Physics PDF

CHAPTER 1
Electric Charges and Fields

Chapter Analysis with respect to Last 3 Years’ Board Exams
List of Topics 2016 2017 2018

Delhi All India Delhi All India Delhi/All India
2Q
Electric Charges (1 mark)
Conservation of charge, 1Q
Coulomb’s law-force be- (1 mark)
tween two point charges,
forces between multiple
charges
Superposition principle
and continuous charge
distribution
Electric field, electric field 1Q
due to a point charge, (3 marks)
electric field lines, elec-
tric dipole, electric field
due to a dipole, torque on
a dipole in uniform elec-
tric field.
Electric flux, statement of 1Q 2Q
Gauss’s theorem (1 mark) (5 marks)
Gauss’s theorems appli- 1Q 1Q 1Q 2Q 2Q
cations to find field due (3 marks) (1 mark), (1 mark) (5 marks) (3 marks)
to infinitely long straight 1Q
wire, uniformly charged (3 marks)
infinite plane sheet and
uniformly charged thin
spherical shell (field in-
side and outside)
On the basis of above analysis, it can be said that from exam point of view Electric Flux, Gauss
theorem and its applications , electric force on a charge and Electric Field due Electric Dipole
are the most important topics of the chapter.
2 CHAPTER 1 : Electric Charges and Fields
[Topic 1] Coulomb’s law, electrostatic field and

electric dipole
Summary Three basic properties of electric

charge
Electric Charge
• Quantization: When the total charge of a body is
• Electrostatic charge is a fundamental property of an integral multiple of a basic quantum of charge,
matter due to which it produces and experiences this is known as quantization of electric charge.
electrical and magnetic effects. i.e., q = ne where
• Properties of atoms, molecules and bulk matter n = ±1, ±2, ±3, .................
are determined by electric and magnetic forces.
• Additivity: It means that the total charge of a
• It can be inferred from simple experiments based system is the algebraic sum (adding taking into
on frictional electricity that there are two type of account negative and positive signs both) of all
charges in nature: negative and positive; and like the charges in the system.
charges repel and unlike charges attract.
• Conservation of charge: Conservation of
• By convention, the charge on electron is electric charges means that there will be no
considered as negative and the charge on proton change in the total charge of the isolated system
is considered as positive and the charge present is with time. There is transfer of the electric charge
equal. The S.I. unit of electric charge is coulomb. from one body to another, but no charge will be
Its C.G.S unit is stat coulomb. created or destroyed.
• The nature and amount of electric charge present
in a charged body is detected by Gold-leaf Coulomb’s law
electroscope. The force between two point charges q1 and q2
• Total charge on a body is expressed as q = ± ne. is directly proportional to the product of the two
charges(q1q2) and inversely proportional to the
Conductors and Insulators square of the distance between them(r2)and it acts
along the straight line joining the two charges.
• Objects that allow charges to flow through them
are called Conductors (metals) and objects that k ( q1 q 2 )
do not allow charges to flow through are called F12 = force on q2 due to q1 = r̂21
r 221
Insulators (rubber, wood, and plastic).
1
• Objects that behave as an intermediate between where k =
conductors and insulators are called semi- 4 πε 0
conductors, for example- silicon. The experimental value of the constant ε0 is
• The process of sharing charges with the earth, 8.854 × 10–12C2N–1m–2
when we bring a charged body in contact with the Therefore, the approximate value of k is
earth is called grounding or earthing. 9 × 109Nm2C–2
F21
Charging by Induction F12
r21 = r2 – r1
q2
q1
• Charging by induction means charging without
contact. r1 r2
• If a plastic comb is rubbed with wool, it becomes o

negatively charged.
Fig. Depiction of Coulomb’s law
CHAPTER 1 : Electric Charges and Fields 3
Facts about Coulomb’s law:  In regions of constant electric field, the field
lines formed are uniformly spaced parallel
• Coulomb’s law is not valid for charges in motion; straight lines.
it should only be used for point charges in vacuum • Field lines are continuous curves. There will be
at rest. no breaks.
• The electrostatic force obeys Newton’s third law
of motion and acts along the line joining the two
charges.
• Presence of other charges in the neighborhood + +
does not affect Coulomb’s force. q q
• The ratio of electric force and gravitational force
between a proton and an electron is represented
k e2 Fig. Electric field lines
by ≅ 2.4 × 1039
G me mp • Field lines are not intersecting. They cannot cross
each other.
Superposition Principle • Electrostatic field lines begin at positive charges
and terminate at negative charges.
The presence of an (or more) additional charge does
not affect the forces with which two charges attract • No closed loop can be formed by them.
or repel each other. Superposition principle states
that the net force on any charge due to n number of Electric Dipole
charges at rest is the vector sum of all the forces on
that charges, taken one at a time. • A pair of equal and opposite charges q and
–q separated by small distance 2a is known
i.e. F0 = F01 + F02 + F03 + ..F0n as electric dipole. The magnitude of its dipole
• The force on a small positive test charge q placed moment vector is 2qa and is in the direction of
at the point divided by the magnitude of the the dipole axis from –q to q.
charge is the electric field E at a point due to 2a
E+q E–q p
charge configuration. p q –q
r
Electric Field Fig. Electric dipole
• The space around a charge up to which its force • Field of an electric dipole in its equatorial plane
can be experienced is called electric field. at a distance r from the center:
• Electric field due to a point charge q has a −p 1
q E=
magnitude E ( r ) = (
4 πε o a + r 2 )
32
rˆ 2
4 πε 0 r 2
 It is radially outwards if q is positive. −p
≅ for r >> a
 It is radially inwards if q is negative. 4 πε o r 3

• Electric field satisfies the superposition principle. • Dipole electric field on the axis at a distance r
 The unit of electric field is N/C. from the center:
 Electric field inside the cavity of a charged 2pr
E=
conductor is zero.
( )
2
4 πε o r 2 − a 2

Electric Field lines 2p
≅ for r >> a
• The tangent at each point on the curve of electric 4 πε o r 3

field line, gives the direction of electric field at
that point. The 1/r3 dependence of dipole electric fields should
be noted in contrast to the 1/r2 dependence of electric
• The relative strength of electric field at different
field due to a point charges.
points is indicated by the relative closeness of
field lines. • In a uniform electric field E, a dipole experiences
a torque t given by
 In regions of strong electric field, they crowd
near each other. t=p×E
 In regions of weak electric field, they are far But no net force will be experienced by it.
apart.
(b) Two charged spherical conductors of radii R1
PREVIOUS YEARS’ and R2 when connected by a conducting wire

acquire charge q1 and q2 respectively. Find
the ratio of their surface charge densities in
EXAMINATION QUESTIONS terms of their radii.
TOPIC 1 [DELHI 2014]
8. A charge is distributed uniformly over a ring of
1 Mark Questions radius ‘a’. Obtain an expression for the electric
1. What is the geometrical shape of equipotential intensity E at a point on the axis of the ring.
surface due to a single isolated charge? Hence show that for points at large distances
[DELHI 2014] from the ring, it behaves like a point charge.
2. Why do the electric field lines never cross each [DELHI 2016]
other?
[ALL INDIA 2014]
5 Marks Question
9. (a) Derive an expression for the electric field
3. A point charge +Q is placed at point O as shown
E due to a dipole of length “2a’ at a point
in the figure. Is the potential difference VA — VB
distant r from the centre of the dipole on the
positive, negative or zero?
+Q
axial line.
O A B (b) Draw a graph of E versus r for r >> a.
[DELHI 2016] (c) If this dipole were kept in a uniform
4. In which orientation, a dipole placed in a external electric field E0,diagrammatically
uniform electric field is in (i) stable, (ii) unstable represent the position of the dipole in stable
equilibrium? and unstable equilibrium and write the
[DELHI 2018] expressions for the torque acting on the
5. Draw a graph to show the variation of E with dipole in both the cases.
perpendicular distance r from the line of charge. [ALL INDIA 2017]
Solutions
[DELHI 2018]
2 Marks Question
1. The equipotential surfaces of an isolated charge
6. An electric dipole of length 4 cm, when placed
are concentric spherical shells(co-centric shells)
with its axis making an angle of 60° with a
and potential will be inversely proportional to
uniform electric field, experiences a torque of
distance. [½]
4√3 Nm. Calculate the potential energy of the
dipole, if it has charge ± 8 nC. [DELHI 2014]
3 Marks Questions +q
7. (a) Obtain the expression for the energy stored
per unit volume in a charged parallel plate
capacitor. [½]
(b) The electric field inside a parallel plate Fig. Equipotential surfaces of an isolated charge
capacitor is E. Find the amount of work
done in moving a charge q over a closed 2. If two electric fields cross each other then there
rectangular loop abcda. would be two different values of electric field
+++++++++++ with individual directions at that location which
a is impossible, hence electric field lines never
b
cross each other. [1]
3. Potential at a distance r from a given point
charge Q is given by,
+q
O A B
d c rA
–––––––––––
rB [½]
OR
(a) Derive the expression for the capacitance of 1 Q
a parallel plate capacitor having plate area V=
4πεo r
A and plate separation d.
Q 2
VA = 1 Q2 1 (eo EA)
4πεo rA U= =
2 C 2 eo A d
Q
VB = 1 [½]
4πεo rB U= eo E 2 ( Ad) J
2
Since, rA < rB ⇒ VA > VB The volume between the plates is Ad metre3.

Therefore, the energy per unit volume is
Hence, VA – VB is positive. [½] given by,
4. A dipole placed in a uniform electric field is in: U 1
(i) Stable Equilibrium: When the electric field U
= = ε E 2 J / m3
Ad 2 o
is directed along the direction of the dipole +Q –Q
+ –
i.e. when E is parallel to p . [½]
+ –
F F
+ –
(ii) Unstable Equilibrium: When the electric
field is directed at an angle of 180° with the + –
+ –
direction of the dipole i.e. when E is anti-
a b [½]

parallel to p . [½] Fig.: Parallel plate capacitor
5. (b) Work done, W = F .d
Here, F is the exerted on the charge (q) due
to electric field (E) and is given by, F = qE
E
[1]
Net displacement, d = 0
Hence, W = 0
r
Fig: graph to show the variation of E with OR
perpendicular distance r from the line of (a) Derivation for the capacitance of parallel
charge. [1] plate capacitor:
6. As τ = pE sin q Surface charge Area A
density
∴ 4 3 = pE sin q [1] 1
++++++++++++++++
3
⇒ pE × =4 3
2 E d
⇒ pE = 8
Potential energy of dipole, ––––––––––––
U = –pE cos q [1] 2
U = –pE cos 60° Surface charge
U = –4 J density – [½]
7. (a) Let us consider a parallel-plate capacitor of Fig. Capacitance of a parallel plate capacitor
plate area A. If separation between plates is A parallel plate capacitor consists of two large
d metre (meter), capacitance C in given by plane parallel conducting plates separated by a
C = eo A F [1] small distance d. The two plates have charges
d q and -1 and distance between them is d.
We know that the magnitude of the electric
field between the charged plates of the
capacitor in q
Plate 1 has charge density σ =
σ A
E=
εo q
Plate 2 has charge density σ = −
Where, σ is the surface density of either A

plate. Therefore, the plate charge in is
In the inner region between the plates 1 and 2,
Q = sA = e0 EA Now, the energy stored in the
capacitor in the electric fields due to the two charged plates
add up
q q q q As the total charge q is uniformly distributed,

E= + = = [½] the charge dq on the element dl is
2ε o 2ε o ε o Aε o
q
dq = ⋅ dl
For this electric field, potential difference 2p a
between the plates in given by,
\ The magnitude of the electric field produced
1 qd by the element dl at the axial point P is
V = Ed =
ε o A dq kq dl
dE = k ⋅ = ⋅
2 2p a r 2
The capacitance C of the parallel plate capacitor r
Q εo A (i) The axial components dEcosθ and
is then, C = = [½]
V d (ii) The perpendicular component dEsinθ. [½]
(b) The surface charge density for a spherical Since the perpendicular component of any two
conductor of radius R1 is given by: diametrically opposite elements are equal and
q1 opposite, they cancel out in pairs. Only the axial
σ= components will add up to produce the resultant
4πR12 field.
Similarly, for spherical conductor R2, the E at point P is given by,
surface charge density is given by: 2π a
q2
σ2 = [½] E= ∫ dE cosθ [½]
4πR 22 0
σ1 q1R 22 (∵ Only the axial components contribute towards

= E)

σ2 q 2 R12

(1 )
2p a
As the spheres are connected so the charges will kq dl x
flow between the spherical conductors till their
E= ∫ ⋅ ⋅
2p a r 2 r
0
potential become equal.
2p a
kq1 kq2 kqx 1  x
= E= ⋅
2p a r 3 ∫ dl
∵ cosθ = 
R R2 0 r
1 [½]
q1 q kqx 1 2p a
= 2 E= ⋅ (l)
R R2 ( 2) 2p a r 3 o
1
Using (2) in (1) kqx 1
E= ⋅ ⋅ 2p a
2p a 32
We have, x 2 + a2( ) [½]
σ1 R1 R 22 R
= ⋅ ⇒ 2
σ2 R 2 R12 R1 ∵ r 2 = x2 + a2

kqx
σ1 R 2 E=
( )
= 32
σ R1 x2 + a2 [½]
2 [½]
8. Suppose we have a ring of radius a that carries If x > > a, then x2 + a2 ≈ x2
a uniformly distributed positive charge q.
1 qx
dl dE Sin E=
++
+ r 4πε o
(x )2 32
++

+++++++
dE
+++ +++++
P

a dE Cos 1 q
x
E=
2
dE 4πε o x [½]
++
+++ r
dE Sin This expression is similar to electric field due to
dl
point charge.
Fig. Uniform distribution of a charge over a ring
[½]
9. –q O +q EA EP (b) Graph of E versus r for r >> a

EB
A l l B
2l (x–l)
E
x
(x+l)
[½]
Electric field intensity due to as electric dipole r [1]
(a) Dipole at a point on the axial wire: we Fig.: E versus r

have to a calculate the field intensity (E) (c) Torque on an electric dipole in uniform
at a point P on the axial line of the dipole electric field :-
and dt a distance OP = π from the centre
O of the dipole. Resultant electric field B
F
+q
intensity at the point P, E P = E A + E B
2l
The vectors EA and EB are collinear at opposite.
\ EP = EA – EB A
–F C
Here,
–q
[½]

1 q 1 q Consider an electric dipole considering of two
EA = ⋅ and EB = ⋅
4πε 0 ( x − l )2 4πε 0 ( x + l )2 changes –q and +q placed is a uniform external
electric field of intensity E. The dipole moment
P makes an angle θ with the direction of the
[½] electric field. The net force is zero. Since, the
two forces are equal in magnitude and opposite
Thus, in direction and act at different points therefore
they constitute a couple. A net torque τ acts
 
1  q q  1 4 qlx on the dipole about an axis passing through
Ep =  − = ⋅
4πε0  ( x − l)2 ( x + l)2  4πε0 2 the mid-point of the couple. Now τ = force ×
  x2 − l 2 ( ) perpendicular distance BC between the parallel
force qE (2l sin q)
1 4 px t = (q × 2l) E sin q or t = pE sin q [½]

Hence, E p = ⋅
2
[ ∴ p = q × 2l ] [½]

4πε0
( x2 − l 2 )
In vector notation,
t=p×E
1 4 px
In vector form, E p = ⋅

4πε0 2 SI unit of torque is newton-meter (N-m) and its
( x2 − l 2 ) dimensional formula is [ML2T2]
Case-I: If q = 0° then t = 0,
If the dipole is short,i.e. , 2l << x, then [½]
T h e d i p o l e i s i n s t a b l e e q u i l i b r i u m
2 P Case-II: If q = 90°, thent = PE (maximum value)
Ep = ⋅
...... (i) The torque acting on dipole will be maximum.
4πε0 x3
Case-III: If q = 180° then t = 0 [½]
The direction of EP is long BP produced clearly,
1 The dipole is in unstable equilibrium
EP ∝ [½]
x3
Topic 2: Electric Flux

Summary Where, r is the radial (perpendicular) distance of the
point from the wire and n̂ is the radial unit vector in
• Electric flux is proportional to number of lines
the plane normal to the wire passing through the
leaving a surface, outgoing lines with positive
point.
sign, incoming lines with negative sign.
E •
Infinite plane sheet (thin) of uniform surface
charge density s
Surface charge
z density
S
E y

E E
Fig. Electric flux 1 2 x
• Through a small area element DS, the flux Df of
electric field E is given by x x
Df = E.DS Fig. Infinite plane sheet (thin)

And the vector area element DS is σ
E= nˆ
∆S =∆Snˆ 2ε o

Where n̂ is a unit vector normal to the plane and
Where DS is the magnetic of the area element and n̂ going away from it.
is normal to the area element, which can be considered •
Thin spherical shell of uniform surface charge
planar for the sufficiently small DS. density s
q
Gauss’s Law and its application= E
4 πε o r 2
r ( r ≥ R)

• The flux of electric field through any closed Gaussian surface
surface S is 1/e0 times the total charge enclosed Surface charge
density
by S. P
R r
q enclosed
= ∫ dA
φ E=
ε0
O

• The law is mainly useful in determining electric
Fig.: Thin uniformly surface charged spherical
field E, when the source distribution has simple
shell (r > R)
symmetry:
(For r > R)
 Thin infinitely long straight wire of uniform
linear charge density l E = 0 (r < R)
Gaussian
+ Surface charge surface
+
+ density
+
+
+ R
+ O
+
+ r
+ P
+
P r +
+ l
E + Fig.: Thin uniformly surface charged spherical
+
+
+ shell (r < R)
+
+
+
r (For r < R)
+
+
Where r is the distance of the point from the center
+ of the shell whose radius is R with the total charge
Fig. Thin infinitely long Straight wire q. The electric field outside the shell is the same as
the total charge is concentrated at the center. A solid
λ
E= n sphere of uniform volume charge density shows the
2πε o r
same result. Inside the shell at all the points, the
field is zero.
spherical shell of radius R at a point outside the
PREVIOUS YEARS’ shell. Draw a graph showing the variation of

electric field with r, for r > R and r < R.
EXAMINATION QUESTIONS [ALL INDIA 2011]
TOPIC 2 5 Marks Questions

8. (a) An electric dipole of dipole moment
1 Mark Questions
1. What is the electric flux through a cube of side p consists of point charges +q and –q
1 cm which encloses an electric dipole? separated by a distance 2a apart. Deduce

[DELHI 2015] the expression for the electric field E due
to the dipole at a distance x from the centre
2. Figure shows three point charges +2q, -q and
of the dipole on ts axial line in terms of the
+3q. Two charges +2q and –q are enclosed within
a surface ‘S’. What is the electric flux due to this dipole moment p . Hence show that in the
configuration through the surface ‘S’? limit

[DELHI 2015] ˆ,
(b) Given the electric field in the region E = 2 xl
+2q
–q find the net electric flux though the cube and
the charge enclosed by it.
+3q S y
3. How does the electric flux due to a point charge

enclosed by a spherical Gaussian surface get
affected when its radius is increased?
[DELHI 2016] x
2 Marks Questions x a
OR
ˆ /C ,
4. Given a uniform electric field E = 5 ×103 iN
find the flux of this field through a square of 10 (a) Explain, using suitable diagrams, the
cm on a side whose plane is parallel to the y-z difference in the behavior of a (i) conductor
plane. What would be the flux through the same and (ii) a dielectric in the presence of
square if the plane makes a 30o angle with the external electric field. Define the terms
x-axis? polarization of a dielectric and write its
relation with susceptibility.
[DELHI 2014]
(b) A thin metallic spherical shell of radius a
5. Given a uniform electric field E = 2 ×103i N C . carries a charge Q on its surface. A point
Find the flux of this field through a square of Q
charge is placed at its centre C and
side 20 cm, whose plane is parallel to the y-z 2
plane. What would be the flux through the same another charge +2Q is placed outside the
square, if the plane makes an angle of 30° with shell at a distance x from the centre as
the x-axis? shown in the figure. Find (i) the force on the
[DELHI 2014] charge at the centre of shell and at the point

6. Given a uniform electric field E = 4 ×103i N C , A, (ii) the electric flux through the shell.
find the flux of this field through a square of 5 [DELHI 2015]
cm on a side whose plane is parallel to the y-z 9. (a) Use Gauss’s theorem to find the electric field
plane. What would be the flux through the same due to a uniformly charged infinitely large
square, if the plane makes an angle of 30°with plane thin sheet.
the x-axis?
(b) An infinitely large thin plane sheet has a
[DELHI 2014]
uniform surface charge density +σ. Find
the amount of work done in bringing a point
3 Marks Question charge q from infinity to a point, distance r,
7. Using Gauss’s law to obtain the expression for in front of the charged plane sheet
the electric field due to a uniformly charged thin [ALL INDIA 2017]
10. (a) Define electric flux. Is it a scalar or a vector 4. When the plane is parallel to the y-z plane:
quantity? A point charge q is at a distance of Electric flux, f = EA
3
d Here, E = 5 ×10 jN / C
directly above the centre of a square of side
2 A = 10 cm2 , i = 10 −2 im2 = 10 −2 im2 [1]
‘d’, as shown in the figure. Use Gauss’s 3 −2
φ = 5 × 10 i10 i ⇒ φ = 50 Weber or Nm2C–1
theorem to obtain the expression for the
electric flux through the square.

When the plane makes a 30o angle with the
q x-axis, the area vector makes 60o with the x-axis.

φ = E. A ⇒ φ = EA cos θ
d/2
f = 5 × 103.10–2 cos 60°
d
50
φ=
2
d ⇒ φ = 25 Weber or Nm2 C–1 [1]
(b) If the point charge is now moved to a distance 5. When the plane is parallel to the y-z plane:

‘d’ from the centre of the square and the f = E. A

side of the square is doubled, explain how the

electric flux will be affected. E = 2 × 103 i
OR 2
A = ( 20 cm) i = 0.04 M 2 i
Use Gauss’ law to derive the expression for the
A = (20 cm)2 i = (20 × 10–2)2 = 0.04m2i
electric field E ( ) due to a straight uniformly
( )(
∴ f = 2 ×103i ⋅ 0.04i)
charged infinite line of charge density lC/m.
⇒ φ = 82
[ALL INDIA 2018]
Weber or 80 Nm2C–1 [1]

When the plane makes an 30°angle with the
Solutions x-axis, the area vector makes an 60° angle with
the x-axis.
1. From Gauss law the net flux passing through
a surface is proportional to the charge enclosed E ⋅ A ⇒ φ = EA cos θ
φ =
within the surface. Since , net charge enclosed
by electric dipole is zero hence flux will be zero.
φ =×
2 103 × 0.04 cos60o
[1] φ = 2 × 103 × 0.04 cos 30°
2. From gauss law net flux is ratio of total charge 1
q f = 2 ×103 × 0.04 ×
enclosed divided by ( S) = from the figure 2
eo
⇒ φ = 40 Weber or 40 Nm2C–1 [1]
total charge enclosed is +2q – q = q. Hence 6. When the plane is parallel to the y-z plane:
q
( S) = [1] Electric flux, f = E ⋅ A
eo

3. According to Gauss’s law, Here, E = 4 ×103i N C
qen 2
φ = ∫ ε ⋅ ds =
εo A = (5 cm) i = 0.25 ×10−2im2

[½]
Where qen is the total charge enclosed by the

( )(
f = 4 ×103i ⋅ 25 ×10−4i )
surface. From above formula it is clear that
⇒ φ = 10 Weber or Nm2C–1 [1]
electric flux does not depend on radius, hence it
remains constant.

When the plane makes an angle of 30° with the
x-axis, the area vector makes an angle of 60°
Flux depends only on the charge enclosed.
with the x-axis.
Hence, the electric flux remains constant.

E ⋅ A ⇒ φ = EA cos θ
φ = For r > R , electric field outside a charged
thin spherical shell is same as if the whole
⇒ φ = (4 × 103) (25 × 10–4) cos 60°
charge Q is concentrated at the centre.
10
⇒f= 8. (a) Electric field at a point on the axial line
2
kq kq
⇒ φ = 5 Weber or Nm2C–1 [1] E+ q = E− q =
2 [1]
7. ( x − a) ( x + a)2
Q EO  1 
R ∵ k = 
rP  4πε o 

∵ P = 2aq
[½]  
Fig.: Spherical Gaussian surface
Consider a spherical Gaussian surface of If x>>>a,
2p
radius r (›R), concentric with given shell. In vector form, E = [1]
If E is electric field outside the shell, then 4πε o x 3
by symmetry, electric field strength has
same magnitude Eo on the Gaussian surface
(b) Since, the electric field is parallel to the faces
and is directed radially outward. Also
parallel to xy and xz planes, the electric flux
the direction of normal at each point is
through them is zero.
radially outward, so angle between Eo and ds
is zero at each point. Hence, electric flux through Electric flux through the left face, [1]
Gaussian surface [½] fL = (EL) (a ) cos 180°
2
= fs Eo ds fL = (0 )(a2) cos 180° = 0

= fs Eo ds cos 0° = Eo 4 pr2ds Electric flux through the right face,
N o w, G a u s s i a n s u r f a c e i s o u t s i d e fR = (ER) (a2) cos 0°
the given charged shell, so charge fR = (2a) (a2) × 2a3
enclosed by the Gaussian surface is Q. q enclosed
Total flux ( φ ) = 2a 3 = [1]
Hence, by Gauss’s theorem εo
1 ∴ q enclosed = 2a 3 ε o
φs Eo ⋅ ds = × charge − enclosed
εo

Add sign of integration in this formula
OR
1
2
⇒ Eo ⋅ 4πr = ×Q Q A

εo 2Q
1 Q
⇒ Eo = [1]
4πεo r 2 Q C
Thus, electric field outside a charged thin 2
spherical shell is same as if the whole [1]
charge Q is concentrated at the centre. (a) (i) Conductor Eo → External field
Graphically,
E m → Internal field created by the
Y
Emax redistribution of electrons inside the metal
E
When a conductor like a metal is subjected
Ex1/r 2 to external electric field, the electrons
experience a force in the opposite direction
x
0 r=R r [½] collecting on the left side. [1]
E is proportional to 1/r not multiple as shown
2 A positive charge is therefore induced on the
in the figure. right hand side. This creates an opposite
electric field (Em) that balances out (Eo).
For r < R , there is no strength of electric
Hence, the net electric field inside the
field inside a charged spherical shell.
conductor becomes zero. [1]
∴Total electric flux over the centre surface of

– + cylinder f = 2EA [1]
– +
Total charge enclosed by the cylinder, q = sA
– Metal +
q
–
Em
+ acc. to Gauss’ law, φE =
– + ε0
σA σ
E0 ∴ 2EA = or E = [1]
(ii) Dielectric ε0 2ε 0
–+ –+ Internal electric (b) Let V0 be the potential on the surface at

field sheet that at a distance r from it
–+ –+
dV = E ⋅ dr
E0 σ
When external electric field is applied, V0 – V = r
2ε 0
dipoles are created (in case of non-polar σ
dielectrics). The placement of dipoles is V = V0 – r
2ε 0
as shown in the given figure. An internal
10. (a) Electric flux is defined as, fe = E. ds
electric field is created which reduces the It is scalar quantity. Electric flux through
external electric field. q
Polarization of dielectric (P) is defined as square is φε = [3]
εo 6
the dipole moment per unit volume of the
q
polarized dielectric. (b) Flux will not change, i.e. φε = [2]
εo 6
P = ce e0 E
OR
Where, cesusceptibility [1]

To calculate the field, imagine a cylindrical
E → Electric field Q
(b) Net force on the charge , placed at the Gaussian surface,as shown in the figure. Since
2 the field is every where radial,flux through the two
centre of the shell, is zero.
Force on charge 2Q kept at a point A [1] ends of the cylindrical Gaussian surface is zero. At
the cylindrical part of the surface, E is normal to
 3Q 
I 2Q the surface at every point,and its magnitude is
 2 
F = E × 2Q = constant, since it depends on r. The surface area
k3Q2 4πε o r 2 of the curve if 2prl, where l is the length of the
F= cylinder. [1]
r2
Q [1] Flux through the Gaussian surface
Electric flux through the shell, φ = ε
o = flux through the curved cylindrical part of the
9. (a) Gaussian surface for a thin infinite plane
surface = E ×2prl
sheet of uniform charge density
>o The surface includes charge equal to ll. Gauss’s
Cylinderical Sheet
+++ n̂ λl
law then gives E × 2πrl = [1]
Gaussian Surface +++ +
+++ ++++
+ + + +
+++
+ ds φo
s +++ ++
S 3 ++ ++++ + ES
E
+++ ++++
E λ
i.e., E = [1]
+

+
+++ ++++
+ +
ds s1 +++ ++++ s2ds
r + + + +
+++
+ + +
+ r 2πrφo
λ ∧
+++ ++++
+ +
+++ ++++
?+ + + +
[1] Vectorially, E at any point is given by E = n
2πrφo
Let σ be the surface charge density of the sheet. +
+
+
From symmetry, E on either side of the sheet +
+
must be perpendicular to the plane of the sheet, +
+
having same magnitude at all points equidistant +
+
from the sheet. We take a cylindrical cross- +
+
sectional area A and length 2r as the Gaussian P r +
+ l
E +
surface. On the curved surface of the cylinder E +
+
and n̂ are perpendicular to each other. Therefore +
+
+ r
flux through curved surface = 0. Flux through +
+ [2]
the flux surface = EA + EA = 2EA +
+
CHAPTER 2
Electrostatic Potential and
Capacitance

Electric potential, poten-
tial difference
electric potential due to a
point charge, a dipole and
system of charges
Equipotential surfaces,
electrical potential en-
ergy of a system of two
point charges and of elec-
tric dipole in an electro-
static field
Conductors and insula-
tors, free charges and
bound charges inside a
conductor
Dielectrics and electric
polarization
capacitors and capaci- 1Q 1Q
tance, combination of ca- (3 marks) (2 marks)
pacitors in series and in
parallel
capacitance of a parallel 1Q 1Q 1Q
plate capacitor with and (3 marks), (3 marks) (3 marks)
without dielectric medi- 1Q
um between the plates, (5 marks)
energy stored in a capac-
itor
On the basis of above analysis, it can be said that from exam point of view, Potential Differ-
ence, Dielectric and Capacitor, Electric Dipole and Parallel Plate Capacitor are the most im-
portant concepts of the chapter.
16 CHAPTER 2 : Electrostatic Potential and Capacitance
Topic 1: Electrostatic Potential and Electrostatic Potential

Energy
Summary q from a point R to a point P is VP – VR , which is

the difference in potential energy of charge q be-
Electrostatic potential: tween the final and initial points.
• The amount of work done by an external force in Potential difference:
moving a unit positive charge from one point to
When the work is done upon a charge to change its
another in electrostatic field is called electrical
potential energy then the difference between the fi-
potential.
nal and the initial location is called electric potential
1 q difference.
• Such that V =
4 πε r
Electric Potential due to a dipole:
here, q = charge causing the field, e = permittiv-
• W • The electrostatic potential at a point with distance
ity, r = separation between centre of charge point. r due dipole at a point making an angle q with
• Electrostatic force is a conservative force. dipole moment p placed at the origin is given by
• Work done by an external force (equal and oppo- 1 p ⋅ rˆ
=
V(r) ⋅ 2 .
site to the electrostatic force) in bringing a charge 4 πε o r
r1
q
+
r
2a r2
p O
–
–q
Fig. Electrical potential due to dipole
• It is a scalar quantity. • The result is true also for a dipole (with charges
• Let A and B be the initial and final location for a –q and q separated by 2a for r >> a.
single charge q then the potential difference be- Dipole and System of charges
tween A and B is given by: • For a charge configuration q1, q2, ......, qn with po-
B B B sition vectors r1, r2, r3, ......, rn, then the potential
∆V = − ∫ E × ds =
VB − VA = − ∫ Eds cos θ =
− ∫ E × ds V1 at point P due to charge q1 will be,
A A A
1 q1
V1 =
Where, E is the field due to a point charge, ds = dr, 4 πε 0 r1

so that
And the sum of potentials due to individual charges
rB r
q dr q 1 
B
q 1 1 is given by the superposition principle,
VB −=
VA ∫
rA
=
4 πε 0 r 2
4 πε 0
=
r  − 
  rA 4 πε 0  rB rA  1  q1 q 2 q 
=V  + + .... + n 
4 πε o  r1P r2P rnP 
CHAPTER 2 : Electrostatic Potential and Capacitance 17
q5 • In this system the two charges q1 and q2 when
separated by distance r, will either repel or
q1 q4 attract each other.
r5P
• Electrical potential of charges q1 and q2 is given by:
r4P
r1P 1 2
U= ∑ q i Vi
2 i =1
r3P
P q3
r2P
Potential Energy in an External Field:
q2
• The potential energy of a charge q in an external
• In a uniformly charged spherical shell, the electric potential V(r) is qV(r). The potential energy of a di-
field outside the shell with outside potential is pole moment p in a uniform electric field E is –p.E.
given by,
• Electric dipole in an electrostatic field: Electric
1 q potential due to a dipole at a point at distance r
V=
4 πε 0 r and making an angle q with the dipole moment p

is given by
Equipotential surfaces 1 p cos θ
V=
• A surface over which potential has a constant val- 4 πε 0 r 2

ue is known as an equipotential surface.
• The amount of work done in moving a charge over Electrostatics of conductors:
an equipotential surface is zero.
• Concentric spheres centered at a location of the • Electrostatic field is zero inside a conductor.
charge act as equipotential surfaces for a point • Electrostatic field at the surface of a charged
charge. conductor must be normal to the surface at every
point.
• The electric field E, at a point and equipotential
surface are mutually perpendicular to each other • In the static situation, there cannot be any excess
through the point. The direction of the steepest charge in the interior of a conductor.
decrease of potential is in E. • Throughout the volume of the conductor, the
• Regions of strong and weak fields are located be- electrostatics potential is constant and has same
cause of the spacing among equipotential surfac- value on its surface.
es. • Electrostatics field E is zero in the interior of a
conductor; just outside the surface of a charged
Potential Energy of a System of conductor, E is normal to the surface given by
Charges: E=
σ
nˆ where n̂ is the unit vector along the
Potential energy stored in a system of charges is εo
the work done by an external agency in assembling outward normal to the surface and σ is the surface
the charges at their locations. Total work done charge density.
in assembling the charges is given by • Electrostatic shielding: A field which is inside
1  q1 q 2 q1 q 3 q 2 q 3  the cavity of a conductor is always zero and it
U= ⋅ + + where r12 is distance remains shielded from the electric field, which is
4 πε o  r12 r13 r23 
known as electrostatic shielding.
between q1 and q2, r13 is distance between q1 & q3 and Dielectrics and Polarization:
r23 is distance between q2 & relabel q3.
r23 • Dielectrics: A non-conducting substance which
q3
q2 has a negligible number of charge carriers unlike
conductors is called dielectrics.
r13 r12 • Electric polarization: The difference between
induced electric field and imposed electric field in
q1 dielectric due to bound and free charges is known
Fig. Potential energy due to System of charges as electric polarization. It is written as:
D−E
Electric potential energy of system
P=
4π
of two point charges Note: Polarisation can also be written as polariza-
• Here the work done doesn’t depend on path. tion (with ‘z’ in place of ‘s’)
6. An electric dipole of length 2 cm, when placed with
PREVIOUS YEARS’ its axis making an angle of 60° with a uniform

electric field, experiences a torque of 8√3 Nm.
EXAMINATION QUESTIONS Calculate the potential energy of the dipole, if it
has a charge of ± 4 nC.
1 Mark Questions 7. (i) Can two equipotential surfaces intersect each

1. A point charge Q is placed at point O as shown other? Give reasons.
in the figure. Is the potential difference VA − VB (ii) Two charges -q and +q are located at points
positive, negative or zero, if Q is (i) positive (ii) A (0, 0, –a) and B (0, 0, + a) respectively. How
negative? much work is done in moving a test charge
Q from point P (97, 0, 0) to Q(–3, 0, 0)?
[DELHI 2017]
O rA
A B
3 Mark Questions
8. (a) Define electric dipole moment. Is it a scalar or
rB
a vector? Derive the expression for the electric
[ALL INDIA 2011]
field of a dipole at a point on the equatorial
2. For any charge configuration, equipotential
plane of the dipole.
surface through a point is normal to the electric
field. Justify. (b) Draw the equipotential surfaces due to an
[DELHI 2014] electric dipole. Locate the points where the
3. What is the amount of work done in moving a potential due to the dipole is zero.
[ALL INDIA 2017]
point charge Q around a circular arc of radius
‘r’ at the centre of which another point charge 9. (i) Draw equipotential surfaces for a system of
‘q’ is located ? two identical positive point charges placed
[ALL INDIA 2016] a distance ‘d’ apart.
2 Mark Questions (ii) Deduce the expression for the potential

4. Two uniformly large parallel thin plates energy of a system of two point charges q1
and q2 brought from infinity to the points
having charge densities +σ and –σ are kept in

the X-Z plane at a distance ‘d’ apart. Sketch r1 and r2 respectively in the presence of

an equipotential surface due to electric field external electric field E .
between the plates. If a particle of mass m [DELHI 2017]
and charge ‘-q’ remains stationary between the
plates, what is the magnitude and direction of 10. Four point charges Q, q, Q and q are place d at
the corners of a square of side ‘a’ as shown in
the field?
the figure.
Or Q q
Two small identical electrical diploes AB and
CD, each of dipole moment ‘p’ are kept an angle
of 120° as shown in the figure. What is the
resultant dipole moment of this combination?
If this system is subjected to electric field E
directed along +X direction, what will be the q
a Q
magnitude and direction of the torque acting on Find the (a) resultant electric force on a charge
this? Q, and (b) Potential energy of the system
[ALL INDIA 2011]
5. An electric dipole of length 2 cm, when placed with OR
its axis making an angle of 60° with a uniform (a) Three point charge sq, –4 q and 2q are place
electric field, experiences a torque of 6√3 Nm. d at the vertices of an equilateral triangle
Calculate the potential energy of the dipole, if it ABC of side ‘l’ as shown in the figure. Obtain
has a charge of ± 2 nC. the expression for the magnitude of the
[DELHI 2014]
resultant electric force acting on the charge q.
A 12. (a) Explain why, for any charge configuration,

the equipotential surface through a point
q
is normal to the electric field at that point.
Draw a sketch of equipotential surfaces due
to a single charge (–q), depicting the electric
field lines due to the charge
(b) Obtained an expression for the work done to
dissociate the system of three charges placed
–4q at the vertices of an equilateral triangle of
–2q
side ‘a’ as shown below.
q
B C
l
(b) Find out the amount of the work done to
separate the charges a t infinite distance.
[ALL INDIA 2018]
a a
5 Mark Questions
11. (a) State Gauss’s law in electrostatics. Show,
with the help of a suitable example along
with the figure, that the outward flux due –4q a +2q
to a point charge ‘q’. in vacuum within a [ALL INDIA 2016]
closed surface, is independent of its size or
shape and is given by q
Solutions
e0
kq
1. Potential at a point: V = ,Given q = Q
(b) Two parallel uniformly charged infinite plane r
1 1
sheets, ‘1’ and ‘2’, have charge densities + σ VA  VB  kQ   
and –2 σ respectively. Give the magnitude r
 A rB

and direction of the net electric field at a Q
point.
A B
(i) in between the two sheets and O rA
(ii) outside near the sheet ‘1’.
Or rB [½]
(a) Define electrostatic potential at a point. Where, rA < rB
Write its S.I. unit. Three point charges q1, 1 1
>
q2 and q3 are kept respectively at points A, rA rB
So,
B and C as shown in the figure, Derive the
expression for the electrostatic potential 1 1
  0
energy of the system. r r
 A B
A
If Q at O is positive, VA – VB will be positive
[½]
r12
r13 If Q at O is negative, VA – VB will be negative.
2. If the electric field were not normal to equipotential
surface, it would have non-zero component along
the surface. To move a charge against this
B C component, work would have to be done. But
r23
no work is needed to move a test charge on an
(b) Depict the equipotential surface due to (i) equipotential surface. Hence electric field must
an electric dipole, (ii) two identical positive be normal to the equipotential surface at every
charges separated by a distance. point. [1]
[DELHI 2015]
3. Given charge ‘q’ is located at centre and charge Direction of resultant dipole moment:
‘Q’ on surface. Then, work done p sin 120
tan  
1 qQ p  p cos 120
W [½]
4  o r 2
t an   3
Work done to move the charge over the circular
q = 60°
arc is zero, because it is moving over an
equipotential surface. [½] That is, 30 degrees with +x axis.
Given applied E is along +x axis,
4.
So torque on resultant dipole will be
pE
  pE sin 30  [½]
2
Direction will be along –Z-axis.

5. As t = PE sin q
o
6 3 = pE sin 60
3
 pE  or PE = 12
2
U= –PE cos q [1]
Fig. Two uniformly charged plates [½]
Potential energy of a dipole,
Here the dotted lines represent the parallel
= –12 cos 60°
equipotential surfaces along X-Z plane.
1
If a charge q has to be held stationary between  12 
the two plates, it will have to be balanced by two 2
forces. U = – 6J [1]
Gravitational force: mg, downwards 6. As t = PE sin q
Electrostatic force = 2qE , acting upwards. 8 3 = pE sin q

This implies, that in X-Z plane, the upper plate is
+ charged plate & lower plate is –charged plate. 8 3 = pE sin 60 o

So, E field lines have to be directed along –y axis.
pE 3
[½] 8 3=
2
Or
⇒ pE = 16 [1]
Also Potential energy of the dipole,
⇒ U= –PE cos q
U= –PE cos 60°
16.1
U  8 J
2
U= –8J [1]
7. (i) Two equipotential surfaces cannot intersect
each other because when they will intersect,
[½] the electric field will have two directions,
Fig. An Electric dipole which is impossible. [1]
Resultant dipole moment, (ii) Charge P moves on the perpendicular
   bisector of the line joining +q and –q. Hence,
Pres = p1 + p2 this perpendicular bisector is equidistant
from both the charges. Thus, the potential
pres  p12  p22  2p1p2 cos 120 will be same everywhere on this line.

Pres = p Therefore, work done will be zero. [1]

8. (a) Electric dipole moment of an electric dipole The net electric field intensity due to the electric
is defined as the product of the magnitude dipole.
of either charge of the electric dipole and the 2 2
∴E= E+ + E− + 2 E+ E− cos 2q
dipole length.
2 2 2
= E+ + E+ + 2 E+ cos 2q ( E+ = E− )

2 2
2l = 2 E+ + 2 E+ cos 2q

[½] 2
= 2 E+ (1 + cos 2q )
Fig. An Electric Dipole

i.e., p  q 2l  
2
= 2 E+ (
× 2 cos2 q 1 + cos 2q = 2 cos2 q )
The magnitude of dipole moment is
1 q
p = q × 2l ∴ E = 2 E+ cos 2θ = 2 × cos θ [½]
Dipole moment is a vector quantity. SI Unit of (
4πεo r + l 2
2
)

 
dipole moment p is coulomb metre (Cm)
[Using equation (iii)]
F
l
cosq =

(r 2 + l2 )
P
q
E=  3 
2  2 
E

(
4πεo r 2
+l )
q
If l << r, E =
4πεo r 3
(b) [1]
A B
–q +q
Fig: Electrical field of a dipole at a point on Dipole equator
equatorial plane of the dipole

Electric field intensity due to +q charge is given
by
1 q
E+ = ……….(i) [½] Fig. Equipotential Surface due to dipole
4πεo BP 2
Electrical potential is zero at all points in the
Electric field intensity due to –q charge is given
by plane passing through the dipole equator.
1 q 9. (i) [1]
E− =
4πεo AP 2

= 1 q ………….(ii)
(
4πεo r 2 + l 2 )
From (i) and (ii),
1 q ………….(iii) [½]
E+ = E− = Fig. Equipotential Surface for System of Charges
(
4πεo r 2 + l 2 )
(ii) The work done in bringing charge q1 from

= F’ 2 F12
= 2 F1
infinity to r1 is q1V r1 .  
Work done on q2 against external field F3 and resultant of F1 and F2 will be in same

direction

 q2 V r2   Net force F = F’ + F3
Work done on q2 against the field due to 1 qQ 1 Q2
F= 2 +
q1 q2 4πεo a2 4πεo 2a2
q1 = [1]
4πεo r12
1 qQ  Q
Where, r12 is the distance between q1 and q2. F=
2  q 2 +  [1]
4πεo a  2
By the superposition principle for fields,
Work done in bringing q2 to (b)


qq 

( )
r2 is  q2 V r2 + 1 2 
4πεo r12 

Thus,
Potential energy of system = The total work
done in assembling the configuration

qq
( )
= q1V r1 + q2 V r2 + 1 2 ( ) [1] 4 KQq Kq2 KQ2
4πεo r12 w  
a 2a 2a
10. ( ) There will be three forces on charge Q 1
K=
4πεo
[½]
Or
F2
A
120°
60°
F1
l

l
1 qQ
F1 =
4πεo a2
–4q 2q
1 qQ
F2 = (a) B l C
4πεo a2

1 (4 q)( q)
1 QQ 1 Q2 F1 =
F3 = =
4πεo l2
4πεo 2 4πεo 2a2

( 2a ) 1 4 q2
F1 =
F1 and F2 are perpendicular to each other so their 4πεo l 2

resultant will be
1 q2
F1 =
F’  F12  F22  2 F1 F2 cos 90 πεo l2

F1 = F2 1 ( q)(2q)
F2 =
F’  F12  F12  2 F1 F1  0
4πεo l2

1 q2 q
F2 = φE =
2πεo l 2 εo

Angle between F1 and F12 is 120° And for fig (b)

F  F12  F22  2 F1 F2 cos 120 q
φE = which is same in both the cases so it is
εo
F1 = 2 F2
independent of size and shape of closed surface.

F 2 F2 2  F22  4 F22 cos 120
[½]
F 4 F22  F22  2 F22

F = 3 F22
P
Q
F = 3 F2

3 q2
F= ^
r

2πεo l2 [1]
(b) Plate 1 Plate 2 [½]
(b) The amount of work done to separate the
Let r̂ be the unit vector directed from left
charges at infinity will be equal to potential
to right
energy.
1  Let P and Q are two points in the
U= q ×(−4 q) + ( q × 2q) + (−4 q × 2q) inner and outer region of two plates
4πεo l 
respectively charge densities on plates
1  are +s and –2s
U= −4 q2 + 2q2 − 8 q2 
4πεo l  
[½]

1  (i) Electric field at point P in the inner region
U= −10 q2 
4πεo l   of the plates
1  E = E1 E2
U =− 10 q2  unit
4πεo l    σ σ
[½] E =  +  r
 2εo εo 
11. (a) Statement: The electric flux linked with a
closed surface is equal to ε times the net
3σ
charge enclosed by a closed surface. E= r
2εo
Mathematical expression: [½]
1 (ii) Electric field at point Q in the outer region
 E ⋅ ds =
φE = ∫
εo
( qnet )
[½] of plate 1
Consider two spherical surfaces of radius r and σ 2σ
2r respectively and a charge 1 is enclosed in it. E1 = (−r) and E2 = r
2εo 2εo
According to gauss theorem the total electric [½]
flux linked with a closed surface depends on the ∴ Net electric field in the outer region of the
charge enclosed in it so For fig (a) [½] plate 1(i.e, at Q) is
σ σ 
E = E1 + E2 =  − r
 εo 2εo 
[½]
(q) 2r r σ
E= r
2εo
(q) [½]
(a)
[½]
(b)
Or 12. (a) If the field were not normal to the

(a) The electrostatic potential (V) at any point equipotential surface, it would have non-
in a region with electrostatic field is the zero component along the surface. To move a
work done in bringing a unit positive charge unit test charge against the direction of the
(without acceleration) from infinity to that component of the field, work would have to
point. Its S.I. unit is Volt. The potential be done. But this is in contradiction to the
energy of a system of three charges q1, q2 definition of an equipotential surface: there
and q3 located at r1, r2 and r3 respectively. To is no potential difference between any two
bring q1 first from infinity to r1 no work is points on the surface and no work is required
required. Next we bring q2 from infinity to to move a test charge on the surface.
r2. Work done is [1] [2]
1 q1 q2
q2 V1 ( r2 ) = … (1)
4πεo r12
The charge q1 and q2 produce at potential,
which at any point P is given by –q
1  q1 q  Spherical
V1⋅2 =  + 2  …. (2) Equipotential
4πεo  r1 P r2 P 
surface
Work done next in bringing q3 from infinity [1]
to the point r3is q3 times V12 at
(b) Total electrostatic potential energy of system
1  q1 q2 q2 q3  U = U12 + U23 + U31
q3V1,2 ( r3 ) =  +  …. (3) [1]
4πεo  r12 r23 
1  q (−4 q) (−4 q)(2q) q (2q) 
=  + + 
The total work done in assembling the 4πεo  a a a 

charges at the given locations is obtained
by adding the work done in different steps 1  10 q2 
[Eq. (1) and Eq. (3)] and gets stored in the =−  
4πεo  a 
form of potential energy.   [1]
1  q1 q2 q1 q3 q2 q3  \ Work done to dissociate the system
U=  + + 
4πεo  r12 r13 r23  W=–U

1  10 q2 
(b) Equipotential surfaces for (a) a dipole and   [1]
(b) two identical positive charges are shown
=
4πεo  a 
 
in Figure.
– +
– +
(a) (b)
Some equipotential surfaces for (a) a dipole, (b)
two identical positive charges. [1]
Topic 2: Capacitance
Summary • The net electric field inside the dielectric and
hence the potential difference between the plates
Capacitor and Capacitance is thus reduced. Consequently, the capacitance
C increases from its value Co when there is no
• Capacitor: The system of two conductors
medium (vacuum),
separated by an insulator is called capacitor. ε
C = KCo where K = is the dielectric constant
The device which is used to store charge is known ε0
as capacitor. The applied voltage and size of
capacitor decides the amount of charge that can of the insulating substance.
be stored i.e., Q = CV
Two similar connecting plates are placed in
Types of capacitor:
capacitor in the front of each other where one
A
plate is connected to the positive terminal and • Parallel plate capacitor: C= Kε 0
other plate is connected to the negative terminal. d
• Capacitance: The ratio of magnitude of charge 1
stored on the plate to potential difference between • Cylindrical capacitor: C = 2πKε 0
the plates is called capacitance. It is written as: ln ( b a )
Q  ab 
C= • Spherical capacitor: C = 4 πKε 0 
∆V  b − a 
Size, shape, medium and other conductors in
surrounding influence the capacitance of a
conductor.
Combination of Capacitors
Its S.I. unit is farad. • For capacitors in the series combination, the total
capacitance C is given by
1F = 1CV–1 For a parallel plate capacitor (with
A 1 1 1 1 1
vacuum between the plates), C = ε o where A = + + + ....
d C C1 C2 C3 Cn
is the area of each plate and d in the separation
between the parallel plates. • In the parallel combination, the total capacitance
Area A C is C = C1 + C2 + C3 ...... Cn, where C1, C2, C3 ......
I are individual capacitances.
1 • Capacitors connected in series have the same
+ + + + + + + + + + + +
charges and when connected in parallel have the
same voltage.
E • Potential across capacitor remains same if the
d battery is connected but if it is disconnected
– – – – – – – – – – –
then charge remains the same which is stored in
capacitor.
2
II Electrical Energy Stored in a
Capacitor:
Fig. Capacitor • The energy U stored in a capacitor of capacitance
C, with charge Q and voltage V is
Effect of Dielectric on Capacitance:
1 1 1 Q2 .
• If the medium between the plates of a capacitor = U = QV CV 2
=
is filled with an insulating substance (dielectric), 2 2 2 C
the electric field due to the charged plates induces • The electric energy density (energy per unit vol-
a net dipole moment in the dielectric. This effect, 1
ume) in a region with electric field is ε o E2 .
called polarization, gives rise to a field in the 2
opposite direction.
• The dielectric is polarised by the field and also • Electric density is alternatively known as
the effect is equivalent to two charged sheets with electrostatic pressure.
surface charge densities sp and –sp.
Van De Graaff Generator: plastic

spherical
charge taken by roller
• A Van de Graaff generator consists of a large metal cover
pointed electrode
spherical conducting shell (a few meters in diam-
eter).
• There are two pulleys, one at ground level and
one at the center of the shell. Both of them are
wounded around by a long and narrow endless
belt of insulating material.
• The motor drives the lower pulley which keeps
moving this belt continuously. pointed electrode
rubber belt
• At ground level to the top, it continuously carries produces charge
the positive charge and sprayed on to it by a brush. by friction or high
Then the positive charge is transferred by it to an- voltage plastic
other conducting brush connected to the large shell. roller
• After the transferring of the positive charge is
done, it spreads out uniformly on the outer sur-
face. It can build the voltage difference of as much
as 6 to 8 million volts. Fig. Yande Graff Generator
How will the (i) charge and (ii) potential
PREVIOUS YEARS’ difference between the plates of the capacitors

be affected after the slabs are inserted?
EXAMINATION QUESTIONS
TOPIC 2
1 Mark Questions
1. Why should electrostatic field be zero inside a
[DELHI 2011]
conductor?
[All INDIA 2012]
6. A slab of material of dielectric constant K has
2. A capacitor has been charged by a dc source. the same area as that of the plates of a parallel
What are the magnitudes of conduction and d
plate capacitor but has the thickness 2 , where
displacement current, when it is fully charged? 3
[All INDIA 2013] d is the separation between the plates. Find out
3. Define dielectric constant of a medium. What is the expression for its capacitance when the slab
its S.I. unit? is inserted between the plates of the capacitor.
[DELHI 2015] [DELHI 2011]
4. Predict the polarity of the capacitor in the 7. A capacitor of unknown capacitance is connected
situation described below: across a battery of V volts. The charge stored in
it is 360 μC. When potential across the capacitor
is reduced by 120 V, the charge stored in it
becomes 120 μC.
A Calculate:
S N S N
B (i) T h e p o t e n t i a l V a n d t h e u n k n o w n
capacitance C.
(ii) What will be the charge stored in the
[All INDIA 2017] capacitor, if the voltage applied had increased
by 120 V?
2 Mark Questions [DELHI 2011]
5. Figure shows two identical capacitors, C1 and C2 8. A parallel plate capacitor of capacitance C is
each of 1µF capacitance connected to a battery of charged to a potential V. It is then connected to
6V. Initially switch ‘S’ is closed. After sometime another uncharged capacitor having the same
‘S’ is left open and dielectric slabs of dielectric capacitance. Find out the ratio of the energy
constant K = 3 are inserted to fill completely the stored in the combined system to that stored
space between the plates of the two capacitors. initially in the single capacitor
[All INDIA 2014]
9. A capacitor ‘C’ a variable resistor ‘R’ and a bulb (ii) Calculate the potential difference between
‘B’ are connected in series to the ac mains in the plates of X and Y.
circuit as shown. The bulb glows with some (iii) Estimate the ratio of electrostatic energy
brightness. How will the glow of the bulb change stored in X and Y.
if (i) a dielectric slab is introduced between the [DELHI 2015]
plates of the capacitor, keeping resistance R to 13. The potential difference applied across a given
be the same; (ii) the resistance R is increased resistor is altered so that the heat produced per
keeping the same capacitance? second increases by a factor of 9. Bywhat factor
B
does the applied potential difference change?
[All INDIA 2017]
C R 14. Two identical parallel plate capacitors A and B
are connected to a battery of V volts with the
switch S closed. The switch is now opened and the
free space between the plates of the capacitors
is filled with a dielectric of dielectric constant K.
Find the ratio of the total electrostatic energy
Mains stored in both capacitors before and after the
introduction of the dielectric.
[DELHI 2014] S
10. Two capacitors of unknown capacitance C1and C2
are connected first in series and then in parallel
across a battery of 100 V. If the energy stored
in the two combinations is 0.045 J and 0.25 J
respectively, determine the value of C1 and C2. A B
Also calculate the charge on each capacitor in E
parallel combination.
[DELHI 2015]
3 Mark Questions [All INDIA 2017]

11. Three circuits, each consisting of a switch ‘S’ and 15. A thin conducting spherical shell of radius R
two capacitors, are initially charged, as shown has charge Q spread uniformly over its surface.
in the figure. After the switch has been closed,
Using Gauss’s law; derive an expression for an
in which circuit will the charge on the left-hand
electric field at a point outside the shell. Draw a
capacitor (i) increase, (ii) decrease and (iii)
graph of electric field E(r) with distance r from
remains same ? Give reasons.
S S S the centre of the shell for 0  r  
+ + 6Q + + 3Q + + [DELHI 2017]
6Q
2C C
3Q C C 6Q
3C C
3Q
16. Three identical capacitors C 1, C 2 and C 3 of
(a) (b) (c) capacitance 6 pF each are connected to a 12 V
[All India 2015] battery as shown.
12. Two parallel plate capacitors X and Y have the C1
same area of plates and same separation between
them. X has air between the plates while Y
contains a dielectric medium of k = 4.
X Y
+
12V C3
–
+ –
C2
15V
Find (i) charge on each capacitor (ii) equivalent
(i) Calculate capacitance of each capacitor if capacitance of the network (iii) energy stored in
equivalent capacitance of the combination the network of capacitors
is 4 µF. [DELHI 2017]
5 Mark Questions 19. A parallel-plate capacitor is charged to a

17. Draw a labeled diagram of Van de Graff potential difference V by a dc source. The
generator. State its working principle to show capacitor is then disconnected from the source.
how by introducing a small charged sphere into If the distance between the plates is doubled, a
a larger sphere, a large amount of charge can be state with reason how the following change:
transferred to the outer sphere. State the use of (i) The electric field between the plates,
this machine and also point out its limitations. (ii) Capacitance and
Or
(iii) Energy stored in the capacitor
(a) Deduce the expression for the torque acting [DELHI 2017]

on a dipole of dipole moment p in the
20. Explain the principle of a device that can build
up high voltage of the order of a few million volts.
presence of a uniform electric field E
Draw a schematic diagram and explain the
(b) Consider two hollow concentric spheres, working of this device.
S1 and S2, enclosing charges 2Q and 4Q Is there any restriction on the upper limit of the
respectively as shown in the figure high voltage set up in this machine? Explain.
(i) Find out the ratio of the electric flux through [DELHI 2012]
them.
(ii) How will the electric flux through the
Solutions
sphere S1 change if a medium of dielectric 1. In a conductor charges reside on its surface,
constant ‘εr’ is introduced in the space inside there are no charges present inside a conductor.
S1 in place of air? Deduce the necessary Hence electric field inside is zero. [1]
expression. 2. When capacitor is fully charged it maintains
4Q a constant voltage and charge will also be
constant. Since current is defined as rate of
change of charge it will be zero.
Conduction current, IC = C dV
2Q S2 dt
Since, V is constant, dV = 0 [½]

dV dt
 IC  C  0
S1 dt
 q
[DELHI 2014] d 
 o 
18. (a) Distinguish, with the help of a suitable Displacement current, ID  o
diagram, the difference in the behavior of dt
a conductor and a dielectric placed in an
external electric field. How does polarized  q
dielectric modify the original external field? Since q is constant, d    0
 q  o 
(b) A capacitor of capacitance C is charged d 
fully by connecting it to a battery of emf  o  [½]
 ID   o 0
E. It is then disconnected from the battery. dt
If the separation between the plates of
the capacitor is now doubled, how will the 3. Dielectric constant (or relative permittivity) of
following change? a dielectric is the ratio of the absolute permittivity
(i) Charge stored by the capacitor. (ε) of a medium to the absolute permittivity
(ii) Field strength between the plates. 
of free space(ε o ) K  , It is unit less
o
(iii) Energy stored by the capacitor.
Justify your answer in each case. quantity. [1]
[All INDIA 2016]
4. According to Lenz law the polarity of induced emf q1 360  10 6

is such that it opposes the cause of its production C 
V1 180
so the polarity of the capacitor is as shown
= 2 × 10–6 F = 2µF [1]
(ii) If the voltage applied had increased by 120
A V, then
S N S N
B Va = 180 + 120 = 300 V
Hence, charge stored in the capacitor,
Qa = CVa = 2 × 10–6 × 300 = 600 µ C [1]
Fig.: Lenz Law 1 1
8. Uinitial  CV 2  0  CV 2
[½] 2 2
5. In C2 : After connecting common potential
Charge Qo = CD VD C V  C2V2   V
Vcommon  1 1
C1C2 2
Where, CD = KC = increases K times
V 2 2
VD = = decreases K times [1] 1 V 1 V 1
U Final  C    C    CV 2 [1]
K 2  2  2  2  4
In C1:
Charge Qo = CD V 1 
CV 2 
U Initial  4 
Potential V remains the same as 6V  
U Final 1 2
Charge QD = KCV = KQ, increases K times [1]  CV 
2
6. 2d/3 [1]
U Initial
U =1:2 [1]
Final
9. (i) As the dielectric slab is introduced between

K the plates of the capacitor, its capacitance
will increase. Hence, the potential drop
across the capacitor will decrease (V= Q/C).
As a result, the potential drop across the
bulb will increase (since both are connected
d in series’). So, its brightness will increase.
[1]
eo A 3eo AK
C= = (ii) As the resistance (R) is increased, the
2d / 3 d / 3 d ( 2 + K ) [1]
+ potential drop across the resistor will
K 1 increase. As a result, the potential drop
7. (i) Initial voltage, V1 = V volts and charge across the bulb will decrease (since both are
stored, Q1 = 360 µC connected in series). So, its brightness will
decrease. [1]
Q1 = CV1 (1)
Charged potential, V2 = V – 120 10. When the capacitors are connected in parallel,
Q2 = 120 µC Equivalent capacitance, Cp = C1 + C2
Q2 = CV2 (2) The energy stored in the combination of the
1
By dividing (2) from (1), we get capacitors, E p = C pV 2
2
Q1 CV1 1
E p  C1  C2  100  0.25 J
= 2
Q2 CV2 2 [½]

360

V  C1  C2   5  10 5
(1)
120 V  120
When the capacitors are connected in series.
V = 180 Volts
C1C2 C1 1
Equivalent Capacitance, Cs  =
C1  C2 C2 er

The energy stored in the combination of the ⇒ C2 = er C1

1 2 C1 = C
capacitors, Es = Cs V
2
1 C1C2 C2 = 4 C ( er = 4) [½]
 Es  1002  0.045J Since two capacitance are connected in series
2 C1  C2

so, equivalent capacitance will be
1 C1C2
 1002  0.045J 1

1

1
2 5  10 5 [½] Ceq C1 C2

 C C  0.045  10 4  5  10 5  2
1 2 C1C2
Ceq 
= 4.5 × 10 –10
C1  C2

 C1  C2   C1  C2   4C1C2
2 2
C × 4C
4µF =
C + 4C
 C1  C2   7  10 10  2.64  10 5
⇒ C = 5µ F

C1 = C2 = 2.64 × 10–5
(2) [½]
So, C1 = 5µF and C2 = 20µF [½]
Solving (1) and (2), we get
(ii) Ceq Vnet = QTotal
C1 = 38.2 µF And C2 = 0.12 µF

QTotal = 60 µC
When the capacitors are connected in parallel,
Since in series configuration charge on each
the charge on each of them can be obtained as
capacitor is equal.
follows:
Hence, Q1 = Q2 = QTotal = 60C [½]
Q1 = C1 V = 382.2 × 10–6 ×100 = 38.2 10–4 C
Using Q = CV
Q2 = C2 V = 0.12 × 10–6 ×100 = 0.12 10–4 C [½]
Q 60µC
11. When charged capacitors are connected to V1 = 1 = = 12V
C1 5µF
each other then the charge will flow from the
capacitor with higher potential towards the Q2 60µ C
capacitor with lower potential untill a common V2 = = = 3V [½]
C2 20µ F
potential is reached.
2
(a) In fig. (a) the potential of both the capacitor 1 Q12 1 (60µ C)
(iii) U1 = = = 360µ J
is same so the charge on left hand capacitor 2 C1 2 5µ F
remains the same [1] 2
(b) In fig. (b) the potential of left hand 1 Q22 1 (60µ C)
U2 = = = 900µ J [½]
capacitor is high so charge from 6Q to 3Q. 2 C2 2 20µ F
Therefore charge on left hand capacitor will
U1 4
decrease. [1]  
U2 1
(c) In fig. (c) the potential of left hand capacitor
is low so charge will flow from 3Q to 6Q.
Therefore charge on left hand capacitor will V1 : V2 : : 4 : 1 [½]
increase. [1] 13. Let the heat dissipated per unit time.
12. (i) Let the capacitance of X be C1 and capacitance
of Y be C2 H
 V 2
R [1]
e A
C1 = o
d
H
122  24 J / sec
e e A 6 [1]
C2 = r o
d The new heat dissipated per unit time (H = H X
9 = 216J/Sec
Let the new voltage be V’ Q

E=
V 2
 216
εo 4π r 2 ( )
R [1]
Where r is the distance of the point from the
(V) = 216 × 6
2
centre of the shell
V = 36 volt q
14. Net capacitance before filling the gap with E=
dielectric slab is given by [½]
εo 4π r 2 ( )
CInitial = A + B ... (i) The graph of electric field E(r) with distance r
Net capacitance at here filling the gap with from the centre of the shell for 0 ≤ r ≤ ∞ [½]
dielectric slab of dielectric constant E
Cfinal = K(A + B) ... (ii) [½]
2
Energy stored by capacitor is given by U = Q
2C
Electric field
[½]
So energy stored in capacitor Combination before E=0
1
introduction of dielectric slab 2
r
Q2
Uinitial  [½]
 A  B r
R
Energy stored in combination after introduction
of dielectric slab Distance [½]
Fig. Valuation of Electric field with respect to distance
Q2
U final  16. The 12 V battery is in parallel with C1, C2 and
K  A  B
C3. C1, C2 are in series with each other while C3
Uinitial K is in parallel with the combination formed by C1
Ratio of energy stored = [½] and C2. Total voltage drop across
U final 1
C3 = 12 V
15. According to Gauss law: q3 = CV
Where, q = Charge on the capacitor C1, C2, C3 =
e E  dA = q
o ∫ 6 μF (Given in the question)
Where, q is the point charge E is electric field q3 = 6 × 12 = 72 µ C [1]
due to the point charge dA is a small area on Voltage drop across C1 and C2 combined will be
the Gaussian surface at any distance and eois 12 V.
the proportionality constant.
Let the voltage drop at C1 = V1
For a spherical shell at distance r from the point
Let the voltage drop at C2 = V2
charge, the integral ∫ dA is merely the sum of Then, V = V1 + V2
all differential of dA on the sphere. [1]
q
2 V2 = 2
Therefore, ∫ dA = 4p r C

(
εo E 4π r 2 = q ) V1 =
q1
C
or,
q1 q2
q   12
E= 6 6 [1]

(
εo 4π r 2 ) [1] As both the capacitors are in series.
Therefore, for a thin conducting spherical shell q1 = q2 = q
of radius R and charge Q. spread uniformly over Then,
its surface, the electric field at any point outside
1 1 
the shell is q     12
 6 6 
1 shell by a wire, the charge will flow to the shell

q  12 because the shell is at a lower potential. [½]
3
Application.
q = 36 micro coulombs Thus. charge on each of
is 36 coulombs. [1] It can be used to accelerate particles like protons,
deuterons, α- particles and other ions. These
17. Theory. Consider a large spherical conducting accelerated particles are called ‘’projectiles’’.
shell A having radius R and charge +Q. Potential These particles are used in nuclear physics for
1 Q collision experiments. [½]
inside the shell, V1 = , Assume that
4πεo R
Limitations of Van de Graff generator: Due
a small conducting sphere B of radius r and to very high electric field at sphere, sparking &
carrying charge +q is placed at the centre of leakage of charge takes place so high pressure
shell A, [1 + 1] gasses are used around sphere. But still leakage
takes place at higher electric field so highest
A potential is limited. [1]
Or
+Q
(a) Consider an electric dipole placed in uniform
+q
R electric field then
B E
r
+q qE
2a
Potential due to A at the surface of B
1 q
V2 = V1 = –q
4πεo Q V2 – V1
Potential due to B at the surface of B, qE
[1]
1 q
V = VB − VA = = Fig. Electric Dipole in Uniform electric field
4πεo r r

t = Force × Perpendicular distance
Total potential at the surface of shell A,
= (qE) (2a sin q)
1  Q q
VA = V1 + V2 =  +  t = PE sin q
4πεo  R r 
t = PE [1]
Total potential at the surface of B, (b) (i) According to Gauss theorem
1 Q q  ∑q ∝
VB = V1 + V3 =  + 
4πεo  R R  ϕnet =
εo εr
∑q

Potential difference, jS1 2Q 1
= =
1 Q q Q q  jS2 2Q + 4Q 3
V = VB − VA =  + − −  [2]
4πεo  R r R R 
(ii) If medium is filled in S1 then
q 1 1 
=  − . [1] ϕs1 =
∑q = 2Q
4πεo  r R  εo εr εo εr
[1]
The potential difference is independent of charge
18. (a) When conductor is placed in an external
Q on the shell. If the sphere is connected to the
electric field then induction phenomena
occur, due to which induced charge get
develop on the conductor surface.
U2 = 2U1 [1]
Energy stored in capacitor gets doubled to
– +
– its initial value.
– + E
– 19. Q = CV
E=0 +
– e A
+
– Q =  o ( Ed)
– +  d 
– +
Q = eo
Fig. Conductor in external field Q
∴E=
When dielectric is placed in an external eo A
[1]
electric field then polarization phenomena
will occur. [1] Therefore, the electric field between the parallel
plates depends only on the charge and the plate
E
area. It does not depend on the distance between
the plates.
– + – + – + Since, the charge as well as the area of the plates
– + – + – + does not change, the electric field between the
plates also does not change. [1]
– + – + – +
Let the initial capacitance be ‘C’ and the final
– + – + – + capacitance be ‘ `’. Accordingly,
e A
– + – + – + C= o
d
E
[1] eo A
Fig. Polarization And C ’ =
d
(b) After disconnection from battery and C C
= 2= or C ’
doubling the separation between two plates C ’ 2 [1]
(i) charge on capacitor remains same. Hence, the capacitance of the capacitor gets
i.e., CV = C′ V′ halved when the distance between plates is
doubled.
 C
 CV    V ’ 1 Q2
 2 (iii) Energy of a capacitor, U =

2 C
⇒ V′ = 2V [1]
Since, Q remains the same but the capacitance
(ii) Q Electric field between the plates decreases,
n’ 2n
E’ = = 1 Q2
d ’ 2d U’ 
2  C
n  
E’ = =E 2 [1]
d
U 1
⇒ Electric field between the two plates =
U’ 2
remains same. [1]
U’ = 2U
(iii) Energy stored in capacitor when
connected from battery The energy stored is in the capacitor gets doubled
when the distance between the plates is doubled.
q2 q2
U1 = = [1]
c c
2× 20. Van de Graff generator is the device used for
2
building up high voltages of the order of a few
 q2  million volts. Such high voltages are used to

U2 = 2   = 2U1 accelerate charged particles such as electrons,
 2c 
protons, ions, etc. It is based on the principle that
charge given to a hollow conductor is transferred down and collects the positive charge from B1,
to outer surface and is distributed uniformly which in turn is collected by B 2 . This is
over it. [1] repeated. Thus, the positive charge of S
Construction: goes on accumulating. In this way, potential
differences of as much as 6 or 8 million volts
C (with respect to the ground) can be built up.
+ S +
+ + The main limiting factor on the value of high
potential is the radii. If the electric field just
+ + outside the sphere is sufficient for dielectric
B2 ion
+ source breakdown, of air, no more charge can be
+ + P2 D + transferred to it. [1 + 1]
+
+ For a conducting sphere, electric field just
+ + +
+ outside sphere
+
+
+ Q
+ E=
+ 4πε R2
+
+
+ And electric potential
P1 + P2
+ Q
+ V=
+ 4πεo R
+
+
+ V
+ Thus, E =
B1
+ R
+ P1
+
+ Now, for E = 3 × 106V/m (dielectric breakdown)
H.T.
Radius of sphere should be 1 m.
Target (T) Thus, the maximum potential of a sphere of
[1]
radius 1 m would be 3 × 106V . [1]
Fig.: Van de Graff Generator
OR
It consists of large spherical conducting shell
(a) Electric Flux: It is the number of electric field
(S) supported over the insulating pillars. A long lines passing through a surface normally.
narrow belt of insulating material is wound
f = E⋅ A
around two pulleys P1 and P2, B1 and B2 are
two sharply pointed metal combs. B1 is called Where E electric field, A = Area
the spray comb and B2 is called the collecting S.I. unit of flux is Nm2C–1 [1]
comb.
++ n
Working: The spray comb is given a positive + + ++
+ + ++ E
potential by high tension source. The positive + + ++
+ ++ +
charge gets sprayed on the belt. As the + ++ +
P
belt moves and reaches the sphere, a negative ^ + + ++
E n + + ++ E
charge is induced on the sharp ends of collecting + + ++
comb B2 and an equal positive charge is + + ++
++
induced on the farther end of B2 . This positive [1]
charge shifts immediately to the outer surface
of S. Due to discharging action of sharp (b) Consider a uniformly infinite plane sheet of
points of B2, the positive charge on the belt charge density s .
is neutralized. The uncharged belt returns
We have to find electric field E at point P as It shows that electric field is uniform due to
shown in figure in the form of cylinder. charged infinite plane sheet. Also, we can
Applying Gauss’s law, say that E is independent of distance from
σ∆s the sheet.
E ⋅ ds = ε
φ=∫ σ
o (c) E = [½]
2εo
σ∆s
⇒ E∆s + E∆s + 0 = (i) Direction of field will be away from the sheet
εo
[1] if sheet is positively charged.
σ∆s
= 2E∆s = (ii) E = −σ
εo 2εo

σ Direction of field will be towards the sheet if sheet
⇒E= is negatively charged. [½]
2εo
[1]
CHAPTER 3
Current Electricity

Electric current; flow of 1Q 1Q
electric charges in a metallic (5 marks) (2 marks)
conductor; drift velocity, 1Q
mobility and their relation
with electric current (5 marks)
Ohm’s law, electrical 1Q 1Q
resistance, V-I (1 mark) (2 marks)
characteristics (linear and
non-linear)
1Q
electrical energy and power, 1Q (3 marks),
electrical resistivity and 1Q
conductivity (3 marks) (4 marks)
Carbon resistors, colour code
for carbon resistors; series
and parallel combinations
of resistors; temperature
dependence of resistance
Internal resistance of a 1Q 1Q 1Q
cell,potential difference and (2 marks) (3 marks) (2 marks)
emf of a cell, combination of
cells in series and in parallel,
measurement of internal
resistance of a cell
Kirchhoff’s laws and simple 1Q 1Q
applications; Wheatstone (5 marks) (3 marks)
bridge, metre bridge
Potentiometer - principle and 1Q 1Q
its applications to measure (5 marks) (3 marks)
potential difference and for
comparing emf of two cells
On the basis of above analysis, it can be said that from exam point of view Kirchhoff’s Rule, Drift
Velocity, Relaxation Time Meter Bridge, Potentiometer, Resistance in series and parallel, emf of a cell
and Power loss are most important concepts of the chapter. This is one of the important chapters in
Class XII Physics from exam point of view.
38 CHAPTER 3 : Current Electricity
Topic 1: Electricity conduction, Ohm’s law and resistance

Summary R=
ρl
, ρ being the resistivity of the material of the
Electric Current: Net charge flowing across a given A
area of conductor per unit time is defined as electric conductor.
current.
l
q
I= , S.I. unit of current is Ampere (A).
t A

A steady current is generated in a closed circuit
where electric charge moves from lower to higher Fig.: Resistance in a conductor
potential. Electromotive force or emf is the work Resistivity: It depends on the nature of the material
done by the source in taking the charge from higher and temperature. It is also termed as specific
to lower potential energy. resistance.
Drift velocity: The free electrons drift with some m gives the relation between resistivity and
velocity towards the positive terminal when a ρ=
ne 2 τ
potential difference is applied across the ends. The
average velocity with which the electrons move is relaxation time.
termed as drift velocity. There is an increasing order of resistivity as we go
eEτ eVτ from metal to insulator.
Drift velocity,=
vd = ρmetals < ρsemiconductors < ρinsulators
m ml
Conductivity and conductance: The reciprocal of
Where e = charge on electron resistivity is conductivity (s).
E = Electric field intensity 1
σ= and its S.I. unit is W–1m–1.
V = Potential difference across the ends of the ρ
conductor The reciprocal of resistance is the conductance of the
t = Relaxation time conductor. Its S.I. unit is mho.
m = Mass of electron Current Density: The amount of charge flowing per
Relation between current and drift velocity: unit area per second is called the current density.
Current is directly proportional to the drift velocity. J = mqvd , where vd is the drift velocity of the charge
I ∝ vd carriers, n is the number of charge carriers and q is
When the number of electrons are less, current is less the charge.
so the drift velocity is small. The relation between current density and conductivity
When the number of electrons are large, high current is
flows so the drift velocity is large. J = sE
Ohm’s law: The voltage across the ends of the Mobility: Mobility is the ratio of drift velocity to the
conductor is directly proportional to the electric applied electric field. Mobility is symbolized by m.
current flowing through the conductor.
v d qτ
V ∝I =
µ =
E m
Or V = IR, where R is the electrical resistance of the
conductor Its S.I. unit is m2s–1V–1.
Resistance: The property that resists the flow of Resistors: The objects which resist the flow of charge
current through any conductor is called the resistance are called resistors which can be of two types, i.e. wire
of the conductor. bound resistors and carbon resistors.
V Resistors can combine in two different ways; either in
R= series or in parallel.
I
• Consider n number of resistors connected in series,
It varies directly with the length of the conductor
then the combined resistance will be as follows:
while depends inversely on the area of cross section
of the conductor. R eqv = R1 + R 2 + R 3 + ....... + R n

CHAPTER 3 : Current Electricity 39
Same amount of current will flow through each resistor Internal resistance: It is the resistance on the cur-
connected in series while the potential difference rent offered by the electrolyte and the electrodes. It
would be different for every resistor. is symbolize by r.
• Consider n number of resistors connected in Let us assume a cell with 2 electrodes connected by
parallel, then the combined resistance will be as ε
an external resistance R. Then current is, I =
follows: R+r
1 1 1 1
R eqv = + + + ...... + where e = emf, r = Internal resistance
R1 R 2 R 3 Rn

The current flowing through each resistor would be
different in this case while the potential difference
would be same for all the resistors.
7. Define the term ‘electrical conductivity’ of a
PREVIOUS YEARS’ metallic wire. Write its S.I. unit.

[DELHI 2014]
EXAMINATION QUESTIONS 8. Show variation of resistivity of copper as a

function of temperature in a graph.
1 Mark Questions 9. Graph showing the variation of current versus

1. When electrons drift in a metal from lower to voltage for a material GaAs is shown in the
higher potential, does it mean that all the free figure, Identify the region of:
electrons of the metal are moving in the same (i) Negative resistance
direction? (ii) Where Ohm’s law is obeyed
[ALL INDIA 2012]
2. Show on a graph, the variation of resistivity with
temperature for a typical semiconductor.
Current I
[ALL INDIA 2012] C D

3. The graph shown in the figure represents
a plot of current versus voltage for a given
semiconductor. Identify the region, if any, E
B
over which the semiconductor has a negative
resistance.
A
Voltage V
Current (mA)
B
[DELHI 2015]
C 10. V-I graph for a metallic wire at two different
A temperature T1 and T2 is as shown in the figure.
Which of the two temperatures is higher and
O Voltage (V) why ?
[ALL INDIA 2013] T1
4. Define the term ‘Mobility’ of charge carriers in
a conductor. Write its S.I. unit. V
[DELHI 2014] T2
5. Plot a graph showing variation of current versus θ1
voltage for the material Ge. θ2
[DELHI 2014]
6. Define the term ‘drift velocity’ of charge carriers I
in a conductor and write its relationship with [ALL INDIA 2015]
the current flowing through it.
[DELHI 2014]
11. Nichrome and copper wires of same length and 18. Define the terms (i) drift velocity, (ii) relaxation
same radius are connected in series. Current I time. A conductor of length L is connected to a
is passed through them. Which wire gets heated dc source of emf 8. If this conductor is replaced
up more? Justify your answer. by another conductor of same material and same
area of cross-section but of length 3L, how will
[ALL INDIA 2017]
the drift velocity change?
2 Marks Questions [ALL INDIA 2011]
12. A wire of resistance 8R is bent in the form of a 19. Derive an expression for drift velocity of free
circle. What is the effective resistance between electrons in a conductor in terms of relaxation
the ends of diameter AB? time.
[DELHI 2017]
20. (a) Define the term ‘conductivity’ of a metallic
wire. Write its SI unit.
A B
(b) Using the concept of free electrons in a
conductor, derive the expression for the
conductivity of a wire in terms of number
density and relaxation time. Hence obtain
the relation between current density and
[DELHI 2018]
the applied electric field E.
13. Explain the term ‘drift velocity’ of electrons in a [ALL INDIA 20168]
conductor. Hence obtain the expression for the
current through a conductor in terms of ‘drift 5 Marks Questions
velocity’. [ALL INDIA 2013]
21. (i) Define the term drift velocity.
14. Estimate the average drift speed of conduction (ii) On the basis of electron drift, derive an
electrons in a copper wire of cross-sectional expression for resistivity of a conductor in
area 2.5 × 10–7 m2 carrying a current of 2.7 A. terms of number density of free electrons
Assume the density of conduction electrons to be and relaxation time. On what factors does
9 × 1028 m–3. resistivity of a conductor depend?
[ALL INDIA 2014]
(iii) Why alloys like constantan and manganin
15. When 5V potential difference is applied across a are used for making standard resistors?
wire of length 0.1 m, the drift speed of electrons [DELHI 2016]
is 2.5 × 10–4 m/s. If the electron density in the
wire is 8 × 1028 m–3, calculate the resistivity of Solutions
the material of wire.
[ALL INDIA 2018] 1. No, when electric field is applied the electrons
will have net drift from lower to higher field but
3 Marks Questions locally electrons may collide with ions and may
16. Define the terms change its direction of motion. [1]
(i) drift velocity, (ii) relaxation time. 2. The following curve shows the variation of
A conductor of length L is connected to a source resistivity with temperature for a typical
of emf e. If this conductor is replaced by another semiconductor.
conductor of same material and same area of
cross-section but of length 3L, how will the drift
velocity change?
[ALL INDIA 2011]
17. A potential difference V is applied across a ρ
conductor of length L and diameter D. How is
the drift velocity, v1 of charge carriers in the
conductor affected when (i) V is halved, (ii) L
is doubled and (iii) D is halved? Justify your T

answer in each case.
[ALL INDIA 2015] Fig.: Variation of resistivity with respect to
temperature. [½]
This is because for a typical semiconductor, Or, the reciprocal of resistivity of a material is
resistivity decreases rapidly with increasing called its electrical conductivity.
temperature. [½]
1 1
3. From ohm’s law: Conductivity = or σ = [½]
resistivity ρ
In the region BC, DV is positive and DI is
negative, hence resistance is negative. [1] 8. The variation of resistivity of copper with
temperature is parabolic in nature. This is
4. Mobility of charge carriers in a conductor is shown in the following graph:
defined as the magnitude of their drift velocity
per unit applied electric field. [½]
Mobility, µ → Drift of electric field
Vd
µ=
Resistivity
E [½]
S.I. unit of mobility is m2V–1 s–1 or ms–1N–1C.
5. Current—Voltage characteristics graph for :
+I (mA) O Temperature
Fig.: Variation of resistivity of Cu with Temperature [1]

Reverse
Bias Forward 9. (i) DE is the region of negative resistance
Zener voltage
Bias because the slope of curve in this part is
negative. [½]
–V (in V) +V (in V) (ii) BC is the region where Ohms law is obeyed
Knee because in this part, the current varies
voltage
linearly with the voltage. [½]
10. The slope of V-I graph gives the resistance of
–I (mA) the metallic wire and the slope is higher at
temperature T1 and we know that on increasing

the temperature of metallic wire resistance of the
Fig.: V-I Characteristics of material Ge [1]
wire increases, so T1 temperature is higher. [½]
6. The net speed achieved by an electron due to
11. Heat dissipated in a wire is
a current carrying conductor is called as drift
velocity. [1] H = I2RT
i = neAV ρl  ρl 
H = I2 ×t ∴ R = 
n is number of electron per unit volume A A [½]
e is charge per unit electron For same current “I” length (l) and Area (A), H
depends on  and t.
A be cross sectional area
rnichrome > rcopper.
V be drift velocity
So, HNichrome > HCopper. [½]
7. The electric conductivity of a metallic wire is
defined as the ratio of the current density to the 12. When it is bent to form circle the ends of
electric field it creates electrical conductivity. diameter separates the circle into semicircles
in such a way that the two semicircles make a
J
σ = , S.I. Unit = ohm m–1. [½] parallel connection in the circuit. Also both the
E semicircles have equal circumference therefore
they have equal resistance. We get 4R as the
resistances of the semi circles {8R = 4R + 4R}
[1] V
= ρ neν d [1]
1 1 1 l
Therefore 1/effective resistance = + = ’
4 4 2 V
=ρ
Effective resistance [1] nelν d

13. Drift velocity: Drift velocity is defined as the
5
average velocity with which free electrons in a = [1]
conductor get drifted in a direction opposite to 0.1 × 8 × 10 28
× 1.6 × 10 −19 × 2.5 × 10 −4
the direction of the applied electric field. Let n
be the number of free electrons per unit volume = 1.56 × 10–5 Ωm
of the conductor. Then, total number of free
16. The average velocity of electrons, independent of
electrons in the conductor = n × volume of the
time, although accelerated through a conductor
conductor = n × Al [1]
is called drift velocity. [1]
E The average time interval between successive

collisions is called relaxation time. [1]
A
Vd Drift velocity
l eτ E eτ V
Vd = = [1]
I m mL
V
Where V is the potential difference applied across
the length of the conductor.
If e is the magnitude of charge on each electron,
then the total charge in the conductor, Keeping V constant, if Length L of the conductor
is made 3 times, the drift velocity will become
Q = (nAl)e ------- (1) 1/3rd.
The time taken by the charge to cross the 17. We know drift velocity is given by
conductor length is given by
eτ
1 νd = m E [1]
t=
νd
V
Also, E =
Where vd is the drift velocity of the electrons. L
eτ  V 
So, ν d =   [1]
According to the definition of electric current, m  L
Q nAle
I= = = neAν d (i) When V is halved drift velocity (vd) gets

t l /ν d halved
14. I = ne Avd [1] (ii) When L is doubled drift velocity (vd) gets
halved
1 2.7
∴ Vd = = (iii) When D is halved drift velocity (vd) remains
neA 9 × 10 × 1.6 × 10 −19 × 2.5 × 10 −7
28
same. [1]
= 7.5 × 10–4 m/s [1] 18. (i) Drift velocity: The average velocity with
15. Given p.dV = 5V which the free electrons drift towards
positive terminal under the influence of an
l = 0.1 m external field is called drift velocity. [1]
vd = 0.1 m (ii) Relaxation time: Average time interval
between two successive collisions of an
n = 8 × 1028
electron with the ions / atoms of the
E = rJ conductor. [1]
The drift velocity will be inversely proportional Then,

1
to L, i.e. ν d ∝ hence it will become one third − eE
L νd = τ
m [1]
of its initial value.
Negative sign shows that electrons drift opposite
ν to the applied field.
νd = d
3 20. (a) Conductivity
19. If there are N electrons and the velocity of the
electron at a given time is v, A
1 N
where, i = (1, 2, 3, ... N), then
N
∑ i
ν =0
=1 i
l
(If there is no external field) l

R= ρ
A
When an external electric field is present, the
electrons will be accelerated due to this field by RA

ρ=
− eE l
a=
m 1 l
=
ρ R⋅ A [1]
Where, – e = Negative charge of the electron
E = External field [1] Conductivity of a wire is defined as reciprocal

of resistivity. S.I. unit mho / m
m = Mass of an electron Let vi be the velocity
immediately after the last collision after which (b) As we know that
external field was experienced by the electron.
If vi is the velocity at any time t then from the ρl
R=
equation V = u + at, we obtain A

V = IR
V = ν − − eE t ……………..(i) [1]
i i
m Iρl
V=
A
For all the electrons in the conductor, average
value of vi is zero. The average of vi is vd or drift
I
velocity. = j= current density
A
This is the average velocity experienced by an
electron in an external electric field. There is
V = j rl
no fixed time after which each collision occurs.
Therefore, we take the average time after which V=El
one collision takes place by an electron.
E, l = j rl
Let this time, also known as relaxation time t
be. E = j r
Substituting this in equation (i) E

j= [1]
ρ
νi = 0
V = u + at
t=t
F
a=
νi = ν d m
− eE V
a= Using, E = −
m I
 eV 
I ⋅ ∆t = neA ( Vd ) ∆t I = neA  τ
 ml 

i
j= ∆t  ne2 Aτ  1
A = V = V [1]
 ml  R
eE
j = neA
m I ∝ V → Ohm’s Law
ml
ne2 E Where, R = is a constant for a
j= [1] nAe2τ
m
21. (i) Drift velocity is defined as the average particular conductor at a particular
velocity which the electrons are drifted temperature and is called the resistance of
towards the positive terminal under the the conductor.
effect of applied electric field. Thermal
velocities are randomly distributed and  m  l ρl
R=  =
2
 ne τ  A A
average thermal velocity is zero. [1]

u1 + u2 + ........ + uN  m 
= 0 [1] ρ=
N  ne2τ 

eEτ Where P is the specific resistance or
vd = −
m resistivity of the material of the wire. It
depends on number of free electron per unit
(ii) We know that the current flowing through volume and temperature.
the conductor is: (iii) They are used to make standard resistors
because: [1]
I = Aevd (a) They have high value of resistivity
(b) Temperature coefficient of resistance is

 eEτ  [1] less.
I = neA  − 
 m
(c) They are least affected by temperature.
E
A
+ –
V
Topic 2: Kirchhoff’s Laws, cells and their combinations

Summary Kirchhoff’s law:
• Junction Rule: The sum of currents entering a
Cells in series and in parallel junction would be equal to the sum of currents
• The equivalent emf of a series combination of n leaving the junction.
cells is just the sum of their individual emfs • Loop Rule: The sum of changes in potential
• The equivalent internal resistance of a series around any loop that is closed should be zero.
combination of n cells is the sum of their internal Wheatstone bridge: It is an arrangement of four
resistances. resistors in a way so that a galvanometer is placed
ε1 ε2 between the two opposite arms.
I
There is a null-point condition in the wheat
A r1 B C stone bridge where current is zero which can be
r2
represented as follows:
e = e1 + e2 R1 R 3
=
R2 R4

• In a parallel connection,
B
1 1 1 ε eq ε1 ε I4 U
= + ...... + and = + ...... + n nk
req r1 rn req r1 rn no
w
I2 R2 R4 n
ε1 A C
I1 I1 I1 R1 R3
r1
m rd
ar nda
A B1 ε2 B2 I C
a
I2
St
I2 I3
r2 D
ε
Fig.: Wheastone bridge
3. Two identical cells, each of emf E, having
PREVIOUS YEARS’ negligible internal resistance, are connected

in parallel with each other across an external
EXAMINATION QUESTIONS resistance R. What is the current through this
resistance?
TOPIC 2 [ALL INDIA 2013]
1 Mark Questions 2 Marks Questions
1. A cell of emf E and internal distance r draws a 4. A cell of emf E and internal resistance r is
current ‘I’. Write the relation between terminal connected to two external resistances R1 and
voltage ‘V’ in terms of E, I, r. and R2 a perfect ammeter. The current in the
[DELHI 2013] circuit is measured in four different situations:
2. A heating element is marked 210 V, 630 W. What (i) without any external resistance in the circuit
is the value of current drawn by the element
(ii) with resistance R1 only
when connected to a 210 V, dc source?
(iii) with R1 and R2 in series combination
[DELHI 2013]
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42 11. An ammeter of resistance 0.80 Ω can measure
A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in current up to 1.0 A.
that order. Identify the currents corresponding (i) What must be the value of shunt resistance
to the four cases mentioned above. to enable the ammeter to measure current
[ALL INDIA 2012] up to 5.0 A?
5. An ammeter of resistance 1 Ω can measure current (ii) What is the combined resistance of the
ammeter and the shunt?
up to 1.0 A. (i) What must be the value of the shunt
resistance to enable the ammeter to measure up [DELHI 2013]
to 5.0 A ? (ii) What is the combined resistance of 12. In the given circuit diagram a voltmeter ‘V’ is
the ammeter and the shunt? connected across a lamp ‘L’. How would (i) the
brightness of the lamp and (ii) voltmeter reading
[DELHI 2013]
‘V’ be affected, if the value of resistance ‘R’ is
6. A 10 V battery of negligible internal resistance is decreased? Justify your answer.
connected across a 200 V battery and a resistance
of 38Ω as shown in the figure. Find the value of
the current in circuit. V L +
10V
9V
38Ω R
200V
[DELHI 2013]
7. The emf of a cell is always greater than its terminal
[DELHI 2013]
voltage. Why?
13. A cell of emf ‘E’ and internal resistance ‘r’ is
[DELHI 2013]
connected across a variable resistor ‘R’. Plot a
8. A cell of emf ‘E’ and internal resistance is graph showing variation of terminal voltage ‘V’
connected across a variable resistor ‘R’. Plot a of the cell versus the current ‘I’. Using the plot,
graph showing the variation of terminal potential show how the emf of the cell and its internal
‘V’ with resistance ‘R’. Predict from the graph the resistance can be determined.
condition under which ‘V’ becomes equal to ‘E’. [ALL INDIA 2014]
[DELHI 2017] 14. Two cell of emfs 1.5 V and 2.0 V having internal
9. A variable resistor R is connected across a cell of resistances 0.2 Ω and 0.3 Ω respectively are
emf ε and internal resistance r as shown in the connected in parallel. Calculate the emf and
figure. Draw a plot showing the variation of: internal resistance of the equivalent cell.
(i) Terminal voltage V [DELHI 2016]
(ii) the current I, as a function of R. 15. A 10 V cell of negligible internal resistance is

R connected in parallel across a battery of emf
200V and internal resistance 38Ω as shown
in the figure. Find the value of current in the
circuit. [ALL INDIA 2018]
I I 10V
+ –
ε, r
[ALL INDIA 2015]
10. State Kirchhoff’s rules. Explain briefly how these
rules are justified. 200V 38Ω
[DELHI 2014]
3 Marks Questions 20. Using Kirchhoff’s rule, determine the value of

16. A cell of emf ‘E’ and internal resistance ‘r’ is unknown resistance R in the circuit so that no
connected across a variable load resistor R. current flows through 4Ω resistance. Also, find
Draw the plots of the terminal voltage V versus the potential difference between A and D.
1Ω
(i) R and (ii) the current i. It is found that when F
E
D
R = 4Ω, the current is 1A, when R is increased to
9Ω, the current reduces to 0.5 A. Find the values 1Ω 4Ω R
of the emf E and internal resistance r.
[DELHI 2016]
I 6V
17. Two identical cells of emf 1.5 V each joined in
parallel supply energy to an external circuit
A B C
consisting of two resistances of 7Ω each joined 9V 3V
in parallel. A very high resistance voltmeter [ALL INDIA 2012]
reads the terminal voltage of cells to be 1.4 V. 21. In the figure shown, an ammeter ‘A’ and a
Calculate the internal resistance of each cell. resistor of 4 Ω are connected to the terminals of
the source. The emf of the source is 12 V having
7Ω an internal resistance of 2 Ω. Calculate the
voltmeter and ammeter reading.
V
7Ω
1.5V 12 V 2 Ω
A r B
1.5V
R=4Ω
r
[ALL INDIA 2017]
V 22. In the circuit shown, R1 = 4 Ω, R2 = R3 = 15 Ω,
[ALL INDIA 2016] R4 = 30Ω and E = 10 V. Calculate the equivalent
18. A cell of emf E and internal resistance r is resistance of the circuit and the current in each
connected to two external resistances R1 and resistor. A
R2 and a perfect ammeter. The current in the
circuit is measured in four different situations: I1 R1 I4 I3
I2
(i) without any external resistance in the circuit
E
(ii) with resistance R1 only
R2 R4 R3
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in a parallel combination
l R
The currents measured in the four cases are 0.42
A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in [ALL INDIA 2011]
the order. Identify the currents corresponding to 5 Marks Questions
the four cases mentioned above. 23. (a) State Kirchhoff’s rules for an electric
[DELHI 2014]
network. Using Kirchhoff’s rules, obtain the
19. In the circuit shown, R1 = 4 R2 = R3 = 15, R4 = 30
balance condition in terms of the resistance
and E = 10V. Calculate the equivalent resistance
of four arms of Wheatstone bridge.
of the circuit and the current in each resistor.
A [DELHI 2013]
I1 R1 I2
I4 I3 Solutions
E 1. When the current I draws from a cell of emf E
I R2 R4 and internal resistance r, then the terminal
R3
voltage is V = E – IR [1]
l 2. In dc source, P = VI
R
[ALL INDIA 2011] Given that, P = 630 Wand V = 210 V
P 630 4x = 1
So, =
I = = 3A [1] x = 0.25
V 210
Thus the shunt resistance is 0.25 Ω
3. [1] (ii) Combined resistance of the ammeter and
the shunt,
R
ε x
R=
1+ x
ε i 0.25
R=
1 + 0.25
i=ε
R 0.25
4. The current relating to corresponding situations R=
1.25
is as follows:
= 0.2 Ω [1]
(i) Without any external resistance in the
6. Since, the positive terminal of the batteries are
circuit: connected together, so the equivalent emf of the
E batteries is given by
I1 =
r E = 200–10= 190 V [1]
The current in this case would be maximum. E 190
=I =
R 38 [1]
So I1 = 4.2A [½]
Hence, the current in the circuit is given by
(ii) With resistance R1 only:
I = 5A
E 7. The emf of a cell is greater than its terminal
I2 =
r + R1 voltage because there is some potential drop across

The current in this case will be second the cell due to its small internal resistance. [2]
smallest value, so I2 = 1.05A [½] 8. V becomes equal to E when no current flows
(iii) With R1 and R2 in series combination through the circuit. [1]
E Y
I3 =
r + ( R1 + R2 ) E

The current in this case will be minimum
as the resistance will be maximum, so I3 =
0.42A [½] V
(iv) With R1 and R2 in parallel combination
E R X
I4 =
 R R  The condition under which V will be equal to E
r+ 1 2 
 R1 + R2  is E = ∞ [1]

The current in this case would be the second 9. (i) Terminal voltage across a cell as a function
largest value so I3 = 0.42A [½] of R [1]
5. We have,
As resistance R increases current (I) in the
Resistance of ammeter, RA = 1 Ω
circuit decreases and terminal voltage (V)
Maximum current across ammeter, IA = 1.0A increases. We know, V = ε–Ir, Where ε is emf
So, voltage across ammeter, of the cell.
V = IR = 1.0 × 1.0 = 1V
(i) Resistance of ammeter with shunt, (ii) Current I as a function of R.
RA x x The current across a cell is given by
R= =
RA + x 1 + x
ε
Current through ammeter, I = 5A I=
 x  R+r [1]
Now,  × 5 = 1 .0 [1]
 1 + x 
x × 5 = 1 + x
I (A) For the closed loop BACB:

ε E1 − E2 = I1 R1 + I2 R2 − I3 R3
r
For the closed loop CADC:
E2 = I3 R3 + I4 R4 + I5 R5
This law is based on the law of conservation of
R (Ω) energy.
Fig.: I versus R 11. We have, resistance of ammeter, RA = 0.80 ohm

Maximum current across ammeter, IA = 1.0 A.
10. Kirchhoff s first Law—Junction Rule: In an
So, voltage across ammeter, V = IR = 1 × 0.80 =
electrical circuit, the algebraic sum of the
0.8 V
currents meeting at a junction is always zero.
Let the value of shunt be x.
(i) Resistance of ammeter with shunt,
I1 RA x 0.8 x
I2 R= =
R A + x 0.8 + x

Current through ammeter, I = 5A
 0.8 x 
∴ × 5 = 0.8
I3  0.8 + x 
[1]
I4
[1] ⇒ 0.8x × 5 = 0.8 (0.8 + x)
I1, I2, I3 and I4 are the currents flowing through 4x = 0.64 + 0.8x
the respective wires. 0.64
x=
Convention: The current flowing towards the 3 .2
junction is taken as positive. The current flowing x = 0.2
away from the junction is taken as negative.
Thus, the shunt resistance is 0.2 Ω
I3 + (–I1) + (–I2) + (–I4) = 0 (ii) Combined resistance of the ammeter and
This law is based on the law of conservation of the shunt,
charge. 0 .8 0.8 × 0.2 0.16
R= = =
Kirchhoff’s Second Law — Loop rule: In a closed 0.8 + x 0.8 + 0.2 1 [1]
loop, the algebraic sum of the emf ‘s is equal
R = 0.16 Ω
to the algebraic sum of the products of the
resistances and the currents flowing through 12. The given figure is Common Emitter (CE)
them. configuration of an n-p-n transistor is shown in
E1 the figure. The input circuit is forward biased
I1 B and collector circuit is reverse biased. [½]
R1 If resistance R decreases, forward biased in the
R2
input circuit will increase, thus the base current
(IB) will decrease and the emitter current (IE)
R3 I2 will increase. This will increase the collector
E2
A ε C current (IC) as IE = IB + IC. [1]
I4 I3 When an IC increase which flows through the
lamp, the voltage across the bulb will also
increase making the lamp brighter and the
R4 R5
voltmeter is-connected in parallel with the lamp,
I5 the reading in the voltmeter will also increase.
D [1] [½]
13. Plot between V and I is straight line of positive 0.06

intercept and negative slope [½] =
0 .5
V
req = 0.12 Ω
38Ω
15. [1]
190
I
[½]
E = 200 – 10 = 190 V
(i) value of potential difference corresponding
V 190
to zero current gives emf of cell [½] =
I = = 20 A
P 30 [1]
(ii) Maximum current is drawn when terminal
OR
voltage is zero so
Answer: Potentiometer at open circuit, l1 = 350
V = E – Ir
R = 9 [1]
0 = E – Imax. R
l2 = 300
E
⇒r=  350 
Imax r = 9 − 1
[½]  300 
[1]
14. Given, E1 = 1.5V, R1 = 0.2 Ω r = 1.5 Ω
= E2 = 2V, R2 = 0.3 Ω [½] 16. (i) Graph between terminal voltage (V) and
1.5V 0.2 resistance (R)
V (in volts)
A B
[½]
2V 0.3
E1 E2 [1]
+ R (Ω)
r1 r2
Equivalent emf =
1 1 (ii) Graph between terminal voltage (V) and
+
r1 r2 current (i)
=
(1.5 × 0.3) + (2 × 0.2)
0.2 + 0.3 [½]
0.45 + 0.4
=
V (in volts)
0 .5
0.85
= = 1.7volt
0.5
r1 r2 [1]
Equivalent internal resistance =
r1 + r2 I (A)

0 .2 × 0 .3 (iii) When R = 4 Ω and I = 1A
= Terminal voltage, V = E – Ir
.2 + 0.3
0 [½]
So, V = IR = r = E – Ir
E – r = 4 (1)
When R = 4 Ω and I = 0.5A E

I1 = [½]
V = IR = 0.5 × 9 = E – 0.5r r
E – 0.5r = 4.5
The current in this case would be maximum.
Subtracting (1) from (2) we get, So,
E – 0.5r = – E + r = 4.5 – 4 I1 = 4.2 A
0.5r = 0.5 (ii) with resistance R1 only:
r=1Ω E
I2 = [½]
Substituting value of r in (1) r + R1
E – 1 = 4
E = 5V The current in this case will be the second
smallest value. So, I2 = 1.05 A
Thus
(iii) with R1 and R2 in series combination:
r = 1 Ω and E = 5V [1]
E
E1 r2 + E2 r1 I3 =
17. Eeq of cells across AB, Eeq = r + ( R1 + R2 ) [1]
r1 + r2
E1 + E2 = 1.5V
The current in this case will be minimum
1.5r + 1.5r
Eeq = as the resistance will be maximum. So, I3
2r = 0.42 A
Eeq = 1.5V [1] (iv) with R1 and R2 in a parallel combination:
Equivalent resistance E
I4 =
r1 r2 r 2
r  R R 
req = ⇒ req = ⇒ req = r+ 1 2  [1]
r1 + r2 2r 2  R1 + R2 

Equivalent circuit The current in this case would be the second
7Ω largest value. So, I4 = 1.4 A
19. R3, R4 and R2:Parallel
Upon calculation comes out to be [1]
7Ω
Rp = 6 Ω
i R1 and Rp are in series.
1.5V r/2
Req = 6 + 4 = 10 Ω
Using Kirchoff’s rules we get the following
equations,
V E – I1 R1 – I2R2 = 0
1 .5 3 – I4 R4 + I2R2 = 0
i= ⇒i=
7 r 7+r E – I1 R1 – I3R3 = 0
+
2 2 [1] – I3 R3 + I2R2 = 0
Reading of voltmeter V = iR
I1 = I2 + I3 + I4 [1]
 7
1 .4 = i   Solving the above equation we get
 2
I1 =1A
 3 7 10
  = 1.4 ⇒ 7 + r = 7.5 I2 = A
7 + r 2
25
r = 0.5 Ω [1]
10
18. The current relating to corresponding situation I3 = A
25
are as follows:
(i) without any external resistance in the 5
circuit: I4 = 25 A [1]
Internal resistance (r) = 2Ω

20.
1Ω C I–I
B
1
D Exterrnal resistance (R) = 4Ω
I1
1Ω 4Ω RΩ F 12 12
Then current (I) = = = 2A [1]
R+r 4+2 6
6V
I So reading of ammeter will be 2A
A 9V E
F 3V We know V = E – Ir
Apply Kirchhoff’s law in loop ABCFA: ⇒ V = E – Ir
I + I + 4I1 = 9 – 6 ⇒ V = 12 – 2 × 2
2I + 4I = 3 ----- (1) ⇒ V = 12 – 4 = 8V [1]
As there is no current flowing through the 4Ω So the reading in voltmeter will be 8V.
resistance,
I1 = 0
22. R2, R3, R4 are in parallel combination
Or 2I = 3
1 1 1 1
= + +
I = 1.5A R234 R2 R3 R4

Thus current through resistance R is 1.5A.
1 1 1
As there is no current through branch CF, so = + +
equivalent circuit will be, 15 15 30
1Ω 2+ 2+1
D =
30
1Ω RΩ
5
=
30
R234 6Ω [1]
A
9V 3 V
Now R234 is in series with R1, so,

Req = 4Ω + 6 Ω = 10 Ω
By applying Kirchhof’s loop law we get,
E
1.5 + 1.5 + R (1.5) = – 3 [1] ∴I =
Req

R = 2Ω Potential Difference between A and D
10
21. = = A 1A
10
V
\ Ii = 1A [1]
12 V 2Ω Current through R1 = 1A
A
P.D. across R1, V = IR1 = 1 × 4 = 4V
R = 4Ω [1] So, P.D. across R234 = 6V
Emf (E) = 12V \ I2 R2 = I3 R3 = I4 R4 = 6V

6 E1 – E2 = I1 R2 – I3 R3
=I2 = A 0 .4 A
15 For closed part CADC,
6 E2 = I3 R3 + I4 R4 + I5 R5 [1]
=I3 = A 0 .4 A
15
Wheatstone Bridge is an arrangement of four
6 resistances as shown in the following figure.
=I4 = A 0.2 A
30 [1] B
I4
23. (a) Kirchhoff’s First Law – Junction Rule: The U
nk
algebraic sum of the currents meeting at a no
I2 R2 R4 w
point in an electrical circuit is always zero. n
A G C
I1 R1 R3
d
d ar
I1 an rm
I2
St a
I3
D
I3
V [1]
I4
Let the currents be I1, I2, I3, and I4 R1, R2, R3 and R4 are the four resistances.
Convention: Current towards the junction - Galvanometer (G) has a current I g flowing
positive [1] through it at balanced condition, Ig = 0
Current away from the junction – negative Applying junction rule at B,
I3 + (– I1) + (– I2) + (– I4) = 0 \ I2 = I4
Kirchhoff’s Second Law – Loop Rule: In a closed Applying loop rule to closed loop ADBA,
loop, the algebraic sum of the emfs is equal to the
–I1 R1 + 0 + I2 R2 = 0
algebraic sum of the products of the resistance
and current flowing through them. I1 R2
∴ =
E1 I2 R1 (1)
I1 B
Applying loop rule to closed loop CBDC,
R1 R2 I2 R4 + 0 – I1 R3 = 0
=∵ I3 I=
1 , I4 I2
I2 [1]
E2 R3
A C I1 R4
I3 ∴ =
I4 I2 R3
I5 (2)
From equations (1) and (2),

R4 R5
R2 R4
I5 =
R1 R3
D
For closed part BACB, This is the required balanced condition of

Wheatstone Bridge.
Topic 3: Electrical devices

Summary Potentiometer: It is a device which is used to
compare potential differences and emf’”s. It also
Meter Bridge: Meter Bridge is the simplest form measures the internal resistance of a cell.
of the Wheatstone bridge which is used for accurate ε1 1
comparison of resistances. 3
In order to find out an unknown resistance R with the ε2
2
help of a standard known resistance S: B A
R S
B G
N1
R
G
A l1 100 - l1 C
D C N2
Metre scale K1
Fig.: Potentiometer
ε K1 ε1 l
= 1
ε 2 l2
Fig.: Meter bridge
l1 Potentiometer does not draw any current from
R=S , l1 being the distance of the jockey from
100 − l1 the voltage source being measured. The internal
resistance of a given cell can be measured by:
end A at the balance point. l 
=r R  1 − 1
 l2 
4. Two electric bulbs P and Q have their resistances
PREVIOUS YEARS’ in the ratio of 1:2. They are connected in series

across a battery. Find the ratio of the power
EXAMINATION QUESTIONS dissipation in these bulbs.
[DELHI 2018]
TOPIC 3 5. In a potentiometer arrangement for determining
1 Mark Questions the emf of a cell, the balance point of the cell in
1. A resistance R is connected across a cell of emf ε open circuit is 350cm.When a resistance of 9Ω
and internal resistance r. A potentiometer now is used in the external circuit of the cell, the
measures the potential difference between the balance point shifts to 300 cm. Determine the
terminals of the cell as V. Write the expression internal resistance of the cell.
for ‘r’ in terms of ε, V and R.
[ALL INDIA 2018]
[ALL INDIA 2011]
2 Marks Questions 3 Marks Questions

2. Use Kirchhoff’s rules to obtain conditions for the 6. A potentiometer wire of length 1.0 m has a
balance condition in a Wheatstone bridge. resistance of 15 Ω. It is connected. o a 5V battery
[DELHI 2015] in series with a resistance of 5 Ω. Determine the
3. Describe briefly, with the help of a circuit diagram, emf of the primary cell which gives a balance
how a potentiometer is used to determine the
point at 60 cm.
internal resistance of a cell.
[ALL INDIA 2013] [DELHI 2016]
7. Answer the following: 10. In a meter bridge, the null point is found at a
(a) Why are the connections between the distance of I1 cm from A. If now a resistance of
resistors in a meter bridge made of thick X is connected in parallel with S, the null point
copper strips ? occurs at I2 cm. Obtain a formula for X in terms
of I1, I2 and S.
(b) Why is it generally preferred to obtain the X
balance point in the middle of the meter
bridge wire? R
S
(c) Which material is used for the meter bridge
wire and why? G
Or A B
A resistance of R Ω draws current from a
potentiometer as shown in the figure. The
potentiometer has a total resistance RoΩ. A
voltage V is supplied to the potentiometer.
Derive an expression for the voltage across R
[DELHI 2018]
when the sliding contact is in the middle of the
11. (a) Write the principle of working of a metre
potentiometer.
V bridge.
(b) In a metre bridge, the balance point is
A R0 formed at a distance I1 unit with resistances
C R and S as shown in the figure
R S
R
[ALL INDIA 2014] G

8. A potentiometer wire of length 1 m has a l1
resistance of 10Ω. It is connected to a 6 V battery B
in series with a resistance of 5Ω. Determine the
emf of the primary cell which gives a balance
An unknown resistance X is now connected in
point at 40 cm
parallel to the resistance S and the balance point
[DELHI 2014]
9. In the figure a long uniform potentiometer wire is found at a distance I2. Obtain a formula for X
AB is having a constant potential gradient along in terms of I1, I2 and S. [ALL INDIA 2017]
its length. The null points for the two primary 5 Marks Questions
cells of emf’s e1 and e2 connected in the manner X Y
shown are obtained at a distance of 120 cm and
300 cm from the end A. Find B
ε
(i) 1 and G
ε2
(ii) position of null point for the cell ε1.
How is the sensitivity of a potentiometer A D C
increased?
12.
300 cm The figure shows experimental set up of a meter
bridge. When the two unknown resistances X
120 cm
and Y are inserted, the null point D is obtained
A B 40 cm from the end A. When a resistance of 10 0
ε1 ε2
is connected in series with X. the null point shifts
ε 1 ε2 by 10 cm. Find the position of the null point
[All India 2012] when the 10 0 resistance is instead connected in
series with resistance ‘Y’. Determine the values
of the resistances X and Y.
[DELHI 2017]
13. (a) State the working principle of a potentiometer. Solutions

With the help of the circuit diagram, explain
how a potentiometer is used to compare 1. e = I (R + r ) ---------- (1)
the emfs of two primary cells. Obtain the V – IR ----------------- (2) [½]
required expression used for comparing the From equations (1) and (2), we have
emfs.
ε R+r
(b) Write two possible causes for one sided =
deflection in a potentiometer experiment. V R
OR ε r
=1+
In the meter bridge experimental set up, shown V R
in the figure, the null point ‘D’ is obtained at a r ε
distance of 40 cm from end A of the meter bridge ⇒ = −1
R V
wire. If a resistance of 10 Ω is connected in series
with R1, null point is obtained at AD = 60 cm. ε 
⇒ r =  − 1
Calculate the values of R1 and R2.  V 
[½]
R1 R2
2. Let us consider a Wheatstone bridge arrangement
B as shown below:
Wheat stone bridge is a special bridge type
G circuit which consists of four resistances,
a galvanometer and a battery. It is used to
D determine unknown resistance. [1]
A C
In figure four resistance P, Q, R and S are
connected in the form of four arms of a
K quadrilateral. Let the current given by battery
[DELHI 2013] in the balanced position be I. This current on
reaching point A is divided into two parts I1 and
14. (i) State the principle of working of a
I2. As there is no current in galvanometer in
potentiometer.
balanced state, therefore, current in resistances
(ii) In the following potentiometer circuit AB is P and Q is Ii and in resistances R and S it is I2.
a uniform wire of length 1 m and resistance
B
10 Ω. Calculate the potential gradient along I1
the wire and balance length AO (=l). P Q
2V 15Ω K 1
I1
A G C
O I2
A B R S
I I2
1.2Ω
D
G
1.5V
E K1
0.3Ω Applying Kirchhoff’s first law at point A
[DELHI 2016] I – I1 – I2 = 0 or I = I1 + I2 (1)
15. Write the principle of working of a potentiometer. Applying Kirchhoff’s second law to closed mesh
Describe briefly, with the help of a circuit ABDA
diagram, how a potentiometer is used to
–I P + I2 R = 0 or I1 P = I2 R (2)
determine the internal resistance of a given cell.
[DELHI 2018]
Applying Kirchhoff’s second law to mesh BCDB
–I Q + I2 S = 0 or I1 Q = I2 S (3)
Dividing equation (2) by (3), we get 5. Potentiometer at open circuit, l1 = 350

I1 P I2 R P R R = 9 [1]
= = or
I Q I2 S Q S l2 = 300
1 [1]
This is condition of balance Wheatstone’s bridge.  350 
r = 9 − 1
3.  300 
[1]
E1 A
+ r = 1.5 Ω
–
Rh
K1 6. Current through potentiometer wire,
V
I=
B RAB + R
A
J' J
E 5
= = 0.25 A
G 15 + 5 [½]
R P.D. across the potentiometer wire,
K2
V = IRAB
Close key K1 keeping key K2 open. Find the point
= 0.25 × 5 = 1.25V [½]
(say J) on the wire AB such that on pressing
the jockey on the wire at J, the galvanometer V
Potential gradient, K = [½]
gives no deflection. At this stage, the potential I
difference across A and J is equal to the e.m.f. V
K =
(E) of the cell. If AJ = l1 then, [1] I
1.25V
E = VAJ = ∝ l1 …….. (i) =
1m [½]
E = Kl1
0.0125V cm –1
Now close key K2 so that a known resistance (R)
is connected across the cell. Find the point (J′, So, unknown e.m.f. balanced against 60 cm of
say) such that on pressing the jockey at J′, the the wire.
galvanometer gives no deflection. The terminal −1
potential difference (V) of cell is equal to the ξ = kl = 0.0125 Vcm × 60 cm
potential difference across A and J′. = 0.75V [½ + ½]
If AJ′ = then V = VAJ = ∝ l2 or V = Kl2 … (ii) Cell
Dividing (i) by (ii), we get,
G
E = l1 …………(iii)
V l2
A B
60 cm
We know, internal resistance of a cell is given by, P
100 cm
r =  E − 1 R …………(iv)
  5V
V
4. RP 1 [1] 5Ω
=
Ro 2 7. (a) For making gaps thick copper strips are used
due to their negligible resistance. [1]
PP I2 R (b) Meter bridge is most sensitive when all four
= P P
PQ IQ2 resistances are of same order. [1]
RQ
[1]
(c) Alloy Magnanin or Constantun are used
for making meter bridge wire due to low
PP 1
= temperature Coefficient of resistance & high
PQ 2 resistivity. [1]

Or
Equivalent circuit V 9.
is as shown
R0/2 B R0/2
A C 300 cm
120 cm C E
A G ε B
1 ε2
R F
So equivalent resistance of circuit
I ε1 ε2 H
RRo
Ro 2 Apply Kirchhoff’s law in loop ACFGA:
+
2 R f(120) = e1 – e2 [1]
R+ o
2 [1] f = potential drop per unit length
V
∴ I circuit = Or, e1 = e2 + f(120) -------- (1)
Req
[1] Loop AEHIA:
\ PD across R
f(300) e1 + e2
 RRo 
 2  = VR Or, ε 2 + (ε 2 + φ (120)) = φ (300)
I
Ro   Ro  [by substituting value of from equation (1)]
 R + 2  2  R + 4  [1]
2 e2 = (300 – 120) f
e2 = 90 f -------- (2) [1]
8. From the figure below: [½]
From equation (1) and (2),
Cell e1 = 210 f------- (3)
ε1 210 7
G Hence, = =
ε2 90 3
As we know, e = f I
A B
40 cm
P Thus, from equation (2) and (3)
100 cm Null point for cell e2 is 90 cm and for cell e1 it
6V is 210 cm. Senstivity of potentiometer can be
increased by:
5Ω Increasing the length of potentiometer wire,
Total resistance of the circuit, Decreasing the resistance in the primary circuit.
R = (RAB + 5) = 15 Ω 10. Initially when X is not connected
[½] R I1
=
Current in the circuit, S 100 − I1 [Condition for balance] ---------- (1)
V 6 The equivalent resistance (Req.) of the combination
=
i = A of X and S is:
R 15
1 1 1
6 = +
VAB = 6 − ×5 Req. X S
15
= 6 – 2 = 4V [½] SX
Req. =
Hence, voltage across AB, X+S [1]
 l R I2
Emf of the cell, e =   VAB [½] =
 L Req. 100 − I2

Here, l = 40m (balance point) [½] R ( X + S) I2
AB = L = 1m = 100cm (total length of the wire) =
SX 100 − I2 --------- (2)
 40  On dividing (1) by (2), we get
∴e =  4
 100 
[½]
e = 1.6V
R SX I (100 − I2 ) R I1
× = 1 = ………………( i )
S R ( X + S ) I2 (100 − I1 )
S 100 − I1
[1]
X I (100 − I2 ) when the p is connected in parallel with s the
= 1 equivalent resistance is

( X + S) I2 (100 − I1 )
XS
XI2 (100 − I1 ) = ( X + S ) I1 (100 − I2 ) Seq =
X+S
[1]

XI2 (100 − I1 ) = XI1 (100 − I2 ) + SI1 (100 − I2 ) for the second balanced bridge

XI2 (100 − I1 ) − XI1 (100 − I2 ) = SI1 (100 − I2 ) R I1
=
Seq 100 − I2
X  I2 (100 − I1 ) − I1 (100 − I2 ) = SI1 (100 − I2 )

R I2
SI1 (100 − I2 ) =
X= XS 100 − I2
 I2 (100 − I1 ) − I1 (100 − I2 ) X+S

This is the expression for X in terms of l1, l2 and R ( X + S) I2
S. [1] = ....... ( ii )
XS 100 − I2

11.
RB R I2
Resistance wire ∴ =
S 100 − I2
(R) D (S)
( X + S) I1 I2
× =
G S 100 − I1 100 − I2

A (P) B (Q) C
X + S I2 (100 − I1 )
cm (100 – ) cm =
X (100 − I2 ) I1 [1]
+ – Rh K
Cell Rheostat 12. For a meter bridge :
Fig.: Meter Bridge X l1
=
Y 100 − l1
Principle: It is constructed on the principle of
balanced wheatstone bridge. i.e., when a Where, it is given that l1 = 40 cm [1]
P R X 40 2
wheatstone bridge is balanced = where = =
Q S Y 100 − 40 3 --------- (1)
the initial value has usual meaning = at When 10 it resistance is added in series to X,
balancing situation of bridge [1] null point shifts by 10 cm.
P R 1 R 100 − 1 X + 10 40 + 10
=
Q
=
S
⇒
100 − 1
=
S
⇒S=
1
×R Y 100 − ( 40 + 10)

R S 50
X + 10 =
50 [1]
G X + 10
= 1 or X + 10 = Y
l1 Y ------------ (3)
B
Substituting the value of X from equation (2),
we obtain
2
Y + 10 = Y
According to question 3
For first balanced bridge situation 2
10 = Y − Y
3
Y fl2 + 0 – E2 = 0 (2)
Or = 10 [1]
3 E1 I
Y = 30 Ω = 1
E2 I2
[1]
Substituting the value of Y from equation (3),
we obtain Thus, we can compare the emfs of any two
sources. Generally, one of the cells is chosen as
X + 10 30
a standard cell whose emf is known to a high
X = 20 Ω degree of accuracy. The emf of the other cell is
Position of null point when 10 Ω resistance is then calculated from equation (3).
put in series with Y, (b) (i) The emf of the cell connected in main
20 li circuit may not be more than the emf of
=
20 + 10 100 − li the primary cells whose emfs are to be

compared. [1]
2000 = 20 l1 = 40 l1
(ii) The positive ends of all cells are not
60l1 = 2000 connected to the same end of the wire.
2000 [1]
l1 =
60 [1] OR
l1 = 33.3 cm Considering both the situations and writing
13. (a) Working principle of Potentiometer: When a them in the form of equations Let R’ be the
constant current is passed through a wire of resistance per unit length of the potential meter
uniform area of cross-section, the potential wire, [1]
drop across any portion of the wire is directly R1 R ’ × 40 40 2
= = =
proportional to the length of that portion. R2 R ’ (100 − 40) 60 3
(1)
Applications of Potentiometer for comparing
R1 + 10 R ’ × 60 60 3
emfs of two cells: The following figure shows = = = (2)
an application of the potentiometer to R2 R ’ (100 − 60) 40 2
compare the emf of two cells of emf E1 and
Putting the value of R1 from equation (1) and
E2
substituting in equation (2)
E1, E2 are the emf of the two cells. [1]
2 10 3
1, 2, 3 form a two way key. + =
3 R2 2

When 1 and 3 are connected, E1 is connected
to the galvanometer (G). R2 = 12 Ω
Jokey is moved to N1, which is at a distance Recalling equation (1) again
Li from A, to find the balancing length. R1 2
=
E1 1 12 3
1 3 R1 = 8 Ω
E2 2
2 14. (i) Principle of potentiometer: The basic
B A
principle of potentiometer is that when
a constant current flows through a wire
G
N1 of uniform cross-section area and the
R composition of the potential drop across any
length of the wire is directly proportional to
C N2 that length. [1]
A potentiometer is a device used to measure
K1
an unknown emf or potential difference and
Fig.: Construction of Potentiometer internal resistance of a cell accurately.
Applying loop rule to AN1G31A, [1] Construction:
fl1 + 0 – E1 = 0 (1) 1. A potentiometer consists of a long uniform
Where, is the potential drop per unit length cross-section of wire generally made of
Similarly, for E2 is balanced against l2(AN2), manganin or constantan.
2. Usually, 1 m long separate pieces of wire Iρl

are fixed on a wooden board parallel to each V=
A
other.
V = kl (If I ; A and P are constants)
3. The wire are joined in series by thick copper
strips. V ∝ l
4. The ends A and B are connected to a battery Where, Iρ
k=
(called driving all), a plug key and rheostat. A
V
5. A jockey J is provided with the help of which Here, = k (Potential gradient i.e. potential
l
contact can be made at any point on the wire.
6. This circuit sends a constant current I per unit length of wire)
through the wire AB. [1] (ii) Total resistance of the primary circuit
0 100 15 v+ 10 = 25 Ω, emf = 2V
\ Current in the wire AB,
2
=I = 0.08 A
B 400 25 [1]
P.D. across the wire AB = Current ×
R Resistance of wire AB = 0.08 × 10 = 0.8V
Potential gradient
P.D. 0.8
K = = = 0.008Vcm−1
Length 100
E 200l Resistance of secondary circuit
1.2 + 0.3 = 1.5 Ω
A e.m.f. =15V
+ –
G Current in the secondary circuit
Construction of Potentiometer 1.5
= = 1.0 A
1.5 [1]
V
The same is the current in 0.3 Ω resistor
P.D. between points A and O
P.D. across 0.3 Ω resistor in the zero-
deflection condition = Current × Resistance
= 1.0 × 0.3 = 0.3V
Potential difference
Length AO =
Potential gradient
O
Potential vs. Length of difference 15. The working principle of a potentiometer is based
on Kirchhoff’s voltage law. According to this rule,
Principle: When a constant current flows the algebraic sum of changes in voltage around
through a wire of uniform cross sectional area any closed loop involving resistors and cells in
and composition the potential drop across any the loop is zero. [1]
length of the wire is directly proportional to that K2
length. R.B.
Let V be the potential difference across the ε

portion of the wire of length l whose resistance B A
is R
N2 G
By Ohm’s law, V = IR = I ⋅ ρ l [1] R
A
C N1
Where, r → resistivity of wire:
A = Area of cross-section K1 [1]
Fig.: Construction of Potentiometer
Let f be the potential drop per unit length in e = l (r + R) [R = Resistance of the resistance box]
the potentiometer wire. When only a cell is V = IR, which gives, [1]
connected, the balance point is N1. [1]
ε r+R
Applying Kirchhoff’s voltage law, =
V R
e = fl1 (l1 = Length at which the balance point is
achieved] l 
r = R  1 − 1
When some current is drawn using the resistance  l2 

box, the balance point is achieved at N2.
The internal resistance of the cell can be
V = fl2 [1] determined by plugging-in the measured values
This gives, of l1 and l2.
ε l
= 1
V l2

CHAPTER 4
Moving Charges and
Magnetism

Concept of magnetic 1Q 1Q
field, Oersted’s experi- (5 mark) (3marks)
ment. Biot‒Savart law
and its application to
current carrying circular 1Q
loop (3 marks)
Ampere’s law & its appli- 1Q 1Q 1Q
cations to infinitely long (3 marks) (3 marks) (2 marks)
straight wire. Straight
and toroidal solenoids
Force on a moving charge 1Q 1Q
in uniform magnetic and (3 marks)
electric fields; Cyclotron (2 marks)
Force on a current-car- 1Q
rying conductor in a uni- (3 marks)
form magnetic field; force
between two parallel
current-carrying conduc-
tors-definition of ampere
Torque experienced by a 1Q
current loop in uniform (3 mark)
magnetic field;
Moving coil galvanome- 1Q 1Q
ter-its current sensitivity (1 mark) (3 marks)
and conversion to amme-
ter and voltmeter
On the basis of above analysis, it can be said that from exam point of view Moving Coil Galva-
nometer, Magnetic Field ,Magnetic Lorentz Force and permeability & susceptibility are most
important concepts of the chapter.
66 CHAPTER 4 : Moving Charges and Magnetism
[Topic 1] Magnetic Field Laws and their

Applications
Summary • Magnetic field at centre of the coil is
µ 0 Ni
• The Oersted’s law states that an electric current= B =
2R
( x 0)
creates a magnetic field.
• The Biot Savart’s law states that, the magnitude • Magnetic field due to current carrying circular
of magnetic field dB is proportional to the current µ i
arc with centre O is B = 0
I, the element length dl and inversely proportion- 4r
al to the square of the distance r. Its direction is
• If we curl the palm of our right hand around the
perpendicular to the plane containing dl and r.
circular wire with the fingers pointing in the di-
µ
Thus in vector notation, dB ∝ Idl × r , where 0 rection of the current, the right hand thumb rule
r3 4π
gives the direction of the magnetic field.
is the constant of proportionality and is equal to • Ampere’s circuital law: The line integral of the
10–7 Tm/A. magnetic field around some closed loop is equal to
Current element the times the algebraic sum of the currents which
Y pass through the circular loop. For some circuital
θ loop, C, ∫ B.dl =
C
µ 0I
Idl
r
P dB
Applications of Ampere’s Law
I
Magnetic field due to current carrying solenoid, B
= m0nI
X
µ 0 nI
At the end of a short solenoid, B =
Fig.: Biot Savart’s law 2
• The magnetic force produced by a Solenoid as

Applications of Biot-Savart’s Law: stated by Ampere’s law is given as F = m0nI,
where n is the number of turns of the wire per
• Magnetic field at a point in circular loop will be
unit length, I is the current flowing through the
µ 0 IR 2 wire and the direction is given using the right
B= 3
hand thumb rule.
(
2 R 2 + x2 ) 2
• Due to a toroid a magnetic field is given as,

Y µ NI
B= 0 where ‘N’ is the number of turns of the
2πr
dl r dB1 toroid coil, I is the amount of current flowing and
dB r is the radius of the toroid.
R
O θ • Antiparallel currents repel and parallel currents
I x X
P dBx attract.
dl • Magnetic moment on a rectangular current loop
in a uniform magnetic field, m = NIA where m
is the magnetic moment and N is the number of
Z closely wounded turns and A is the area vector.
Fig.: Magnetic field at a point in circular loop
CHAPTER 4 : Moving Charges and Magnetism 67
5. (a) State Ampere’s Circuital law, expressing it
PREVIOUS YEARS’ in the integral form.

(b) Two long coaxial insulated solenoids, S1
and S2 of equal lengths are wound one over
EXAMINATION QUESTIONS the other as shown in the figure. A steady
TOPIC 1 current “I” flows through the inner solenoid
S1 to the other end B, which is connected
1 Mark Questions to the outer solenoid S 2 through which
1. Magnetic field lines can be entirely confined the same current “I” flows in the opposite
within the core of a toroid, but not within a direction so as to come out at end A. If n1 and
straight solenoid. Why? n2 are the number of turns per unit length,
[DELHI 2017] find the magnitude and direction of the net
2 Mark Questions magnetic field at a point (i) inside on the axis
2. State Biot-Savart law. A current I flows in a and (ii) outside the combined system.
conductor placed perpendicular to the plane of 1
the paper. Indicate the direction
 of the magnetic r1 B
field due to a small element dl at point P situated
at distance r from the element as shown in the A
figure. r1
Z-axis I S1
n1 turns
n2 turns S2
[DELHI 2014]
dl O Y 6. (a) State Ampere’s Circuital law. Use this law to
r P obtain the expression for the magnetic field
inside an air cored toroid of average radius r,
having ‘n’ turns per unit length and carrying
X
a steady current I.
[DELHI 2017]
(b) An observer to the left of a solenoid of N
3 Mark Questions turns each of cross section area A observes
3. Two identical circular wires P and Q each of that a steady current I in it flows in the
radius R and carrying current ‘I’ are kept in clockwise direction. Depict the magnetic
perpendicular planes such that they have a field lines due to the solenoid specifying
common centre as shown in the figure. Find the its polarity and show that it acts as a bar
magnitude and direction of the net magnetic magnet of magnetic momentum M = NIA.
field at the common centre of the two coils. A
Q
I
N
P OR
I (a) Define mutual inductance and write its S.I.
units.
(b) Derive an expression for the mutual
[DELHI 2012] inductance of two long co-axial solenoids of
4. (a) State Biot – Savart law and express this law same length wound one over the other.
in the vector form. [DELHI 2015]
(b) Two identical circular coils, P and Q each of 7. A long straight wire of a circular cross-section of
radius R, carrying currents 1A and 3 A radius ‘a’ carries a steady current ‘I’. The current
respectively, are placed concentrically and is uniformly distributed across the cross-section.
perpendicular to each other lying in the XY Apply Ampere’s circuital law to calculate the
and YZ planes. Find the magnitude and magnetic field at a point ‘r’ in the region for (i)
direction of the net magnetic field at the r < a and (ii) r > a.
centre of the coils. [DELHI 2018]
[DELHI 2017]
5 Mark Questions Net field at centre,

2 2
8. (a) Using Biot-Savart’s law, derive the expression BN  BP  BQ
for the magnetic field in the vector form at a
point on the axis of a circular current loop. 2 2
 I  I
(b) What does a toroid consist of? BN   o    o 
 2R   2R 
Find out the expression for the magnetic field
inside a toroid for N turns of the coil having o I
BN 
the average radius r and carrying a current 2R [½]
I. Show that the magnetic field in the open
Direction of net magnetic field,
space inside an exterior to the toroid is zero.
B
[DELHI 2013] tan  P  1
BQ
9. State Biot-Savart law, giving the mathematical 
expression for it. Use this law to derive the tan   tan
4
expression for the magnetic field due to a
circular coil carrying current at a point along its 
 
axis. How does a circular loop carrying current 4 [½]
behave as a magnet? [DELHI 2011] 4. (a) Biot-savart’s law: This law states that the
magnetic field (dB) at point P due to small
Solutions current element Idl of current carrying
1. Magnetic field lines form closed loops around a conductor is
current-carrying wire. The geometry of a straight
solenoid is such that magnetic field lines cannot
loop around circular wires without spilling over
P
to the outside of the solenoid. The geometry of a θ
toroid is such that magnetic field lines can loop Id
r
around electric wires without spilling over to
the outside. Hence, magnetic field lines can be
entirely confined within the core of a toroid, but

not within a straight solenoid. [1]
2. Biot-Savart’s law states that the magnitude of the Fig.: Biot-Savart’s law [½]
magnetic field dB is proportional to the current (i) directly proportional to the Idl (current
element idl and inversely proportional to the element of the conductor) dB ∝ idl
square of the distance. The direction of magnetic (ii) directly proportional to sin q ∝ dBsinq,
field is along the negative X-direction. [1] q is the angle b/w dl and r
3. Magnetic field at the centre of circular loop (iii) inversely proportional to the square of
carrying current ‘I’ is given by, the distance of point p from the current
 I 1
B o element dB ∝ 2 [½]
2a r
Here, a = R now magnetic field due to loop Q: Combining all the inequalities
 I Idlsin  Idlsin
BQ  Bx  o dB   o
2R r 2 4 r2

Magnetic field due to loop P: o
 I where  107 Tm / A for free space
BP  By  o 4
2R [½]
Q The direction of magnetic field can be
obtained using right band thumb rule
I  Idl  rˆ
dB  o
4 r 2
P
I In vector form Biot-Savart’s law can be
written as
 Idl  r
dB  o
Fig.: Two identical circular wires 4 r 3 [½]
placed in perpendicular planes. [½]
(b) B = µo Ni [½]
P Where, N = number of turns per unit
I1 length
I2 R i = current through the solenoid.
Q Now, the magnetic field due to solenoid
R
yz plane S1 will be in the upward direction and
the magnetic field due to S2 will be in
[½] the downward direction (by right hand
I1 = 1A screw rule)
I2 = 3 A Bnet  BS1  BS2

Magnetic field due to coil Q at its centre is Bnet  o n1 I  o n2 I  o I  n1  n2 

 I [½]
Bq  0 1 along X-axis
2R (ii) The magnetic field is zero outside the
solenoid. [1]
B 6. (a) Amperes circuital law in electro magnetism
x-axis Q
B is analogous to Gauss law in electrostatics.
This law states that “The line integral of
resultant magnetic field along a closed
plane curve is equal to µ0 times the total
current crossing the area bounded by the
BP z-axis closed curve provided the electric field

inside the loop remains constant. Thus
[½]
Resultant magnetic field is  B.dl  o Ienclosed , where permeability
of free space and Ienclosed is the net current
Bp2 + Bq2 enclosed by the loop.

2 2 2 A toroid is a hollow circular ring on which a
 0 I1   0 I2   0 
 2 R    2 R    2 R   I12  I22  large number of turns of a wire are closely
wound. Consider an air-cored toroid (as

shown below) with centre O.
0 0
 3 
2
 12  2  0 
2R 2R 2R
and its direction is in X – Z plane [½] r l
5. (a) Ampere’s Circuital law states that the O
circulation of the resultant magnetic field
along a closed, plane curve is equal to po
times the total current crossing the area
bounded by the closed curve, provided
l l
the electric field inside the loop remains

constant. [½]
Fig.: Ampere circuital law
B Given:
i3
r = Average radius of the toroid
I = Current through the solenoid.
n = Number of turns per unit length
i1 To determine the magnetic field inside the
i2 toroid, we consider three amperian loops
(loop 1, loop 2 and loop 3) as show in the
figure below.
In the above illustration, the Ampere’s
Circuital
Law can be written as follows:
 B  dl  o i
[½]
Where, i = i1 – i2
(b) (i) The magnetic field due to a current
carrying solenoid:
Magnetic moment of single current carrying

loop is given by
m = LA
r l Where,
O Loop 1
I = Current flowing through the loop A
Loop 2
= Area of the loop
Loop 3 So, Magnetic moment of the whole solenoid
is given by
l l
M = Nw’ = N (IA) [1]
According to Ampere;s Circuital Law, we OR
have (a) Mutual inductance is the property of two
 B.dl  o Ienclosed [½] coils by the virtue of which each opposes
Total current for loop 1 is zero because no any change in the value of current flowing
through the other by developing an induced
current is passing through this loop.
emf. The SI unit of mutual inductance is
So, for loop 1,  B . dl  0  o  Total current  henry and its symbol is H. [1]
For loop 3, according to Ampere;s Circuital (b) Consider two long solenoids S 1 and S 2

Law, we have  B . dl  o  Total current  of same length l such that solenoid S 2
Total current for loop 3 is zero because net surrounds solenoid S1 completely.
current coming out of this loop is equal to
the net current going inside the loop. S1 I1
For Loop 2: The total current flowing
through the toroid is NI, where N is the total
number of turns

 B.dl 0  o  NI  ...(1)
I2
Now, B and dl are in the same direction l

S2
[½]
 B.dl  B  dl Let:

  B.dl  B 2 r  ...(2) n1 → Number of turns per unit length of S1
Comparing and we get, n2 → Number of turns per unit length of S2
B 2 r   o NI I1 → Current passed through solenoid S1
f21 → Flux linked with S2 due to current
 NI flowing through S1
B o
2 r f21 ∝ I1
Number of turns per unit length is given by f21 = M21 I1
N When current is passed through solenoid S1
n an emf is induced in solenoid S2
2 r
Magnetic field produced inside solenoid S1
B = µ0 nI on passing current through it is given by B
This is the expression for magnetic field = µ0 nI1
inside air-cored toroid. [1] Magnetic flux linked with each turn of
solenoid S2 will be equal to B1 times the area
(b) Given that the current flows in the clockwise
of cross-section of solenoid S1
direction for an observer on the left side of Magnetic flux linked with each turn of the
the solenoid. This means that left face of solenoid f21 = B1 A
the solenoid acts as South Pole and right Therefore, total magnetic flux linked with
face acts as North Pole. Inside a bar magnet the solenoid S2 is given by f21 = B1 A f21 A =
the magnetic field lines are directed from µo n1 I1 A [½]
N221
south to north. Therefore, the magnetic field M21 
lines are directed from left to right in the I1

solenoid. [½] N  n I A
M21  2 o 1 1  o n1 N2 A
I1

Where N2 is total number of turns wound

over the secondary coil.  B.dl  o Ienclosed

M21 = µo n1 N2 A B.dl  Bdl cos 

Similarly the mutual inductance between
q = 0°
the two solenoids when current is passed  

through solenoid S2 and induced emf is B.dl = Bdl
produced in solenoid S1 is given by I enclosed = I

M12 = µo n2 I1 A [1]
 Bdl  o I
7. (i) For r < a,
I1
B  dl  o I
B 2 r   o I

l o I
B
2 r [2]
I
8.
l
dl
X CY r = R2 dBcosθ
[½] 90° + k2 θ dB

R
 B.dl  o Ienclosed O x
θ dBsinθ
I
I enclosed I
 θ
 a2  r2 dB dBcosθ
X' Y'
r2 [1]
I enclosed = I
a2 (a) According to Biot-Savart’s law, magnetic

field due to a small element XY at point P is
B  dl  Bdl cos  Idl sin 
dB  o

4 r2
B  dl  Bdl  cos  1
o Idl sin 90 Idl
dB    Q sin 90  1
r2 4 r2 r2
 Bdl  o I

a 2 Resolving dB into two components:

2
(i) dB cos q, which is perpendicular to the
r axis of the coil
 dl  o I
B
a2 (ii) dB sin q which is along the axis of the

coil and away from the centre of the coil.
r2
B 2 r   o I  I sin 90
a2 B   dB sin  or B  o  dl
4  r2

o 1
B r o I sin   2 R
2  a2 [2] B 
4 2 r 2

(ii) For r > a,
o  2 IR2
B 

4 2 4 r 2  x2

3/2
[2]
r
(b)
Winding
Core
dl O r
2a B
[½] I I
From Ampere’s Circuital law,

For entire closed circular loop;
 B  dl  B  2 r …(i)
According to Ampere’s circuital law,

×  B  dl  o net current enclosed by the r
Q P
circle of radius r
= µo total number of turns × I
µo (n × 2pr) …(ii)
Comparing equation (i) and (ii), we get
I I
B × 2pr = µo (n × 2pr)I
B = µo nI [2] 2 r o Idl sin 90
For any point inside the empty space
dB  0 4 r2

surrounded by toroid and outside the toroid,
magnetic field B is zero because the net o I 2 r o  2 l 
current enclosed in these spaces is zero. But
B
4 r 2 0 dl  
4  r 

[1]
magnetic field is not exactly zero.
o  2 nl 
9. For any general shape of the wire, magnetic field For n turns of a coil; B   
due to current carrying wire is given by Biot 4  r 
Savart Law: The magnetic field lines due to a circular wire
form closed loops.
d
The direction of the magnetic field is given by
r right hand thumb rule. [1]
I P
 
  0 I dl  r
4  r3
B(P) 
[1]
Where dl is a vector tangent to the current wire The current carrying loop produces a magnetic
equal to dy j for a straight wire along y-axis. field around it, whose magnetic moment is
Consider a circular loop of radius r carrying a given as I × A (here, I is the current through the
current I. loop and A is the area of cross-section; hence it
Since dl ^ r behaves like a magnet. [1]
→ q = 90°
Applying Biot Savart law:
 Idl sin 90
dB  o
4 r2 [1]
Topic 2: Lorentz Force and Cyclotron
Summary Magnetic field out

of the paper
Deflection plate
• The electric field, E produced by the source of the

Qrˆ Exit Port
field Q, is given as E = , where r̂ is the
( 4π ∈0 ) r 2
unit vector and the field E is a vector field. A Charged
charge ‘q’ interacts with this field and experiences particle
a force F, expressed as
qQrˆ
= =
F qE

( 4π ∈0 ) r 2 D1 D2
• In the presence of both electric field E(r) and mag-

netic field B(r) there is a point charge q (moving OSCILLATOR
with a velocity v and located at ‘r’ at a given time
t ). The force on an electric charge ‘q’ due to both Fig.: Cyclotron
of them is written as • The charged particle gets accelerated and moves
F q [ E(r) + v × B(r)
= = ] Felectric + Fmagnetic . This force in a circular path whose radius is given by
mv
r =
is called the Lorentz force. q0B
• We can calculate the Lorentz force for a straight • The frequency of the cyclotron is given by
rod, if B is the external magnetic field by consid- 1 Bq
=
v =
ering the straight rod as a collection of linear T 2π m
strips dlj , where l is the length of the rod, j is the
current density. Hence, the force can be calculat- • A charge of any type in uniform circular motion
would have an associated magnetic moment given
ed=
as F ∑ Idl
j
j ×B.
−e
by µL = l , where l is the magnitude of
2 me
Cyclotron: angular momentum of electron.
• It consists of two D’s which are placed in a strong µL e
= = 8.8 × 1010 C / kg . , and this ratio is
magnetic field. An oscillating electric field is ap- l 2 me
plied from the oscillator which is parallel to the called Gyro magnetic ratio.
magnetic field.
3 Mark Questions
PREVIOUS YEARS’ 6. A metallic rod of length L is rotated with a
frequency v with one end hinged at the centre
EXAMINATION QUESTIONS and the other end at the circumference of a
TOPIC 2 circular metallic ring of radius r, about an axis
passing through the centre and perpendicular
1 Mark Questions to the plane of the ring. A constant uniform
1. Write the expression, in a vector form, for the magnetic field B parallel to the axis is present
everywhere. Using Lorentz force, explain how
Lorentz magnetic force F on a charge –q moving
emf is induced between the centre and the
with velocity V in a magnetic field B . What is
metallic ring and hence obtain the expression
the direction of the magnetic force?
for it.
[DELHI 2014]
[DELHI 2013]
2. A proton and an electron travelling along parallel
paths enter a region of uniform magnetic field, 7. Figure shows a rectangular conducting loop
acting perpendicular to their paths. Which of conducting PQRS in which the arm PQ is free
them will move in a circular path with higher to move. A uniform magnetic field acts in the
frequency? direction perpendicular to the plane of the loop.
[DELHI 2018] Arm PQ is moving with a velocity v towards the
arm RS Assuming that the arms QR, RS and SP
2 Mark Questions have negligible resistances and the moving arm
3. (a) Write the expression for the magnetic force PQ has the resistance r, obtain the expression
acting on a charged particle moving with for (i) the current in the loop (ii) the force and

velocity v in the presence of magnetic field (iii) the power required to move the arm PQ.

B.
S
(b) A neutron, an electron and an alpha particle P
V
moving with equal velocities, enter a
uniform magnetic field going into the plane Q
R
of the paper as shown. Trace their paths in
the field and justify your answer.
α [DELHI 2013]
8. A charge ‘q’ moving along the X-axis with a

n velocity c is subjected to a uniform magnetic field
B along the Z-axis as it crosses the origin 0.
e B Z-axis
[DELHI 2016]
O Y
4. State the underlying principle of a cyclotron.
Write briefly how this machine is used to q
accelerate charged particles to high energies.
[DELHI 2018] X
5. Find the condition under which the charged (i) Trace its trajectory.
particles moving with different speeds in the (ii) Does the charge gain kinetic energy as it
presence of electric and magnetic field vectors enters the magnetic field? Justify your
can be used to select charged particles of a
answer.
particular speed.
[DELHI 2017]
5 Mark Questions Write the expression for the magnetic moment



9. (a) Draw a schematic sketch of a cyclotron.  
m due to a planar square loop of side ‘l’ carrying
Explain clearly the role of crossed electric a steady current ‘I’ in a vector form. In the given
and magnetic field in accelerating the figure this loop is placed in a horizontal plane
charge. Hence derive the expression for the near a straight long conductor carrying a steady
kinetic energy acquired by the particles. current ‘I1’ at a distance of ‘l’ as shown. Give
(b) An α-particle and a proton are released reason to explain that the loop will experience
from the centre of the cyclotron and made a net force but no torque. Write the expression
to accelerate. for this force acting on the loop.
(i) Can both be accelerated at the same [DELHI 2018]
cyclotron frequency? Give reason to justify
your answer. Solutions
(ii) When they are accelerated in turn, which of 1. The Lorentz magnetic force is given by the
the two will have higher velocity at the exit following relation:
slit of the dees? 
 
 
[DELHI 2013]
F q VB   [1/2]
10. With the help of a labelled diagram, state the Here, q is the magnitude of the moving charge.
underlying principle of a cyclotron.
The direction of the magnetic force is
Explain clearly how it works to accelerate the perpendicular to the plane containing the
charged particles.
velocity vector V and the magnetic field vector
Show that cyclotron frequency is independent of
B. [1/2]
energy of the particle. Is there an upper limit on
 qB 
the energy acquired by the particle? Give reason. 2. As freq. of rev   here both charged
 m 
[DELHI 2011] particles will move in circular tracks. As charge
11. (a) Deduce an expression for the frequency of on electron and proton is same and both are
revolution of a charged particle in a magnetic subjected to same magnetic field so electron will
field and show that it is independent of show higher frequency due to less mass. [1]
velocity or energy of the particle. 3. (a) A charge particle having charge q is moving
(b) Draw a schematic sketch of a cyclotron. with velocity ‘v’ in a magnetic field of field
Explain, giving the essential details of its strength ‘B’ then the force acting on it is

construction, how it is used to accelerate the  
given by the formula F  q V  B and F =
charged particles.
[DELHI 2014] qvB sin q (Where q is the angle between
12. Explain the principle and working of a cyclotron velocity vector of magnetic field).
with the help of a schematic diagram. Write the Direction of force is given by the cross
expression for cyclotron frequency. product of velocity and magnetic field. [1]
[DELHI 2017] (b)
13.
α
I1
n
I
e
I

I

Fig.: Magnetic field
a Particle will trace circular path in

clockwise direction as it’s deviation will be E

 
in the direction of v  B i.e. perpendicular
to the velocity of particle, neutron will pass

without any deviation as magnetic field does O
t
not exert force on neutral particle.
Electron will trace circular path in
anticlockwise direction as its deviation will

 
be in the direction opposite to v  B with

Fig.: Square Wave
a smaller radius due to large charge/mass [1/2]
mv The accelerating electric field reverses just at
ratio as r = [1]
qB the time the change particle finishes its half
circle so that it gets accelerated across the gap
4. The underlying principle of a cyclotron is that an
between the Ds.
oscillating electric field can be used to accelerate
a charge particle to high energy. The particle gets accelerated again and again,
and its velocity increases. Therefore, it attains
A cyclotron involves the use of an electric field
high kinetic energy.
to accelerate charge particles across the gap
between the two D-shaped magnetic field The positively charged ion adopts a circular path
regions. The magnetic field is perpendicular to with a constant speed v, under the action of
the paths of the charged particles that makes magnetic field B, which is perpendicular to the
them follow in circular paths with in the two Ds. planes of D’s of radius r.
An alternating voltage accelerates the charged mv
r= [1/2]
particles each time they cross the Ds. The radius qB
of each particles path increases with its speed.
5. Moving uniformly charged particles can be
So, the accelerated particles spiral toward the
selected under electric and magnetic field
outer wall of the cyclotron. [1/2]
E
Magnetic field out vectors form the equation v = , where E is
of the paper Deflection plate B
electric field and B is magnetic field. [2]
Exit port
6. Suppose the length of the rod is greater than the
radius of the circle and rod rotates anticlockwise
Charged and suppose the direction of electrons in the rod
particle at any instant be along +y direction.
P
Suppose the direction of the magnetic field is
D1 D2
along +z direction.
Then, using Lorentz law, we get the following:


F  e v  B 
Oscillator   
F   e  v j  Bk
Fig.: Cyclotron  

[1/2]

[The Ds are the semi-circular structures (D1 F   eBi [1]
and D2) between which the charges move. The Thus, the direction of force on the electrons
accelerating voltage is maintained across the is along – x-axis. So, the electrons will move
opposite halves of the Ds.] Square wave electric towards the centre i.e., the fixed end of the
fields are used to accelerate the charged particles rod. This movement of electrons will effect in
in a cyclotron. current and thus it will generate an emf in the
rod between the fixed end and the point touching
the ring.
Let q be the angle between the rod and radius of Y

B is the magnetic field
the circle at any time t.
Velocity
Then, area swept by the rod inside the circle
1
  r 2
2 [1]
Induced emf
d 1 2 
 B   r   Force
dt  2 [1/2]
1 d 1 2 (ii) No, the charge does not gain kinetic
  r2 B   r B
2 dt 2 energy because the force and velocity are
1 perpendicular to each other.
  r 2 B 2 v
2 W  F    F cos 

Induced emf = p2r2Bv [1] q = 90° [1]
7. Let the length RQ = x and RS = l. Thus, force does not bring out any change
in the velocity. [1/2]
Let the magnitude of the uniform magnetic field 9. (a)
be B.
S
(i) The magnetic flux f enclosed by the loop
PQRS is given by, f = Blx Deflecting High frequency
plate
oscillator
As, x is changing with time, the rate of
dφ Source of positively
change of flux will induce an emf given
dt Target charged particle
by:
d d  Blx  dx
E    Bl  Blv
dt dt dt N
 dx  Fig.: Cyclotron
 as, dt  v [1]
 
Electric field accelerate the charged particle
Current in the loop is given by, whereas magnetic field makes its path
E Blv circular.
=
I = [1]
r r 1 m
(ii) The magnetic force on the PQ is, K . E.  m 2 , r 
2 qB
 Blv 
F  BIl  B  l 2
 r  1  qBr 
 m 
2  m 
B2 l 2 v
F= q2 B2 r 2
r [1] K . E. =
2m
(iii) Power emitted to move the arm PQ is,
qB 2e B eB
B2 l 2 v (b) (i) f   
P  F v
r
v eB 
2 m 2 4 m p 
4 m p
fp 
4 m p
B2 l2 v2
P=  f  f p
r [1] [2]
8. The direction of magnetic field is along the So both can’t be accelerated by same
negative X-direction. Hence the magnetic force frequency.
will act in such a way that this particle describes
a circular motion as shown below. [1]
m Since frequency of circular motion, given by

r
(ii)
qB  qB 
fc   is independent of r, the frequency
 2 m 
qBr
 
m of alternating voltage is made twice fc, so that
q when the particle reaches point Q, V1 – V2 flip
 
m
and D2 is at higher potential than D1, so that
particle is accelerated again. This goes on till
p q p m
 the kinetic energy is high enough and
 q m p

consequently the radius is large enough so that
 e   4m p  the particle is ejected out tangentially from the
  
 2e   m p 
magnetic field with high energy.
p
2 Frequency [1]
v
 qB 
fc   is independent of the energy or the

 p  2  2 m 
 p   speed of the charged particle.

[2]
Upper limit is relativistic speeds. The cyclotron
10. The principle behind the working of a cyclotron
is limited by relativistic effects due to which the
is that in a uniform magnetic field, charged
mass of the accelerating particle increases with
particle executes uniform circular motion with
energy and so fc changes after each cycle.
a frequency that is independent of its radius
(which depends on the energy of the particle) Also, for light charged particles, fc is enormously
high and difficult to maintain.
D1
Also, it is not easy to maintain uniformity of the
P magnetic field.
~ To add to that, a charged particle in circular
Q
motion (even if it is uniform) is accelerating and
all accelerated charges radiate electromagnetic
D2 energy, thereby losing energy. [1]
Fig.: Cyclotron [1] 11. (a) If particle is performing circular motion due
A cyclotron has a uniform magnetic field spread to magnetic force then
over a circular region split in two halves by two Centripetal force = Magnetic force [1/2]
semi-circular D’s, D1 and D2.
mv2
Let us assume for definiteness that the particle  qvB sin 90
r
has positive charge. Then, as it comes around
in D1 clockwise to point P, D1 and D2 are so mv
r
connected to an alternating voltage that at this qB [1]
point, D1 is at a higher potential than D2. This
2 r
accelerates the particles, so that it gains kinetic So, time period 
v
energy by T = q(V1 – V2) where v1 and v2 where 2 mv 2 m
T 
D1 and D2 are potentials of respectively. [1] v qB qB
This increases the radius of circular motion since 2 m
T  vo
 m  qB
r and is proportional to the square root
 qB 
1 qB
∴ \Frequency f    vo [1]
T 2m
of kinetic energy.
(b) Cyclotron is a device to accelerate ions to The particles move in two semi-circular
extremely high velocities, by accelerating containers D1 and D2, called Dees. Inside the
them repeatedly through high voltages. metal box, the charged particle is shielded from
D.P external electric fields. [1]
When the particle moves from one dee to another,
electric field is acted on the particle.
S
The sign of the electric field is changed
alternately, in tune with the circular motion
of the particle. Hence, the particle is always
accelerated by the electric field. As the energy of
D1 D2 the particle increases, the radius of the circular
Oscillator path increases.

Fig.: Cyclotron [1/2] mv2
qvB =
Principle – A positive ion can acquire r
sufficiently large energy with a small mv
alternating potential difference by making r=
qB [1]
the ion cross the same electric field time and
again by making use of a strong magnetic Time taken for a particle for one complete
field. [1] revolution = T
Construction- It consists of a pair of hollow 2 r
metal cylindrical chambers shaped like T

D, and called the Dees; Both the Dees are
placed under the circular pole pieces of a 2 m
T
very strong electromagnet. The two Dees qB
are connected to the terminals of a very high
1
frequency and high voltage oscillator whose T
frequency is of the order of a few million c

cycles per second.
Where v, is the cyclotron frequency
Working -Charged ion is passed through
Then,
electric field again & again to be energized.
Inside Dee’s strong perpendicular magnetic 1 2 m

field turns the particle towards gap. So c qB

radius of semi-circular path increases
continuously. [1] 2 m
c 
qB
12.
Magnetic field out The above expression is the expression for
of the paper Deflection plate cyclotron frequency.
The oscillator applies an ac voltage across the Ds
Exit port
and this voltage must have a frequency equal to
that of cyclotron frequency. [1]


Charged
particle 13. The expression for the magnetic moment m  
D2 due to a planar square loop of side ‘l’ carrying
D1
steady current ‘l’ in a vector form is given as

 

 
m  IA

 
 
Oscillator 2
Therefore, m  I  l  n
Fig.: Cyclotron [1]
∧
Cyclotron is a machine used to accelerate charged Where, n is the unit vector along the normal to
particles or ions to high energies. It uses both the surface of the loop. The attractive force per
electrical and magnetic fields in combination to unit length on the loop is, [5]
increase the speed of the charged particles.
Topic 3: Magnetic Force and Torque between

Two Parallel Currents
Summary Moving coil galvanometer
Force on a current-carrying • Its main use is to detect and measure small
electric currents.
conductor in a uniform magnetic • The current carrying coil is suspended in a
field: uniform magnetic field, so it produces a torque
• The force on a current carrying conductor of which is responsible for rotating the coil.
length l in a uniform magnetic field B when q is Scale
the angle between current and magnetic field can
be calculated by F = IBl sin q
• Fleming’s Left-Hand Rule is used to find the Pointer Permanent magnet
direction of the magnetic force which is right
angled to the plane containing conductor and Coil
magnetic field.
Force between two parallel current-
carrying conductors: N Sp S
Two parallel conductors carrying current Pivot

experiences a force. When current flows in same Soft-iron
core
direction, wire B experiences magnetic field due
µ II
to wire A which is: B 1 = o 1 Uniform radial magnetic field
2π d
Force per unit length in the given wire is Fig.: Moving coil galvanometer
F 2 µo I 1I 2 • The torque is given by t = F × b = nBIl × b = BInA
= sinq
l 2π d d
Current sensitivity of galvanometer
a • When a galvanometer produces a large deflection
for a small amount of current, it is said to be
sensitive.
b
• The voltage sensitivity of galvanometer is
Fba
deflection per unit voltage and is given as
θ nBA
=
L I C
Ia Conversion of galvanometer into
ammeter
Ba
Ib A small resistance called a Shunt resistance is
Fig.: Force between two parallel currents attached with the galvanometer coil in parallel so
carrying conductores. that most of the current passes through the shunt
resistance.
Torque experienced by a current
loop in uniform magnetic field: Conversion of galvanometer into
The torque experienced by a rectangular loop in
voltmeter
uniform magnetic field B of length l, breadth b A high resistance is connected in series with the
with current I lowing through it is: galvanometer coil so that the galvanometer acts
t = nBIAsinq as a voltmeter.
PREVIOUS YEARS’ I = 5A
4 cm
TOPIC 3 2A 2A
1 Mark Questions 10 cm
1. Using the concept of force between two infinitely
long parallel current carrying conductors, define
one ampere of current.
[DELHI 2014] 1 cm
2. Write the underlying principle of a moving coil (i) the torque acting on the loop and
galvanometer. (ii) the magnitude and direction of the force on
[DELHI 2016] the loop due to the current carrying wire.
7. A cyclotron’s oscillator frequency is 10 MHz.
3. Two long straight parallel conductors carry What should be the operating magnetic field for
steady current I1 and I2 separated by a distance accelerating protons? If the radius of its ‘dees’ is
d. If the currents are flowing in the same 60 cm, calculate the kinetic energy (in MeV) of
direction, show how the magnetic field set up the proton beam produced by the accelerator.
in one produces an attractive force on the other. [DELHI 2015]
Obtain the expression for this force. Hence
8. Deduce the expression for the torque τ acting on
define one ampere.
a planar loop of area Ar and carrying current I
[DELHI 2016]
placed in a uniform magnetic field B. If the loop
4. A wire AB is carrying a steady current of 6A and
is free to rotate, what would be its orientation
is lying on the table. Another wire CD carrying
in stable equilibrium?
4A is held directly above AB at a height of 1mm.
[DELHI 2015]
Find the mass per unit length of the wire CD so
that it remains suspended at its position when 9. Derive the expression for force per unit length
left free. Give the direction of the current flowing between two long straight parallel current
in CD with respect to that in AB. [Take the value carrying conductors. Hence define one ampere.
of g = 10ms–2] 5 Mark Questions
[DELHI 2013]
5. State the principle of working of a galvanometer. 10.
(a) Draw a labeled diagram of a moving coil
A galvanometer of resistance G is converted galvanometer. Describe briefly its principle
into a voltmeter to measure up to V volts by and working.
connecting a resistance R1 in series with the (b) Answer the following:
coil. If a resistance R2 is connected in series with (i) Why is it necessary to introduce a cylindrical
V soft iron core inside the coil of a galvanometer?
it, then it can measure up to volts. Find the (ii) Increasing the current sensitivity of a
2
resistance, in terms of R1 and R2, required to be galvanometer may not necessarily increase
connected to convert it into a voltmeter that can its voltage sensitivity. Explain, giving reason
read up to 2 V. Also find the resistance G of the [DELHI 2014]
galvanometer in terms of R1 and R2. 11. State the underlying principle of working of a
[DELHI 2015] moving coil galvanometer. Write two reasons
3 Mark Questions why a galvanometer cannot be used as such to
6. A rectangular loop of wire 4 cm × 10 cm carries measure current in a given circuit. Name any
a steady current of 2A. A straight long wire two factors on which the current sensitivity of
carrying 5A current is kept near the loop as a galvanometer depends
shown. If the loop and the wire are coplanar, find [DELHI 2018]
Solutions
(l → length of CD)
for balance of CD
1. One ampere is that current which if passed
in each of two parallel conductors of infinite fm = mg [1/2]
length and one meter apart in vacuum, causes  o i1i2 
each conductor to experience a force of 2 × 10–7  2 h  l  mg
Newton per meter of length of conductor. [1]
2. When a current carrying coil is placed in 7
m 2  10  6  4
magnetic field then it experiences a torque. [1/2 
NIAB = ka
l 103  10
k m
I   4.8  104 kg / m
NAB l [1/2]
N ⇒ The number of turns
I ⇒ Current Direction of current in CD will be opposite to AB.

A ⇒ Area of the loop 5. Principle: When a current-carrying coil is
B ⇒ Magnetic field placed in a magnetic field, it experiences a
k ⇒ Torsional constant of the wire torque. From the measurement of the deflection
a ⇒ Angle of deflection [1/2] of the coil, the strength of the current can be
3. Magnetic field produced on the wire (carrying computed. A high resistance is connected in
current I2) due to I1 will be. series with the galvanometer to convert it into
voltmeter. The value of the resistance is given
by R  V  Rg Where, V → Potential difference
Ig
across the terminals of the voltmeter [1/2]

I1 I2
F l Vg → Current through the galvanometer
G → Resistance of the galvanometer
d When the resistance R1 is connected in series
with the galvanometer. [1/2]
V = I (G + R1)
Fig.: Two long parallel conductor separated by When the resistance R2 is connected in series
 I distance d
B o 1 with the galvanometer.
2 d
V
Force acting at l length is F = I2 lB  I G  R2 
o I1 I2 l 2
F  2 d towards I1 [1] G  R1
2
Attractive force between the wires G  R2

If l = 1m, d = 1m, I1 = I2 = I and F = 2 × 10–7 N
⇒ I = 1A G = R1 – 2 R2 [1/2]
So one ampere is defined as the current, which
when maintained in two parallel infinite length Let R3 be the resistance required for conversion
conductors, held at a separation of one meter into voltmeter of range 2V
will produce of a force of 2 × 10–7 N per metre
of each conductor. [1] \ 2V = Ig (G + R3)
4. Also, V = Ig (G + R1)
Fm G  R3
4A 2
C D G  R1
i
mg h \ R3 = G + R1 = R1 – 2R2 + 2R1
R3 = 3R1 – 2R2 [1/2]
A B
6A

[1]
Fig.: Current carrying wires Ab & CD
6. 1 1 
 2  10  107  10   
I = 5A
FB 1 5 
4 cm
D C = 160 × 10–7 N
r’
7. Frequency of oscillators
F F' (v) = 10 MHz = 107 Hz
10 cm Mass of proton, m = 1.67 × 10–27 kg
2A
Charge of proton = 1.6 × 10–19 C
Operating magnetic field is given by the relation
r
2 m 2  3.14  1.67  1027  107
1 cm A B B 
FB q 1.6  1019
Fig.: Current carrying loop [1/2] = 0.65T [1]

(i)   M  B  MBsin  Radius of dees = 60cm = 0.6m
Here M and B have the same direction,
so q = 90° KE 
q2 B2 r 2


1.6  1019 0.65 0.6  MeV
2 2
 2m 2  1.67  1027  1.6  1013

  MBsin   0
= 7.28 MeV [1]

(ii) We know FB  il  B 8.
On line AB and CD magnetic forces are equal C
and opposite. So they cancel out each other. B l
Magnetic force on line AD. N S
B
θ
l m
FB  il  B (Attractive) D
= ilB (i = 10 cm = 0.1m) A
 I
B o
[1]
2 r
F2
 iIl
F  (Attractive)
2 r
Magnetic force on line CB,

α/2
FB  il  B (Repulsive) a/2sinθ
θ B
 F ’  F  ilB ’ a/2 m

o I
B
2 r ’
o iIl (Repulsive)
F’ 
2 r ’ F1
So, net force, Fn = F – F’ Fig. (a) & (b): Planar loop
 iIl  1 1  [1/2]
Fn  o   

2  r r ’  (a) The area vector of the loop ABCD makes an

arbitrary angle q with the magnetic field.
Given, i = 5A, I = 2A, r = 1cm = 0.01 m,
(b) Top view of the loop. The forces F1r and F2r
r = (1 + 4) = 5 cm and l = 10cm = 0.1m
acting on the arms AB and CD are indicated.
Substituting these values in above equations,
We consider the case when the plane of the loop,
we get
is not along the magnetic field, but makes an

   1
Fn  2  107 520.1 
 0 . 01

0
1 
. 05 
angle q with it. Fig.(a) illustrates this general
case. The forces on the arms BC and DA are By Ampere’s circuital law, we have
equal, opposite, and act along the axis of the o Io
coil, which connects the centres of mass of BC Ba 
2 d
and DA. Being collinear along the axis they
Conductor b will experience a sideways force
cancel each other, resulting in no net force or
because of conductor a. Let this force be Fba
torque. The forces on arms AB and CD are F1
and F2. They too are equal and opposite, with Fba = Ib LBa (F = ILB) [1/2]
magnitude, F1 = F2 = I b B But they are not  o I a Ib L

collinear! This results in a couple Fig.(b) is a 2 d
view of the arrangement from the AD end and it By symmetry,
illustrates these two forces constituting a couple. Fba = –Fab
The magnitude of the torque on the loop is, 1 ampere is the value of that steady current
which when maintained in each of the two very
a a
  F1 sin   F2 sin  long, straight, parallel conductors of negligible
2 2 cross section and placed one metre apart
= Iab sin q in vacuum, would produce on each of these
IAB sin q …(i) [1] conductors a force equal to 2 × 10–7 Newton per
As q → 0 , the perpendicular distance between metre of length.
the forces of the couple also approaches zero. 10. (a) Principle: When a current carrying coil is
This makes the forces collinear and the net force placed in magnetic field, it experiences a
and torque zero. The torques in above equation torque.
can be expressed as vector product of the Construction: It consists of a narrow
magnetic moment of the coil and the magnetic rectangular coil PQRS consisting of a large
field. We define the magnetic moment of the number of turns of fine insulated copper
current loop as, m = I A where the direction wire wound over a frame made of light, non-
magnetic metal. A soft iron cylinder known
of the area vector A is given by the right-hand
as the core is placed symmetrically within
thumb rule and is directed into the plane of the
the coil and detached from it. The coil is
paper in Fig.(a). Then as the angle between m suspended between the two cylindrical pole
and B is q, equation (i) can be expressed by one pieces (N and S of a strong permanent horse
expression shoe magnet) by a thin flat phosphor bronze
t=m×B strip the upper end of which is connected to
a movable torsion head T. The lower end
we see that the torque tr vanishes when mr is
of the coil is connected to a hair spring s of
either parallel or antiparallel to the magnetic
phosphor bronze having only a few turns.
field Br . This indicates a state of equilibrium as Radial magnetic field. The magnetic field
there is no torque on the coil (this also applies to in the small air gap between the cylindrical
any object with a magnetic moment m r). When pole pieces is radial. The magnetic lines of
mr and Br are parallel the equilibrium is a stable force within the air gap are along the radii.
one. [1] On account of this, the plane of the coil
9. Two long parallel conductors a and b separated remains always parallel to the direction of
by a distance I and carrying currents Ia and Ib the magnetic field. [1]
respectively are shown below. T
d
a T1 T2
b m
Fba
P S
I N core S
Ia Q R
Ib Ba s
Fig.: Two parallel long conductors separated by ‘d’
[1/2] Fig.: Moving coil Galvanometer [1/2]

The magnetic field in the small air gap (ii)

Voltage sensitivity
between the cylindrical pole-pieces is radial. Charge Sensitivity
On account of this, the plane of the coil = Resistance of Coil
remains always parallel to the direction of
[1/2]
the magnetic field CS
VS =
Theory : Let I = current flowing through the Rcoil
coil
B = magnetic field induction If Rcoil = constant V S ∝ C S
l = length of the coil; It means that VS increases if CS is

increased but if resistance of coil is also
b = breadth of the coil increases in same ratio then VS may be
N = number of turns in the coil constant [1]
A (= l × b) = area of the coil 11. The underlying principle for the working of
Moment of deflecting couple = NBIl × b a moving coil galvanometer is that when a
= NBIA current-carrying conductor is placed inside a
When the coil deflects, the suspension fiber magnetic field, it experiences a magnetic force.
gets twisted. On account of elasticity, a The two reasons why a galvanometer cannot be
restoring couple is set up in the fiber. This used for measuring current are:
couple is proportional to the twist. (i) The high resistance of galvanometer can
If α be the angular twist then Moment of disturb the original current flowing through
restoring couple = kα the circuit. [2]
For equilibrium of the coil, NBIA = kα or (ii) The high current present in the circuit can
 k  destroy the coil windings present in the
I 
 NBA  galvanometer. [2]
The factors on which the current sensitivity
or I = Kα of a galvanometer depends are:
 k  (i) Number of turns in the coil [1/2]
where K   is the galvanometer
 NBA  (ii) Torsional spring constant [1/2]
constant
Now, I ∝ α or α ∝ I
[1]
(b) (i) By using soft iron core, magnetic field
is increased so sensitivity increases and
magnetic field becomes radial So angle
between plane of coil & magnetic line of
force is zero in all orientations of coil. [1]

CHAPTER 5
Magnetism and Matter

Current loop as a mag-
netic dipole and its
magnetic dipole moment
magnetic dipole moment
of a revolving electron
Magnetic field intensity
1Q
due to a magnetic dipole 1Q
1Q (1marks),
(bar magnet) along its
(4 marks) 1Q
axis and perpendicular to (2 marks)
(3 marks)
its axis
Torque on a magnetic
dipole (bar magnet) in a
uniform magnetic field;
bar magnet as an equiv-
alent solenoid,magnetic
field lines
Earth’s magnetic field 1Q
and magnetic elements (3 marks),
1Q
(2 marks)
Para-, dia- and ferro - mag-
netic substances, with ex-
amples. Electromagnets 1Q
and factors affecting their (2 marks)
strengths,permanent
magnets
On the basis of above analysis, it can be said that from exam point of view Magnetic Dipole and
Earth magnetic field are most important concepts of the chapter.
88 CHAPTER 5 : Magnetism and Matter
[Topic 1] Magnetic Dipole and Magnetic Field

Lines
Summary • Bar magnet as an equivalent solenoid: The
magnetic field B due to bar magnet of size l and
Magnetism: magnetic moment m which is at a distance r from
the mid-point when r >> l, is given by
• Magnetic phenomena are universal in nature.
Magnetism is a physical phenomenon produced
µ0 2m
by the motion of electric charge, which results in B = (Along axis)
attractive and repulsive forces between objects. 4π r 3
• The magnetic field of the Earth points from
x
geographical south to the north.
• A bar magnet always points in the north-south dx
direction when suspended freely.
• When same poles of two magnets are brought close a
P
to each other, a repulsive force is experienced.
When Opposite poles of two magnets are brought O r
close, then an attractive force is experienced.
Bar Magnet: l
2l
Iron fillings sprinkled on a glass plate kept over
a short bar magnet arrange themselves in a
pattern. It shows that the magnet has two poles Fig: Bar magnet as an equivalent solenoid
in the same way as the positive and negative
charge of an electric dipole called as the North • Dipole in a uniform magnetic field: When a bar
and the South pole. magnet is having a dipole moment m and it is
Magnetic field lines: The magnetic field lines of placed in uniform magnetic field B,
a bar magnet form continuous closed loops. The The force acting on it is equal to 0.
direction of net magnetic field at any point is
determined by the tangent to the field line at that The torque acting on the magnet is m × B
point. The magnitude of the magnetic field will be It has a potential energy of –m.B
stronger for the area from which more number of
field lines are passing. The magnetic field lines Gauss’s law for magnetic fields:
never intersect each other.
It states that the magnetic flux through any
Axis
B closed loop is equal to zero.

φB = ∑∆φB = ∑B .∆S = 0
all all
H
N
i
Fig: Magnetic field lines in a bar magnet
CHAPTER 5 : Magnetism and Matter 89
 3 3  0.44  2.836
PREVIOUS YEARS’ 90
mB sin  d   mB  cos 90  cos 60
dw 
 60
TOPIC 1  1
 6  0.44    
3 Mark Questions  2 [1]
1. A bar magnet of magnetic moment 6J/T is aligned = 3 × 0.44
at 60° with a uniform external magnetic field of = 1.32 J
0.44T. Calculate (a) the work done in turning
(ii)
the magnet to align its magnetic moment (i)
180
normal to the magnetic field, (ii) opposite to the
dw   mB sin  d   mB  cos 180  cos 60
magnetic field, and (b) the torque on the magnet
 60
in the final orientation in case (ii).
[DELHI 2018]  1
 6  0.44  1  
Solutions  2 [1]
1. M = 6J/T  3
 6  0.44   
q = 60°  2
B = 0.44T = 9 × 0.44 39.6J
6 × 0.44 sin 60° (b) W = m × B
3 W = m B sin q
 6  0.44 
2 = 6 × 0.44 sin 180° [1]
W=0
[Topic 2] Earth’s Magnetism and Magnetic

Properties of Material
SUMMARY Magnetization and magnetic field:
• The magnetization M is equal to its magnetic
Earth’s Magnetism: moment per unit volume
• The earth’s magnetism is of the order of 10–5 T. Its m
M = net
strength is different at different place. The pole V
near to geographic north pole is called the north • The magnetic intensity H is defined as the amount
magnetic pole and the pole near to geographic of magnetic flux in a unit area perpendicular to
south pole is called south magnetic pole. The the direction of magnetic fow.
magnetic of the field on the earth’s surface is B
4 × 10–5 T. H = 0
µ0
• There are three elements of the earth’s magnetic
field which are used to specify the magnetic field • The magnetic field B in the material is given by,
of earth’s surface – the horizontal component, the B = m0(H + M)
magnetic declination and the magnetic dip. • The degree of magnetization of a material in
• The magnetic field of a bar magnet tilted 11° from response to an applied magnetic field is denoted
the spin axis of Earth is in the same direction as as magnetic susceptibility. It is given by
the Earth’s magnetic field. M
χ =
H
So, m = m0mr
Where mr = 1 + 
Magnetic properties of materials:
Magnetic Materials are broadly classified as paramagnetic, diamagnetic and ferromagnetic materials. For
paramagnetic materials  is positive and is small, for diamagnetic materials  is negative and lies between
0 and -1 and for ferromagnetic materials  is positive and large.
Property Ferromagnetic Diamagnetic Paramagnetic
Effect of magnets They are strongly They are feebly repelled by They are feebly attracted by
attracted by magnets. magnets. magnets.
Susceptibility value Large and positive Small and negative Small and positive
m m > 1000
Permeability value µ >> µ0 µ < µ0 µ < µ0
In a uniform Freely suspended rod Freely suspended rod Freely suspended rod aligns
magnetic field aligns itself parallel to aligns itself perpendicular itself parallel to the field.
the field. to the field.
Relative It is greater than 1000. Slightly less than 1. Slightly greater than 1.
permeability value
Effect of temperature Susceptibility Susceptibility is Susceptibility varies inversely
decreases with independent of with temperature.
temperature. temperature.
Physical state of the Solids only. Solid, liquid or gas. Solid, liquid or gas.
material
Hysteresis effect Shows hysteresis Does not show hysteresis. Does not show hysteresis.
Removal of magnetic Magnetization retain Magnetization is only for Magnetization is only for the
field even on removal of the time magnetic field is time magnetic field is applied.
magnetic field. applied.
Examples Fe, Ni, Gd, Co Bi, Si, Cu, Pb Al, Ca, Na
Ferromagnetic materials show the property of Permanent magnets:

hysteresis.
• Permanent magnets are those substances which
B
(Tesla) at room temperature retain their ferromagnetic
1.5 a property.
b • An iron rod held in north-south direction and
1.0 if it is hammered repeatedly it will become a
0.5 permanent magnet.
c • It can also be made by placing a ferromagnetic
–200 –100 O 100 200 H rod in a solenoid and passing current through it.
–0.5
A/m The rod gets magnetized by the magnetic field of
–1.0 the solenoid.
d e • A material having high permeability, high
–1.5 coercivity, and high retentivity could be suitable
for permanent magnets.
Fig: Magnetic hysteresis loop
The magnetic hysteresis loop is the B-H curve for
Electromagnets:
ferromagnetic materials • A solenoid having a core of iron with wire wrapped
around it is called an electromagnet.
Curie’s law: • Ferromagnetic materials are used for core of
The intensity of magnetization I of a paramagnetic electromagnets.
material varies directly to the strength of the external • Some of the applications of electromagnets
magnetic field H, called magnetizing field and is are loudspeakers, electric bells, telephone
inversely proportional to absolute temperature of the diaphragms.
material.
C
χ = where C is Curie constant.
T
5. Which of the following substances are

PREVIOUS YEARS’
paramagnetic?
Bi, Al, Cu, Ca, Pb, Ni
EXAMINATION QUESTIONS [DELHI 2013]
2 Mark Questions
TOPIC 2 6. A magnetic needle free to rotate in a vertical
1 Mark Questions plane parallel to the magnetic meridian has its
1. The permeability of a magnetic material is north tip down in 60° with the horizontal. The
0.9983. Name the type of magnetic materials it horizontal component of the earth’s magnetic
represents. field at the place is known to be 0.4 G. Determine
[DELHI 2011] the magnitude of the earth’s magnetic field at
2. The horizontal component of earth’s magnetic the place.
field at a place is B and the angle of dip is 60°. [DELHI 2011]
What is the value of vertical component of the 7. The susceptibility of a magnetic material is –2.6
earth’s magnetic field at equator? × 10–5. Identify the type of magnetic material
[DELHI 2012] and state its two properties.
3. What are permanent magnets? Give one [DELHI 2013]
example. 8. Show diagrammatically the behaviour of
[DELHI 2013] magnetic field lines in the presence of (i)
4. Which of the following substances are paramagnetic and (ii) diamagnetic substances.
diamagnetic? How does one explain this distinguishing
Bi, Al, Na, Cu, Ca and Ni feature? [DELHI 2014]
[DELHI 2013]
9. Out of the two magnetic materials, ‘A’ Solutions

has relative permeability slightly greater
than unity while ‘B’ has less than unity. 1. It represents a diamagnetic substance since its
Identify the nature of the materials ‘A’ and permeability (0.9983) is less than 1. [1]
‘B’. Will their susceptibilities be positive or 2. On the equator, the values of both angle of dip (d)
negative? and the vertical component of earth’s magnetic
[DELHI 2014]
field is zero. So, Bv = 0 [1]
10. Write two properties of a material suitable for 3. The magnets which have high retentivity
making and high coercivity are known as permanent
magnets. For example: Steel [1]
(a) a permanent magnet, and
4. Bi and Cu are diamagnetic substances.Since
(b) an electromagnet. their net magnetic moment is zero. [1]
[DELHI 2017]
5. Paramagnetic substances are Aluminium (Al) and
11. (i) Write two characteristics of a material used Calcium (Ca).They have unpaired electrons. [1]
for making permanent magnets. 6. Horizontal component of earth’s magnetic field,
(ii) Why is the core of an electromagnet made BH = 0.4G
of ferromagnetic materials? Angle made by the needle with the horizontal
Or plane = Angle of dip d = 60° [1]
Draw magnetic field lines when a (i) diamagnetic, Earth’s magnetic field strength = B
(ii) paramagnetic substance is placed in external We can relate B and BH as:
magnetic field. Which magnetic property
BH = B cos q
distinguishes this behaviour of the field due to
 B  BH  0.4  0.8G
the substances?
cos  0.5
[DELHI 2018]
12. (a) An iron ring of relative permeability µr, has Hence, the strength of earth’s magnetic field at
windings of insulated copper wire of n turns the given location is 0.8G. [1]
per metre. When the current in the windings 7. D i a m a g n e t i c m a t e r i a l s h a v e n e g a t i v e
is I, find the expression for the magnetic field susceptibility. So, the given material is
in the ring. diamagnetic. Properties of magnetic material
(b) The susceptibility of a magnetic material are as mentioned below: [1]
is0.9853.Identify the type of magnetic i. They do not obey Curie’s law. [1/2]
material. Draw the modification of the field ii. They are feebly repelled by a magnet. [1/2]
pattern on keeping a piece of this material 8.
in a uniform magnetic field.
N S N S
[DELHI 2018]
5 Mark Questions Fig.: (Diamagnetic Material) Fig.: (Paramagnetic Material)

13. (a) A small compass needle of magnetic moment
[1]
tn is free to turn about an axis perpendicular
to the direction of uniform magnetic field ‘B’. Magnetic Permeability of paramagnetic’s is more
The moment of inertia of the needle about than air so it allows more lines to pass through
the axis is I. The needle is slightly disturbed it while permeability of diamagnetic is less than
from its stable position and then released. air so it does not allow lines to pass through it.
Prove that it executes simple harmonic n2 o
motion. Hence deduce the expression for its r 0.53 A
z [1]
time period.
(b) A compass needle, free to turn in a vertical 9. For a paramagnetic material, the relative
plane orients itself with its axis vertical at permeability lies between 1 < µr < 1 + e and its
a certain place on the earth. Find out the susceptibility lies between 0 < c < e
values of (i) horizontal component of earth Hence, ‘A’ is a paramagnetic material and
s magnetic field and (ii) angle of dip at the its susceptibility is positive. This is because
place. its relative permeability is slightly greater
than unity. For a diamagnetic material, the
[DELHI 2013]
relative permeability lies between 0  r  1
and its susceptibility lies between –1 < c < 0. (ii) The magnetic field lines, when a paramagnetic
Hence, ‘B’ is a diamagnetic material and its material is placed in an external magnetic
susceptibility is negative. This is because its field, can be diagrammatically represented
relative permeability is less than unity. Here as:
µr, and c refer to the relative permeability and
susceptibility [1+1]
10. (a) Permanent magnet [1]
1. High coercivity.
2. High retentivity [1]
(b) Electromagnet [1]
1. High permeability Fig: The magnetic field lines, when a
2. Low coercivity. paramagnetic material is placed in an external
11. (i) The material used for making permanent magnetic field
magnets should have the following Diamagnetic and paramagnetic materials
characteristics: are distinguished by the magnetic property
(a) High retentivity: It ensures that the called magnetic susceptibility. For diamagnetic
magnet remains strong even after materials, magnetic susceptibility is negative,
removal of the magnetising field. [1/2] whereas for paramagnetic materials, magnetic
(b) High coercivity: It ensures that the susceptibility is slightly positive. [1]
magnetism of the material does not get
12. (a) There will be two magnetic field one due to
easily lost. [1/2]
current and another due to magnetics.h
Apart from these two criteria, the
material should have high permeability.
(ii) The core of an electromagnet should have high
permeability and low retentivity. The high
permeability of the core of an electromagnet
ensures that the electromagnet is strong.

On the other hand, low retentivity of the
core ensures that the magnetism of the core
material gets lost as soon as the current is
switched off. Ferromagnetic materials have
both high permeability and low retentivity.
B = Bo + Bm
Hence, ferromagnetic materials are the
most suitable for making the core of an B = m0 nI + m0 M
electromagnet. [1]
Or B = m0 H + m0 mr H

(i) The magnetic field lines, when a diamagnetic
B = m0 (1 + mr ) H
material is placed in an external magnetic
field, can be diagrammatically represented B = m0 mr H [1]
as:
(b) The susceptibility of this material is
between 0 and 1 so it’s a paramagnetic
material.
N S N S
B=0
Fig: Magnetic field lines, when a diamagnetic
material is placed in an external magnetic
field
[1] [1]
13. (a) Consider a rectangular loop-ABCD carrying If there is V such turns the torque will be
current I. nIAB
a
Rotation axis Where, b → Breadth of the rectangular coil
a → Length of the rectangular coil
C
A = ab → Area of the coil.
N B I
b
I S Case II: Plane of the loop is not along
b
the magnetic field, but makes angle with
it.
A
Brush m
D
C
I I
I B
+ – B
N S
I
Fig: A small compass D
[1]
Case I: The rectangular loop is placed such A
that the uniform magnetic field B is in the [1]
plane of loop. [1] Angle between the field and the normal
No force is exerted by the magnetic field on is q. Forces on BC and DA are equal and
the arms AD and BC. Magnetic field exerts opposite and they cancel each as they are
a force F1 on arm AB. collinear.
∴ F1 = IbB Force on AB is F1 and force on CD is F2
Magnetic field exerts a force F2 on arm CD F1 = F2 = IbB
F2 = IbB = F1 Magnitude of torque on the loop as in the
Net force on the loop is zero. figure:
The torque on the loop rotates the loop in F1
anti-clockwise direction.
F2
a/2
a/2 a/2
B
a/2
m
F1
Torque, F1
a a [1]
  F1  F2
2 2 a a
  F1 sin   F2 sin 
a a 2 2
 IbB  IbB
2 2 t = labB sin q
= l (ab) B t = lAB sin q
t = BIA Where, A = ab
(b) We know, Lorentz force, F = Bqv sin q (a) The torque on the needle is t = m × B
Where, q = angle between velocity of particle In magnitude, t = m × B sin q
and magnetic field = 90° Here t is restoring torque and q is the angle
So, Lorentz force, F = Bqv between m and B.
Thus, the particles will move in circular Therefore, in equilibrium,
path.
d2
I   mB sin 
mv2 mv dt2
Bqv  r m
r Bq p
Negative sign with mB sin q implies that
Let mp = mass of the proton, md = mass restoring torque is in opposition to deflecting
of deuteron, vp = velocity of proton and torque. For small values of in radians, we
vd = velocity of deuteron approximate sin q = q and get
The charge of proton and deuteron are equal. d2
I   mB sin 
Given that, mp vp = md vd [1] dt2

m pv p 2
rp = d  mB
Bq ...(1) Or,  
2 I
dt
m v
rd = d d
Bq This represents a simple harmonic motion.
...(2)
The square of the angular frequency is
As (1) and (2) are equal, so rp = rd = r w2 = mB/I and the time period is,
Thus, the trajectory of both the particles will
I
be same. T  2
mB
Or
(b) (i) As, Horizontal component of earth’s
B magnetic field, BH = B cos d
Putting d = 90°, BH = 0
(ii) For a compass needle align vertical at
a certain place, angle of dip, d = 90°
r
O

CHAPTER 6
Electromagnetic Induction

Electromagnetic induc-
tion
Faraday’s laws, induced 1Q 1Q
emf and current (1 mark) (1 mark)
Lenz’s Law, Eddy cur-
1Q 1Q
rents. Self and mutual
(3 marks) (1 mark)
induction
On the basis of above analysis, it can be said that from exam point of view Electromagnetic
induction, Self and Mutual Inductance are most important concepts of the chapter.
98 CHAPTER 6 : Electromagnetic Induction
[Topic 1] Electromagnetic Induction Laws
Summary Lenz’s law and Conservation of

• Electromagnetic Induction is the one in which by Energy
which electric current is generated with the help The Lenz’s law states that the polarity of
of a magnetic field. induced emf is such that it tends to produce
• The Experiments of Faraday and Henry a current which opposes the change in
The observations from the experiments of magnetic flux that produced it.
Faraday and Henry concluded that it is the
relative motion between the magnet and the coil
that is responsible for generation or induction of
the electric current in the coil.
• Magnetic Flux N N

It is the amount of field lines cutting through a
surface area A defined by unit area vector. The Fig. Lenz’s law
magnetic flux that passes through a plane of area
A and has a uniform magnetic field B, is given by,
Motional Electromotive Force
fB = B.A = BA cosθ where θ is the angle between The relationship between induced emf and a
magnetic field B and Area A. Magnetic flux is a wire moving at a constant speed v is given by
scalar quantity and its SI unit is weber. e = Blv
Energy Consideration: A
B Quantitative Study
A • ‘r’ is the resistance of the movable arm PQ of the
rectangular conductor. Assume that remaining
arms QR, RS, SP have negligible resistance
compared to r. In the presence of magnetic field
there will be a force on the arm AB. This force I(l
Fig. Field lines in a magnetic field × B) is outwards directed in a direction opposite
to the velocity of rod.
B 2 l 2v
Faraday’s Law of Induction • Magnitude of force is = F I= lB .
r
• Faraday’s First Law: Whenever a conductor is B 2 l 2v 2
• Magnitude to push arm PQ = Fv =
placed in a varying magnetic field, there is an r
induced emf and if the conductor circuit is closed,
I
there is an induced current. S
• Faraday’s Second Law: This law of electromagnetic P M
induction states that the magnitude of the
induced emf in a circuit is equal to the time rate l v
of change of magnetic flux through the circuit.
Mathematically, the induced emf is given by Q N
−d φB
ε= , the negative sign indicates direction R
dt
of the induced emf and hence the direction in a x
closed loop. Fig. Energy Consideration in a Magnetic field
CHAPTER 6 : Electromagnetic Induction 99
is present everywhere. Deduce the expression
PREVIOUS YEARS’ for the emf between the centre and the metallic
ring.
[DELHI 2012]
3 Mark Questions
TOPIC 1 7. A rectangular conductor LMNO is placed in
1 Mark Questions a uniform magnetic field of 0.5 T. The field
1. Predict the directions of induced currents in is directed perpendicular to the plane of the
metal rings 1 and 2 lying in the same plane conductor. When the arm MN of length of 20 cm
where current I in the wire is increasing steadily. is moved towards left with a velocity of 10 m/s,
calculate the emf induced in the arm. Given the
1 resistance of the arm to be 5Ω (assuming that
other arms are of negligible resistance) find the
value of the current in the arm.
l
2 B
L M
[ALL INDIA 2012]
2. The motion of copper plates is damped when it V
is allowed to oscillate between the two poles of l
a magnet. If slots are cut in the plate, how will
the damping be affected?
[ALL INDIA 2013] O
N
3. Two spherical bobs, one metallic and the other of
glass, of the same size are allowed to fall freely
from the same height above the ground. Which Or
of the two would reach earlier and why? A wheel with 8 metallic spokes each 50 cm
[DELHI 2014] long is rotated with a speed of 120rev/min in
4. A conducting loop is held below a current a plane normal to the horizontal component of
carrying wire PQ as shown. Predict the direction the Earth’s magnetic field. The Earth’s magnetic
of the induced current in the loop when the field at the place is 0.4 G and the angle of dip is
current in the wire is constantly increasing. 60°. Calculate the emf induced between the axle
and the rim of the wheel. How will the value of
P Q
emf be affected if the number of spokes were
increased?
[ALL INDIA 2011]
8. While travelling back to his residence in the car,
Dr. Pathak was caught up in a thunderstorm. It
[ALL INDIA 2014] became very dark. He stopped driving the car
5. A planar loop of rectangular shape is moved and waited for thunderstorm to stop. Suddenly
within the region of a uniform magnetic field he noticed a child walking alone on the road.
acting perpendicular to its plane. What is the He asked the boy to come inside the car till
direction and magnitude of the current induced the thunderstorm stopped. Dr. Pathak dropped
in it? the boy at his residence. The boy insisted that
[ALL INDIA 2015] Dr. Pathak should meet his parents. The parents
expressed their gratitude to Dr. Pathak for his
2 Mark Questions concern for safety of the child.
6. A metallic rod of ‘L’ length is rotated with angular
Answer the following questions based on the
frequency of ω with one end hinged at the centre
above information:
and the other end at the circumference of a
(a) Why is it safer to sit inside a car during a
circular metallic ring of radius L about an axis
thunderstorm?
passing through the centre and perpendicular
(b) Which two values are displayed by Dr. Pathak
to the plane of the ring. A constant and
in his action?
uniform magnetic field B parallel to the axis
(c) Which values are reflected in parent’s 3. Glass bob will reach the ground earlier than
response to Dr. Pathak? the metallic bob. As the metallic bob falls, it
(d) Give an example of similar action on your intercepts earth’s magnetic field and induced
part in the part from everyday life. currents are set up in it which oppose its
[DELHI 2013] downward motion. But no such currents are
induced in the glass. [1]
9. (a) State Faraday’s law of electromagnetic
4. When current in wire is increased, inward flux
induction.
with loop increases. According to Lenz Law, loop
(b) Explain, with the help of a suitable example, induces outward magnetic flux so anti clockwise
how we can show that Lenz’s law is a current is induced in loop. [1]
consequence of the principle of conservation 5. The magnetic flux linked with a circuit is not
of energy. changing with time so there will be no induced
(c) Use the expression for Lorentz force acting current in the loop. [1]
on the charge carriers of a conductor to dφ B
6. The induced emf =
obtain the expression for the induced emf dt
d
across the conductor of length l moving with e= ( BA)
velocity v through a magnetic field B acting dt
perpendicular to its length. dA
=B
[ALL INDIA 2015] dt
5 Mark Questions \ fB = BA cos f
10. (a) Describe a simple experiment (or activity) to  φ = 0°
[½]
show that the polarity of emf induced in a dA
Where = Rate of change of area of loop
coil is always such that it tends to produce dt
a current which opposes the change of formed by the sector OPQ. Let q be the angle
magnetic flux that produces it. between the rod and the radius of the circle at
(b) The current flowing through an inductor of P at time t.
self-inductance L is continuously increasing. The area of the sector OPQ
Plot a graph showing the variation. θ 1
= π R2 × = R2θ [½]
(i) Magnetic flux versus the current 2π 2
di Where R = Radius of the circle
(ii) Induced emf versus
dt d 1 2  1 2 dθ
Hence e = B ×  R θ  = BR
(iii) Magnetic potential energy stored versus the dt  2 2 dt
Bω R2
current. [DELHI 2015] e=
2 [½]
Solutions Q
1. Using Lenz law, we can predict the direction
of induced current in both the rings. Induced R
current oppose the cause of increase of magnetic
flux. So, [½] P
O
1
[½]
l
2 7. e = vB = 10 × 0.5 × 0.2 = 1 volt
[½] ε 1
i= =
It will be clockwise in ring 1 and anticlockwise R 5
in ring 2. i = 0.2A [1 + 1 + 1]
2. When slots are made in the plate, path length Or
of induced current increases hence resistance BH = Be cosq = 0.4 × 10–4 cos 60° = 2 × 10–5T
increased so eddy current minimized and that’s
why it is less damped. [1] BH ω R2
ε=
2
2 Suppose that the induced current was in the

2 × 10−5 × 4π (0.5)
= direction opposite to the one depicted in Fig.(a).
2 In that case, the South-pole due to the induced
120 current will face the approaching North-pole
ω = 2π × = 4π
60 [1 + 1 + 1] of the magnet. The bar magnet will then be
On increasing the number of spokes, emf attracted towards the coil at an ever increasing
will remain same because they form parallel acceleration. A gentle push on the magnet will
combination. initiate the process and its velocity and kinetic
8. (a) It is safer to be inside a car during energy will continuously increase without
thunderstorm because the car acts like a expending any energy. If this can happen, one
Faraday cage. [1] could construct a perpetual-motion machine
(b) Awareness and humanity [½] by a suitable arrangement. This violates the
law of conservation of energy and hence cannot
(c) Gratitude and obliged [½]
happen. Now consider the correct case shown
(d) Once I came across a situation where a in Fig.(a). In this situation, the bar magnet
puppy was struck in the middle of a busy experiences a repulsive force due to the induced
road during rain and was not able to cross current. Therefore, a person has to do work in
due to heavy flow, so I quickly rushed and moving the magnet. Energy spent by the person
helped him. [1] is dissipated by Joule heating produced by the
9. (a) Faraday gave laws for relating induced emf induced current. [1]
to the flux. These are given as under: (c) I
(i) Whenever there is a change of magnetic flux S
through a circuit, there will be an induced P M
emf and this will last as long as the change
l v
persists.
(ii) The magnitude of the induced emf in a Q N
circuit is equal to the time rate of change of
R
magnetic flux through the circuit.
Mathematically, the induced emf is given by x
dφ The arm PQ is moved to the left side, thus
ε =− B
dt [1] decreasing the area of the rectangular loop. This
(b) Lenz’s law states that the polarity of induced movement induces a current I as shown.
emf is such that it tends to produce a current Let us consider a straight conductor moving in
which opposes the change in magnetic flux a uniform and time independent magnetic field.
that produced it. The negative sign shown Figure shows a rectangular conductor PQRS in
dφ B which the conductor PQ is free to move. The rod
in equation ε = − represents this effect.
dt PQ is moved towards the left with a constant
Conservation of energy: velocity v as shown in the figure. Assume that
there is no loss of energy due to friction. PQRS
forms a closed circuit enclosing an area that
changes as PQ moves. It is placed in a uniform
magnetic field B which is perpendicular to the
N
(a)
plane of this system. If the length RQ = x and
RS = l, the magnetic flux B enclosed by the loop
PQRS will be fB = Blx Since x is changing with
time, the rate of change of flux jB will induce
an emf given by:
dx
N ε = − Bl = Blν
(b) dt
dx
Illustration of Lenz's law. where we have used = – v which is the
dt
speed of the conductor PQ. The induced emf
Fig. Illustration of Lenz’s Law
Blv is called motional emf. Thus, we are able to anti-clockwise direction, with respect to server.
produce induced emf by moving a conductor instead The magnetic moment M associated with this
of varying the magnetic field, that is, by changing induced emf has north polarity, towards the north
the magnetic flux enclosed by the circuit. It is also pole of the approaching bar magnet. Similarly,
possible to explain the motional emf expression by when the north pole of the bar magnet is moved
invoking the Lorentz force acting on the free charge away from the coil, the magnetic flux linked with
carriers of conductor PQ. Consider any arbitrary the coil decreases. To counter this decrease in
charge q in the conductor PQ. When the rod moves magnetic flux, current is induced in the coil in
with speed v, the charge will also be moving with clockwise direction so that its south pole faces
speed v in the magnetic field B. The Lorentz force the receding north pole of the bat magnet. This
on this charge is qvB in magnitude, and its direction would result in an attractive force which opposes
is towards Q. All charges experience the same force, the motion of the magnet and the corresponding
in magnitude and direction, irrespective of their decrease in magnetic flux. [1]
position in the rod PQ. The work done in moving the (b) (i) Since f = LI
charge from P to Q is,
Where, I → strength of current through the coil at
W = qvBl [1] any time
Since emf is the work done per unit charge,
f→ Amount of magnetic flux linked with all turns
W of the coil at that time and, L → Constant of
ε=
q proportionally called coefficient of self-induction. [1]
= Blv
This equation gives emf induced across the rod PQ
The total force on the charge at P is given by
F = q(E + v × B)
10. (a) Lenz law: According to Lenz s law, the
polarity of the induced emf is such that
it opposes a change in magnetic flux
responsible for its production l
− dφ d ( LI )
(ii) Induced emf, e = =−
dt dt
 dl 
i.e. e = − L  
S N  dt 
N S Or
dl
dl
dt
dt
Induced
e.m.f.
Induced
e.m.f.
N S When dl increases When I increases
N S dt
at constant rate at constant rate [1]
(iii) Since magnetic potential energy is given by

[1] 1 2
U= LI
Fig. Effect of Movement of bar magnet on 2
Magnetic flux.
When the north pole of a bar magnet is pushed U
towards the coil, the amount of magnetic flux
linked with the coil, increase. Current is reduced
in the coil from a direction such that it opposes
the increase in magnetic flux. This is possible
only when the current induced in the coil is in
l [1]
[Topic 2] Eddy currents, self and mutual inductance
Summary Self – Inductance

• The production of induced emf in a circuit when
Eddy Currents the current changes in the same circuit is called
• When bulk pieces of conductors are subjected to self-induction.
changing magnetic flux then induced currents are
dI
produced in them which are called as eddy currents. • The induced emf is given by ε = −L , where is
dt
• Eddy currents create a significant drag known as
magnetic damping. the coefficient of self-induction.
• The applications of eddy currents are in magnetic • The direction of induced emf is given by Lenz’s
braking in trains, electromagnetic damping, Law.
electric power meters and induction furnace.
AC Generator
Inductance
• The electromagnetic induction has its applications
• Flux change produced by another coil in the close
proximity of a coil or flux exchange produced by in an AC generator, where mechanical energy is
the same coil induces electric current. converted to electrical energy.
• The inductance in series is given by Ls = L1 + L2 +
L3 + .......... Coil Axle
• The inductance in parallel is given by
1 1 1 1
= + + + ...........
L p L1 L 2 L 3
Mutual- Inductance
• When the emf is induced into the adjacent coil
situated within the same magnetic field, the emf N S
is said to be induced magnetically or by mutual
induction. Slip
rings Alternating emf
• Mutual inductance of a pair of coils, solenoids etc.
depends on their relative orientation as well as
their separation.
dI 2 Carbon
ε1 = −M
dt brushes
• Mutual Inductance of two coils is given by Fig. A.C. Generator
µ0 µr N p N s As where A , A are the cross • The motional emf is of a coil with N turns and area
M = p s
Ip A, rotated at v revolutions per second in a uniform
d
sectional areas of primary and secondary coil in magnetic field B is given as, ε = −NBA (cos ωt )
dt
m2, I is the coil current and Ns, Np are the number
of turns of secondary and primary coils
respectively.
6. (a) Define self inductance. Write its S.I. units.
PREVIOUS YEARS’ (b) Derive an expression for self inductance of

a long solenoid of length I, cross-sectional
area A having N number of turns.
TOPIC 2 3 Mark Questions
1 Mark Questions 7. Define mutual inductance between a pair of coils.
1. A plot of magnetic flux (f) versus current (I) Derive an expression for the mutual inductance
is shown in the figure for two inductors A and of two long coaxial solenoids of same length
B. Which of the two has larger value of self wound one over the other.
inductance? Or
A Define self-inductance of a coil. Obtain the
expression for the energy stored in an inductor
L connected across a source of emf.
B [ALL INDIA 2017]
8. State the principle of an ac generator and explain
its working with the help of a labeled diagram.
Obtain the expression for the emf induced in a
coil having N turns each of cross-sectional area,
I [DELHI 2018]
rotating with a constant angular speed in a
2 Mark Questions magnetic field B, directed perpendicular to the
2. A current is in induced in coil C1 due to the axis of rotation.
motion of current carrying coil C2. [ALL INDIA 2018]
(a) Write any two ways by which a large deflection 5 Mark Questions
can be obtained in the galvanometer G,
9. (a) Draw a schematic sketch of an ac generator
(b) S u g g e s t a n a l t e r n a t i v e d e v i c e t o describing its basic elements. State briefly
demonstrate the induced current in place its working principle. Show a plot of
of a galvanometer. variation of
C1 (i) Magnetic flux and
C1
(ii) Alternating emf versus time generated by a
loop of wire rotating in a magnetic field.
(b) Why is choke coil needed in the use of
fluorescent tubes with ac mains?
G K [DELHI 2014]

[ALL INDIA 2011] Solutions
3. Define self-inductance of a coil. Show that
1. Inductor A Self inductance is given by L = NΦ/I,
magnetic energy required to build up the current
where I is current . From the plot it is clear that
1 2
I in a coil of self-inductance L is given by LI A has a smaller value of I for given φ hence larger
2 value of self inductance. [1]
. [ALL INDIA 2012]
4. How does the mutual inductance of a pair of coils 2. (a) Two ways by which a large deflection can be
change when obtained in the galvanometer are :
(i) distance between the coils is decreased and (i) By increasing the relative motion between
(ii) number of turns in the coils is decreased? the coils. [1]
[ALL INDIA 2013] (ii) By inserting an iron rod into the coils along
5. (i) Define mutual inductance. their axes. [½]
(ii) A pair of adjacent coils has a mutual (b) An LED. (Light Emitting Diode) [½]
inductance of 1.3 H. If the current in one coil 3. Self-inductance is the inherent inductance of a
changes from 0 to 20 A in 0.5 s, what is the circuit, given by the ratio of the electromotive
change of flux linkage with the other coil? force produced in the circuit by self-induction to
[ALL INDIA 2016] the rate of change of current producing it. It is
also called coefficient of self-induction. dI = 20A, Dt = 20s

Suppose I = Current flowing in the coil at any − MdI
e=
time dt [1]
f = Amount of magnetic flux linked 20
It is found that f ∝ I e = −1.5 ×
0.5
f = LI
e = –60V
Where, L is the constant of proportionality and
So the flux linked with the other coil is given by
is called coefficient of self induction.
Df = eDt = –60 × 0.5
SI unit of self-inductance is Henry.
= – 30Wb [1]
Let at t = 0 the current in the inductor is zero.
So at any instant t, the current in the inductor 6. The phenomenon in which emf is induced in a
dI single isolated coil due to change of flux through
is I and the rate of growth of I is . the coil by means of varying the current through
dt
dI the same coil is called self inductance. S.I. unit
Then, the induced emf is e = L × [1] of inductance is Henry. [1]
dt
If the source is sending a constant current I (b) Magnetic field B inside a solenoid carrying
through the inductor for a small time dt, then a current i is µoni
small amount of work done by the source is given B = µoni
by
Let n be the number of turns per unit length.
 LdI 
dW = eIdt =  Idt = LIdI Where, Nϕ B = nlBA
 dt 
N is total number of turns [½]
The total amount of work done by the source of l is the length of the solenoid
e.m.f., till the current increases from its initial
Nϕ B
value I = 0 to its final value I is given by Inductance, L =
I I
i
 I2 
W = ∫ LIdI = L∫ IdI = L   Substituting, we obtain
0 0  2 
nlBA
L=
1 i [1]
= LI 2
2 [1] Substituting the value of B. we obtain
This work done by the source of emf is used in nl µo niA
L=
building up current from zero to Io is stored in i
the inductor in energy form. Therefore, energy
2
stored in the inductor is L = n l µo A
1 2 Inductance L of a solenoid is: [½]
U= LI
2 L = n2 lµoA
4. (i) Mutual inductance increased on decreasing 7. Mutual inductance: The phenomenon
distance. [1] according to which an opposing emf is produced
(ii) Mutual inductance decreased on decreasing as result of change in current or magnetic
the number of turns. [1] flux linked with a neighboring coil. Mutual
5. (i) Mutual induction is the phenomenon of inductance of two long wareial solenoids:
production of induced emf in one coil due to
change of current in the neighboring coil. S1
The coil in which the current changes is
called primary coil and the coil in which emf
is induced is called the secondary coil. [1]
l
(ii) M = 1.5 H
Ii = OA S2
Ib = 20A
Fig. Mutual Inductance in a coil
Let n1 be the no. of turns per unit length of S1, The induced emf is also called back emf when
n2 be the number of turns per unit length of the current in a coil is switched on. The self-
S2. I1 be current passed through S1, f21 be the induction opposes the growth of the current
flux linked with S2 due to charge flowing in and when the current is switched off, the
S1. [1] self-induction opposes the decay of the current
f21 ∝ S1 so, self-induction is also called the inertia of
electricity. [1]
f 21 = M 21 I 1where M 21 coefficient of mutual
induction of two solenoid. When current is (ii) Self-inductance of long solenoid: A long
passed through , an emf is induced in solenoid solenoid is one whose length is very large
S 2 . Magnetic field produced inside S 1 on as compared to its area of cross section.
passing current I1, B1 = µo n1 I1 Magnetic field (B) at any point inside B =
Magnetic flux linked with each turn of the µ NI
B= 0 …………. ( i )
solenoid S2 will be equal to B1 times the area l
of cross section of solenoid S1. So, magnetic Magnetic flux through each two of the solenoid
flux linked with each turn of the solenoid  N 
S2 = B1 A Therefore, total magnetic flux linked f = B × Area of each turn φ =  µ0 I  A
 l 
with solenoid S2 will be
Where A = area of each turn of the solenoid
φ21 = B1 A × n 2 l = µ0 n1 l1 × A × n 2 l
Total magnetic flux linked with the solenoid
φ21 = µ0 n1n 2 AI1 l = flux through each turn total no. of turns [1]
M21 = µ0 n1 n2 Al ………… ( i ) N f = µ oIA × N ..........(ii)
[1]
If L is coefficient of self-inductance of the
Similarly, the mutual inductance between the solenoid then
two solenoids, when current is passed through
N f = LI .......(iii)
S2 and induced emf is produced in solenoid S1
and is given by from (ii) & (iii) we get
M12 = µo n1n2AI1l N
LI = µ0 I×N
M12 = M12 = M = (say) l
µ N2 A
Hence coefficient of mutual induction between or L = 0 ……….. ( iv )
the two long solenoid M12 = µo n1n2Al l
We can write equation (i) as The magnitude of emf is given by
N N  dE
M = µ0  1   2  π r12 × l Let e = L
 l  l  dt
Multiplying (I) to the both sides we get
µ µ N N A eIdt = LIdt ..........(v)
M= 0 r 1 2
l [1] dq
OR =
But I =
dt
or Idt dq
Self-inductance: Self-inductance is the Also, work done

property of a coil by virtue of which, the coil (dw) = voltage (e) × charge (dq)
opposes any change in the strength of current Or dq = e × dq = eldt ………….. (vi)
flowing through it by inducing an emf in itself. So, from (v) & (vi)
dw = LIdI ……….. (vii)
L B Total work done in increasing current from

zero to I 0, we have By integrating both sides
of equation (vii) we get
w I0
+ –
∫ dw = ∫ LIdI
0 0
B R
1 2 The basic elements of an ac generator are

w= LI shown in figure. It consists of a coil mounted
2 0
on a rotor shaft. The axis of rotation of the
This work done through inductor is stored as coil is perpendicular to the direction of the
the potential energy (u) in the magnetic field magnetic field. The coil (called armature)
of inductor is mechanically rotated in the uniform
1 2 magnetic field by some external means. The
u= LI
2 0 [1] rotation of the coil causes the magnetic flux
through it to change, so an emf is induced in
8. Principle-Electromagnetic Induction
the coil. The ends of the coil are connected
AC Generator: The phenomenon of to an external circuit by means of slip rings
electromagnetic induction has been and brushes.
technologically exploited in many ways. An When the coil is rotated with a constant
exceptionally important application is the angular speed w the q angle between the
generation of alternating currents (ac). The magnetic field vector B and the area vector A
modern ac generator with a typical output of the coil at any instant is w = qt (assuming
capacity of 100 MW is a highly evolved q = 0° at t = 0).
machine. In this section, we shall describe As a result, the effective area of the coil
the basic principles behind this machine. The exposed to the magnetic field lines changes
Yugoslav inventor Nicola Tesla is credited with time, and from equation the flux at any
with the development of the machine. As was time t is
pointed out in, one method to induce an emf f B = BA cos q = BA cos wt
or current in a loop is through a change in the From Faraday’s law, the induced emf for the
loop’s orientation or a change in its effective rotating coil of N turns is then,
area. As the coil rotates in a magnetic field dφ d
B, the effective area of the loop (the face e = − N B = − NBA (cos ω t )
dt dt
perpendicular to the field) is A cosq,where q
Thus, the instantaneous value of the emf is
is the angle between A and B. This method
\ e = NBAw sin wt -------- (1)
of producing a flux change is the principle
Where NBAw is the maximum value of the
of operation of a simple ac generator. An ac
emf, which occurs when sin wt = ±1. If we
generator converts mechanical energy into
denote NBAw as
electrical energy.
f = f o sin wt ------------ (2)
Coil Axle Since the value of the sine function varies
between +1 and – 1, the sign, or polarity of
the emf changes with time. Note from figure
that the emf has has its extreme value when
q = 90° or q = 270°, as the change of flux is
greatest at these points.
The direction of the current changes
N S periodically and therefore the current is
called alternating current (a.c.)
Slip Since f = 2pv, equation (2) can be written as
rings Alternating emf f = f o sin 2pvt ------------------- (3)
Where f is the frequency of revolution of the
generator’s coil.
Note that equation 2 and 3 give the
Carbon instantaneous value of the emf and q varies
brushes between +q o varies between –q o periodically.
Fig. An A. C. Generator We shall learn how to determine the time-
averaged value for the alternating voltage
and current. [1 + 1]
Fig. Operating/Working of an a.c. generator {1]

9. (a) Principle is “Based on the phenomenon of electromagnetic induction.
Construction:
B C C B
S N S N
I I
A D D A
B1 B1
R1 R1
B2 B2
R2 R2
R R
(i) (ii)
Fig. An A.C. Generator
[1]
Main parts of an ac generator:
• Armature: The rectangular coil ABCD
• Field Magnets Two pole pieces of a strong electromagnet
• Slip Rings. The ends of the coil ABCD are connected to two hollow metallic rings R1 and R2.
• Brushes: B1 and B2 are two flexible metal plates or carbon rods. They are fixed and are kept in tight
contact with R1 and R2, respectively. [1]
Working: As the armature coil is rotated in the As the coil rotates, angle q changes. Therefore,
magnetic field, angle q between the field and magnetic flux linked with the coil changes and
the normal to the coil changes continuously. an emf is induced in the coil. At this instant, if
Therefore, magnetic flux linked with the coil e is the emf induced in the coil, then [1]
changes and an emf is induced in the coil. dθ d
According to Flemings right hand rule, current e=− = − ( NAB cos ω t )
dt dt
is induced from A to B in AB and from C to D in
CD. In the external circuit, current flows from B2 d
= − NAB (cosω t )
to B1. To calculate the magnitude of emf induced: dt
Suppose, = NBA(–sinwt) w
A → Area of each turn of the coil e = NBA wsin wt
N → Number of turns in the coil The graph between alternating emf versus time

B → Strength of the magnetic field is shown below
y
q → Angle which normal to the coil makes with
at any instant t [1]
Normal Coil e
S N
2T/4 T
x
T/4 3T/4
Fig. A.C. Generator
Hence, magnetic flux linked with the coil in this
position is given by,

( )
–e
φ = N B⋅ A

NBA cosq = NBA cos wt (1)

Where, “w” is angular velocity of the coil.
Fig. Graph Between Alternating emf versus time
Graph between magnetic flux and time, according
(b) A choke coil is an electrical appliance used
to equation (1) is given below:
for controlling current in an a.c. circuit.
Therefore, if we use a resistance R for the
same purpose, a lot of energy would be
wasted in the form of heat etc. [1]
Fig. Graph Between Magnetic flux & time

CHAPTER 7
Alternating Current

Alternating currents,
peak and RMS value
of alternating current/
voltage
Reactance and Imped- 1Q
ance (5 marks)
1Q 1Q 1Q
LC oscillations (qualita- (1 mark), (5 marks) (5 marks)
tive treatment only); LCR 1Q 1Q
series circuit,Resonance (5 marks) (3 marks)
Power in AC circuits, 1Q 1Q 1Q
power factor wattless cur- (5 marks) (5 marks), (3 marks)
rent. AC generator and 1Q
transformer (5 marks)
On the basis of above analysis, it can be said that from exam point of view AC Source, AC
Circuit, AC Generator, L-C-R Circuit and Phasor Diagram are most important concepts of the
chapter.
112 CHAPTER 7 : Alternating Current
[Topic 1] Introduction to Alternating Current
Summary Mean value of Alternating Current

The value of alternating current that would give
Alternating Current same amount of charge in to a circuit at half cycle
The electric main supply that varies like a sine that is sent for steady current in the same duration.
function with time is called alternating voltage 2I o
and the current drawn by it in the circuit is called =
I avg = 0.637I o
π
Alternating current. Alternating current is the
current which varies on two factors i.e. magnitude
and the direction periodically and alternatively.
Alternating Voltage
Mathematically alternating current can be expressed Alternating voltage is the voltage which varies on two
as: factors i.e. magnitude and the directions periodically
I = I0sinwt and alternatively.
Where I0 , is the peak value of alternating current. Alternating Voltage is expressed mathematically as,
V = V o sin ωt

Vo
=
V r ms = 0.707V o or V r ms
= 70.7% of V o
2
2V o
=
V avg = 0.637V o or V avg
= 63.7% of V o
π
The alternating current and alternating voltage is
illustrated in the following diagram:
Fig.: Alternating Current in an electrical circuit.

RMS value of Alternating Current
The value of alternating current over a complete
cycle which would generate same amount of heat in
a given resistors that is generated by steady current
in the same resistor and in the same time during a
complete cycle.
Io
= =
I r ms 0.707I o
2
Fig.: Variation of V0, I0 w.r.t wt.
Fig.: Variation of Current with respect to wt.

CHAPTER 7 : Alternating Current 113
3 Mark Questions
PREVIOUS YEARS’ 8. The figure shows a series LCR circuit with
L = 75 H, C = 80 μF, R = 40Ω connected to a
variable frequency 240 V source, calculate
EXAMINATION QUESTIONS (i) the angular frequency of the source which
TOPIC 1 drives the circuit at resonance,
(ii) the current at the resonating frequency,
1 Mark Questions (iii) the rms potential drop across the inductor
1. Define the term ‘wattless current’. at resonance.
[ALL INDIA 2011]
2. Why is the use of a.c. voltage preferred over d.c.
voltage? Give two reasons.
[ALL INDIA 2014]
3. Define capacitive reactance. Write its S.I. units.
[DELHI 2015]
4. Define ‘quality factor ‘of resonance in series LCR
circuit. What is its SI unit? [ALL INDIA 2012]
[DELHI 2016] 9. The figure shows a series LCR circuit with L
= 10.0H, C = 40 µF, R = 60W connected to a
2 Mark Questions variable frequency 240 V source. Calculate:
5. When an ideal capacitor is charged by a dc
battery, no current flows. However, when an ac
source is used the current flows continuously.
How does one explain this based on the concept
of displacement current?
[DELHI 2012]
6. (a) For a given a.c, i  im sin  t show that the
average power dissipated in a resistor R (i) the angular frequency of the source which
1 2 drives the circuit at resonance,
over a complete cycle is im R. (ii) the current at the resonating frequency,
2
(iii) the rms potential drop across the inductor
(b) A light bulb is rated at 125 W for a 250 V at resonance. [DELHI 2012]
a.c. supply. Calculate the resistance of the 10. In a series LCR circuit connected to an ac source
bulb. of variable frequency and voltage v = vm sin ωt,
[ALL INDIA 2013] draw a plot showing the variation of current
7. (i) When an AC source is connected to an ideal (I) with angular frequency (ω) for two different
capacitor, show that the average power values of resistance R1 and R2 (R1> R2). Write
supplied by the source over a complete the condition under which the phenomenon
cycle is zero. of resonance occurs. For which value of the
(ii) A bulb is connected in series with a resistance out of the two curves, a sharper
variable capacitor and an A.C. source as resonance is produced? Define Q-factor of the
shown. What happens to the brightness of circuit and give its significance.
the bulb when the key is plugged in and [DELHI 2015]
11. An inductor L of inductance XL is connected in
capacitance of the capacitor is gradually
series with a bulb B and an ac source. How would
reduced .
brightness of the bulb change when (i) number of
turns in the inductor is reduced, (ii) an iron rod
is inserted in the inductor and (iii) a capacitor
of reactance XC = XL is inserted in series in the
circuit. Justify your answer in each case.
4 Mark Questions 5 Mark Questions

12. A group of students while coming from the school 15. (a) When a bar magnet is pushed towards
noticed a box marked “Danger H.T. 2200 V” at (or away) from the coil connected
a substation in the main street. They did not to a galvanometer, the pointer in the
understand the utility of such a high voltage, galvanometer deflects. Identify the
while they argued the supply was only 220 V.
phenomenon causing this deflection and
They asked their teacher this question the next
write the factors on which the amount and
day. The, teacher thought it to be an important
direction of the deflection depends. State the
question and therefore, explained to the whole
laws describing this phenomenon.
class.
Answer the following questions: (b) Sketch the change in flux, emf and force
(i) What device is used to bring the high voltage when a conducting rod PQ of resistance R
down to low voltage of a.c. current and what and length l moves freely to and fro between
is the principle of its working? A and C with speed υ on a rectangular
(ii) Is it possible to use this device for bringing conductor placed in uniform magnetic field
down the high dc voltage to the low voltage? as shown in the figure.
Explain.
(iii) Write the values displayed by the students
and the teacher.
[DELHI 2015]
13. (a) Using phasor diagram, derive the expression
for the current flowing in an ideal inductor
connected to an a.c. source of voltage, v = v0
sin wt. Hence plot graphs showing variation
of (i) applied voltage and (ii) the current as
a function of ωt. Or
(b) Derive an expression for the average power In a series LCR circuit connected to an ac source
dissipated in a series LCR circuit.
of voltage transformation ratio v = vm sin wt,
[ALL INDIA 2015]
use phasor diagram to derive an expression
14. The teachers of Geeta’s school took the students
for the current in the circuit. hence obtaine
on a study trip to a power generating station,
the expression for the power dissipated in the
located nearly 200km away from the city. The
circuit. Show that power dissipated at resonance
teacher explained that electrical energy is
is maximum.
transmitted over such a long distance to the city,
in the form of alternating current (ac) raised to a [ALL INDIA 2016]
high voltage. At the receiving end in the city, the 16. (i) An a. c. source of voltage V = V0 sin wt is
voltage is reduced to operate the devices. As a connected to a series combination of L, C
result, the power loss is reduced. Geeta listened and R. Use the phasor diagram to obtain
to the teacher and asked questions about how expressions for impedance of the circuit and
the ac is converted to a higher or lower voltage. phase angle between voltage and current.
(a) Name the device used to change the Find the condition when current will be in
alternating voltage to a higher or lower phase with the voltage. What is the circuit
value. State one cause for power dissipation in this condition called? (ii) In a series LR
in this device. circuit XL = R and power factor of the circuit
(b) Explain with an example, how power loss is P1. When capacitor with capacitance C
is reduced if the energy is transmitted over such that XL = XC is put in series, the power
long distances as an alternating current P
rather than a direct current. factor becomes P1. Calculate 1 .
P2
(c) Write two values each shown by the teachers
and Geeta. [DELHI 2016]
[ALL INDIA 2018]
17. A device ‘X’ is connected to an ac source v = v0 sin 19. A device X is connected across an ac source of
wt. The variation of voltage, current and power voltage V = V0 sin wt. The current through X is
in one cycle is shown in the following graph: given as I = I0coswt
(a) Identify the device X and write the
expression for its reactance.
(b) Draw graphs showing variation of voltage
and current with time over one cycle ac for
X.
(c) How does the reactance of the device X vary
with frequency of the ac? Show this variation
graphically.
(a) Identify the device ‘X’.
(d) Draw the phasor diagram for the device X.
(b) Which of the curves A, B and C represent the
[ALL INDIA 2018]
voltage, current and the power consumed in
the circuit? Justify your answer.
(c) How does its impedance vary with frequency Solutions
of the ac source? Show graphically.
(d) Obtain an expression for the current in the 1. An AC circuit containing only capacitor or
circuit and its phase relation with ac voltage. inductor will have zero power dissipation even
Or though the current is flowing through it. Such
(a) Draw a labelled diagram of an ac generator. current is called wattless current. [1]
Obtain the expression for the emf induced 2. (i) AC generator are simpler & cheaper than
in the rotating coil of N turns each of DC generator as commutator is not used in
cross-sectional area A, in the presence of a AC generator. [½]
magnetic field B (ii) AC can be stepped up or down using
(b) A horizontal conducting rod 10 m long transformer so its transmission is cheaper
extending from east to west is falling
and efficient. [½]
with a speed 5.0 ms–1 at right angles to
the horizontal component of the Earth’s 3. Capacitor reactance is the resistance offered by
magnetic field, 0.3 × 10Wbm–2. Find the a capacitor to the flow of a.c.
instantaneous value of the emf induced in
It is given by, XC  1 [½]
the rod. C
[ALL INDIA 2017]
18. (a) Derive an expression for the average power Where, w = 2pf
consumed in a series LCR circuit connected
to a.c. source in which the phase difference f → Frequency of the source
between the voltage and the current in the 1
XC 
circuit is 0. 2 fC
[½]
(b) Define the quality factor in an a.c. circuit.
Why should the quality factor have high w → Angular frequency of the source
value in receiving circuits? Name the factors C → Capacitance of the capacitor
on which it depends. The SI unit of capacitor reactance is ohm (W).
Or
4. The Q factor of series resonance circuit is defined
(a) Derive the relationship between the peak
as the ratio of the voltage developed across
and the rms value of current in an a.c.
circuit. the inductor or capacitor at resonance to the
(b) Describe briefly, with the help of labelled impressed voltage, which is the voltage across
diagram, working of a step-up transformer. R. [½]
A step-up transformer converts a low voltage 1 L
into high voltage. Does it not violate the Q=
R C
principle of conservation of energy? Explain.
[DELHI 2017] It is dimensionless hence, it has no units. [½]
5. When an ideal capacitor is charged by dc battery, 1

charge flows till the capacitor gets fully charged. 
When an ac source is connected then conduction 400  10 6
dq 1 1000
current ic = flows in the connecting wire.  =
dt 20  10 3 20 [1]

Due to charging current, charge deposited on
w = 50rads–1
the plates of the capacitor changes with time.
Changing charge causes electric field between (ii) At resonant frequency, we know that
the plates of capacitor to be varying, giving rise the inductive reactance cancels out the
d capacitive reactance.
to displacement current id   o c . [As Impedance, Z = R = 40W
dt
displacement current is proportional to the rate The current at resonant frequency
of flux variation] [1] V 240
=
Irms = = 6A
Between the plates, electric field R 40 [1]
 q (iii) For rms potential drop across inductor
E 
 o A o VL  Irms  X L
q
Electric flux,  c  E A  A  Irms   L
A o
= 6 × 50 × 5
 d d  qA  dq = VL = 1500V [1]
So, id  o c    ic
dt dt  A o  dt 9. (i) Resonant angular frequency: Given L = 10H
and C = 40µF = × 10–6 F
Displacement current brings continuity in
1 1
the flow of current between the plates of the o  
capacitor. [1] LC 10  40  106

2 2 2

6. P  i R  im sin  t R  
1

1
3
400  10 6 20  10
2
 
2 im R 1000
(a) P  im R sin 2  t = [1] =
2 20

2
V 2  250 wo = 50rads–1 [1]
(b) R   [1] (ii) At resonant frequency, we know that
P 125
R = 500 W the inductive reactance cancels out the
7. (i) Power dissipation in AC circuit is capacitive reactance.
R The impedance = Z = 60W the value of
P  Vrms Irms cos where cos 
Z resistance.
for an ideal capacitor R = 0
The current amplitude at resonant frequency,
So cos f = 0, So P = 0
Eo 2 Ev 2  240 339.60
Hence power dissipated is minimum. [1] Io    
Z R 60 60
(ii) When AC source is connected, the capacitor
1 Io = 5.66A [1]
offers capacitive reactance XC  . The
C (iii) The R.M.S value of current
current flows in the circuit and the lamp I 5.66
I  o 
glows. On reducing C, XC increases. 2 2

Therefore, glow of the bulb reduces. [1]
Iv = 4A
8. Given, L = 5. H, C = 80 μF, R = 40Ω V = 240V
For R.M.S potential drop across inductor
(i) Resonant angular frequenc
VL = Iv × XL
1 1
   = Iv × wL =4 × 50 × 10
LC 5  80  106
VL = 2000V [1]
10. Figure shows the variation of wo M rad/sh in a LCR (ii) When an iron rod is inserted in the inductor,
circuit for two values of resistance R1 and R2(R1 > R2), the self-inductance increases. Consequently,
the inductive reactance XL = wL increases.
This decreases the current in the circuit and
the bulb glows dimmer. [1]
(iii) With capacitor of reactance XC = XL, the
R2   X L  X C   R ,
impedance Z  2
becomes minimum, the current in circuit

becomes maximum. Hence the bulb glows
with maximum brightness. [1]
12. (i) The device that is used to bring high voltage
The condition for resonance in the LCR circuit down to low voltage of an a.c. current is a
1 transformer. It works on the principle of
is,  o  [1] mutual induction of two windings or circuits.
LC When current in one circuit changes, emf is
We can observe that the current amplitude is induced in the neighbouring circuit. [1]
maximum at the resonant frequency wo. Since,
(ii) The transformer cannot convert d.c. voltages
V
im = m at resonance, the current amplitude because it works on the principle of mutual
R induction. When the current linked with
for case R2 is sharper to that for case R1.
the primary coil changes the magnetic flux
Quality factor or simply the Q-factor of a linked with the secondary coil also changes.
resonant LCR circuit is defined as the ratio of This change in flux induces emf in the
voltage drop across the capacitor (or inductor) secondary coil. If we apply a direct current
to that of applied voltage. to the primary coil the current will remain
1 L constant. Thus, there is no mutual induction
It is given by Q = [1]
R C and hence no emf is induced. [1 + 1]
The Q factor determines the sharpness of the (iii) The value of gaining knowledge and curiosity
resonance curve and if the resonance is less about learning new things is being displayed
sharp, the maximum current decreases and by the students. The value of providing
also the circuit is close to the resonance for a good education and undertaking the doubts
larger range of frequencies and the regulation of students has been displayed by the
of the circuit will not be good. So, less sharp the teacher. [1]
resonance, less is the selectivity of the circuit 13. (a)
while higher is the Q, sharper is the resonance
curve and lesser will be the loss in energy of the
circuit.
When XL = XC or VC , the LCR circuit is said to be
in resonance condition. [1]
11.
Fig.: An ac source connected to an inductor

Figure (a) shows an ac source connected to an
inductor. Usually, inductors have appreciable
resistance in their windings, but we shall
assume that this inductor has negligible
resistance. Thus, the circuit is a purely inductive
ac circuit. Let the voltage across the source be
ig.: Inductor Connected in an Electrical Circuit.
F v = vm sin wt. Using the Kirchhoff’s loop rule,
Se(t) = 0, and since there is no resistor in the
(i) When the number of turns in the inductor circuit,
is reduced, its reactance XL decreases. The
current in the circuit increases and hence
brightness of the bulb increases. [1]
The inductive reactance limits the current in

  L di  0 ... (i) a purely inductive circuit in the same way as
dt
the resistance limits the current in a purely
where the second term is the self-induced resistive circuit. The inductive reactance is
Faraday emf in the inductor; and L is the self- directly proportional to the inductance and to
inductance of the inductor. The negative sign the frequency of the current. [1]
follows from Lenz’s law.
From equation (i) we have
di
 L  L m sin  t ... (ii)
dt
di
– Lv  Lvm sin t (b) (c)
dt
Fig.: (b) A phasor diagram for the circuit in fig. (a)
Equation (ii) implies that the equation for i(t),
the current as a function of time, must be such Fig.: (c) Graph of V and I versus.
di
that its slope is a sinusoidally varying (b) We have seen that a voltage v = vm sin wt applied
dt
to a series RLC circuit drives a current in the
quantity, with the same phase as the source circuit given by i = im sin (wt + j)where
voltage and an amplitude given by vm/L. To
m  X  XL 
di im  and   tan 1  C 
obtain the current, we integrate with respect z  R
to time: dt
Therefore, the instantaneous power p supplied
di 
 dt  m  sin  t dt by the source is
dt L
p  vi   m sin  t   im sin  t   

i   m cos  t   constant [1]
L  m im
 cos   cos 2 t   
2  [1 + 1]
The integration constant has the dimension
of current and is time independent. Since the The average power over a cycle is given by the
source has an emf which oscillates symmetrically average of the two terms in R.H.S. of above
about zero, the current it sustains also oscillates equation. It is only the second term which is
symmetrically about zero, so that no constant time-dependent. Its average is zero (the positive
or time-independent component of the current half of the cosine cancels the negative half).
exists. Therefore, the integration constant is Therefore,
zero.  i  i
P  m m cos   m m cos 
  2 2 2
Using  cos  t   sin  t   , we have
 2 = VI cos f
This can also be written as
 
i  im sin  t   P = I2 Z cos f
 2
So, the average power dissipated depends not
m only on the voltage and current but also on the
Where, im  is the amplitude of the current.
L cosine of the phase angle f between them. The
quantity cosφ is called the power factor.
The quantity wL is analogous to the resistance
14. (a) Step up or step down transformer, Eddy
and is called inductive reactance, denoted by XL:
current losses. [1]
XL = wL
(b) With higher voltage, power losses are less,
The amplitude of the current is, then so voltage can be increased by step up
 transformer and transformer works on A/c
im  m only. [1 + 1]
XL

(c) Both are interested towards technical
The dimension of inductive reactance is the same knowledge and both are having sufficient
as that of resistance and its SI unit is ohm (Ω). ideas about power transmission. [1]
15. (a) When a bar magnet pushed towards or away Forward journey
from coil, magnetic flux passing through coil Thus, for b > x ≥ 0
change with time and cause induced emf flux j = BLx
hence induced current according to faraday’s
⇒ j ∝ = BLx
law of induction. Induced emf in the coil is
For 2b ≥ x ≥ b Flux, j = Bbl [Constant]
given as
Return journey
Nd
 For b ≤×≤ 2b,
dt [½]
f = constant = Bbl
Induced emf and hence current depends on For 0 ≤×≤ b,
(i) no. of turns in the coil j = BLx [Decreasing] [½]
(ii) motion of magnet [½] Graphical representation
Direction of current depends on the motion of
magnet whether moving towards coil or away
from the coil.
Faraday’s Laws of Electromagnetic Induction
(i) Whenever there is a change in magnetic flux
linked with a coil, an emf is induced in the
coil. The induced emf is proportional to the
rate of change of magnetic flux linked with [½]
the coil
Case II For b > x ≥ 0
 As, j = BLx
i.e.   [½]
t
d dx  dx 
  B1  B1 v 
(ii) emf induced in the coil opposes the change
dt dt  dt 
in flux, i.e.,
d
 Induced emf e   vB1
 dt1
t
For 2b ≥ x ≥ b,
where k is a constant of proportionality Negative
sign represents opposition to change in flux. In As, j = Bbl
SI system φ is in weber, t in second, ε in volt, d ’
0
 dt1
when k = 1  
t ⇒e=0

If the coil has N-turns, then    k [½] Forward journey
t
For b > x ≥ 0
(b) Case I When PQ moves forward. e = –vBl
(i) For 0 ≤ x < b For 2b ≥ x ≥ b, Backward journey
Magnetic field, B exists in the region. For b > x ≥ 0
\ Area of loop PQRS = lx e = –vBl
\ Magnetic flux linked with loop PQRS, For 2b ≥ x ≥ b, e = 0 [½]
B Variation of induced emf
  BA 
x Or
B
  ---------- (i) [b > x ≥ 0] [½]
x
(ii) For, 2b ≥ x ≥ b
B=0
∴ Flux linked with loop PQRS is uniform and
given by
B
   x  b
b
Figure shows a series LCR circuit connected to 16. (i) Let a, series LCR circuit is connected to an
an ac source ε. As usual, we take the voltage of ac source V (Fig). We take the voltage of the
the source to be v = vm sinwt. source to be V = V0 sin ωt.
Assuming, XL > XC ⇒ VL = VC
So, Net voltage V  VR 2  VL  VC 2
where, VL , VC and VR are PD across L, C and R
[1]
respectively.
The AC current in each element is the same at
But, VR = IR, VL = IXL, VC = IX any time, having the same amplitude and phase.
V It is given by, I = I0 sin (ωt + f)
R2   X L  X C 
2

I
R2   IX L  IXC 
2
V

V
R2   X L  X C 
2
Impedance  Z  
I
VL  VC
tan 
VR

X L  XC
tan 
R [½]
So I = Im sin (wt + j), where φ is the phase
difference between voltage and current source. [1]
Power dissipated in AC circuit: Fig.: Phasor diagram for LCR circuit.
We have seen that a voltage v = vm sinwt applied Let VL , VR , VC and V represent the voltage across
to a series RLC circuit drives a current in the the inductor , resistor, capacitor and the source
circuit given by i = im sin (wt + j) where respectively.
vm  X  XC  VC > VL
im  and tan   tan 1  L 
Z  R  Vo2  VR2   VC  VL 
2

Therefore, the instantaneous power p supplied
Vo2   I o R   I o XC  I o X L 
2 2
by the source is
p  vi  vm sin  t   im sin  t   
Vo2  I o2  R2   XC  X L  
2

  [1]

vm im
 cos   cos  2 t    Vo Vo
2 And I o   Io 
Z
R2   X C  X L 
2
The average power over a cycle is given by the
average of the two terms in R.H.S of above
R2   X C  X L 
2
equation. It is only the second term which is Where, Z 
time-dependent. Its average is zero (the positive
half of the cosine cancels the negative half). It is called the impedance in an AC circuit.
Therefore,
Condition: The current will be in phase with the
v i v i voltage at resonance condition.
P  m m cos   m m cos   VI cos 
2 2 2 At resonance condition.
This can also be written as XL = XC
P = I2 Z cos f [½] 1
L 
Power dissipated at resonance in LCR circuit: C
At resonance, XL – XC = 0. Therefore, cosφ = 1 1 1
and P = I2 Z = I2 R. That is, maximum power is   , 2  [1]
LC LC
dissipated in a circuit (through R) at resonance.
R E = Eo sin wt…. (i)

(ii) As cos 
Z Due to this emf, charge will be produced and it
will charge the plates of capacitor with positive
In L.R. circuit
and negative charge. If potential difference
P1 = cos f across the plates of capacitor is V then
R R
P1   ∵ X l  R = V
C
= or q CV
R2  X L2
2 R2 q

1 The instantaneous value of current in the circuit
P1 =
2 dq d CE  d
In LCR when, Xl = XC
I   V  E   dt CE0 sin t 
dt dt
R \ E = Eo sin wt = CEo sin wt × w
P2 
R2   X L  X C 
2
E0   
 cos  t  cos t  sin  2   t  
1 / c  
R
P=
2 = 1 ∵ X  X 
R L C E0  
I sin    t  .  ii 
P1 1 1 / c 2 
  [1]
P2 2 [1]
I will be maximum when
17. (a) Since current leads the voltage by a plane  
Π sin    t   1 so that I  I0
angle of device X is a capacitor. [1] 2 
2
(b) Curve A shows power consumption over a E0
where, peak value of current I0 
full cycle, curve ‘B’ shows voltage and curve 1 / c

‘C’ show current as in a perfect capacitor the I  I0 sin    t  …. (iii)
current leads the voltage by a plane angle 2 
π
of [1] From (i) & (iii) it is clear current leads the voltage
2
1  
(c) z  xc 
c
by a phase angle of
2
 a0  of .
2
1
xc 
2 fc

1
xc ∝
f [1]
[1]
Or
(a) AC generator :-
(d) AC through capacitor: Let on consider a

capacitor with capacitance C be connected
to an AC source with an emf having
instantaneous value
[½]
Let at any instant total magnetic flux linked with

Power p     m sin  t im sin  t   
the armature will is given G
v i
(Where, q = wt, is the angle made by area vector  m m cos   cos 2 t   
of coil with magnetic field) 2
f = NBA cosq = NBA cos wt [½] Calculating the average power, it is observed that
d
the average of the term cos [2wt + j) is equal to
  – NBA sin  t zero.
dt
Thus,
d Average power,
By Faraday’s law of EMI, e   [½]
dt
v i
p  m m cos   VI cos 
Induced emf is given by NBA w sin wt 2 [1½]
e = eo sin wt where, eo = NBA w peak value of ωL
induced emf [½] (b) The ratio is called the quality factor or
R
Q-factor.
L
Q
R [1]
The quality factor has high value in receiving
circuits in order to get a sharp gain for the
desired channel frequency. The quality factor
[½] depends on the values of the following:
The mechanical energy spent in rotating the coil Inductance
in magnetic field appears in form of electrical
energy. [½] Resistance
Capacitance [1½]
(b)
Or
[½]
(a) The instantaneous power dissipated in the
resistor is P  i2 R  im
2
sin 2  tR
The average value of p over a cycle is:
Given velocity of straight rod v = 5m/s horizontal
component of the earth’s magnetic field P  i2 R  im
2
sin 2  tR
B = 0.30 × 10–4 wb/m2, length of wire l =10m 2
im and R are constants. Therefore,
So the emf induced in the wire is given by
e = Blv sin q (q = 90°)
2
P  im 
R sin 2  t

 [1]
e = 0.30 ×10–4 × 10 × 5 [½] By trigonometric identity,
(∴ Wire is falling at right angle to the earth’s 1
horizontal magnetic field component sin 2  t  1  cos 2 t 
2
e = 1.5 ×10–3 v [½]
The average value of cos 2wt is zero.
Air from west to east (According to Fleming right
We have:
hand rule) [½]
1
18. (a) Power in ac circuit sin 2  t  1  0
2
Voltage v in an ac circuit is:
v = vm sin wt which drives through the circuit The rms value in the ac power is expressed in
a current i the same form as dc power root mean square
or effective current and is denoted by Irms. Peak

i = im sin (w + j) where im  m and current is im therefore,
Z im
=I = 0.707im
 X  XL 
  tan 1  c  2
 R  [1]
2
im
I2 R = R
2
im Thus, combining the above equations,

I=
2 [1½] ip s N
  s
is  p Np

(b) [½]
N   Np 
 s   s   p and I s    Ip
 Np   Ns 
[½]
If Ns > Np, then the transformer is said to be step-
up transformer because the voltage is stepped
up in the secondary coil. No, the transformer
does not violate the principal of conservation
of energies. This can be easily observed by the
following equation:
Vp Ip = Vs Is = P
Power consumed in both the coils is the same
as even if the voltage increases or current
increases, their product at any instant remains
the same. [1]
19. (a) X - Capacitor
1
Xc 
2 fC [1]
(b)
[2]
1
[1] (c) X c ∝
f
In a transformer with Ns secondary turns and
Np primary turns, induced emf or voltage Es is:
d
Es   N s
dt
d
back emf E p   N p ---------- (i)
dt
Ep = Vp
Es = Vs [1]
(d)
d
Thus, Es   N s …………….(ii)
dt
Dividing (i) and (ii), we obtain
Vs N
= s
Vp N p

If the transformer is 100% efficient, then
[1]
ip vp = is vs = Power (p)
[Topic 2] AC Devices
Summary Mathematically it can be expressed as:

1 1
X=C =
ωC 2π fC
Inductive Reactance (XL) Where C is capacitance.
Instantaneous power supplied to the capacitor is
When the current flows in the circuit, the inductor i v
opposes its motion, this opposing nature of the pc = m m sin (2ωt )
2
inductor is termed as Inductive Reactance. π
In case of capacitor, the current leads the voltage by
Mathematically it can be expressed as: 2
XL = wL = 2pfL
Where L is self-inductance.
Instantaneous power supplied to an inductor
i v
p l = − m m sin (2ωt )
2
Average power supplied to an inductor over one
complete cycle is zero.
π
In case of inductor, the current lags the voltage by
2 Fig.: Phasor Diagram of Capacitor
Capacitive reactance can be graphically expressed as
follows:
Fig.: Phasor Diagram of inductor

Inductive reactance can be graphically expressed as follows:
Fig.: Capacitive Reactance Vs f

• For a series LCR circuit driven by voltage v =
vmsin(wt), the current is given by
i = imsin(wt + f).
Fig.: Inductive Reactance Vs f

Capacitive Reactance (XC)
When the current flows in the circuit, the capacitor Fig.: LCR Circuit
opposes its motion, this opposing nature of the
capacitor is termed as capacitive Reactance.
vm • LC Oscillations: When an inductor is connected to

Where i m = an initially charged capacitor, the charge on the
R 2 + (X c − X L )2 capacitor and the current in the circuit exhibit the
phenomenon of electrical oscillations. When the
XC − X L
And φ = t an −1 circuit has no ac source and no resistor then the
R
d 2q 1
charge q of the capacitor is given by + q =
0
2 LC
dt
Z = R 2 + (X C − X L )2 is called the impedance of the
Where
1
= ω0 is the frequency of free
LC
circuit.
oscillation.
Power
In an alternating circuit, the voltage and the current
both keep on changing with respect to time. Hence the
rate at which the electric energy is transferred in a
circuit is called as it’s power. The SI unit is Watt.
• Electric Power: The product of direct current flowing
through a circuit and the voltage across the circuit.
P = IV Fig.: LC Oscillalions
• Instantaneous Power: The product of current and • Idle Current: If the average power consumed in
voltage as a function of time. an alternating current circuit is zero because of the
current flowing through it, this current is called as
Pinst = Einst × Iinst
Idle Current.
• Average Power: Average of instantaneous power
• Pure Inductor circuit and pure capacitor circuit are
can be called as average power.
the two circuits whose average power consumed is
Mathematically it is expressed as, zero as the phase difference is 90°.
Pavg = VrmsIrmscos • In generators and motors, the roles of input and
where cos is power factor. output are reversed. In a motor, electric energy is
• Power factor: It is the ratio of true power to the the input and mechanical energy is the output. In
apparent power. a generator, mechanical energy is the input and
electric energy is the output. Both devices simply
• The phenomenon of resonance is an interesting
transform energy from one form to another.
characteristic of a series LCR circuit. The amplitude
of the current is maximum at the resonant frequency
1
Transformer
and the circuit thus exhibits resonance, ω 0 = .
LC • They convert an alternating voltage from one to
ω0L 1 another of greater or smaller value by using the
The quality factor Q is defined by
= Q = principle of mutual induction.
R ω 0CR
and it tells about the sharpness of the resonance.
Fig.: Transformer Showing Primary & Secondary coils.

• A step-up transformer changes a low voltage in to E sI s

high voltage. η=
E pI p
0
• A step-down transformer changes high voltage to • Energy losses in transformers may be due
low voltage. to Flux leakage, resistance of windings, Eddy
• The primary and secondary voltage and currents currents and Hysteresis.
are given by • The choice of whether the description of an
N  N p  oscillatory motion is by means of sine or cosine
V S =  S V p and I s =  I p
N p  Ns  or by their linear combinations is unimportant,
since changing the zero-time position transforms
• Efficiency of the transformer is the ratio of the one to the other.
output power to the input power. It is usually for
a real one.
transformation ratio is also 100. The input

voltage and power are respectively 220V and
PREVIOUS YEARS’
1100 W Calculate:
(a) Number of turns in secondary.
(b) Current in primary.

(c) Voltage across secondary.
TOPIC 2
(d) Current in secondary.
5 Mark Questions
(e) Power in secondary.
1. (i) With the help of a labelled diagram, describe [CBSE 2016]
briefly the underlying principle and working Solutions
of a step up transformer.
(ii) Write any two sources of energy loss in a 1. (i) Principle underlying the working of
transformer. transformer: The principle is of Mutual
(iii) A step up transformer converts a low input Inductance. When a changing source of
voltage into a high output voltage. voltage is introduced across a coil (which
Does it violate law of conservation of energy? is physically coupled to another coil), the
Explain. changing current through it induces an EMF
2. (i) Write the function of a transformer. State across the second coil.
its principle of working with the help of a
A transformer consists of two sets of coils,
diagram. Mention various energy losses in insulated from each other. They are wound
this device. on a soft-iron core, either one on top of the
(ii) The primary coil of an ideal step up other, or on separate limbs of the core.
transformer has 100 turns and
Fig.: Principle & Working of Transformer


One of the coils is called the primary coil, and
If the transformer is 100% efficient, that is, all
has N2 turns. The other coil, the secondary the input power is transferred to the secondary
coil, has Np turns. The relative numbers without any leakage or losses, then
depend on whether the voltage needs to be I pVp = I s Vs
stepped up or stepped down.

By definition, the voltage to be transformed
This implies that
is introduced across the primary coil. When I p Vs Ns
the alternating voltage is applied across the = =
Is Vp Np
primary, the resulting alternating current
through it produces a changing magnetic
It is clear from that if Ns > Np, the voltage will
field, whose flux through the secondary coil be stepped up, and if Ns < Np it will be stepped
changes. down.

From Faraday’s law, this changing flux
However, in a step down transformer, there will be
induces an EMF across the secondary, a greater current in the secondary as compared to
whose magnitude depends on the amount the primary and vice-versa. [1 + 1]
of coupling of the two coils, numerically
(ii) The possible sources of power losses in
measured as mutual inductance. The
practical transformers can be
more the coupling or association of the two
coils, the more is mutual inductance, and (1) Flux Leakage: Not all flux of the primary can
therefore the induced EMF. be associated with the secondary.

If f is the flux through each turn of the core,
There is always some flux which due to lack
then through N turns around the core,the of absolute coupling, can leak. To avoid this,
total flux is N f the coils are wound over each other again and
again.

So, the EMF induced in the secondary coil
is (2) Resistance of windings: The transformer coil
wires cannot have absolutely zero resistance,
d so some Joule loss is inevitable.
Es   N s
dt
(3) Core eddy currents: Since the core is a very

Similarly, there will also be an EMF good conductor itself, currents are induced in
induced in the primary coil itself, due to it due to changing magnetic fields, called eddy
self-inductance, given by currents. These also result in losses.
d (4) Hysteresis: Some part of energy is frozen into
Ep   N p

dt the core permanently in the form of a residual
magnetic field due to its ferromagnetic

If the voltage applied across the primary is
character. [1]
Vp, then if its resistance is R, the current
V  Ep (iii) No, it does not violate the energy conservation.
through it will be I p  p When low voltage is converted to high volatge,
R the current is lowered, thereby conserving the

However, assuming negligible resistance, total energy dissipated across the primary &
since we cannot have an infinite current secondary coil. [1]
through the coil, then 2. (i) A transformer is an electrical device for
E p ≈ Vp
converting an alternating current at low

voltage into that at high voltage or vice-versa.

If the secondary is an open circuit, no current
1. If it increases the input ac voltage, it is called
is drawn from it then, voltage across it will
step up transformer.
be [1]
2. If it decreases the input ac voltage, it is called
d
Vs  Es   N s step down transformer. [½]

dt

Principle: It works on the principle of mutual

From equation it is clear that induction i.e., When a changing current is passed
Vs N through one of the two inductively coupled coils,
= s
Vp N p an induced emf is set up in the other coil. [½]

[½]
Fig.: Step up transformer & Step down transformer

Working Theory: As the AC flows through the This is called hysteresis loss and can be
primary, it generate an alternating Magnetic magnetized by using core material having
flux in the core which passes through the narrow hysteresis loop.
secondary coil.
• Flux leakage: The magnetic flux produced

Let N1 No. of turns in primary coils by the primary may not fully pass through
N2 = No. of turns in secondary coils the secondary. Some of the flux may leak into

This changing flux sets up an induced emf in air. This loss can be minimized by winding
the secondary, also a self-induced emf in the the primary and secondary coils over one
primary. another. [½]

If there is no leakage of magnetic flux, then flux (ii) Given, N1 = 100
linked with each turn of the primary will be

K = 100
equal to that linked with each of the secondary. V1 = 220V
According to Faraday’s law of induction.
P1 = 1100 W
d
Induced emf in the primary coil, 1   N1 N2
dt
(a) As, K =
d N1
Induced emf in the secondary coil,  2   N2
dt N2 = KN1 = 100 × 100
Where, dφ = rate of change magnetic flux N2 = 10000 [½]
dt
(b) P1 = V1 I1
associated with each turn.
P1 1100
 2 N2 =
I1 =
 V1 220
 N1

1 [½]
Energy Losses in transformer.
I1 = 5V [½]
• Copper loss: Some energy is lost due to the V2
(c)
=K
heating of copper wires used in the primary V1
and secondary windings. This power loss
(P = 12R) can be minimized by using thick
V1 = 5V1; V2 = 100 × 200
copper wires of low resistance.
V2 = 5V [½]
• Eddy current loss: The alternating I1
(d)
=K
magnetic flux induces eddy current in the I2
iron core which leads to some energy loss in
the form of heat. This loss can be reduced by
I2 = 0.05A
using laminated iron core. [½]
P2 = V2 I2
• Hysteresis loss: The alternating current 5
carries the iron core through cycles of P2  22000 

100
magnetization and demagnetization. Work
done in each of these cycles is lost as heat.

P2 = 1100W [½]
CHAPTER 8
Electromagnetic Waves

Basic idea of displace-
1Q
ment current, Electro-
1Q 1Q (1 mark),
magnetic waves, their
(3 marks) (1 mark) 1Q
characteristics, their
(1 mark)
transverse nature
Electromagnetic spec-
1Q
trum (radio waves, micro-
(1 mark),
waves, infrared, visible,
1Q 1Q
ultraviolet, X-rays, gam-
(1 mark) (1 mark),
ma rays) including ele-
1Q
mentary facts about their
(2mark)
use
On the basis of above analysis, it can be said that from exam point of view Characteristics and
Nature of Electromagnetic Spectrum, Applications of Different Parts of Electromagnetic and
Electromagnetic Wave are most important concepts of the chapter.
132 CHAPTER 8 : Electromagnetic Waves
[Topic 1] Electromagnetic Waves, its Types &

Properties
Summary • Faraday’s law:
∫ E .dl =
−d φB
dt
Displacement Current
d φE
• It is defined as the rate of change of electric • Ampere-Maxwell law: ∫ B=
.dl µ0 I c + µ0ε 0
dt
displacement.
An electric charge oscillating harmonically with a
d φε
• It is given by I d = ε 0 where e0 is the frequency, produces electromagnetic waves of the
dt same frequency . The frequency of the electromag-
permittivity of the free space and fs is the amount netic wave naturally equals the frequency ofoscilla-
of electric flux. tion of the charge.
An electric dipole is a basic source of electromagnetic
Properties of EM Waves waves.
• The electric and magnetic fields Ex and By are From Maxwell’s equations it can be seen that the
always perpendicular to each other, and also to magnitude of the electric and the magnetic fields in
the direction z of propagation. Ex and By are given E0
an electromagnetic wave are related as B 0 =
by: c
E x = E 0 sin (kz-ω t )
Properties of EM Waves

• Oscillations of electric and magnetic fields sustain
B = B 0 sin (kz-ω t ) in free space, or vacuum. So, the electromagnetic
y
x waves can travel in vacuum.
E • An electromagnetic wave carries momentum
and energy. Since an electromagnetic wave
z carries momentum, it also exerts pressure, called
radiation pressure.
B • Let the total energy transferred to a surface in time
y
t is U, so the magnitude of the total momentum of
Fig. Electromagnetic Waves
an electromagnetic wave delivered to the surface
Where, U
(for complete absorption) is, P =
“k” is the magnitude of the wave vector (or c
propagation vector) and can be calculated as; • The energy of electromagnetic waves is shared
equally by the electric and magnetic fields.
2π
k =
λ

• w is the angular frequency,
Types of EM Waves
• “k” is direction describes the direction of • Radio waves are produced by the accelerated
propagation of the wave. The speed of propagation motion of charges in conducting wires. They
of the wave is ω . are used in radio and television communication
k systems.The radio waves generally lie in the
• The frequency of EM waves can be from 0 to  . frequency range from 500 kHz to about 1000
Ampere Circuital Law is given by: ∫ B ⋅ dl µ0 i (t )
= MHz
The four Maxwell’s equations are given as: • Microwaves have frequency in the range of
Q gigahertz and are used in aircraft navigation.
∫ E .dA =
• Gauss’s law of electricity:
ε0 • Infrared waves are also referred to as heat waves
as they are produced by hot bodies and molecules.
• Gauss’s law of magnetism: ∫ B .dA = 0
CHAPTER 8 : Electromagnetic Waves 133
• Visible rays can be detected by the human eye. They lie between frequency range of about
4 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 –400 nm.
• Ultraviolet radiation or the UV radiation is produced by special lamps and very hot bodies.
• X-rays lie beyond the UV region and are used as a diagnostic tool in medicine and for treating various kinds
of cancer.
• Gamma rays are emitted by radioactive nuclei and also are produced in nuclear reactions and are used in
destroying the cancer cells.
The properties of different types of EM Waves are:
Type Wavelength range Production Detection
Radio > 0.1 m Rapid acceleration and Receiver’s aerials
decelerations of electrons
in aerials
Microwave 0.1 m to 1 mm Klystron valve or magne- Point contact diodes
tron valve
Infra-red 1 mm to 700 nm Vibration of atoms and Thermopiles, Bolometer,
molecules Infrared photographic film
Light 700 nm to 400 nm Electrons in atoms emit The eye, Photocells, Photo-
light when they move graphic film
from one energy level to a
lower energy level
Ultraviolet 400 nm to 1 nm Inner shell electrons in Photocells, Photographic
atoms moving from one film
energy level to a lower
level
X-rays 1 nm to 10–3 nm X-ray tubes or inner shell Photographic film, Geiger
electrons tubes Ionisation chamber
Gamma rays <10–3 nm Radioactive decay of the Photographic film, Geiger
nucleus tubes Ionisation chamber
5. Write the expression for the de Broglie
PREVIOUS YEARS’ wavelength associated with a charged particle

having charge ‘q’ and mass ‘m’, when it is
EXAMINATION QUESTIONS accelerated by a potential V.
[ALL INDIA 2013]
TOPIC 1 6. Welders wear special goggles or face masks
1 Mark Questions with glass windows to protect their eyes from
1. A plane electromagnetic wave travels in vacuum electromagnetic radiations. Name the radiations
along z-direction. What can you say about the and write the range of their frequency.
direction of electric and magnetic field vectors? [ALL INDIA 2013]
[ALL INDIA 2011] 7. Why are microwaves considered suitable for
2. Name of physical quantity which remains same radar systems, used in aircraft navigation?
for microwaves of wavelength 1 mm and UV [DELHI 2016]
radiations of 1600 A° in vacuum. 8. Write the following radiations in ascending
[ALL INDIA 2012] order with respect to their frequencies: X-rays.
3. State De-Broglie hypothesis. Microwaves. UV rays and radio waves.
4. Which of the following waves can be polarized 9. Do electromagnetic waves carry energy and
(i) Heat waves (ii) Sound waves? Give reason to momentum?
support your answer. [DELHI 2017]
[ALL INDIA 2013]
10. Name the phenomenon which shows the 20. When Sunita, a class XII student, came to know
quantum nature of electromagnetic radiation. that her parents are planning to rent out the
[ALL INDIA 2017] top floor of their house to a mobile company
11. Identify the electromagnetic waves whose she protested. She tried hard to convince her
wavelength vary as parents that this move would be a health hazard.
(a) 10–12 m < 1 < 10–8 m Ultimately her parents agreed:
(b) 10–3 m < 1 < 10–1 m (1) In what way can the setting up of transmission
Write one use of each. tower by a mobile company in a residential
[ALL INDIA 2017] colony prove to be injurious to health?
12. Which part of electromagnetic spectrum is used
(2) By objecting to this move of her parents,
in radar systems?
what value did Sunita display?
[DELHI 2018]
(3) Estimate the range of e.m. waves which can
2 Mark Questions be transmitted by an antenna of height 20
13. Name the electromagnetic radiation’s used for(a)
m. (Given radius of the earth = 6400 km)
water purification, and(b)eye surgery.
[DELHI 2014]
[ALL INDIA 2018]
21. Answer the following:
14. How are infrared waves produced? Why are
these referred to as ‘heat waves’? Write their (a) Name the em waves which are used for the
one important use. treatment of certain forms of cancer. Write
[ALL INDIA 2011] their frequency range.
15. (a) An EM wave is, travelling in a medium with (b) Welders wear special glass goggles while
a certain velocity. Draw a sketch showing working. Why? Explain.
the propagation of the EM wave, indicating (c) Why infrared waves are often called as heat
the direction of the oscillating electric and waves? Give their one application.
magnetic fields. [DELHI 2014]
(b) How are the magnitudes of the electric and 22. Answer the following:
magnetic fields related to velocity of the EM (a) Name the em waves which are used for the
wave ? treatment of certain forms of cancer. Write
[DELHI 2017] their frequency range.
16. How does a charge q oscillating at certain (b) Thin ozone layer on top of stratosphere is
frequency produce electromagnetic waves? crucial for human survival. Why?
Sketch a schematic diagram depicting electric
(c) Why is the amount of the momentum
and magnetic fields for an electromagnetic wave
transferred by the em waves incident on the
propagating along the Z-direction.
surface so small? [DELHI 2014]
[DELHI 2017]
23. Answer the following:
17. (a) Why are infra-red waves often called heat
waves? Explain. (a) Name the em waves which are suitable for
(b) What do you understand by the statement, radar systems used in aircraft navigation.
“Electromagnetic waves transport Write the range of frequency of these waves.
momentum”? [ALL INDIA 2018] (b) If the earth did not have atmosphere, would
its average surface temperature be higher
3 Mark Questions or lower than what it is now? Explain.
18. Write Einstein’s photoelectric equation and point
out any two characteristic properties of photons (c) An em wave exerts pressure on the surface
on which this equation is based. on which it is incident. Justify.
Briefly explain the three observed features
which can be explained by this equation. [DELHI 2014]
[ALL INDIA 2013] 24. Name the parts of the electromagnetic spectrum
19. Write Einstein’s photoelectric equation. State which is
clearly how this equation is obtained using the (a) Suitable for radar systems used in aircraft
photon picture of electromagnetic radiation. navigation
Write the three salient features observed in (b) Used to treat muscular strain
photoelectric effect which can be explained using (c) Used as a diagnostic tool in medicine
this equation. [ALL INDIA 2012]
Write in brief, how these waves can be 2. Both microwaves and UV rays are a part of the
produced. electromagnetic spectrum. Thus, the physical
[DELHI 2015] quantity that remains same for both types of
25. Define the terms “stopping potential’ and radiation will be their speeds, equal to c. [1]
‘threshold frequency’ in relation to photoelectric 3. It states that, “Moving object sometimes acts as
effect. How does one determine these physical a wave and sometimes as a particle” or a wave
quantities using Einstein’s equation? is associated with the moving particle, which
[ALL INDIA 2015] controls the particle in every respect. This wave
26. How are electromagnetic waves produced? associated with the moving particle is called
What is the source of the energy carried by a matter wave or de Broglie wave. Its wave length
propagating electromagnetic wave? Identify the is given by:
electromagnetic radiations used
h
(i) In remote switches of household electronic 
m [1]
device; and
Where, h = planck’s constant
(ii) as diagnostic tool in medicine
m = mass of the object
[ALL INDIA 2015]
27. How are e.m. waves produced by oscillating v = velocity of the object
charges? 4. Heat waves can be polarized as they are transverse
Draw a sketch of linearly polarized e.m. waves waves whereas sound waves cannot be polarized
propagating in the Z-direction. Indicate the as they are longitudinal waves. [½]
directions of the oscillating electric and magnetic Transverse waves can oscillate in the direction
fields. perpendicular to the direction of its transmission
OR but longitudinal waves oscillate only along the
Write Maxwell’s generalization of Ampere’s direction of its transmission. So, longitudinal
Circuital Law. Show that in the process of waves cannot be polarized. [½]
charging a capacitor, the current produced h
5.   [1]
d 2mqV
within the plates of the capacitor is i   o E
dt
6. Ultraviolet rays frequency range
where fE is the electric flux produced during
(7.5 × 1014 – 5 × 1015 Hz) [1]
charging of the capacitor plates.
[DELHI 2016] 7. Microwaves of frequency 1 GHz to 300 GHz
28. State two important properties of photon bounces from even the smallest aircraft so
which are used to write Einstein’s photoelectric that they are suitable to avoid getting bombed.
equation. Define Microwaves can penetrate through clouds
also. [1]
(i) Stopping potential and
8. The given radiations can be arranged in
(ii) Threshold frequency, using Einstein’s
ascending order with respect to their frequencies
equation and drawing necessary plot
as: Radio waves < Microwaves < UV rays <
between relevant quantities.
X-rays [1]
[DELHI 2016]
29. How are electromagnetic waves produce? 9. Yes, electromagnetic waves carry energy (E) and
What is the source of energy of these waves? momentum (P) is given by
Write mathematical expressions for electric hc
E
and magnetic fields of an electromagnetic  [½]
wave propagating along the z-axis. Write any P = mC [½]
two important properties of electromagnetic Here C is speed of EM(Electromagnetic) wave in
waves. vaccum λ is wavelength of EM wave.
[ALL INDIA 2016]
10. “Photoelectric effect” shows the quantum nature
Solutions of electromagnetic radiation. [1]
1. The electromagnetic wave travels in a vacuum 11. (a) X-ray
along the z-direction. The electric field (E) and Uses- These are used in surgery to detect
the magnetic field (B) are in the x-y plane. They fracture, damaged organs, stones in the
are mutually perpendicular. [1] body, etc. [1]
(b) Gamma rays [1] 18. If energy of photon = E (= hv)

Uses- These are used in radio therapy for work function of metallic surface = φ = (hv0)
the treatment of tumor and cancer. 1
kinetic energy of emitted electron = mv2
12. The microwave range [1] 2
1
13. (a) UV rays [1] E    mv2
2
(b) Laser [1]
1
14. Infrared waves are produced by hot bodies or due mv2  h  –  o 
2 [½]
to vibrations of atoms and molecules. They are
called heat waves because they cause heating Einstein’s equation
effect or increase the temperature. [1] Two characteristics of photons
Use: Infrared lamps,play important role in (i) This equation is based on particle nature of
maintaining warmth through greenhouse light [½]
effect. [1]
(ii) Total energy of photon is transferred
15. (a) Given, completely to single electron. [½]

Velocity, v = vi and electric field, E along y-axis (iii) If incident frequency ν< threshold frequency
and magnetic field, B along z-axis, [½] 1
(νo), then kinetic energy mv2 will be
The propagation of EM wave is shown below: 2
y negative which is not possible because
kinetic energy cannot be negative. This
E
shows that photoelectric emission is not
possible if frequency (ν) of incident light is
x less than the threshold frequency (νo) of the
metal. [1
z B (iv) One photon can emit only one electron
from the metal surface, so the number of
Fig.: Electromagnetic Waves [½]
photoelectrons emitted per second is directly
(b) Speed of EM wave can be given as the ratio proportional to the intensity of incident light
of magnitude of electric field (Eo) to the which depends upon number of photons
Eo present in the incident light. [½]
magnitude of magnetic field (Bo), c = B [1]
o (v) From equation it is clear that kinetic energy,
16. An oscillating charge is an example of accelerated
1
charge. We know from Maxwell’s theory that mv2 increases with the increase in the
2
accelerated charge radiates electromagnetic
waves. These electromagnetic waves are frequency(ν) of the incident light. [½]
produced because oscillating charge produces 19. Einstein’s Photoelectric Equation,
oscillating magnetic field which in turn produces
hν, = Kmax + f
an oscillating electric field. This process goes on,
giving rise to an electromagnetic wave. [1.5] Where, h = Planck’s constant
x v = frequency
E
f = Work function
1 2
z K max  m max  eVo
2
B (Where vmax = maximum velocity of emitted
y
photoelectron and Vo = Stopping potential)
Fig.: Electromagnetic Waves [½] According to Planck’s quantum theory, light
17. (a) Infrared produce heat or emitted from hot radiations consist of small packets of energy.
bodies. [1] Einstein postulated that a photon of energy ‘hv’
(b) When electromagnetic waves hit body the is absorbed by the electron of the metal surface,
mass is lost but the momentum is conserved then the energy equal to f is used to liberate
.i.e., transferred from EMW to body. electron from the surface and rest of the energy
Therefore, EMW transports momentum. [1] i.e. hv – f becomes the kinetic energy. [1]
∴ Energy of photon is given by, E = hv (where h = 22. (a) X-ray, Gamma(y) rays are used for the
Planck’s constant and v = frequency of light) treatment of certain forms of cancer. Their
The minimum energy required by the electron of frequency range is 1018m to 1022m. [1]
a material to escape out of it, is work function f. (b) The thin ozone layer on top of stratosphere
The additional energy acquired by the electron absorb most of the harmful ultraviolet
appears as the maximum kinetic energy Kmax of rays coming from the Sun towards the
the electron. i.e. Kmax = hv – f Earth. They include UVA, UVB and UVC
Or hv = K max + f [Einstein’s Photoelectron radiations, which can destroy the life system
Equation] [1] on the Earth. Hence, this layer is crucial for
human survival. [1]
Kmax = eVo
(c) Momentum transferred = Energy Speed of
Salient features observed in photoelectric effect:
light = hvc = 1022 (for v – 1020 Hz)
The stopping potential and hence the maximum
Thus, the amount of the momentum
kinetic energy of emitted electrons varies
transferred by the em waves incident on the
linearly with the frequency of incident radiation.
surface is very small. [1]
There exists a minimum cut off frequency vo,
23. (a) Microwaves are suitable for radar systems
for which the stopping potential is zero.
used in aircrafts navigation. The range of
Photoelectric emission is instantaneous. [1] frequency for these waves is 109 Hz to 1012
20. (1) A transmitting tower makes use of Hz. [1]
electromagnetic waves such as microwaves, (b) In the absence of atmosphere, there would
exposure to which can cause severe health be no greenhouse effect on the surface of
hazards like, giddiness, headache, tumour and the Earth. As a result, the temperature of
cancer. Also, the transmitting antenna operates the Earth would decrease rapidly, making
on a very high power, so the risk of someone it difficult for human survival. [1]
getting severely burnt in a residential area
(c) An em wave carries a linear momentum
increases. [1]
with it. The linear momentum carried by a
(2) By objecting to this move of her parents, Sunita portion of wave having energy U is given by
has displayed awareness towards the health and p = UC [½]
environment of her society. [1]
Thus, if the wave incident on a material
(3) Range of the transmitting antenna. surface is completely absorbed, it delivers
d = 2hR energy U and momentum p = UC to the
surface. If the wave is totally reflected, the
Here, h is the height of the transmitting antenna
momentum delivered is p = 2UC because
and R is the radius of the Earth.
the momentum of the wave changes from
R = 6400 km = 64 × 105 m p to –p. Therefore, it follows that em waves
5 incident on a surface exert a force and hence
d  2  20  64  10 a pressure on the surface. [½]
d = 16000m [1] 24. (a) Microwaves are suitable for radar systems
21. (a) Gamma rays are used for the treatment that are used in aircraft navigation.
of certain forms of cancer. Their frequency These rays are produced by special vacuum
ranges. [1] tubes, namely -klystrons, magnetrons and
(b) Welders wear special glass goggles while Gunn diodes. [1]
working so that they can protect their eyes (b) Infrared waves are used to treat muscular
from harmful electromagnetic radiation. [1] strain. [1]
(c) Infrared waves are often called as heat These rays are produced by hot bodies and
waves because they induce resonance in molecules.
molecules and increase internal energy in
(c) X-rays are used as a diagnostic tool in
a substance. [½]
medicine. [1]
Infrared waves are used in burglar alarms,
These rays are produced when high energy
security lights and remote controls for television
electrons are stopped suddenly on a metal of
and DVD players. [½]
high atomic number.
25. Stopping potential: For a particular frequency absorbing the energy quanta and therefore, the
of incident radiation, the minimum negative number of electrons coming out of the metal
(retarding) potential Vo given to the anode plate (for    o ) is more and so is photoelectric
for which the photocurrent stops or becomes current. [½]
zero is called the cut-off or stopping potential. 26. According to Maxwell’s theory accelerated
Threshold frequency: There exists a certain charges radiate electromagnetic waves. Consider
minimum cut-off frequency Vo for which the a charge oscillating with some frequency. (An
stopping potential is zero and below Vo the oscillating charge is an example of accelerating
electron emission is not possible. This cut-off charge.) This produces an oscillating electric
frequency is known as threshold frequency, field in space, which produces an oscillating
Vo which is different for different metal. In magnetic field, which in turn, is a source
photoelectric effect, an electron absorbs a of oscillating electric field, and so on. The
quantum of energy (hν) of radiation. If this oscillating electric and magnetic fields thus
quantum of energy absorbed by electron exceeds regenerate each other, so to speak, as the wave
the minimum energy required to come out of the propagates through the space. The frequency
metal surface by electron, the kinetic energy of of the electromagnetic wave naturally equals
the emitted electron is the frequency of oscillation of the charge. The
K = hν – f... (1) [1.5] energy associated with the propagating wave
where, f is the minimum energy for electron comes at the expense of the energy of the source
to come out of the metal, and is different for – the accelerated charge. The electromagnetic
different electrons in the metal. The maximum radiation used are: [1 + 1]
kinetic energy of photo electrons is given by ... (1) Infrared rays
Kmax = hν – fo (2) (2) X-rays [1]
where, fo – work function or least value of f 27. These waves are constituted by varying or
equation (2) is known as Einstein’s photoelectric oscillating electric and magnetic fields. The
equation. electric and magnetic fields are perpendicular
Explanation of photoelectric effect with the help to each other and are also perpendicular to the
of Einstein’s photoelectric equation direction of propagation of the wave E is the
(i) According to equation (2), kmax depends linearly envelope of electric intensity vector and B is the
on ν, and is independent of intensity of radiation. envelope of magnetic intensity vector. [1 + 1]
x
This happens because, here, photoelectric effect E
arises from the absorption of a single quantum
of radiation by a single electron. The intensity
z
of radiation (that is proportional to the number
of energy quanta per unit area per unit time) is
irrelevant to this basic process. [½] B
Fig.: Electromagnetic Waves [1]
(ii) Since kmax must be non-negative, equation (2) OR
Correction in Amperes Circuital law (Modified
implies that photoelectric emission is possible Ampere’s law) : Maxwell removed the problem
 of current continuity and inconsistency observed
only if h   o or    o , where  o  o in Ampere’s Circuital law by introducing the
h
concept of displacement current, Displacement
o current arises due to change in electric flux with
Thus, there exists a threshold frequency  o  d E
h time and is given by id   o [1]
dt
exists, below which photoelectric emission is not Electric flux through the loop fE = EA
 Q A Q
possible, and is independent of intensity. [½] E  A 
o A o o
(iii) As intensity of radiation is proportional to the
number of energy quanta per unit area per unit (Q = charge on either plates)
time. The greater the number of energy quanta Q
available, the greater is the number of electrons E 
o
d E

1 d eV0  h – h 0
dt  o dt
h h
Vo    o
d E dQ e e [1]
o 
dt dt [1] V0
dQ h
is called conduction current which is equal Slope
dt e
d E
to  o , which is displacement current.
dt V0 V
Hence, ic = id
– 0/e
Generalization of Ampere’s circuital law is:


B.dl  o ic  id 
Fig.: Graph of V0 Vs v
Conduction current is because of How of charges 29. Electromagnetic waves are produced by
but displacement current is not because of How accelerated or oscillating charges and do not
of charges but because of change in electric need any medium for propagation.
flux. [1] The source of energy of these waves is energy of
28. Characteristic properties of photons: oscillating charge particle.
(i) Energy of photon is directly proportional to The mathematical expressions for electric and
the frequency (or inversely proportional to the magnetic fields of an electromagnetic wave
wavelength). [1] propagating along the z-axis are:
(ii) In photon-electron collision, total energy and Ex  Eo sin  kz   t 
momentum of the system of two constituents [1]
remains constant. [1] By  Bo sin  kz   t 

(iii) In the interaction of photons with the free
Properties of electromagnetic waves:
electrons, the entire energy of photon is
absorbed. Cut off or stopping potential is that 1. The direction of E, B and direction of propagation
minimum value of negative potential at anode of waves are mutually perpendicular to one
which just stops the photo electric current. For a another.
given material, there is a minimum frequency of 2. The amplitude ratio of E and B gives velocity of
light frequency is called as threshold frequency. light.
By Einstein’s photo electric equation 3. Electromagnetic waves are not deflected by
hc electric and magnetic fields. [1 + 1]
KEmax    o  h  h o where  0  h 0

CHAPTER 9
Ray Optics and Optical
Instruments

Reflection of light,
1Q
spherical mirrors, mirror
(3 marks)
formula
Refraction of light, total
1Q 1Q
internal reflection and its
(5 marks) (3 marks)
applications Optical fiber
Refraction at spherical
surfaces
Lenses; thin lens formu-
la, lens maker’s formula;
1Q 1Q 1Q
magnification, power of a
(3 marks) (4 marks) (5 marks)
lens, combination of thin
lenses in contact
Refraction and dispersion 1Q
of light through a prism 1Q 1Q (1 mark), 1Q
(3 marks) (3 marks) 1Q (3 marks)
(5 marks)
Scattering of light - blue
color of sky and reddish 1Q
appearance of the sun at (1 mark)
sunrise and sunset
Optical instruments: Mi-
croscopes and astronom-
1Q
ical telescopes (reflecting
(3 marks)
and refracting) and their
magnifying powers.
On the basis of above analysis, it can be said that from exam point of view Biconvex lens,
Dispersion of Prism, Reflecting Index of Glass, Prism Formula, Lens Maker’s Formula and
reflecting telescope are most important concepts of the chapter.
142 CHAPTER 9 : Ray Optics and Optical Instruments
[Topic 1] Reflection, refraction and dispersion of

light
Summary Object on left Mirror
The speed of light in vacuum is given by c = 3 × Heights Incident light

upwards
108ms–1, which is the highest speed that can be at- positive
tained in nature. x-axis
A light wave travels along a straight line from one Distances
point to another. This path is called a ray of light, Heights
against
and bundle of such rays together form a beam of light. incident
downwards light Distances
negative negative along incident
Laws of reflection states that light positive
Fig.: Cartesian sign convention
• The angle of reflection (i.e., the angle between
If the distance of the object and the image is given by
reflected ray and the normal to the reflecting u and v, respectively and f is the focal length of the
surface or the mirror) equals the angle of mirror. Then the mirror formula is given by,
incidence (angle between incident ray and the 1 1 1
normal), i.e. ∠i = ∠r + =
v u f
• The incident ray, the normal to the mirror at the Focal length f for a concave mirroris negative and is
point of incidence and the reflected ray, they all positive for a convex mirror.
lie in the same plane. The magnification produced by a mirror is given by
sin i h' v
Snell’s law for refraction is given by =n, m = = − where h′ is the height of the image and
sin r h u
where the angle of incidence, angle of refraction and h is the height of the object.
refractive index of the medium is given by i, r and n
Total Internal Reflection:
respectively.
The angle of incidence at which a ray travelling from • When light travels from an optically denser
a denser to rarer medium makes an angle of refrac- medium to a rarer medium at the interface, it
tion of 90° is a critical angle and is denoted by ic. is partly reflected back into the same medium
and partly refracted to the second medium. This
Cartesian sign convention: reflection is called internal reflection when all
light is reflected back, it is called total internal
• Positive sign is used for distances measured in
reflection.
the same direction as the incident light, whereas
negative sign is used for those measured in the B
Rarer
direction opposite to the direction of incident medium
Water-air
light. (Air)
O1 r O2 r O3 D O4 interface
• All distances are measured from the pole of the ic
i i N i>ic
mirror or the optical centre of the lens. The
N N Totally
heights measured upwards with respect to x-axis Denser reflected ray
and normal to the principal axis of the mirror/ medium Partially
(water)
lens is taken as positive. A C reflected rays
• The heights measured downwards with respect to Fig.: Total Internal Reflection
x-axis are taken as negative.
• The applications of total internal reflection
include mirage, diamond, prism and optical
fibers.
CHAPTER 9 : Ray Optics and Optical Instruments 143
Refraction through glass slab: Focal length f is positive for a converging lens and
The emergent ray through a glass slab is parallel to is negative for a diverging lens.
the incident ray but it is laterally displaced. • The magnification produced by a mirror is given
Also, ∠Angle of incidence = ∠Angle of emer gence h′ v
= =
by m where h′ is the height of the image
Medium al h u
(Glass) a ter t
L hif
s and h is the height of the object.
r3
Medium i2 • The power (P) of a lens is given by, P = 1 .
Medium f
(Air) r1 (Air)
i1 Where f is the focal length of the lens and the SI
unit of power is dioptre (D): 1 D = 1 m–1
Fig.: Reflection through glass slab • The effective focal length of a combination of thin
lenses of focal length f1, f2, f3 ..... is given by
Refraction at spherical surfaces 1 1 1 1
= + + + ..........
If the rays are incident from a medium of refractive f f1 f 2 f3

index n1 to another of refractive index n2, then
And the effective power of the same combination
n 2 n1 n 2 − n1
− = is given by
v u R
P = P1 + P2 + P3 ......
N Dispersion:
n1 i n2
• Splitting of light into its constituent colors is
r
known as dispersion of light.
O C I
M R • When a white light is incident on a prism, the
white light is split into seven components, violet,
u v
indigo, blue, green, yellow, orange and red (given
by the acronym VIBGYOR)
Fig.: Refraction at spherical surface
Some natural phenomenon due to sunlight are rainbow
For a prism of the angle A, of refractive index n2 and scattering of light.
placed in a medium of refractive index n1 and Dm be-
The Eye: It has a convex lens of focal length about
ing the angle of minimum deviation.
2.5 cm. This focal length can be varied somewhat by
A
the help of ciliary muscle so that the image is always
formed on the retina. This ability of the eye of adjusting
M the muscle to form a clear image is called accommo-

i r e dation.
Q r1 2 R
N In a defective eye, if the image is focused before the
retina, it is called myopia. For correction of myopia, a
P S diverging corrective lens is needed.
In a defective eye, if the image is focused beyond the
B C
retina, it is called hypermetropia. For correction of hy-
Fig.: Prism permetropia, a converging corrective lens is needed.
n 2 sin ( A + D m ) / 2 Astigmatism: A refractive error in which the vision is
n=
21 = blurred at all distances, is corrected by using cylindri-
n1 sin ( A / 2)

cal lenses.
• If the distance of the object and the image is given
by u and v, respectively and f is the focal length of
the lens. So, the lens formula is,
1 1 1
− =
v u f
prism inside it. Find the angle of incidence for

this ray.
[ALL INDIA 2012]
PREVIOUS YEARS’ 11. A convex lens of focal length 25 m is placed co
axially in contact with a concave lens of focal
EXAMINATION QUESTIONS length 20 cm. Determine the power of the
TOPIC 1 combination. Will the system be converging
or diverging in nature?
1 Mark Questions
[DELHI 2013]
1. When monochromatic light travels from one
medium to another its wavelength changes but 12. (a) Write the necessary conditions for the
frequency remains the same. Explain. phenomenon of total internal reflection to
occur.
[ALL INDIA 2011]
(b) Write the relation between the refractive
2. Under what condition does a biconvex lens of
index and critical angle for a given pair of
glass having a certain refractive index act as a
optical media.
plane glass sheet when immersed in a liquid?
[DELHI 2013]
[ALL INDIA 2012]
13. A convex lens of focal length f1 is kept in contact
3. Write the relationship between angle of
with a concave lens of focal length f2. Find the
incidence ‘i’, angle of prism ‘A’ and angle of
focal length of the combination.
minimum deviations Dm for a triangular prism.
[DELHI 2013]
[DELHI 2013]
14. A convex lens of focal length 30 cm is placed
4. A convex lens is placed in contact with a plane
coaxially in contact with a concave lens of
mirror. A point object at a distance of 20 cm
focal length 40 cm. Determine the power of the
on the axis of this combination has its image
combination. Will the system be converging or
coinciding with itself. What is the focal length
diverging in nature?
of the lens? [DELHI 2014]
5. A biconcave lens made of a transparent material [ALL INDIA 2013]
of refractive index 1.25 is immersed in water of 15. A convex lens of focal length 20 cm is placed co
refractive index 1.33. Will the lens behave as a axially in contact with a concave lens of focal
converging or a diverging lens? Give reason. length 25 cm. Determine the power of the
[ALL INDIA 2014] combination. Will the system be converging
or diverging in nature?
6. A concave lens of refractive index 1.5 is immersed
in a medium of refractive index 1.65. What is the [DELHI 2013]
16. Two monochromatic rays of light are incident
nature of the lens? [DELHI 2015]
normally on the face AB of an isosceles right-
7. When light travels from an optically denser angled prism ABC. The refractive indices of
medium to a rarer medium, how does the the glass prism for the two rays ‘1’ and ‘2’ are
critical angle of incidence depend on the colour respectively 1.38 and 1.52. Trace the path of
of light? these rays after entering through the prism.
[ALL INDIA 2015] A
8. Why can’t we see clearly through fog? Name the
phenomenon responsible for it.
45
°
[ALL INDIA 2016] '1'

9. How does the angle of minimum deviation of a '2'
glass prism vary, if the incident violet light is
replaced by red light? Give reason. 45°
B C
[ALL INDIA 2017]
[ALL INDIA 2014]
2 Mark Questions 17. Use the mirror equation to show that an object
10. A ray of light, incident on an equilateral prism placed between f and 2f of a concave mirror
produces a real image beyond 2f.
( µ g = 3 ) moves parallel to the base line of the [DELHI 2015]
18. A ray PQ incident on the refracting face BA is 22. (a) A ray of light is incident normally on the
refracted in the prism BAC as shown in the face AB of a right angled glass prism of
figure and emerges from the other refracting refractive index. aµg = 1.5. The prism is
face AC as RS such that AQ = AR. If the angle partly immersed in a liquid of unknown
of prism A = 60º and refractive index of material refractive index. Find the value of refractive
of prism is 3 , calculate angle θ. index of the liquid so that the ray grazes
A along the face BC after refraction through
the prism.
Q R
A B
60°
S
P
B C
[ALL INDIA 2016]
3 Mark Questions
19. A convex lens of focal length 20 cm is placed
coaxially with a convex mirror of radius of C
curvature 20 cm. The two are kept 15 cm apart. A (b) Trace the path of the rays if it were incident
point object is placed 40 cm in front of the convex normally on the face AC.
lens. Find the position of the image formed by [ALL INDIA 2015]
this combination. Draw the ray diagram showing
23. (a) Calculate the distance of an object of height h
the image formation. [ALL INDIA 2014]
from a concave mirror of radius of curvature
20. One day Chetan’s mother developed a severe 20 cm, so as to obtain a real image of
stomach ache all of a sudden. She was rushed magnification 2. Find the location of image.
to the doctor who suggested for an immediate
(b) Using mirror formula, explain why a convex
endoscopy test and gave an estimate of
mirror always produces a virtual image.
expenditure for the same. Chetan immediately
[DELHI 2016]
contacted his class teacher and shared the
information with her. The class teacher arranged 24. (i) A screen is placed at a distance of 100 cm
for the money and rushed to the hospital. On from an object. The image of the object is
realizing that Chetan belonged to a below formed on the screen by a convex lens for
average income group family, even the doctor two different location of the lens separated
offered concession for test fee. The test was by 20 cm. Calculate the focal length of the
conducted successfully. lens used.
Answer the following questions based on the (ii) A converging lens is kept coaxially in contact
above information with a diverging lens-both the lenses being
of equal focal length. What is the focal length
(a) Which principle in optics is made use of in
of the combination?
endoscopy?
[ALL INDIA 2016]
(b) Briefly explain the values reflected in the
25. (a) Monochromatic light of wavelength 589 nm
action taken by the teacher.
is incident from air on a water surface. If
(c) In what way do you appreciate the response µw =1.33, find the wavelength, frequency and
of the doctor on the given situation? speed of the refracted light.
[ALL INDIA 2013]
(b) A double convex lens is made of a glass of
21. (a) A mobile phone lies along the principal axis
refractive index 1.55, with both faces of the
of a concave mirror. Show, with the help of a
same radius of curvature. Find the radius
suitable diagram, the formation of its image.
of curvature required, if the focal length is
Explain why magnification is not uniform.
20 cm.
(b) Suppose the lower half of the concave
[ALL INDIA 2017]
mirrors reflecting surface is covered with an
opaque material. What effect this will have 26. A symmetric biconvex lens of radius of curvature
on the image of the object? Explain. R and made of glass of refractive index 1.5,is
[DELHI 2014] placed on a layer of liquid placed on top of a
plane mirror as shown in the figure. An optical (b) Draw a ray diagram to show the image
needle with its tip on the principal axis of the formation by a concave mirror when the
lens is moved along the axis until its real, object is kept between its focus and the pole.
inverted image coincides with the needle itself. Using this diagram, derive the magnification
The distance of the needle from the lens is formula for the image found.
measured to be x. On removing the liquid layer [ALL INDIA 2011]
and repeating the experiment, the distance is 29. (a) A point-object is placed on the principal
found to be y. Obtain the expression for the axis of convex spherical surface of radius
refractive index of the liquid in terms of x and y. of curvature R, which separates the two
media of refractive indices n1 and n2 (n2 >
n1). Draw the ray diagram and deduce the
relation between the distance of the object
(u), distance of the image (v) and the radius
of curvature (R for refraction to take place
at the convex spherical surface from rarer
[ALL INDIA 2018] to denser medium.
27. The figure shows a ray of light falling normally (b) Use the above relation to obtain the condition
on the face AB of an equilateral glass prism on the position of the object and the radius
3 of curvature in terms of n1 and n2 when the
having refractive index , placed in water of real image is formed.
2
4 [ALL INDIA 2015]
refractive index 3 . Will this ray suffer total
internal reflection on striking the face AC? 30. (a) Draw a ray diagram to show image formation
Justify your answer. when the concave mirror produces a real,
inverted and magnified image of the object.
A
(b) Obtain the mirror formula and write the
expression for the linear magnification.
(c) Explain two advantages of a reflecting
telescope over a refracting telescope.
[ALL INDIA 2018]
Solutions
B C 1. Refraction occurs when the energy of an incoming
[ALL INDIA 2018] light wave matches the natural vibration
5 Mark Questions frequency of the electrons in a material. The
light wave penetrates deeply into the material,
28. (a) Draw a ray diagram to show refraction of a
and causes small vibrations in the electrons. The
ray of monochromatic light passing through
electrons pass these vibrations on to the atoms
a glass prism. Deduce the expression for the
in the material, and they send out light waves
refractive index of glass in terms of angle of
of the same frequency as the incoming wave. [1]
prism and angle of minimum deviation.
2. A biconvex lens will act like a plane sheet of
(b) Explain briefly how the phenomenon of total
glass if it is immersed in a liquid having the
internal reflection is used in fibre optics.
same index of refraction as itself. In this case
OR the focal length. [1]
(a) Obtain Lens Maker formula using the 3. The relation between the angle of incidence i,
expression angle of prism A, and the angle of minimum
n2 n1 ( n2 − n1 ) deviation ∆m, for a triangular prism is given by
− = ( A + ∆m)
ν u R i=
2 [1]
Here the ray of light propagates from a rarer
medium of refractive index (n1) to a denser 4. The figure shows a convex lens L placed in
medium of refractive index (n2) is incident contact with a plane mirror M. P is the point
on the convex side of spherical refracting object, kept in front of this combination at a
surface of radius of curvature R. distance of 20 cm, from it. [½]
M So, (δ m )violet > (δ m ) Red [½]

L
X 10. It is given that the prism is equilateral in shape.
So, all the angles are equal to 60°.
P Thus, the angle of prism, A = 60° The angle of
A
refraction in case of a prism, r = = 30° [1]
X 2
Using Snell’s law,
[½] µa sin i = µg sin r
Fig.: Convex lens with a mirror Here, µa → refreactive index of air, n1 = 1
As the image coincides with itself, the rays from µg → refreactive index of air, n2 = 3
the object, after refraction from lens, should fall
i = angle of incidence
normally on the mirror M, so that they retrace
their path. For this, the rays from P, after Thus,
refraction from the lens must from a parallel  µg   µg 
beam perpendicular to M. For clarity, M has been sin i =   sin r =   sin 30°
 µa   µa 
shown at a small distance from L (in diagram).
As the rays from P, form a parallel beam after 3
refraction, P must be at the focus of the lens. sin i = = sin 60°
2
Hence the focal length of the lens is 20 cm.
5. Biconcave lens will change its nature as So, the angle of incidence, i = 60° [1]
refractive index of outside medium is greater 11. We have focal length of convex lens,
than that of lens material so it will behave as f1 = +25cm = + 0.25
converging lens. [1] Focal length of the concave lens,
6. Since µ g( )lens < ( µm )surroundings [1] f2 = +25cm = + 0.25
It behaves like a converging lens. Equivalent focal length,
7. The critical angle for a given pair of medium is 1 1 1 1 1 1
= + = + =−
1 F f1 f2 25 −20 100
given by sin ic = where µ is refractive index
µ \ F = – 100 cm [1]
of denser medium. The refractive index of a Power of convex lens,
medium depends on the wavelength given by
1 1
B P=
1 =
the relation µ = A + 2 , and each colour of light
f1 0.25
λ
Power of concave lens,
is associated with specific wavelength, so as
the wavelength of light increases critical angle 1 1
P2 = =
increases. [1] f2 −0.20

8. Due to scattering phenomenon, we can’t see Power of the combination,
clearly through fog. [1]
P = P1 + P2
9. lred > lviolet
1 1 100 100
1 P= + = −
we know λ ∝ 0.25 −0.20 25 20
µ
So, 400 − 500 100
= =− = −1
µred > µviolet [½] 100 100
For minimum deviation
The focal length of the combination = –1m =
δ + A 
Sin  m –100cm As the focal length is negative, the
 2 
µ= system will be diverging in nature. [1]
 A
sin  
2

12. (a) Necessary conditions for total internal \ F 100cm

reflection to occur are:
Power of convex lens,
(1) The incident ray on the interface should
travel in optically denser medium. 1 1
P=
1 =
f1 0.20
(2) The angle of incidence should be greater
than the critical angle for the given pair of Power of concave lens,
optical media. [1]
1 1
1 P2 = =
(b) a µb = f2 −0.25
sin C [1]
Where a and b are the rarer and denser Power of the combination lens,
media respectively. C is the critical angle P = P1 + P2
for the given pair of optical media. [1]
1 1
1 1 1 = +
= + 0.20 −0.25
13. f + f1 − f2
1 f2 − f1 100 100
= = +
f f1 f2 20 −25

500 − 400
f f =
f = 1 2 100
f2 − f1
[2]
100
14. We have, = = 1D
100
Focal length of convex lens,
The focal length of the combination = 1 m = 100
f1 = +30cm = + 0.30m
cm. As the focal length is positive, the system
Focal length of concave lens, will be converging in nature. [1]
f2 = –40cm = – 0.40m 1 1
sin i = sin 45° = =
Equivalent focal length, 16. 2 1.414
1 1 1 1 1 1 1
⇒ = + = + sin i = >
F f1 f2 30 −40 1 . 414 1 . 52
40 − 30 1 \ sin i > sin ic for ray (2)
= =
1200 120
⇒ i > ic for ray (2) [1]
\ F = 120cm = 1.2m [1] So, ray (2) will suffer T.I.R.
Power of the combination lens,
1 1
<
1 1 As 1.414 1.38
=
P = = 0.83 D
F 1.2 ⇒ i > ic (1)
The focal length of combination So, ray (1) got refracted. [½]
= 1.2 m = 120 m
As the focal length is positive the system will be
converging in nature. [1]
(1)
15. We have, (2)
Focal length of convex lens,
f1 = +20cm = + 0.20m
Focal length of concave lens, (2)
[½]
f2 = –25cm = – 0.25m
Fig.: Trace of ray path through a prism.
Equivalent focal length,
1 1 1 19.
17. Mirror equation is = +
f u v
Where, u is distance of object from the mirror v
is the distance of image from the mirror and f
is the focal length of the mirror.
For a concave mirror f is negative i.e. f < 0, v < 0
For a real object (on the left of mirror)
\ 2f < v < f [1] [1]
1 1 1 1 1 1 ‘O’ is at 2f of lens so it will form image at 2f i.e.
⇒ > > − > −
2f u v f f f 40 cm from lens so position of object for mirror
is at (40 – 15) cm = 25cm behind the mirror For
1 1
⇒ > >0 mirror [1]
2f v
f = + 10 cm
1 1 1 u = + 25 cm
∵ > >−
v f v u=?
1 1 1 1 1
⇒ > <0 + =
2f v v u f
⇒ v < 0 Also, v > 2f 1 1 1
+ =
Image implies that v is negative and greater v 25 +10
than 2f. This means that the image lies beyond 50 50
2f and it is real. [1] ∴v = + cm i.e. behind the mirror [1]
3 3
18. Given AQ = AR hence in ∆AQR 20. (a) Total internal reflection.
∠Q = ∠R = j (Let) (b) Teacher plays a good role to teach a moral in
A + 2 j = 180° (for ∆AQR) a perfect way to student how to support for
humanity by presenting him as an example
⇒ j = 60 ⇒ ∠r1 = ∠r2 = 830°
for students. [1]
This is the condition of minimum deviation
(c) As Chetan belongs to below average income
So using relation group family, keeping this in his/her mind,
  A + δ min   doctor offered concession for the test fee and
 sin  2   served his oath for humanity as a doctor in
ν =  a best possible way. [1]
 A 
 sin 
2 21. (a)
[1]
Given ν = 3 & A = 60°
  A + δ min  
 sin  2  
ν = 
 A 
 sin 
2

 60 + δ min  3
sin   = [1]
 2  2
The image of the mobile phone formed by
 60 + δ min  the concave mirror is shown in the above
 2  = 60 figure. The part of the mobile phone that
is at C will form an image of the same size
dmin = 60° [1] only at C. In the figure, we can see that B’C
= BC. The part of the mobile phone that lies
between C and F will form enlarged image ic = 41°

beyond C as shown in the figure. It can be Now since the angle of incidence is smaller than
observed that the magnification of each part the critical angle refraction will take place as
of the mobile phone cannot be uniform on shown in figure
account of different locations. That is why
For ray 2
the image formed is not uniform. [1]
The critical angle for glass-liquid interface is
(b) As the laws of reflection are true for all
points of the mirror, the height of the whole µ l 1.3
sin ic = = = 0.866
image will be produced. However, as the area µ g 1.5

of the reflecting surface has been reduced,
the image intensity will be reduced. In ic = 60° [½]
other words, the image produced will be less Now since the angle of incidence is smaller than
bright. [1] the critical angle refraction will take place as
shown in figure [½]
A B 23. (a) Given, Height of object = ho
60° Radius of curvature = 20cm
30° Magnification, m = 2
i
Object distance, u =?
Image distance, u =?
Magnification,
υ h
C M= = i
22. (a) u ho [1]

Given the refractive index of prism µg = 1.5 the
υ
ray will grazes along the face BC when the angle 2= ⇒ υ = 2u
u (i)
of incidence i is equal to the critical angle for the
glass and liquid interface here i = 60° (from the Using mirror formula,
fig.) [½] 1 1 1
+ =
µ υ u f
and sin i = l
µg 1 1 −2
− =
µi = µg sin i −2u u 20
µl = 1.5 sin 60° +3 +2
=
µl = 1.3 2u 20
So the refractive index of the liquid is 1.3 [½] 60
=
u = 15cm
B 4
A
i1 60° Putting in (i) v = 1 × 15cm = 30 cm in front of the
ray 1
i2 mirror.
ray 2 hi υ
=
h
o u
C hi 2u
(b) =
[½] ho u
[1]
The angle of incidence for ray 1 and ray 2 on face
Height of image, h1 = 2ho
BC is equal to 30°
For convex mirror, f = +ve (always)
For ray 1
(b) Mirror formula,
The critical angle for glass-air interface is
1 1 1
µ 1 + =
sin ic = a = = 0.66 υ u f
µ g 1.5

1 1 1 (ii)
= −
υ f u
As u = -ve(for real object)

1 1  1 
= − 
υ f  −u 
1 1 1
= +
υ f u (both are of equal focal length = f )
equivalent power of combination
u = +ve [1]
PeqL = P1 = P2
Hence, it will form virtual image.
 1 1 
24. (i) PeqL =  +
 + f − f 

Position Position
PeqL = 0 ⇒ feqL = ∞
(1) (2) [1]
Focal length of the combination is infinite.
25. (a) Given: Monochromatic light of wavelength
I
O l1 = 589 nm = 589 × 10–9 m
20cm Speed of light in air,
Screen
v1 = 3 × 108 ms−1
x (100–x)
[1]
Refractive index of water,
Lens at position (1)
µw = µ2 = 1.33

Object distance, u = –x
Refractive index of air, [½]
Image distance, u = (10 – x)
µa = µ1 = 1

Using lens formula
Find: wavelength of reflected light (l2) = ?
1 1 1
= − Frequency of refracted light (f2) = ?
f ν u Speed of refracted light (v2) = ?
1 1 1 λ1 µ2 µw
f = 100 − x + x -------------- (i) We know, = = [½]
λ2 µ1 µa

Lens at position (2) 589 × 10 −9

= 442.85 × 10 −9
1.33
Object distance, u = –(x + 20)
= 442.85 nm
Image distance, v = –(20 – x) [1]
For speed
Using lens formula
µw µ2 ν 1
1 1 1 ν2, = =
= − µa µ1 ν 2
f ν u
1.33 3 × 108
1 1 1 =
f = 80 − x + x + 20 ---------- (ii)
1 ν2
Solving equation (1) & (2) = 2.25 × 108 m/s

f = 24 cm For frequency, v2 = fl2
v 2.25 × 108
f = 2 =

λ2 442.85 × 10−9 [½]
f = 5.08 × 1014 m/s

(b) Given the refractive index of glass with 1 0.5 × 2

respect to air aµg = 1.55 =
y R
aµg = 1.55
R=y
For double lens
Now for liquid [1]
R1 = R, R2 = – R
1 1
= ( µ ’ − 1)  
f2  R

R1 R2 ( y − x) 1 1
− = ( µ ’ − 1)  − 
xy  R ∞

[½]
( y − x) ( µ ’ − 1)
− =
xy y
 \both faces have same radius of curvature
[for double convex lens, one radius is taken ( x − y)
∴ = ( µ ’ − 1)
as positive and other negative] focal length x
of lens,
y
f = + 20 cm 1+1− = µ’
x
Using lens formula
y
1  1 1  2− = µ’
f
( )
= aµ g − 1  −
 R1 R2 
x [1]
27. (a)
in this formula put for a
Incident AIR Reflected
1  1 1
⇒ = (1.55 − 1)  + 
20  R R
[½]
1 2
⇒ = 0.55 ×
20 R Refracted
R = 0.55 × 2 × 20 = 22cm
Medium
Thus the required radius of curvature is [1]
22 cm [½] (b) No it will not suffer total internal reflection
26. f = focal length liquid + lens A
f1 = focal length of lens only
60°
f2 =focal length of liquid mirror
1 1 1 30°
= +
f f1 f2 60°

60°
1 1 1
= −
f2 f f1 C [1]
B
1 1 1 µw 4 / 3 8
= − µ v= = =
f2 x y µg 3/2 9

xy  8
f2 = =i sin −1 µ= sin −1  = 62.73°
y − x [1] v
 9 

1 2 So if I is less than 60° then TIR will not
Now, = ( µ − 1)   (Convex lens)
f1  R happen. [1]
1 2
= (1.5 − 1)
y R
28. (a) (a)

A N1 N2
M
n1 n1
i r1 r2 e n2
Q R
N
P O P1 C P2 C1 I I1
S R2 v
u v1
B C R1

[1]
Fig.: Lens Maker Formula
Fig.: Ray diagram to depict refraction of
monochromatic ray through a glass For refraction at the first surface
prisms. n2 n1 ( n2 − n1 )
− = ------------- (1)
From ∆MQR, (i − r1 ) + ( e − r2 ) = δ ν1 u R1
So (i + e) − (r1 + r2 ) = δ For the second surface, I1 acts as a virtual
From ∆RQN , r1 + r2 + ∠QNR = 180° object (located in the denser medium) So, for
Also, A + ∠QNR = 180° refraction at this surface, we have
Thus A = r1 + r2 n1 n2 ( n1 − n2 ) ---------- (2)
− =
So i = e – A = d ν ν1 R2
At minimum deviation, i = e, r1 = r and d = Adding (1) and (2)
dm
n1 n2 ( n2 − n1 ) ( n1 − n2 )
A + δm A − = +
⇒i= and r = ν u R1 R2
2 2
sin i n1 n2  1 1 
Also µ = [1] − = ( n2 − n1 )  −
sin r ν u  R1 R2 

 A + δm 
 2  From the above two equations,
Hence µ = sin [1]
 A 1 1  n2  1 1 
sin   − = − 1  −
2 ν u  n1   R1 R2 
(b) [1]
Low n The point, where image of an object, located
at infinity is formed, is called the focus F,
of the lens and the distance f gives its focal
length.
So for u = ∞, n = +f
High n
Fig.: Total internal reflection in fibre optics
1 1  n2  1 1 
⇒ − = − 1  − 
ν u  n1  1R R2
Each optical fibre consists of a core and
cladding, refractive index of the material (b)
of the core is higher than that of cladding. A
When a signal, in the form of light, is M
directed into the optical fibre, at an angle A
greater than the critical angle, it undergoes C B P B
F
repeated total internal reflections along the
length of the fibre and comes out of it at
the other end with almost negligible loss of DABP is similar to DA’ B’ P
intensity. [1] A’ B’ B’ P
=
OR So, AB BP
Now, A’ B’ = I
AB = O Now, for DNOC, i is the exterior angle. Therefore

B’ P’ = n i = ∠NOM + ∠NCM,
BP = –u MN MN
i= + -------- (1) [½]
I ν OM MC
Magnification, m = = − [2]
O u
29. (a) Similarly,
r = ∠NCM + ∠NIM
MN MN
N i.e, r = MC − MI -------- (2)
n1 i n1
Now, by Snell’s law
r
O I n1 sin i = n2 sin r
C
M R Or for small angles
u v n1 i = n2 i
Substituting i and r from Eqs.(1) and (2), we get
Refraction at spherical surface separating two n1 n n − n1
+ 2 = 2
media. OM MI MC ---------- (3)
Figure shows the geometry of formation of Here, OM, MI and MC represent magnitudes
image I of an object O on the principal axis of of distances. Applying the Cartesian sign
a spherical surface with centre of curvature C, convention
and radius of curvature R. The rays are incident OM = –u, MI = +v, MC + R
from a medium of refractive index n1, to another Substituting these in Eq. (3), we get
of refractive index n2. As before, we take the
n2 n1 n2 − n1
aperture (or the lateral size) of the surface to be − = ------------ (4)
ν u R
small compared to other distances involved, so
that small angle approximation can be made. In Equation (4) gives us a relation between object
particular, NM will be taken to be nearly equal and image distance in terms of refractive index
to the length of the perpendicular from the point of the medium and the radius of curvature of the
N on the principal axis. curved spherical surface. It holds for any curved
We have, for small angles, [2] spherical surface. [½]
MN
tan ∠NOM = (b)
OM
MN A
tan ∠NCM = N1 N2
MC
MN
tan ∠NIM =
MI O u B D v I
C
(a)
N1 A
i1 A
i1 r2
r2 N1
C1 I1 C1
O B R1 D I I1
n1 n2 R1
n2 n1
C
C (b)
(a) The position of object, and the image formed

by a double convex lens
 n2 
(b) Refraction at the first spherical surface and ∴ n21 = n 
2
(c) Refraction at the second spherical surface.
Equation (5) is known as the lens maker’s
Figure (a) shows the geometry of image formation formula. It is useful to design lenses of desired
by a double convex lens. The image formation focal length using surfaces of suitable radii of
can be seen in terms of two steps: (i) The first curvature. Note that the formula is true for a
refracting surface forms the image I1 of the object concave lens also. In that case R1 is negative, R2
O [Fig. (b)]. The image I1 acts as a virtual object is positive and therefore, f is negative.
for the second surface that forms the image at I
[Fig. (c)]. Applying From Eqs. (3) and (4), we get
n n n − n1 n1 n n
Equation 2 − 1 = 2 to the first interface + 1 = 1
ν u R OB DI f ..(6)
ABC, [1]
We get, Again, in the thin lens approximation, B and D

are both close to the optical centre of the lens.
n1 n n − n1 Applying the sign convention,
+ 2 = 2
OB BI1 BC1
..(1) BO = –u, DI = +v, we get
A similar procedure applied to the second 1 1 1
interface ADC gives, − = ..(7)
ν u f
n2 n1 n − n1
− + = 2
D1 D1 DC2 Equation (7) is the familiar thin lens formula.
..(2)
For a thin lens, BI1 = DI1 30. (a)
Adding eqs.(1) and (2), we get B

n1 n  1 1 
+ 1 = ( n2 − n1 )  + ..(3)
OB DI  BC1 DC2 
A F l
A i
Suppose the object is at infinity, i.e, OB → ∞ and
DF = f, Eq.(3) gives
B
n1  1 1 
= ( n2 − n1 )  + 
[1]
f  BC1 DC2
…(4) (b)
The point where image of an object placed at
infinity is formed is called the focus F, of the A
lens and the distance f gives its focal length. By M
the sign convention,
BC1 = +R1, B P
B C F
DC1 = –R2 f
A
So, eq 4 can be written as v
u N
1  1 1 
= ( n21 − 1)  − D
f  R1 R2  ..(5) [½]
Ray diagram for image formation by a Using these in equations, we get

concave mirror
−ν + f −ν
=
Figure shows the ray diagram considering three
−f −u
rays. It shows the image A’B’(in this case, real) of
an object AB formed by a concave mirror. It does ν−f ν
not mean that only three rays emanate from the Or =
f u
point A. An infinite number of rays emanate
from any source, in all directions. Thus, point
This relation is known as the mirror equation.
A’ is image point of A if every ray originating at
The size of the image relative to the size of the
point A and falling on the concave mirror after
object is another important quantity to consider.
reflection passes through the point A. [½]
We define linear magnification (m) as the ratio
We now derive the mirror equation or the relation of the height of the image h’ to the height of the
between the object distance (u), image distance object(h): [½]
(v) and the focal length (f). From figure, the
h’
two right-angled triangles A’B’F and MPF are m=
similar. (For paraxial rays, MP can be considered h
to be a straight line perpendicular to CP.)
h and h’ will be taken positive or negative in
Therefore,
accordance with the accepted sign convention.
B’ A’ B’ F ’ In triangles A’ B’ P and ABP, we have,
=
PM FP B’ A’ B’ P
=
B’ A’ B’ F ’ BA BP
Or = (∵ PM = AB) ----------- (i)
BA FP With the sign convention, this becomes
Since ∠APB = ∠ A’ PB’, the right angles triangle
A’B’P and ABP are also similar. Therefore, [½] − h ’ −ν
=
h −u
B’ A’ B’ P
BA = BP ---------- (ii) h’ ν
So that, m = =− [½]
Comparing equation (i) and (ii), we get h u
(c)
B ’ F B ’ P − FP B ’ P
= = ------------ (iii) 1. Spherical and chromatic abbreviation
FP FP BP
eliminated.
Equation (iii) is a relation involving magnitude
of distances. We now apply the sign convention. 2. Objective lenses are large and expensive
We note that light travels from the object to the in refracting telescope, whereas reflecting
mirror MPN. Hence this is taken as the positive telescope is economical. [1]
direction. To reach the object AB, image A’B’as
well as the focus F from the pole P, we have to
travel opposite to the direction of incident light.
Hence, all the three will have negative signs.
Thus,
B ’ P = −ν , FP = − f , BP = −u [½]
[Topic 2] Optical Instrument
Summary =
the objective. Also, m
L
×
D
, where f0 and fe are
fo fe
Simple microscope the focal length of the objective and the eyepiece,
• A simple magnifier or microscope is a converging respectively, and L is the distance between their
lens of small focal length. focal points.
u fo
fc
A
B Eyepiece
Eye focussed h
B B O h E
on near point
Objective A
u
D
A
Fig.: Compound microscope
Telescope
h • The telescope is used to provide angular

i
magnification of distant objects. It also has an
f i
objective and an eyepiece
• Magnifying power (m) of a telescope is the ratio of
Eye focussed the angle b subtended at the eye by the image to the
at infinity angle a subtended at the eye by the object.
Fig.: Magnifier β fo
=
m = , f0 and fe are the focal length of the
D  α fe
• The magnifying power (m) is given by m= 1 +   ,
f
objective and the eyepiece, respectively.
where D = 25 cm is the least distance of distinct vision Objective fo Eyepiece
and f is the focal length of the convex lens. If the image
fe
is at infinity, magnifying power (m) is given by,
O B E
D h
m = .
f A
• For a compound microscope, the magnifying

power m is given by m = me × m0 where,
D 
me = 1 +   is the magnification due to the
f  Fig.: Telescope
eyepiece and mo is the magnification produced by
6. Draw a schematic ray diagram of reflecting

telescope showing how rays coming from a
PREVIOUS YEARS’ distant object are received at the eyepiece. Write

its two important advantages over a refracting
telescope.
TOPIC 2 7. (a) Draw a ray diagram depicting the formation
of the image by an astronomical telescope in
2 Mark Questions normal adjustment
1. Two convex lenses of same focal length but of
aperture A1 and A2 where A1 < A2 are used as the (b) You are given the following three lenses.
objective lenses in two astronomical telescopes Which two lenses will you use as an
having identical eyepieces. What is the ratio of eyepiece and as an objective to construct an
their resolving power? Which telescope will you astronomical telescope? Give reason.
prefer and why? Give reason. Lenses Power Aperture
[ALL INDIA 2011]
2. (i) A giant refracting telescope has an objective L1 3 8
lens of focal length 15 m. If an eye-piece of L2 6 1
focal length 1.0 cm is used, what is the angular L3 10 1
magnification of the telescope? (ii) If this [ALL INDIA 2017]
telescope is used to view the moon. What is the
diameter of the image of the moon formed by
the objective lens? The diameter of the moon is
5 Mark Questions
3.48 × 108 m and the radius of lunar orbit is 8. Define magnifying power of a telescope. Write
3.8 × 108 m. its expression.
[DELHI 2015] A small telescope has an objective lens of focal
length 150 cm and an eye piece of focal length
3 Mark Questions 5 cm. if this telescope is used to view a 100 m
3. Draw a ray diagram showing the image high tower 3 km away, find the height of the
formation by a compound microscope. Hence final image when it is formed 25 cm away from
obtain expression for total magnification when the eye piece.
the image is formed at infinity. Or
[DELHI 2013] How is the working of a telescope different from
4. Draw a labeled ray diagram of a refracting that of microscope?
telescope. Define its magnifying power and write The focal lengths of the objective and eye piece of
the expression for it. a microscope are 1.25 cm and 5 cm respectively.
Write two important limitations of a refractive Find the position of the object relative to
telescope over a reflecting type telescope. the objective in order to obtain an angular
[DELHI 2013] magnification of 30 in normal adjustment.
5. (a) Draw a labelled ray diagram showing the [ALL INDIA 2012]
formation of a final image by a compound 9. (a) Draw a labelled ray diagram showing
microscope at least distance of distinct the formation of image by a compound
vision. microscope in normal adjustment. Derive
(b) The total magnification produced by the expression for its magnifying power.
a compound microscope is 20. The (b) How does the resolving power of a microscope
magnification produced by the eye piece is change when
5. The microscope is focussed on a certain (i) the diameter of the objective lens is
object. The distance between the objective decreased.
and eye-piece is observed to be 14 cm. If least
(ii) the wavelength of the incident light is
distance of distinct vision is 20 cm, calculate
increased ?
the focal length of the objective and the eye-
piece. [DELHI 2014] Justify your answer in each case.
[ALL INDIA 2015]
Solutions u fo
fe
A
a K B Eyepiece
1. Resolving power =
1.22λ B B O K E
Objective A
( R.P )1 A
⇒ = 1

( R.P )2 A2
[1]
The telescope with the objective of aperture A1 D
should be preferred for viewing as because A1 >
A2, this would: A [1]
Fig.: Ray Diagram showing image formation
(i) Give a better resolution
by a compound microscope.
(ii) Have a higher light gathering power of
If uo is the distance of the object from the
telescope. [1]
objective and vo is the distance of the image
2. (i) Let, fo → Focal length of the objective lens from the objective, then the magnifying
= 15m = 1500cm power of the objective is
fe → Focal length of the eye lens = .0m h’ L  h   h’  
Angular magnification of the giant refracting
Mo = =  Using, tan β =   =   
h fo   fo   L  
telescope is given by, [1]
Where, h, h are object and image heights
fo
mo = respectively and fo is the focal length of the
fe objective.

(ii) Diameter of the image of the moon formed L is the tube length i.e. the distance between
by the objective lens, d = afo the second focal point of the objective and
the first focal point of the eyepiece.
Diameter of the moon
⇒d= × fo D
Radius 0 f the lunar orbit When the final image is at infinity, Me = f
e
3.48 × 10 8 Magnifying power of compound microscope,
⇒d= × 1500 × 10 −2 L D
3.8 × 108 M = Mo × Me = ×
fo fe
d = 1373.68 cm = 13.73m [1] [1]
4.
3. A compound microscope consists of two
convex lenses parallel separated by some A Objective Eyepece
distance. The lens nearer to the object Rays fo
distanfrom fe
is called the objective. The lens through t obje
ct Fe Fc Fe
which the final image is viewed is called the B B E
O
eyepiece. At infinity B
A Eye
Magnifying power, when final image is at
infinity: The magnification produced by the ue
compound microscope is the product of the A
D
magnifications produced by the eyepiece and

objective.
[1]
\ M = Me × Mo [1]
Fig.: Astronominal Telescope
Where, Me and Mo are the magnifying powers
Magnifying power of an astronomical telescope
of the eyepiece and objective respectively.
is defined as the ratio of the angle subtended by
the final image at the eye to the angle subtended ⇒ fe = = 5 cm [1]

by the object at the eye. If α and β, be the angles Substituting the value of m and me in equation
subtended by the object and image with the eye (1), we get:
respectively, then
m = mo me
β
M.P. = 20 = mo × 5
α
mo = 4
(a) For distinct vision Now, we have ue + υ o = 14
−f  fe 
M.P. = o  1 + D  (i.e. distance between objective and eye piece)
fc
u + vo = 14
(b) For normal vision, ue = fe vo = 10cm
 fo 
M.P. = −  f  [1] υ
c Now, mo = 1 − o
Limitations of refracting telescope: fo
(i) The refracting telescope suffers from 10
spherical & chromatic aberrations. −4 = 1 −
fo
(ii) Objective lens of refracting telescope of
very large apertureis very difficult to
fo = 2cm [1]
manufacture. [1]
6. Reflecting Telescope: The reflecting telescope
5. (a)
makes us of a concave mirror as objective. The
E
rays of light coming from distant object are
incident on the objective (parabolic reflective).
O v0
C After reflection the rays of light meet at a point
A 0
B2 where another convex mirror is placed. This
F OL B1 L Oe mirror focusses light inside the telescope tube.
u0 v
AL The final image is seen through the eye piece.
The images produced by the reflecting telescope
uo are very bright and its resolving power is
high. [1]
A2
D
[1] Objective
mirror
\\\
Fig.: Ray Diagram in a Compound Microscope Secondary

\\\\\\\\\\
(b) For the least distance of clear vision, the mirror

total magnification is given by:
L  D
m=
fo  1 + f  = mo me Eye plece
e
Where, L is the separation between the eyepiece
and the objective
[1]
fo is the focal length of the objective
Fig.: Reflecting Telescope
fe is the focal length of the eyepiece
Advantages:
D is the least distance for clear vision
(i) The resolving power (the ability to observe
Also, the given distance is for the eyepiece:
two objects distinctly) is high, due to the
D large diameter of the objective.
me = 5 = 1 +
fe (ii) There is no chromatic aberration as the

20 objective is a mirror. [1]
⇒ 5 = 1+
fe

7. Ray diagram of Astronomical telescope : Height of image H’

tan β = =
O Distance of image formation D
Eye lens −36 × 25
parallel lays
fo Thus, H ’ = = −30 cm
from objecs fe 30
at infinity Fo F Negative sign indicates that image formed will

e
be inverted. [3]
C1
Or
B A microscope is used to look into smaller details
Objective lens i.e. structures of cells etc. on the other hand, a
telescope is used to see larger objects that are
too much far like stars, moon etc.
[1] Telescope mainly focuses on collecting the light
Fig.: Ray diagram of astronomical telescope into the objective lens, which should thus
(b) We use L1 as an objective lens because it has be large, while the microscope already has a
higher aperture. That is 8 cm. So that it has focus and the rest is blurred out. There is a big
high resolving power and lens L3 use as an difference in their magnification factor.
eye piece because it has high power. So that For telescope the angular magnification is given
magnification is more. [2] fo
by, M = −
8. Magnifying power of a telescope is defined as the fe
Where fo = is the focal length of the objective
ratio of the angle subtended at the eye by the
lens and fe is focal length of the eye-piece. For
image formed at the least distance of distinct
microscope the angular magnification is given
vision to the angle subtended at the eye by the
by, [2]
object lying at the infinity, when seen directly.
fo  f   D
M =  1 +  ; When image is formed at distance
 1+ e
M=−  fo 
Magnifying power, fe  D
Where, fo = Focal length of the objective = 150 of least distinct
cm D
fe = Focal length of the eye-piece = 5 cm M= ; When image is formed at infinity
fo
D = Least distance of distinct vision = 25 cm
Where D is the distance of least distinct vision
150  5
M=− ×  1 +  = −36 and f o is the focal length of objective lens.
5  25 
Magnifying power of compound microscope
β Given: fo = 1.25, fe = 5 cm and M = – 30
M=
α (magnifying power is negative)
tan β ν  D
M= (As angles α and β are small) We know that, M = o 1 + 
tan α uo  fe 

Height of the object Where, v o = distance of image from the
tanα = objective [2]
Distance of object from objective [2]
uo = distance of object from the objective
H 100 1
= = = D = distance of least distinct vision
u 3000 30
νo  25 
tan β −30 = 1 + 
M= uo 5 
1
30 vo = –5 uo
36 Using lens formula,
tan β = −
30 1 1 1
=− +
fo uo ν o
1 1 1 Where we have used the result

=− −
1.25 uo 5uo  h   h’ 
tan β =   =  
 fo   L 
Uo = –1.5 cm
Thus, the difference of object from objective is Here h′ is the size of the first image, the object
1.5 cm. [1] size being h and fo being the focal length of the
9. (a) objective. The first image is formed near the focal
point of the eyepiece. The distance L, i.e., the
u fo distance between the second focal point of the
fc
A
objective and the first focal point of the eyepiece
h B Eyepiece
(focal length fe) is called the tube length of the
O h
B B E compound microscope. As the first inverted
Objective A image is near the focal point of the eyepiece,
we use the result for the simple microscope to
obtain the (angular) magnification me due to it
 D
equation m =  1 +  , when the final image is
D  f
formed at the near point, is [1]
A  D
me =  1 + 
 fe 
[1]
Fig.: Compound microscope When the final image is formed at infinity, the
Ray diagram for the formation of image by angular magnification due to the eyepiece is
a compound microscope.  D
me =  
A simple microscope has a limited maximum  fe 
magnification (≤ 9) for realistic focal lengths.
For much larger magnifications, one uses two Thus, the total magnification, when the image
lenses, one compounding the effect of the other. is formed at infinity, is
This is known as a compound microscope. A  L   D
schematic diagram of a compound microscope m = mo me =    
 fo   fe 
is shown in Fig.. The lens nearest the object, [1]
called the objective, forms a real, inverted, Clearly, to achieve a large magnification of a
magnified image of the object. This serves as the small object (hence the name microscope), the
object for the second lens, the eyepiece, which objective and eyepiece should have small focal
functions essentially like a simple microscope lengths.
or magnifier, produces the final image, which is
(b) The resolving power of a microscope is given
enlarged and virtual. The first inverted image
by the relation
is thus near (at or within) the focal plane of
the eyepiece, at a distance appropriate for final 1 2n sin β
RP = =
image formation at infinity, or a little closer dmin 1.22λ

for image formation at the near point. Clearly,
(i) When the diameter of the objective lens is
the final image is inverted with respect to the
decreased β decreases so resolving power
original object. We now obtain the magnification
decreases.
due to a compound microscope. The ray diagram
of Figure shows that the (linear) magnification (ii) When the wavelength of the incident light is
h’ increased resolving power decreases. [1]
due to the objective, namely h , equals
h’ L
m=o =
h fo
[1]
CHAPTER 10
Wave Optics

Wave front and Huygen's 1Q 1Q 1Q
principle (3 marks) (5 marks) (2 marks)
Reflection and refraction
of plane wave at a plane
surface using wave fronts
Proof of laws of reflection
and refraction using Huy-
gens’s principle. Interfer-
ence
Young's double slit ex- 1Q 1Q 1Q 1Q 1Q
periment and expression (5 marks) (5 marks) (5 marks) (2 marks) (3 marks),
for fringe width, coherent 1Q
sources and sustained in- (3 marks)
terference of light
Diffraction due to a single
slit; width of central max-
imum
Resolving power of micro- 1Q 1Q
scopes and astronomical (3 marks) (3 marks)
telescopes
Polarisation, plane pola- 1Q 1Q 1Q
rised light, Brewster's (2 marks) (5 marks) (5 marks)
law; uses of plane polar-
ised light and Polaroids
On the basis of above analysis, it can be said that from exam point of view Young’s Double Slit
Experiment, Wavefront and Huygen’s Principle are most important concepts of the chapter.
166 CHAPTER 10 : Wave Optics
[Topic 1] Huygens Principle
Summary • Doppler Effect is defined as the change in

wavelength or frequency of a wave in relation
• The effects which depend on wave nature of light to observer who is moving relative to the wave
are included under wave optics. Interference and source. The Doppler shift is expressed as:
diffraction of light shows that light behaves as
∆ν υ r adial
wave and not as a stream of particles. = −
ν c
• Huygens principle: It states that each point
of the wavefront is the source of a secondary
disturbance. Also, the wavelets emanating from
these points spread out in all directions with
the speed of the wave which are referred to as
PREVIOUS YEARS’
secondary wavelets and if we draw a common EXAMINATION QUESTIONS
tangent to all these spheres, a new position of the TOPIC 1
wavefront is obtained at a later time.
1 Mark Questions
• When a wave gets refracted into a denser medium 1
. In a transistor, doping level in base is increased
the wavelength and the speed of propagation slightly. How will it affect
decrease but the frequency remains the same. (1) Collector current and
(2) Base current? [ALL INDIA 2011]
Incident wavefront 2. What type of wave front will emerge from a
(i) point source, and (ii) distant light source?
A´ B [DELHI 2017]
v1 2 Mark Questions

v1 i 3. Define a wavefront. Using ‘Huygens’ principle,
draw the shape of a refracted wavefront, when
Medium 1 i a plane wavefront is incident on a convex lens.
P A r C P´
v 2 [ALL INDIA 2015]
r
Medium 2
Refracted 3 Mark Questions
v2 < v1 E wavefront 4. Use Huygen’s principle to verify the laws of
refraction. [ALL INDIA 2011]
Incident wavefront 4 Mark Questions

5. Use huygen’s principle to explain the formation
Medium 1 of diffraction pattern due to a single slit
B illuminated by a monochromatic source of light.
i When the width of the slit is made double the
v1 v 1
original width, how would this affect the size
i and intensity of the central diffraction band?
Medium 2 A r C [ALL INDIA 2012]
r v 2
v2
5 Mark Questions
Refracted 6. (a) Define wave front. Use Huygens’ principle
v2 < v1 wavefront to verify the laws of refraction.
E (b) How is linearly polarised light obtained by
the process of scattering of light? Find the
Fig.: Huggen’s Principle
Brewster angle for air – glass interface,
n1 sin i = n2 sin r is the Snell’s law of refraction. when the refractive index of glass = 1.5
[ALL INDIA 2017]
CHAPTER 10 : Wave Optics 167
7. (a) Define a wavefront. Using Huygens’ Refraction of wavefront

principle, verify the laws of reflection at a
plane surface.
(b) In a single slit diffraction experiment, the C
width of the slit is made double the original B H C1
width. How does this affect the size and D
intensity of the central diffraction band? A
E
Explain. F C2
(c) When a tiny circular obstacle is placed in the
path of light from a distant source, a bright
[1]
spot is seen at the centre of the obstacle.
Explain why. Fig.: Law of refraction
[ALL INDIA 2018] Let the time taken by to reach from C to D be
t and now in same time point A would be at F
Solutions hence we can write,
1. I f doping level of base is increased, the resistivity CD = C1t and AF = C2t
of the base will reduce. Also the angle CAD= angle i
(1) The collector current (I C) will decrease. [Angle of incidence for wavefront]
Consequently the current gain of the Similarly, angle ADF= angle r [1]
IC CD t
transistor, will decrease. [½] Now, sin i  sin CAD   C1
IB AC AC
(2) As a result the base current (IB) will increase. AF t
Since, IE = IB + IC [½] And, sin r  sin FDA   C2
FD FD
sin i c1 2
2. (i) For point source, wave front will be
We get the ratio sin r = c = 1 µ replace 1µ2
spherical. [½] 2
This verifies the law of refraction or Snell’s
(ii) For a distant light source, the wave fronts law. [1]
will be plane wave fronts. [½]
5. Consider a parallel beam of monochromatic
3. The locus of points, which oscillate in phase is light is incident normally in a slit of width b
called a wavefront; thus a wavefront is defined (as shown in figure). According to Huygens’s
as a surface of constant phase. [1] principle every point of slit acts as a source of
secondary wavelets spreading in all directions.
Incident Screen is placed at a larger distance.
planewave Consider a particular point P on the screen
F receives waves from all the secondary sources.
All these waves start from different point of the
slit and interface at point P to give resultant
intensity. [1]
Spherical wavefront
of radius f
Fig.: Reflected wavefront when a plane wave P
front is incident on convex lens A
Given above, we consider a plane wave incident
on a thin convex lens, the emerging wave front
is spherical and converges to the point F which b P0
is known as the focus. [1]
4. Laws of refraction: Proof using Huygen’s
principle
If AC is the incident wavefront and FD is
refracted wavefront then we can show that

refraction of wavefront obeys Snell’s law.
Fig.: Single Slit Experiment
[1]
Point Po is at bisector plane of the slit. At Po, all Ak sin i KP ’ sin r

waves are traveling in equal optical path. So are t= +
c n
in phase the waves thus interfere constructively
with each other and maximum intensity is Ak sin i ( AP ’− Ak) sin r
or t = + …(iv)
observed. As we move from Po, the wave arrives c n
with different phase and intensity is changed. The rays from different points or the president
Intensity at point P is given by: [1] wave front will take the same time to reach the
corresponding points on the refracted wave front
sin 2 a
I = Io i.e., given by equation (iv) is independent of Ak.
a will happens so, if
π sin i sinr
Where, α = 2 bsin θ  0
c v
For central maxima a = 0; thus, I = Io
sin i c
When the width of slit is doubled the original = =
width intensity will get four times of its original sin r v [1]
value. Width of central maxima is given by, c
However, =n
2Dλ v
β= This is the shell’s law for refraction of light.
b
Where, D = Distance between screen and slit B
l = wavelength of the light, F
i C
b = size of slit i D B
1
P r Q
So, with the increase in size of the slit the width A
of central maxima decreases. Hence, double the Secondary r (>
2 1 )
Wavelent E
size of slit would result in half the width of the A
central maxima. [1]
6. (a) Wave front: where light is emitted from a A”¯ [1]
source, then the particles present around it Fig.: Huygen’s Principle
begins to vibrate, the locus at all such particles
(b) Polarization by scattering: Polarization
which are vibrating in the same phase is termed
also occurs when light is scattered while
as wave front.
travelling though of medium. When light
Laws of refraction: Suppose when distribution strikes the atoms of a material if will
from point P on incident wave front reaches point often set the electrons of those atoms into
P on the refracted wave front the disturbance vibrations the vibrating electrons then
from point Q reaches the point Q or the produce their own electromagnetic wave
refracting surface XY. Since, A ‘Q’ P’ represents that is radiated outward in all directions.
the refracted wave front the time takes by light These vibrating electrons produce another
to travel from a point on incident wave front to electron magnetic wave that is once more
the corresponding point on refracted wave front radiated outward in all directions. This
would always be the same. Now, time taken by absorption and refraction of light waves
light to go from r to Q’ will be [1] causes the light to be scattered about the
Qk kQ  medium.
t  … (i)
c v This process of scattering contributes to the
(Where c and v are the velocities of light in two blueness of our sky. This scattered light is
medium) partially moralized. [2]
In right angled DAQk, DQAk = I, Here, refractive index m = 1.5 iB = ?
Qk = Ak sin i … (ii) From Brewster’s law tan iB = m = 1.5
In right angleDP’Q’,DQP’ Q’ k = r, iB = tan–1 (1.5) = 56.3°
KQ’= KP’ sin r … (iii)
Substituting eq. (ii) and (iii) in eq. (i), we get
7. (a) B’ A’ B’ P
= ---------------- (ii)
B BA BP
Comparing equation (i) and (ii), we get [½]
B ’ F B ’ P  FP B ’ P
 
FP FP BP --------------- (iii)
F Equation (iii) is involving magnitude of distances.
A’ i
We now apply the sign convention. We note that
A l light travels from the object to the mirror MPN.
Hence this is taken as the positive direction.
To reach the object AB, image A’B’ as well as
the focus F from the pole P, we have to travel
B’ opposite to the direction of incident light. Hence,
[½] all the three will have negative signs. Thus,
B ’ P = −n , FP = − f , BP = −u

(b)
Using these in equations, we get
A M −n + f −n
=
−f −u

n−f n
=
B’ P Or f u [½]
B C F This relation is known as the mirror equation.
The size of the image relative to the size of the
f
A’ object is another important quantity to consider.
v We define linear magnification (m) as the ratio
u of the height of the image ‘h’ to the height of the
N
object (h):
D h’
m=
[½] h
Figure shows the ray diagram considering three h and h’ will be taken positive or negative in
rays. It shows the image A’ ‘(in this case, real) of accordance with the accepted sign convention.
an object AB formed by a concave mirror. It does In triangles A’ B’ P and ABP, we have,
not mean that only three rays emanate from the
point A. An infinite number of rays emanate B’ A’ B’ P
=
from any source, in all directions. Thus, point BA BP [1]
A’ is image point of A if every ray originating at With the sign convention, this becomes
point A and falling on the concave mirror after
−h ’ −n
reflection passes through the point A. [1] =
h −u
We now derive the mirror equation or the relation
between the object distance (u), image distance h’ n
m= =−
(v) and the focal length (f). From Figure, the So that, h u
two right-angled triangles A’B’F and MPF (c)
are similar. (For par axial rays, MP can be 1. Spherical and chromatic abbreviation
considered to be a straight line perpendicular eliminated.
to CP.) Therefore,
2. Objective lenses are large and expensive
B’ A’ B’ F ’ in refracting telescope, whereas reflecting
=
PM FP telescope is economical. [1]
B’ A’ B’ F ’
Or  ∵ PM  AB ----------- (i)
BA FP
Since APB  A ’ PB ’ , the right angles triangle
A’B’P and ABP are also similar. Therefore,
[Topic 2] Interference of Light
Summary S1P – S2P = nl(n = 0, 1, 2, ....) and resultant

intensity is 4I0
• Superposition principle states that at a
• Destructive interference: It is observed in cases
particular point in the medium, the resultant
when two coherent sources are vibrating in phase
displacement produced by a number of waves is
having path difference for a point P as
the vector sum of the displacements produced by
each of the waves.  1
S1P − S 2 P = n +  λ (n =0 , 1, 2, .....) and
• The resultant displacement at a point from two 2
coherent sources will be equal to the sum of the resultant intensity is zero.
individual displacement at that point. • Young’s double slit of length d gives equally
y = 2a cos wt spaced fringes which are at angular separation
Resultant intensity is four times the intensity λ . The midway-point of the slits, the central
produced by each source. d
I = 4I0 and I0  a2 bright fringe and the source, all lie in a straight
• Constructive interference: It is observed in line. But this fringe gets destroyed by an extended
cases when two coherent sources are vibrating in λ
source, if the angle subtended is more than at
phase having path difference for a point P as d
the slits.
P G
P
O
S1 x x
S1° d Z z
O y
S°2 D
S2
D
S
G
Fig.: Young’s Double Slit Experiment
n λD
• Path difference, y =
d
• Fringe width: Distance between two consecutive bright and dark fringes represented by
λD
d
(b) Two wavelengths of sodium light 590 nm

and 596 nm are used, in turn, to study the
diffraction taking place due to a single slit
PREVIOUS YEARS’ of aperture 1 × 10–4 m. The distance between
the slit and the screen is 1.8 m. Calculate the
EXAMINATION QUESTIONS separation between the positions of the first
maxima of the diffraction pattern obtained
TOPIC 2 in the two cases.
1. When monochromatic light travels from one 8. (a) In what way is diffraction from each slit
medium to another its wavelength changes but related to the interference pattern in a
frequency remains the same, Explain. double slit experiment.
[DELHI 2012] (b) Two wavelengths of sodium light 590 nm
and 596 nm are used, in turn to study the
2 Mark Questions diffraction taking place at a single slit of
2. A parallel beam of light of 450 nm falls on a aperture 2 × 10–4m. The distance between
narrow slit and resulting diffraction pattern is the slit and the screen is 1.5 m. Calculate the
observed on a screen 1.5 m away. It is observed separation between the positions of the first
that the first minimum is at a distance of 3 mm maxima of the diffraction pattern obtained
from the centre of the screen. Calculate the in the two cases.
width of the slit. [DELHI 2013]
[ALL INDIA 2013] 9. Answer the following questions:
3. For a single slit of width a, the first minimum of (a) In a double slit experiment using light
the diffraction pattern of a monochromatic light of wavelength 600 nm, the angular width of
of wavelength k occurs at an angle of k/a. At the the fringe formed on a distant screen is 0.1°.
same angle of k/a, we get a maximum for two Find the spacing between the two slits.
narrow slits separated by a distance ‘a’. Explain. o
(b) Light of wavelength 5000 A propagating
[DELHI 2014]
in air gets partly reflected from the surface
4. Draw the intensity pattern for single slit
of water. How will the wavelengths and
diffraction and double slit interference. Hence,
frequencies of the reflected and refracted
state two differences between interference and
light be affected?
diffraction patterns.
[DELHI 2015]
[ALL INDIA 2017] 10. In Young’s double slit experiment, the two slits
4 Mark Questions are separated by a distance of 1.5 mm and
5. Describe Young’s double slit experiment the screen is placed 1 m away from the plane
to produce interference pattern due to a of the slits. A beam of light consisting of two
monochromatic source of light. Deduce the wavelengths 650 nm and 520 nm is used to
expression for the fringe width. obtain interference fringes. Find :
[ALL INDIA 2011] (a) the distance of the third bright fringe for
6. (a) Why are coherent sources necessary to λ = 520 nm on the screen from the central
produce a sustained interference pattern? maximum.
(b) In Young’s double slit experiment using (b) the least distance from the central maximum
monochromatic light wavelength ‘l’, the where the bright fringes due to both the
intensity of light at a point on the screen wavelengths coincide.
where the path differences is ‘l’, is K units. [ALL INDIA 2015]
Find out the intensity of light at a point 11. In a single slit diffraction experiment, when tiny
circular obstacle is placed in path of light from
l
where path difference is . a distance source, a bright spot is seen at the
3 centre of the shadow of the obstacle. Explain
[ALL INDIA 2012]
7. ( a ) Wr i t e t w o c h a r a c t e r i s t i c f e a t u r e s why? State two points of difference between the
distinguishing the diffraction pattern from interference patterns obtained in Young’s double
the interference pattern fringes obtained in slit experiment and the diffraction pattern due
Young’s double slit experiment. to a single slit. [DELHI 2017]
12. (a) If one of two identical slits producing

interference in Young’s experiment is Solutions
covered with glass, so that the light intensity 1. Atoms (of the second medium) oscillate with the
passing through it is reduced to 50%, find same (incident light) frequency and in turn, emit
the ratio of the maximum and minimum light of the same frequency. [1]
intensity of the fringes in the interference 2. l = 450 × 10–9 m
pattern. D = 1.58 m
(b) What kind of fringes do you except to for 1st minima
observe if white light is used instead of
Dl
monochromatic light? [ ALL INDIA 2018] y=
a [½]
5 Mark Questions 1.5  450  10 9
13. (i) ‘Two independent mono-chromatic sources of 3  10 3 
a
light cannot produce a sustained interference
pattern. Give reason. 1.5  450  10 9
a
(ii) Light wave each of amplitude ‘a’ and 3  10 3
frequency ‘ω ’, emanating from two coherent a = 0.225 m [½]
light sources superpose at a point. If the
displacements due to these waves is given by 3. The path difference of two secondary wavelets
y1 = a cos wt and y2 = a cos (ω t + f) where f is is given by nl = a sinq. Since, q is very small sin
the phase difference between the two, obtain q = q. So, for the first order diffraction n = 1, the
the expression for the resultant intensity at l
the point. [DELHI 2014] angle is a . Now, we know that q must be very
small, because of which the diffraction pattern
14. (a) In Young’s double slit experiment, describe is minimum. [1]
briefly how bright and dark fringes are
Now for interference case, for two interfering
obtained on the screen kept in front of a
waves of intensity I1 and I2 we must have two
double slit. Hence obtain the expression for
slits separated by a distance. We have the
the fringe width.
resultant intensity, I = I1 + 2I2 + I1 I2 cos q Since,
(b) The ratio of the intensities at minima to λ
the maxima in the Young’s double slit q = 0 (nearly) corresponding to angle cos θ = 1
a
experiment is 9 : 25. Find the ratio of the (nearly). So,
widths of the two slits. I = I1 + 2I2 + I1 I2 cos q
[ALL INDIA 2014] ⇒ I = I1 + 2I2 + I1 I2
15. (i) In Young’s double slit experiment, deduce We see the resultant intensity is sum of the two
the condition for (a) constructive, and (b) intensities, so there is maxima corresponding
destructive interference at a point on the l
screen. Draw a graph showing variation of to the angle a . This is why, at the same angle
intensity in the interference pattern against of a, we get a maximum for two narrow slits
position y on the screen. separated by a distance a. [1]
(ii) Compare the interference pattern observed
in Young’s double slit experiment with single
slit diffraction pattern, pointing out three 4. Intensity distribution graph for double slit
distinguishing features. experiment is shown below.
[DELHI 2016] Intensity
Central bright
16. (a) Derive an expression for path difference Ist bright
in Young’s double slit experiment and

obtains the conditions for constructive and
destructive interference at a point on the
screen. B2 D2 B1 D1 O D1 B1 D2 B2 D3
[1]
[DELHI 2016] Fig.: Intensity Pattern for Double slit diffraction
Intensity Now consider a point P on the screen. The phase

difference between the waves at P is , where
(Where DPo is optical path difference, DPo = DPg;
DP being the geometrical path difference.)
P
–3 –2 O 2 3 S1
Path difference (d sin) y
Fig.: “Intensity distribution graph for single
d
slit diffraction” is given below. O
Difference between interference pattern and the
S2 d sin
diffraction pattern [1]
Interference D
1. All bright and dark fringes are of equal width.
2. All bright fringes are of same intensity. 2π
= [ S2 P − S1 P ] (here l = 1 in air)
Diffraction λ
As D > > d,
1. The central bright fringe has got double
S P − S1 P ≈ λ d sin θ
width to that of width of secondary maxima 2
and minima.  y
sin θλ ≈ tan θ = 
2. Central fringe is the brightest and intensity  D 

of secondary maxima decreases with
[for very small q]
increase of order of secondary maxima on
2π  dy 
 
either side of central maxima.
Thus, θ = λ  D 
5. Young double slit experiment:
For consecutive interference,
A train of plane light waves is incident on a
barrier containing two narrow slits separated by q = 2n l (n = 0, 1, 2, .....)
a distance d. The widths of the slits are small 2π  dy  D
compared with wavelength of the light used, so
⇒   = 2nπ ⇒ y = nλ
λ D   d [1]
the interference occurs in the region where the
light from s1 overlaps that from s2 [½] Similarly for destructive interference,
D
P

y = (2n + 1)l
2d
( )
n = 1, 2, 3......
S
d Fringe Width W
O It is the separation of two consecutive maxima
S1 or two consecutive minima.
Screen Near the centre O [where q is very small],
double D yn+1 – yn [yn gives the position of nth maxima on
Fig.: Young double slit experiment screen]
A series of alternately bright and dark bands can D
be observed on a screen placed in this region of W =l
d [½]
overlap.
6. (a) Coherent sources produce coherent light
The variation in light intensity along the screen
near the centre O shown in the figure which have wavelength restricted to a very
small range. So the sources have almost
same wavelength, hence produce stable and
sustained interference pattern. [1]
 2π 
O (b) We know, phase difference =  × path
 λ 
difference
a bright fringe fringe width a dark fringe
At path difference l, phase difference
Fig.: Pattern Distribution [1]
 2π  l2 = 596 nm = 5.96 ×10−7 m

φ =  × λ = 2π
 λ 
f  [1] Distance of the slits from the screen = D = 1.5 m
2 
Intensity, I = 4 Io cos  2 
Distance between the two slits, a = 2 × 10–4m [1]
2  2p 
Or, K = 4 Io cos   [given, I = K at path For the first secondary maxima,
2
difference K = 4Io] 3λ1 x
sin θ = = 1
Or, K = 4Io 2a D
K 3l1 D 3l D
 Io  x1 = And x2 = 2
4 [1] 2a 2a
l
Now, at path difference =
3 3 × 590 ×10−9 ×15
2π λ 2π x1 =
φ’ = × = 2 × 2 ×10−4 And
λ 3 3
3 × 596 ×10−9 ×1.5
Intensity, x2 =
2 1  2p  2 × 2 ×10−4
Or, I ’ = 4 Io cos   [1]
2 3  x1 = 663.75 × 10–5 And x2 = 670.50 × 10–5
7. (a) Two characteristic features distinguishing
the diffraction pattern from the interference Spacing between the positions of first secondary
fringes obtained in Young’s double slit maxima of two sodium lines
experiment are: [1] x2 – x1 6.75 × 10–5 M = 0.0675 mm [1]
(i) The interference fringes may or may not 9. (a) Angular width (q) of fringe in double-slit
be of the same width whereas the fringes λ
experiment is given by θ =
of diffraction pattern are always of varying d
width. [1] Where, d → spacing between the slits [1]
(ii) In interference the bright fringes are of same Given, wavelength of light A = 600nm
intensity whereas in diffraction pattern the Angular width of fringe, q = 0.1° = ....... = 0,0018
intensity falls as we go to successive maxima rad
away from the centre, on either side. [1] λ
∴d=
(b) Wavelength of the light beam, λ1 = 590 nm θ
= 5.9 × 10–7 m
600 ×10−9
Wavelength of another light beam, l2 = 596 d=
nm = 5.96 × 10–7 m 18 ×10−1
Distance of the slits from the screen = D = 1.8 m 3
 d  0.33  10 m [1]
Distance between the two slits =1 × 10 m –4
(b) The frequency and wavelength of reflected
For the first secondary maxima, wave will not change.
3l1 D 3l2 D The refracted wave will have same frequency,
x1 = and x2 = 2a
2a only wavelength will change.
Spacing between the positions of first
The velocity of light in water is given by v = lf
secondary maxima of two sodium lines
Where, v → Velocity of light [1]
3D
= (l2 − l1 ) = 1.62 ×10−4 m f → Frequency of light
2a [1]
l → Wavelength of light
8. (a) If the width of each slit is comparable to the
As light ray in travelling from rare (air) medium
wavelength of light used, the interference
to denser medium, its speed will decrease. Hence
pattern thus obtained in the double-slit
wave length (l) will also decrease. [1]
experiment is modified by diffraction from
10. (a) Third bright fringe for l1 = 520 nm is given
each of the two slits. [2]
by
(b) Given that: Wavelength of the light beam,
3l D 3 × 520 ×10−9 ×1
−7
l1 = 590 nm = 5.9 ×10 m
x3 = = = 1.04 ×10−3 m
d 1.5 ×10−3
Wavelength of another light beam, x3 = 1.04 mm [1]
(b) Let n1 bright band of l1 = 520 nm coincides (b) Central fringe will be white and remaining
with n2 bright band of l2 = 650 nm will be coloured in VIBGYOR sequence. [1]
n1l1 D n2l2 D 13. (i) The condition for the sustained interference is
So, =
d d that both the sources must be coherent (i.e.
n1 l1 = n2 l2 they must have the same wavelength and the
n1 l2 650 5 same frequency, and they must have the same
= = =
n2 l1 520 4 phase or constant phase difference). [1]
[1]
So the least distance from the central maximum Two sources are monochromatic if they have
where the bright fringes due to both the the same frequency and wavelength. Since they
wavelengths coincide is are independent, i.e. they have different phases
with irregular difference, they are not coherent
n1l1 D 5 × 520 ×10−9 ×1
x= = sources. [1]
d 1.5 ×10−3 [2] (ii)
= 1.73 × 10–3 m P
= 1.73 mm [1]
11. A bright spot is observed when a tiny circular
object is placed in path of light from a distant
source in a single slit diffraction experiment
because light rays flare into the shadow region S1
of the circular object as they pass the edge of
the tiny circular object. The lights from all the
edges of the tiny circular object are in phase O O
with each other. Thus, they form a bright spot
at the centre of the shadow of the the tiny
circular object. The two differences between the
S2
interference patterns obtained in Young’s double
Screen
slit experiment and the diffraction pattern due
to a single slit are as follows: [2] [1]
(i) The fringes in the interference pattern Let the displacement of the waves from the
obtained from diffraction are of varying width, sources and at point P on the screen at any
while in case of interference, all are of the same time be given by:
width. y1 = a cos wt and y2 = a cos (wt + f)
(ii) The bright fringes in the interference Where, f is the constant phase difference
pattern obtained from diffraction have a central between the two waves.
maximum followed by fringes of decreasing By the superposition principle, the resultant
intensity, whereas in case of interference, all displacement at point P is given by:
the bright fringes are of equal width. [2] y = y1 + y2
12. We know that intensity is directly proportional y = a cos wt a cos (wt + f) [1]
to square of amplitude. [1]
  ωt + ωt + φ  
I ∝ a2 y = 2a  cos   cos  ωt − ωt − φ 
  2 
 
 2 

I 
I =
If 1 2 
 φ   φ 

If intensity reduced to 50%, amplitude will be y = 2a cos ωt +  cos   (1)
a 2 2
then r = 2  f 

2 Let 2a cos   = A (2)
2
 
2
2
Imax  r  1 2 1 Then equation (1) becomes:
 
Imin  r  12
 
2  φ
2 1 y = A cos ωt + 
[1]  2

Imax  2.414  2 2

   5.83 Now, we have:
Imin 0.414
[1] f 
Imax = A2 = 4 a2 cos2  
= 33.98 34 2

Imin

The intensity of light is directly proportional to Then

the square of the amplitude of the wave. The
nl D (n + 1)l D
intensity of light at point on the screen is given xn and xn+1 =
d d
by:
---------------[From eq. (2)]
f
I = 4 a2 cos2 Now the fringe width is
2 [1]
λD
β = xn+1 − xn =
14. (a) The path difference between two rays coming d

from holes S1 and S2 is (S2 P – S1P). If point P Thus, the expression for fringwidth is
corresponds to a maximum.
λD
X β=
d
Imin 9
P (b) =
Imax 25
 
d 2
x d  I1  I2 
  9
S1
2 x 
2  I1  I2  25
d O  
 3
S2
D I1  I2
[1]
I1  I2 5
Now
 5 I1  5 I2  3 I1  3 I2
d   d 
2 2
 
 S2 P  2
  S1 P 
2
  D2   x      D2   x   
 2    2 
    2 I1 = 8 I2

= 2xd, where S1 S2 = d and OP = x I1 4
=
(S2 P + S1P) (S2 P – S1P) = 2xd I2 1

2 xd
 S2 P  S1 P   I1 16

 2  S1 P 
S P
[1] I2
=
1
[1]
For maximum, S2 P + S1P = nl
Ratio of intensity
xd
Thus, nl = 15. (i) Let the two waves arising from the slits A
D and B have the amplitudes a and b and the
nl D
Or, x = xn = d [1] phase difference f. Such that y1 = a sin wt
and y2 = b sin (wt + f).
n = 0, ± 1, ±2, ±3 …….. [For maximum]
The resultant displacement is given as:
Now, for minimum,
G
l
S2 P − S1 P = (2n − 1)
2
y1
l xd
Thus, (2n − 1) 2 = D x
S1 E
lD
Or, x = xn = (2n − 1) 2d
y2

O C
n = 0, ± 1, ±2, ±3………….. [For minima]
d
Thus, bright and dark bands appear on the
screen, as shown in Figure. Such bands are
called ‘fringes’. These dark and bright fringes
are equally spaced. [1] S2 E
Expression for fringewidth (β).
D
Let nth order bright fringe is at a distance xn and
G [1]
(n + 1)th order bright fringe is at xn + 1 from O,
y = y1 + y2
y = a sin wt + b (wt + f) S. Interference Diffraction
y = a sin ωt + b sin ωt cos φ + b cos ωt sin φ No.

1 Interference is the Diffraction is the
y = ( a + b cos φ) sin ωt + b sin φ cos ωt result of superpo- result of superposition

sition of secondary of secondary waves
Let a + b cos φ = A cos δ wavesstarting from starting from differ-
two different wave ent part of same wave
And b sin φ = A sin δ
fronts originating front.
Hence, y = A sin ωt cos δ + A cos ωt sin δ from two coherent
sources.
y = A sin (ωt + δ )
2 All bright and dark The width of central
Where the amplitude A of the resultant fringes are of equal bright fringe is twice
width. the width of any sec-
Wave can be given as:
ondary maximum.
A = a2 + b2 + 2ab cosf 3. All bright fringes Intensity of bright
[1]
are of same inten- fringes decreases as
b sin φ
And tan δ =
a + b cos φ
sity. we move away from
Constructive Interference: Intensity I ∝ A2 and
central bright fringes
for A to be maximum cos f = 1 on either side.
[1]
Or cos φ = cos 2nπ, n = 0, 1, 2, 3.....
16. (a) To observe interference fringe pattern,
Or, φ = 2nπ. there is need to have coherent sources of
And path difference Dx = nl light which can produce light of constant
phase difference. Let two coherent sources
Amax = a + b
of light, S1 and S2 (narrow slits) are derived
(a) Destructive Interference: For I to be minima, from a source S. The two slits, S1 and S2 are
cos f = –1 equidistant from source, S. Now suppose S1
Phase difference: ∆φ = (2n + 1) π and S2 are separated by distance d. The slits
and screen are distance D apart. [2]
And path difference:
l P
∆x = (2n + 1)
2 [1]
S1 yn
Amin = a – b A
d/2
Graph showing interference pattern against
S Od
position ‘x’ on the screen. C M d/2
I B
S2 Screen
D
Imax
[1]
Considering any arbitrary point P on the screen
at a distance yn from the centre O
The path difference between interfering waves is
given by S2 P – S1 P i.e. Path difference S2 P – S1
P [2]
O
x
(ii) Compose of interference pattern observed
in Young’s double slits and the single slits
diffraction:
[Topic 3] Diffraction and Polarisation of Light
Summary • Resolving power of telescope: For two stars to

be just resolved,
• Diffraction: Bending of light around corners
0.61λ f
of an obstacle into the region where shadow of f ∆θ ≈
a
obstacle is expected.
0.61λ
So, ∆θ ≈
To P a
Telescope will have better resolving power if a is
large.
L
M1 • A diffraction pattern with a central maximum is

From S M given by a single slit of width a. At angles of
M2 Q M2
λ 2λ
N ± ,± , etc., along with successively weaker
a a
secondary maxima in between, the intensity
reduces to zero. The angular resolution of a
λ
telescope is limited to , due to diffraction
D
Fig.: Diffraction Phenomenon where D is the diameter. Strongly overlapping
• Light energy is redistributed in interference images are formed when two stars are closer than
and diffraction. When it reduces in one region, this angle. Similarly, in a medium of refractive
emitting a dark fringe, it increases in another index n, a microscope objective subtending angle
region, emitting a bright fringe. In this process 2b at the focus, will just separate two objects
the energy remains constant i.e. neither energy is spaced at a distance
gained nor lost, with the principle of conservation λ
, which is the resolution limit of a
of light. (2 n sin β )
• The resolving power of the microscope is given microscope.
by the reciprocal of the minimum separation of
• The Fresnel distance is given by the formula
two points seen as distant. The resolving power
can be increased by choosing a medium of higher a 2 , where a is the size of the aperture and
ZP =
refractive index. λ
1.22 λ l is the wavelength.
d min =
2 sin β
Image
D
1.22 f
D
Object f
Object
v
plane
Objective Image plane
lens
Fig.: Resolving power of the microscope
• Polarized wave: A long string is held

horizontally, the other end of which is assumed to
PREVIOUS YEARS’
be fixed. If the end of the string is moved up and
down in a periodic manner, a wave propagating
in the +x direction will be generated. Each point
on the string moves on a straight line, the wave
is also referred to as linearly polarised wave. TOPIC 3
The linearly polarized waves are transverse 1 Mark Questions
waves; i.e., the displacement of each point of the 1. If the angle between the pass axis of polarizer
string is always at right angles to the direction of and the analyser is 45°, write the ratio of the
propagation of the wave. intensities of original light and the transmitted
light after passing through the analyzer.
• Unpolarized wave: When the plane of vibration
[DELHI 2017]
of the string is changed randomly in very short
intervals of time, then we have what is known as 2 Mark Questions
an unpolarised wave. Thus, for an unpolarised 2. (a) When a wave is propagating from a rarer to
a denser medium, which characteristic of the
wave the displacement will be randomly changing
wave does not change and why ?
with time though it will always be perpendicular
(b) What is the ratio of the velocity of the wave in
to the direction of propagation. the two media of refractive indices µ1 and µ2?
• A Polaroid consists of long chain molecules [ALL INDIA 2015]
aligned in a particular direction. The electric 3. Unpolarised light is passed through a Polaroid
vectors along the direction of the aligned molecules P1. When this polarized beam passes through
another Polaroid P2 which makes an angle q with
get absorbed. Thus if an unpolarised light wave
the pass axis of P1, then write the expression for
is incident on such a Polaroid then the light wave
the polarized beam passing through P2. Draw a
will get linearly polarized with the electric vector plot showing the variation of intensity when q
oscillating along a direction perpendicular to the varies from 0 to 2p. [ALL INDIA 2017]
aligned molecules; this direction is known as the
pass-axis of the Polaroid. 3 Mark Questions
4. (a) Describe briefly, with the help of suitable
• If I is the intensity of polarized light after passing
diagram, how the transverse nature of light
through the first polariser P1 then the intensity can be demonstrated by the phenomenon of
of the light after passing through the second polarization.
polarizer P2 will be I = Icosθ. This is called Malus’ (b) When unpolarized light passes from air to a
Law. transparent medium, under what condition
does the reflected light get polarized?
• Natural light from the sun is unpolarised which
means that the electric vector takes all possible
[ALL INDIA 2011]
random directions in the transverse plane. A
polaroid transmits only one component of these 4 Mark Questions
vectors, which is parallel to a special axis. 5. (a) What is linearly polarized light? Describe
Therefore the light wave is called plane polarised. briefly using a diagram how sunlight is
When this kind of light is viewed through another polarized.
(b) Unpolarized light is incident on a Polaroid.
polaroid which is rotated through an angle 2p,
How would the intensity of would transmitted
we can see two maxima and minima of same
light change when the Polaroid is rotated?
intensity. [ALL INDIA 2013]
• Plane polarised light can also be producedby 6. What is an unpolarized light? Explain with the
reflection at a special angle known as the help of suitable ray diagram how an unpolarized
π light can be polarized by reflection from a
Brewster angle and by scattering through in transparent medium. Write the expression for
2
Brewster angle in terms of the refractive index
the earth’s atmosphere. of the denser medium. [DELHI 2018]
I
5 Mark Questions = cos2 q
7. (a) How does one demonstrate, using a suitable Im
[½]
diagram, that unpolarised light when passed 2
through a Polaroid gets polarized? I  1 
 cos2 45  
(b) A beam of unpolarised light is incident on a Im  2 

glass-air interface. Show, using a suitable
I 1
ray diagram, that light reflected from the =
interface is totally polarised, when µ = tan Im 2
[½]
iB, where µ is the refractive index of glass 2. (a) Frequency of a wave does not change when
with respect to air and iB is the Brewster’s the wave is propagating from a rarer to a
angle. [ALL INDIA 2014] denser medium because frequency (n) is
8. Describe briefly how a diffraction pattern is given by the relation [1]
obtained on a screen due to a single narrow
slit illuminated by a monochromatic source v
υ = , As the medium changes velocity (v)
of light. Hence obtain the conditions for the λ
and wavelength (l) changes such that ratio
angular width of secondary maxima and remains constant.
secondary minima. [ALL INDIA 2014]
ν1 µ1
9. (a) Why does unpolarised light from a source (b) ν = µ , Here V1 and V2 are the velocity
show a variation in intensity when viewed 2 2
through a polaroid which is rotated ? Show
of the wave in medium 1 and medium 2 and
with the help of a diagram, how unpolarised
µ1 and µ2 are the refractive index of medium
light from sun gets linearly polarised by
1 and medium 2. [1]
scattering. [DELHI 2016]
(b) Three identical polaroid sheets P1, P2 and 3. According to law of Malus, when a beam of
P3 are oriented so that the pass axis of P1, completely plane polarized light is incident
P2 and P3 are inclined at angles of 60º and on an analyzer resultant intensity of light (I)
90º respectively with the pass axis of P1. A transmitted from the analyzer varies directly
monochromatic source S of unpolarized light as the square of cosine of angle θ between the
of intensity I0 is kept in front of the polaroid plane of analyzer and polarizer
sheet P1 as shown in the figure. Determine i.e I ∝ cos2 q
the intensities of light as observed by the ⇒ I = I0 cos2 q
observer at O, when polaroid P3 is rotated
When polarizer and analyzer are parallel
with respect to P2 at angles θ = 30º and 60º.
q = 0° or 180°
So,
cos q = + 1, –1
I = I0 cosq = cos 90° = 0
S O
I=0
In unpolarised light, vibrations are probable
in all direction in a plane perpendicular to the
direction of propagation. [1]
P1 P2 P3
Therefore, q can have any value from 0 to 2p
[DELHI 2018]
2π
1
Solutions ∴ [ cos2θ ]av =
2∫
cos2θdθ
1. I = Im cos2q 0
2π
Where, I is the transmitted intensity 1 1 + cos2θ
2π ∫
= dθ
Im is the maximum value of the transmitted 2
0
intensity
2π
q is the angle between the two polarising 1  sin 2θ 
= 0 + 
directions 2π × 2  2  0

1  cos i p  sin r
=
2  i p  r  90
[1]
If we use malus law then,
5. (a) When vibrations of light wave are confined
I = I0 cos2 q
to only one direction then light is called
1 1 linearly polarized. [1]
I  I0   I
2 2 0 [1] unploarised polarised
The required graph would have the form as
shown in figure.
I
I0/2
[1]
When sunlight passes through Polaroid
then components parallel to axis passes
in unaffected way and components

/2 [1] perpendicular to axis are absorbed so
transmitted light is polarized. [1]
4. (a) When a polaroid P1 is rotated in the path
of an unpolarised light, there is no change in (b) On rotating the Polaroid, intensity remains
transmitted intensity, v. The light transmitted unchanged as half of the incident intensity.
through polaroid P1 is made to pass through 6.
polaroid P2. On rotating polaroid P2, in path of Incident ray
light transmitted from P1 we notice a change in (Unpolarised)
Completely polarised
intensity of transmitted light. This shows the
light transmitted from is polarized. Since light AIR
can be polarized, it has transverse nature. [1]
P A
Partially polarised
P A No Light Medium

[1]
[1} An unpolarized light is one in which the
(b) Whenever unpolarised light is incident from vibration of electric field vector is not restricted
air to a transparent medium at an angle of in one particular plane. When an unpolarized
incidence equal to polarizing angle, the reflected light falls on the surface, the reflected light
light gets fully polarized. is such that the vibration of its electric field
vector is confined to one particular plane. The
According to Brewster’s law
direction of this plane is parallel to the surface
n = tan ip of reflection. A component of electric field vector
Where i p in the polarizing angle and n is is absent from the refracted light. Therefore, the
refractive index of the transparent material refracted light is partially polarized. [2]
By Snell’s law, we have The expression for Brewster angle in terms of
sin i p the refractive index of denser medium is taniB =
n= µ, Where ‘μ’ is the refractive index of the denser
sin r
medium with respect to the rarer medium. [1]
sin i p
tan i p =
sin r
 cos i p  cos 90  r 

7. (a) Where, r is angle of refraction

Unpolarised light A
D According to the Snell’s law:
K N sin iB sin iB
µ= =
sin r sin (90 − iB )

sini B
L M µ=
cos iB

C
B µ = tan iB [1]
Polarised light
Polaroid Hence proved
[1] 8.
The phenomenon of restricting the vibration
of light (electric vector) in a particular
directionperpendicular to the direction of the
wave propagation is called polarization of light.

When unpolarised light is passed through a O
Polaroid, only those vibrations of light pass S
O
through the crystal, which are parallel to the
axis of the crystal (AB). All other vibrations
are absorbed and that is why intensity of the
emerging light is reduced. [1]
The plane ABCD in which the vibrations of [1]
the polarised light are confined is called the
plane of vibration. The plane KLMN that is point source S is placed at the focus of a
A
perpendicular to the plane of vibration is defined converging lens. The source-lens arrangement
as the plane of polarization. provides a plane wavefront which is then
When unpolarised light is incident on the glass- diffracted. Another converging lens is introduced
air interference at Brewster angle iB, then between the diffracting slit and the observation
reflected light is totally polarised. This is called screen such that the screen is in the focal plane
Brewsters Law. of the lens. Plane wavefront emerging from the
When light is incident at Brewster angle, the slit at different angles are brought to focus on
reflected component OB and the refracted the screen using the second lens, as shown in
component OC are mutually perpendicular to the Fig. [1]
each other. [1] Condition for minima
From the figure,
Divide the slit into two equal halves AC and CB,
We have: BOY  YOC  90o a a

   
90 o  iB  90 o  r  90 o
each of size
2
. For every point M1 in AC, there
a
exists a point M2 in CB such that M1 M2 = .
A 2
B The path difference between secondary waves
from M1 and M2 reaching P is
a
Rarer M2 P − M1 P = sin q
2
iB r = iB Point P on the screen would be a first minimum
if this path difference is λ between the secondary
X Y waves from extreme points A and B. Thus,
O
Denser path difference between waves from A and C or

r l
between waves from will be . [1]
2
a λ
Hence, sin θ = or a sin θ = λ for P to be
2 2
first minimum. P is a second minimum if Path
C difference, a sin = 2λ Proceeding in the same
[1]
manner, we can show that the intensity at P is of the sun. Clearly, charges accelerating
zero if Path difference, a sin θ = nλ (condition parallel to the double arrows do not radiate
for minima) where n = 1, 2, 3,…… Condition for energy towards this observer since their
secondary maxima Imagine the slit to be divided acceleration has no transverse component.
into three parts AM2, M1, M2 and M2 B. Let the The radiation scattered by the molecule is
secondary waves reaching P from the extreme therefore represented by dots. It is polarised
3l perpendicular to the plane of the figure. This
points A and B be . The secondary waves explains the polarisation of scattered light
2
reaching P from the corresponding points of the from the sky. [1]
l Incident Sunlight
parts AM1, M1, M2 will have path difference of
2 (Unpolarised)
and interfere destructively. The secondary waves
reaching P from points in the third part M2 B will
contribute to the intensity at P. Therefore, only
one-third of the slit contributes to the intensity at
point P between two minima. This will be much
weaker than the central maximum. This is the Scattered Light
first secondary maximum. The condition for first (Polarised)
secondary maximum is [1 + 1]
3λ
Path difference, asinθ = ,The condition for To Observer
2
second secondary maximum is Path difference, (b) On rotating the polaroid, intensity remains
5λ unchanged as half of the incident intensity.
asinθ = .Proceeding in the same manner,
2 60° 30°
we can show that the condition for a secondary
 λ 
maxima is Path difference, a sin θ = (2n + 1) 
 2
where n = 1, 2, 3, … [1]
9. (a) When vibrations of light wave are confined Io Io
to only one direction then light is called S 2 O
linearly polarised. 8
Unploarised Polarised
P1 P2 P3
When sunlight passes through polaroid 90°

then components parallel to axis passes
Intensity of unpolarized light on P1 is Io
unaffected and components perpendicular
to axis are absorbed so transmitted light is After passing through P 1 the intensity is
polarised. [1]  Io 
 2  [1]
As Fig. shows, the incident sunlight is
unpolarised. The dots stand for polarisation After passing through P2 polaroid P = I cos2 q
perpendicular to the plane of the figure.  Io 
The double arrows show polarisation in where  I  2 
the plane of the figure. (There is no phase I
I ’  o cos2 60
relation between these two in unpolarised 2
light). Under the influence of the electric
Io
field of the incident wave the electrons in I’ =
the molecules acquire components of motion 8
in both these directions. We have drawn
an observer looking at 90° to the direction
( )
Case 1st I ’’ = I ’ cos2 q here q = 0° Io  3 
2
So, I ’’  8  2 
I” = I’  
3 Io
Io I ’’ =
I= 32
8 [1] Case IVth
Case II nd
P3 rotated away from pass axis of P2 by q = 60°
P3 rotated away (q = 30°) from pass axis of P2 Now angle between pass axis of and P2 and P3
Now angle between pass axis of P2 and P3 will be will be
f = 60° f = 90°
I” = I’ cos 60° [½]
2 I” = I’ cos2 90°
Io  I  2 I” = 0 [½]
So, I ’’   
8  2
Io
I ’’ =
32
Case IIIrd:
P3 rotated towards pass axis of P2 by
Now again angle between pass axis of P2 and P3
will be
f = 30°
I” = I’ cos2 30°
CHAPTER 11
Dual Nature of Radiation
and Matter

Dual nature of radia-
tion, Photoelectric effect;
Hertz and Lenard’s 1Q
observations (3 marks)
Einstein’s photoelectric 1Q 1Q 1Q
equation-particle nature (3 marks) (3 marks) (3 marks)
of light
Matter waves-wave na-
ture of particles
de Broglie relation; Da- 1Q 1Q 1Q
visson-Germer experi- (1 mark) (2 marks) (2 marks)
ment
On the basis of above analysis, it can be said that from exam point of view Phenomenon of
Photoelectric Effect, Wave Theory, Intensity of Light and Einstein’s Photoelectric Equation
are most important concepts of the chapter.
188 CHAPTER 11 : Dual Nature of Radiation and Matter
[Topic 1] Photoelectric Effect
Summary light, potential difference between both the plates

and the material of the plate.
• Work Function: The minimum energy which
• Stopping Potential: Stopping potential or cut-
is necessary for an electron to get away from the
off potential is the minimum retarding (negative)
surface of metal is called the work function of
potential for which the photoelectric current stops
the metal which is denoted by 0. The unit for
at a particular frequency of incident light. It is
measuring work function is electron volt (eV). This
denoted by V0.
minimum energy can be provided by thermionic
emission, field emission or photo-electric emission. • Saturation Current: At a certain potential
difference, the photoelectric current stops increasing
Thermionic emission: When a metal is heated,
further. This maximum value of photocurrent is
thermal energy is imparted to free the electrons
known as the saturation current.
from the surface of the metal.
• Maximum Kinetic Energy: The maximum kinetic
Field emission: Electrons can be pulled out of
energy of the photoelectric electrons is denoted by
metal by applying a very strong electric field (of the
Kmax and it depends directly on the frequency of the
order of 108 Vm–1) to it, as in a tesla coil.
incident light. It is independent of the intensity of
Photo-electric emission: Electrons are emitted the light.
when a light of suitable frequency hits a metal
The maximum kinetic energy Kmax = eV0
surface. This can be seen in a photodiode.
• Threshold Frequency: The minimum cut-off
• 1eV is the energy attained by an electron when it
frequency which is required for the emission of
has been accelerated by a potential difference of 1,
electrons is called the threshold frequency which is
so that 1eV = 1.602 × 10–19J.
denoted by ν 0 . No emission is possible for the
• Photoelectric Effect: When metals are irradiated
frequency lower than the cut-off frequency.
by light of suitable frequency, electrons start
emitting from the metal surface. This phenomenon • In the photoelectric effect, the light energy
is known as photoelectric effect. is converted into the electrical energy. The
photoelectric emission is a quick process having
Quartz S
window very less time lag.
Evacuated • Effect of intensity of light on photocurrent:
Photosensitive glass tube
plate Number of photoelectrons emitted per second varies
directly with the intensity of incident radiation.
Electrons
C A • Effect of potential on photoelectric current: The
stopping potential is independent of its intensity for
a given frequency of the incident radiation.
Commutator
• Effect of frequency of incident radiation on
A stopping potential:
V
The stopping potential V0 varies linearly with
the frequency of incident radiation for a given
+ – photosensitive material.
Fig.: Depiction of Photoelectric effect There exists a certain minimum cut-off frequency v0
• Some metals are sensible to ultraviolet light and for which the stopping potential is zero.
some to visible light also. Photocurrent depends • Einstein’s Photoelectric Equation: Einstein
upon the intensity of light, frequency of incident proposed that light is comprised of small discrete
CHAPTER 11 : Dual Nature of Radiation and Matter 189
energy packets known as photons or quanta and 2. The graph shows variation of stopping potential
energy carried by each photon is hv, where v is the versus frequency of incident radiation v for two
frequency of light and Planck’s constant. The photosensitive metals A and B. Which of the two
h metals has higher threshold frequency?
momentum carried by each photon is . In
λ Metal B Metal A
photoelectric effect, emission is possible because of Stopping
the absorption of a photon by an electron. The potential
(v0)
maximum kinetic energy of the emitted electron is: O v0 v0
= hν − φ0 , where f0 is the work function.
K max Frequency of incident
–
Radiation (v)
= h( ν − ν 0 )
The photoelectric emission is possible only when –
hν > φ0 as Kmax must be non-negative.
φ0 [ALL INDIA 2014]
⇒ ν > ν 0 where ν 0 =
h 3. Draw graphs showing variation of photoelectric
• From the photoelectric equation, current with applied voltage for two incident
eV= hν − φ0 , for ν ≥ ν 0 (as K max = eV 0 ) radiations of equal frequency and different
0
intensities. Mark the graph for the radiation of
h  φ
or=V 0  ν − 0 higher intensity.
e  e [DELHI 2014]
According to this result, the graph of V0 versus  is
4. The figure shows a plot of three curves a, b,
a straight line having the slope equal to  h  . c, showing the variation of photocurrent vs
e  collector plate potential for three different
intensities I1, I2 and I3 having frequencies v1, v2
and v3 respectively incident of a photosensitive
surface. Point out the two curves for which the
PREVIOUS YEARS’ incident radiations have same frequency but
different intensities.
Photoelectric
TOPIC 1 current
l1
l2
1 Mark Questions c l3
b
1. The given graph shows the variation of photo-
electric current (I) versus applied voltage (V) a
for two different photosensitive materials and Collector plate potential
for two different intensities of the incident
radiations. Identify the pairs of curves that
[DELHI 2017]
correspond to different materials but same
intensity of incident radiation. 2 Mark Questions
5. Using Bohr’s postulates, obtain the expressions
for (i) kinetic energy and (ii) potential energy
I of the electron in stationary state of hydrogen
atom.
Draw the energy level diagram showing how the
1 transitions between energy levels result in the
3 appearance of Lyman series.
2 [DELHI 2013]
6. (i) Monochromatic light of frequency 6.0 × 1014
Hz is produced by a laser. The power emitted
is 2.0 × 10–3 W. Estimate the number of
4
photons emitted per second on an average
by the source.
V
[DELHI 2013]
(ii) Draw a plot showing the variation of (ii) Is the nucleus formed in the decay of the
photoelectric current versus the intensity of nucleus 22
11 Na ,
an isotope or isobar?
incident radiation on a given photosensitive [DELHI 2014]
device. 14. For the past some time, Aarti had been observing
[DELHI 2014] some erratic body movement, unsteadiness and
7. A 12.5 eV electron beam is used to excite a gaseous lack of coordination in the activities of her sister
hydrogen atom at room temperature. Determine Radha, who also used to complain of severe
the wavelengths and the corresponding series of headache occasionally. Aarti suggested to her
the lines emitted. [ALL INDIA 2017] parents to get a medical check-up of Radha.
8. Plot a graph showing the variation of stopping The doctor thoroughly examined Radha and
potential with the frequency of incident diagnosed that she has a brain tumour.
radiation for two different photo sensitive (a) What, according to you, are the values
materials having work functions W 1 and displayed by Aarti?
W 2 (W 1 > W 2 ). On what factors does the (b) How can radioisotopes help a doctor to
(i) slope and (ii) intercept of the line lines diagnose brain tumour?
depend? [ALL INDIA 2014]
[DELHI 2018] 15. Write Einstein’s photoelectric equation
9. If light of wavelength 412.5nm is incident on and mention which important features in
each of the metals given below, which ones will photoelectric effect can be explained with the
show photoelectric emission and why? help of this equation.
Metal Work function (eV) The maximum kinetic energy of the photo
Na 1.92 electrons gets doubled when the wavelength of
light incident on the surface changes from λ1
K 2.15 to λ2. Derive the expressions for the threshold
Ca 3.20 wavelength λ0 and work function for the metal
Mo 4.17 surface.
[DELHI 2015]
[ALL INDIA 2018]
16. (i) State Bohr’s quantization condition for
3 Mark Questions defining stationary orbits. How does de-
10. An electron and a photon each have a wavelength Broglie hypothesis explain the stationary
1.00 nm. Find orbits?
(a) their momenta, (ii) Find the relation between the three
(b) the energy of the photon and wavelengths from the energy level diagram
(c) the kinetic energy [ALL INDIA 2011] shown below.
11. (a) What is the significance of negative sign in C
the expression for the energy? 1 3
(b) Draw the energy level diagram showing how
the line spectra corresponding to Paschen B
2
series occur due to transition between
energy levels. A
12. (a) Why photoelectric effect cannot be explained 17. Write three characteristic features in
on the basis of wave nature of light? Give photoelectric effect which cannot be explained
reasons. on the basis of wave theory of light, but can be
(b) Write the basic features of photon picture explained only using Einstein’s equation.
of electro-magnetic radiation on which [DELHI 2016]
Einstein’s photoelectric equation is based. 18. Using photon picture of light, show how Einstein’s
[DELHI 2013] photoelectric equation can be established. Write
two features of photoelectric effect which cannot
13. (a) Deduce the expression, N  N o e t for the
be explained by wave theory.
[ALL INDIA 2017]
law of radioactive decay.
(b) (i) Write symbolically the process expressing
22
the b+ decay of 11 Na . Also write the basic
nuclear process underlying this decay.
Solutions 2 ke2
mv = (1)
r
1. Curves 1 and 2 correspond to similar materials
Where, m = mass of the electron
while curves 3 and 4 represent different
materials, since the value of stopping potential r =radius of electronic orbit
for 1, 2 and 3, 4 are the same. For the given v = velocity of electron.
frequency of the incident radiation, the stopping
potential is independent of its intensity. [½] nh
Again, mvr 
2
So, the pairs of curves (1 and 3) and (2 and 4)
correspond to different materials but same
intensity of incident radiation. [½] From equation (1), we get,
2
2.  nh  ke2
m  
V0  2 mr  r
Metal B Metal A
 n2 h2  ke2
m 
 4 2 m2 r 2  r
[½]
P Q
V0 V0
V n2 h2 ke2
O 
2 2 r
4 mr

n2 h2
[½] k
4 2 mre2
As OP > OQ
Using equation (2), we get
 ’ o   o
ke2 4 2 kme2
∴Threshold frequency of A > Threshold Ek 
2n2 h2
frequency of B [½]
3. (ii) Potential energy,
Ip [1]  k  e   e ke2
I1 Ep  
r r
I2 I 1 > I2
4 2 kme 2
Ep   ke 2 
V n2 h2

Fig.: Graph Showing plot of stopping Potential
4 2 k2 me4
vs Frequency Ep  
n2 h2
4. Curves a and b have the same frequency but
different intensities. [1] Hence, total energy of the electron in the nth orbit,
E = Ep + Ek
5. According to Bohr’s postulates, in a hydrogen
atom, a single electron revolves around a 4 2 k2 me4 2 2 k2 me4
nucleus of charge +ke. For an electron moving E 
n2 h2 n2 h2
with a uniform speed in a circular orbit or a
given radius, the centripetal force is provided by
2 2 k2 me4 13.6
Coulomb force of attraction between the electron E  eV
2 2
and the nucleus. The gravitational attraction n h n2 [½]
may be neglected as the mass of electron and When the electron in a hydrogen atom jumps
proton is very small. [½] from higher energy level to the lower energy
2 2 level, the difference of energies of the two energy
So, mv = ke levels is emitted as a radiation of particular
r r2
wavelength. It is called a spectral line.

(ii)
In H-atom, when an electron jumps from the orbit
to orbit , the wavelength of the emitted radiation
is given by,
1  1 1 
 R  
  n2f ni2 

Photoelectric
Where, R → Rydberg’s constant
= 1.09678 × 10–7 m–1
current
For Balmer series, nf = 2 and ni = 3, 4, 5…..
1  1 1 
 R  
2
 2 ni2 

O Intensity
Where, ni = 3, 4, 5……… [½] [1]
Total energy, E(ev) Fig.: Graph Showing variation of Photoelectric
n=5 current with respect to Intensity
n=4
7. The energy of gaseous hydrogen at room
Excited
states
–0.85 n = 3 temperature are

–1.51
E1 = 13.6 eV
n=2
–3.40 E2 = 3.4 eV
E3 = 1.51 eV
E4 = 0.85 eV
E3 – E1 = –1.51 (–13.6) = 12.09 eV and
E4 – E1 = –0.85 (–13.6) = 12.75 eV
As, both the values does not match the given
value but it is nearest to E4 – E1 upto E4 energy
n=1
level the hydrogen atoms would be excited. [1]

Ground state
–13.6 Lyman series:

Layman 1 1 1 
 R  
2
Series

 1 n2 
6. (i) The energy of a proton of frequency u is .
For first member n = 2

  
E    6.63  1034 Js  6  1014 s1  1 1 1  4  1
  R    1.097  107  
1 2 2  4 
19 1 2 
 3.98  10 J [½]
7
If n be the number of photons emitted by the  1  1.215  10 m [½]
source per second, then the power P transmitted
in the beam is given by, P = nE Balmer series:
P 1 1 1 
n   R  
2
E
 2 n2 
2  10 3 For first member n = 3
n
4  10 19
1 1 1
 R  
n = 5 × 1015 photons/sec [½]  22
32 
1
1 1 hc
 1.097  107    (b) Energy of photon, E 
4 9 
3  108
 1  6.56  10
7
m [½]

 
 6.63  10 34 
10 9
8. 17
 19.89  10 J [1]
p2
Stopping (c) Kinetic energy of electron =
potential = W2 2m
 
(v0) 34 2
= W1
v > v0 1 6.63  10
  J
2 9.1  10 31
v0 v 0 v > v 0
O Frequency of incident 19
 2.42  10 J [1]
Radiation (v) [1]
Fig.: Variation of Stopping Potential with respect to 11. (a) Negative sign indicates that revolving
frequency of incident radiation. electron is bound to the positive nucleus. [1]
(i) The slope of the graph is constant and equals to (b) For paschen series, n = 3 and ni = 4, 5……
h
. Therefore, the slope does not depend on any 1  1 1 
e  r  
2
factor. [1]
 3 ni2 
[2]
(ii) The intercept of the lines depends on the work Where, ni = 4, 5………
function f of the metals. Total energy, E(ev)
n=5
n=4
hc
9. E  h  (J)
Excited
states
 –0.85 n = 3
–1.51

hc
 (eV) [1] n=2 Paschen
e –3.40 Series
6.6  10 34  3  108

412.5  10 9  1.6  10 19 [½]
E = 3eV
Na → emission E > lo
K → emission E > lo
Ka → No emission E > lo
n=1 Ground state
Mo → No emission E < lo [½] –13.6
9 12. (a) Wave nature of radiation cannot explain the
10. c   photon  1.00nm  10 m photoelectric effect because of:
(a) For electron or photon, momentum (i) The immediate ejection of photo electrons.
(ii) The presence of threshold frequency for a
h
p  pe  pr  metal surface.
 [1]
(iii) The fact -that kinetic energy of the emitted
34 electrons is independent of the intensity of
6.63  10
p light and depends upon its frequency.
10 9
Thus, the photoelectric effect cannot be
= 6.63 × 10 –25
kg m/s explained on the basis of wave nature of
light. [1½]
(b) Photon picture of electromagnetic radiation On integrating both sides, we get:

on which Einstein’s photoelectric equation N t
dN
is based on particle nature of light. Its basic  N
   dt
features are: No to

(i) In interaction with matter, radiation
behaves as if it is made up of particles called In N  ln N o     t  to 
photons. At to = 0
(ii) Each photon has energy (E = hv), momentum
N
 hv  , and speedc, the speed of light. In   t
 p   No
c 
 N  t   N o e  t
(iii) All photons of light of a particular frequency [1]
v, or wavelength l, have the same energy 22
(b) (i) The b decay for 11 Na is given below:
 hc   hv 
 E  hv   and momentum  p  , 22 22
  c  11 Na   11 Ne  10 

may be.
Or 22
11 Na  
22
 10 Ne  e 
(iv) By increasing the intensity of light of given
If the unstable nucleus has excess protons than
wavelength, there is only an increase in
required for stability, a proton converts itself
the number of photons per second crossing
into a neutron. In the process, a positron e+ (or
a given area, with each photon having
a b+) and a neutrino u are created and emitted
the same energy. Thus, photon energy is
from the nucleus.
independent of intensity of radiation.
(v) Photons are electrically neutral and are not p  n    

deflected by electric and magnetic fields.

(vi) In a photon—particle collision (such Or p    e   [½]
as photon-electron collision), the total This process is called beta plus decay
energy and total momentum are conserved. 22
(ii) The nucleus so formed is an isobar of 11 Na
However, number of photons may not be
because the mass number is same, but the
observed. [1½]
atomic numbers are different. [½]
13. (a) According to the law of radioactive decay, we
14. (a) Aarti shows good awareness towards health
have:
and care for her sister. [1½]
N
N (b) Certain radio isotopes are injected to body
t and they are absorbed by brain tumour and
Where, by detecting intensity of radiations we can
N→ Number of nuclei in the sample measure location and severity of tumour. [2½]
DN→ Amount undergoing decay 15. Einstein’s photoelectric equation is given by,
DtTime 1 2
K max = mvmax
N 2
 N
t K max     o
Where, l = Decay constant or disintegration 1 2
constant Or,    o  m max
2
Dt→ A
Where, Kmax = Maximum kinetic energy of the
dN
  N photoelectron
dt u max = Maximum velocity of the emitted
dN photoelectron
  dt
N [1] m =Mass of the photoelectron
u = Frequency of the light radiation    

fo= Work function  o   1 2 

 22  1 
= Plank’s constant
Work function is the energy required to eject a
If uo is the threshold frequency, then the work photoelectron from the metal.
function can be written as,  o   o
c
1 W
 K max  2
mvmax     o      o  o
2 [1]
The above equations explain the following c 22  1 

W 
results. 1 2 [1]
(1) If v < v0 , then the maximum kinetic 16. (i) Quantization condition: Of all possible
energy is negative, which is impossible. Hence, circular orbits allowed by the classical theory,
photoelectric emission does not take place for the electrons are permitted to circulate only in
the incident radiation below the threshold those orbits in which the angular momentum
frequency. Thus, the photoelectric emission can
h
take place if v > v0. [1] of an electron is an integral multiple of , h
2π
(2) The maximum kinetic energy of emitted being Plank’s constant.
photoelectrons is directly proportional to the Therefore, for any permitted orbit,
frequency of the incident radiation. This means n
that maximum kinetic energy of photo electron L  m r  ; n  1, 2, 3,....
2
depends only on the frequency of incident light.
Where L, m and u is the angular momentum,
According to the photoelectric equation,
mass and speed of the electron, r is the radius
1 2 of the permitted orbit and n is positive integer
K max  mvmax     o
2 called principle quantum number.
c The above equation is Bohr’s famous quantum
K max   o condition. When an electron of mass m is confined
1
to move on a line of length l with velocity v, the
Let the maximum kinetic energy for the de-Broglie wavelength l associated with electron
wavelength of the incident l2 be k2. is:
c  
k2   o  
1 mv p [1]
k2 = 2k1 Or, Linear momentum
From (1) and (2), we have   n
 p  
c  c   2l n 2l
 o  2   o 
2  1  When electron revolves in a circular orbit of

radius ‘r’ then 2l = 2pn
2 1 
  o  c  
–
e
  

 1 2
2 1 
  o  c  
 1 2 

c 2 1 
  c 
o  1 2 

1 2 1 
  
o  1 2 

n n (ii) Dependence of kinetic energy on

P  Or P  r  frequency of incident light: According to
2 r 2
wave theory, the maximum kinetic energy of

Or angular momentum L  p  r is integral emitted electrons should depend on intensity
 of incident light and not on frequency whereas
multiple of which is Bohr’s quantization of
2π Einstein’s equation explains that it depends
on frequency and not on intensity. [1]
angular momentum.
(iii) Instantaneous emission of electrons:

(ii) ECB  C According to wave theory there should be time
1 lag between emission of electrons and incident
of light whereas Einstein’s equation explains
C why there is no time lag between incident of
EBA  light and emission of electrons. [1]
2
[1]
18. Einstein photoelectric equation:
C
When a photon of energy hv falls on a metal
1 surface the energy of the photon is absorbed by
3 the electrons and is used in following two ways.
B (i) A part of energy is used to overcome the

surface barrier and come out of the metal
2 surface. This part of energy is called “work
function”. It is expressed as fo= h vo [1]
A
The remaining part of energy is used in giving
C
ECA  velocity v to the emitted photoelectron.
3

This is equal to maximum kinetic energy of photo
Where, E CA = E CB + E BA (E CB = Energy gap 1 2
electron Vmax
between level B and C) 2
C C C According to the law of conservation of energy
 
3 1  2
1 1
2 2
h   o  mvmax  h o  mvmax
(EBA = Energy gap between level A and B) 2 2 [1]
1 1 1 1
  2
 mvmax  kmax  h    o   h   o
3 1 2 2
(ECA = Energy gap between level A and C)  kmax  h   o

  2 This equation is called Einstein Photoelectric
3  1
1  2 [1] Equation. [1]
17. (i) Existence of threshold frequency:

According to wave theory, there should not
exist any threshold frequency but Einstein’s
theory explains the existence of Threshold
frequency. [1]
[Topic 2] Matter Wave
Summary
Dual Nature of matter PREVIOUS YEARS’
• Particle Nature of matter: EXAMINATION QUESTIONS
Radiation behaves as if it is made up of particles
in interaction of radiation with matter, called
TOPIC 2
photons. 1 Mark Questions
1. Show graphically, the variation of the de-Broglie
Each photon has energy E = hv and momentum wavelength (λ) with the potential (V) through
hυ which an electron is accelerated from rest.
p = , and speed c that is the speed of light.
c [DELHI 2011]
• Wave Nature of Matter: 2. State de-Broglie hypothesis.
De Broglie proposed that the moving particles are [DELHI 2012]
associated with the waves. If a particle is having 2 Mark Questions
a momentum p, then the associated wavelength 3. A proton and a deuteron are accelerated through
h h the same accelerating potential. Which one of
λ =
= , where  is the speed of the moving
p mv the two has,
(a) Greater value of de-Broglie wavelength
particle and its mass. The wavelength l is known associated with it, and
as the de Broglie wavelength and the above (b) Less momentum?
relation as the de Broglie relation. Give reasons to justify your Answer.
4. A proton and an α-particle have the same de
The wavelength of an electron accelerated with
Broglie wavelength. Determine the ratio of (i)
the potential V is:
their accelerating potentials (ii) their speeds.
1.227 [DELHI 2015]
λ= nm
V 5. Plot a graph showing variation of de-Broglie

1
• Heisenberg’s uncertainty principle: This wavelength l versus , where V is accelerating
V
principle states that, “it is not possible to measure
potential for two particles A and B carrying same
both the position and momentum of an electron
charge but of masses m1, m2 (m1 > m2). Which
at the same time exactly. There is always some
one of the two represents a particle of smaller
uncertainty in the position and in momentum. mass and why?
h [DELHI 2016]
∆x ∆p ≈ , where = 6. A proton and an α particle are accelerated
2π
through the same potential difference. Which
• The wave nature of electron was verified and one of the two has (i) greater de-Broglie
confirmed by the electron diffraction experiments wavelength, and (ii) less kinetic energy? Justify
performed by Davisson and Germer, and G.P. your answer.
[ALL INDIA 2016]
Thomson. Many other experiments later also
7. An electron is accelerated through a potential
confirmed the wave nature of electron. difference of 64 volts. What is the De-Broglie
wavelength associated with it? To which part of
the electromagnetic spectrum does this value of
wavelength correspond?
[DELHI 2018]
3 Mark Questions 
12.27

12.27
8. An electron microscope uses electrons accelerated Vproton Valpha
by a voltage of 50 kV. Determine the de-Broglie
wavelength associated with the electrons. Thus,
Taking other factors, such as numerical aperture Vproton
etc. to be same, how does the resolving power of =1
Valpha
an electron microscope compare with that of an [½]
optical microscope which uses yellow light? (ii) We can also write de-Broglie wavelength as,
[ALL INDIA 2014]

Solutions

mv
1.
Where, → plank’s constant, m → mass of the
particle and speed of the particle.
It is given that,  proton  alpha
We know, malpha = 4 m proton

 alpha 
4 m proton valpha
[½]

 
V m proton v proton 4 m proton valpha
[1]
2. De Broglie postulated that the material particles v proton
may exhibit wave aspect. Accordingly a moving =4
valpha
material particle behaves as wave and the [½]
wavelength associated with material particle
1
is [½] 5. qV = mv2
2
h p2
 qV =
m [½] 2m [½]
Where h = Planck’s constant  p  2mqV

m = mass of the object
h
n = velocity of the object p

3. (a) de-Broglie wavelength,
1 h
 (For same accelerating potential) 
mass 2mqV

Mass of a proton is less as compared to a 1
 Slope 
deuteron. So, proton will have greater value of m [1]
de-Broglie wavelength associated with it. [1]
(m2)
(b) Momentum, p ∝ mass (for same accelerating
potential). Mass of deuteron is more as compared
to a proton. So, it will have a greater value of (m1)
momentum. [1] 2
1
4. (i) The de-Broglie wavelength of a particle is 1
12.27 o V
given by,   A [½]
V 6. When a charge particle is accelerated through V
Where, V is the accelerating potential of the potential difference then its kinetic energy
K. E. = qV (q = charge; V = potential difference)
particle. It is given that,
Proton and α-particle accelerated through same
 proton  alpha potential difference so [½]
[½]

because q  q proton q  2e and q proton   e  
6.626  1034
K . E  K . E proton 1863.68  1025

Their Debroglie wavelength 6.626
  109
h 43.17

2mqV l = 0.15 × 10–9 m = 1.5 × 10–10 m [1]

This value of wavelength corresponds to the X-ray
V = same [½]
region of the electromagnetic spectrum. [½]
m q  m p q p  m  4 m p  o
12.27 A
8.  
So    p (la = Debroglie lp wavelength of Vo
α-particle; Debroglie wavelength of proton) 12.27 A
(1) Proton have greater De Broglie wavelength 
50000
(2) Proton have lesser kinetic energy [1]
o
h 12.27 A
7. De-Broglie wavelength,   Where, m 
2meV 233
(mass of electron) = 9.1 × 10–31 kg [½] l = 0.00526 Å [1]
6.626  10 34 1
 Resolving Power (RP)  as wavelength of

2  9.1  1031  1.6  1019  64
moving electron is very small as compared to
6.626  1034 that of yellow light so it has high Resolving

Power than optical microscope. [2]
1863.68  1050
CHAPTER 12
Atoms

Alpha-particle scattering
experiment
Rutherford’s model of 1Q
atom (2 marks)
Bohr model, energy lev- 1Q 1Q 1Q 1Q
els, hydrogen spectrum (1mark) (2 marks) (2 marks) (3 marks)
From analysis of previous years’ papers, it is clear that Bohr Model and Rutherford’s model are
most important topics of the chapter from exam point of view.
202 CHAPTER 12 : Atoms
Topic 1: Rutherford’s Atomic Model, Bohr’s Model & Energy

Level Diagram
Summary • The alpha particles are fast moving and positive-

ly charged Helium nuclei with two protons and
Introduction two neutrons.
• Atoms in simple terms are defined as the smallest Rutherford observed the deflection of alpha particles
unit of matter. after passing through metal sheet and proposed his
• Atoms are electrically neutral because they con- atomic model
tain same number of electrons and protons. • After passing through the metal sheet, the alpha
particles strike on fluorescent screen which was
Plum-Pudding Model coated with zinc sulphide and produced a visible
• In 1898, J. J. Thomson proposed the first model flash of light
of atom. • He concluded that an atom consists of a minute
• He stated, there is a uniform distribution of the positively charged body at its center called as nu-
positive charge of the atom throughout the vol- cleus. The nucleus, though small, contains all the
ume of the atom and like seeds in a watermelon, protons and neutrons.
the negatively charged electrons are embedded in
it. This model was picturesquely called plum pud- Alpha-Particle Trajectory
ding model of the atom. • The trajectory traced by an a particle depends on
the impact parameter, b of collision.
Alpha-Particle Scattering • The particle near to the nucleus suffers large
• Rutherford used a “Gold foil experiment” scattering.
• Rutherford only identified one of type of radiation • Only a small fraction of the number of incident
given off by radioactive elements like polonium, particles rebound back indicating that the num-
uranium and named them as alpha particles. ber of a-particles undergoing head on collision is
small.

b
Target nucleus
Fig.: Alpha-Particle Trajectory
Rutherford’s nuclear model of Atom • Drawbacks of Rutherford’s model: There were
two major drawbacks in Rutherford nuclear model
• According to Rutherford’s model, the entire pos-
in explaining the structure of atom:
itive charge and most of the mass of the atom is
It cannot explain the characteristic line spectra of at-
concentrated in a small volume called the nucleus
oms of different elements.
with electrons revolving around the nucleus just
It contradicts the stability of matter because it specu-
as planets revolve around the sun.
lates that atoms are unstable because the accelerated
• Rutherford scattering is a powerful way to deter- electrons revolving around the nucleus must spiral
mine an upper limit to the size of the nucleus. into the nucleus.
CHAPTER 12 : Atoms 203
Electron Orbits • Bohr’s second postulate defines these stable orbits.

This postulate states that the electron revolves
• The electrostatic force of attraction, Fe between
around the nucleus only in those orbits for which
the revolving electrons and the nucleus provides
the angular momentum is some integral multiple
the requisite centripetal force (Fc) to keep them in
of h/2p where h is the Planck’s constant (= 6.6 ×
their orbits. Hence, for a dynamically states orbit 10– 34 Js). Thus the angular momentum (L) of the
in a hydrogen atom Fe = Fc orbiting electron is quantised. That is L = nh/2p.
• The total energy of the electron is negative. It is • Bohr’s third postulate incorporated into atomic
e2 theory the early quantum concepts that had been
given by E = − .
8πε 0 r developed by Planck and Einstein. It states that
an electron might make a transition from one of its
specified non-radiating orbits to another of lower
Atomic Spectra energy. When it does so, a photon is emitted hav-
• Each element has a characteristic spectrum of ra- ing energy equal to the energy difference between
diation, which it emits. the initial and final states. The frequency of the
• Study of emission line spectra of a material can emitted photon is then given by
therefore serve as a type of “fingerprint” for iden- hv = Ei – Ef, where Ei and Ef are the energies of the
tification of the gas. initial and final states and Ei > Ef.
• The atomic hydrogen emits a line spectrum con- • Bohr radius is represented by the symbol a0, is
sisting of various series as: h 2ε 0
given by a0 = .

1
Lyman series: v = Rc  −
1  π me 2
2  : n = 2, 3 , 4 ,...
1 n2  • The total energy of the electron in the stationary
states of the hydrogen atom is given by
 1 1  13.6
Balmer series: v = Rc  −  : n =3 , 4 , 5,...
2 2
n2  En = − eV
n2
 1 1 
De Broglie’s Explanation of Bohr’s
Paschen series: v = Rc  − : n = 4 , 5, 6,...
 32 n 2 
Second Postulate of Quantisation
 1 1  • De Broglie hypothesis provided an explanation
Brackett series: v = Rc  − : n = 5, 6, 7,...
 4 2 n 2  for Bohr’s second postulate for the quantisation
of angular momentum of the orbiting electron. The
quantised electron orbits and energy states are
 1 1  due to the wave nature of the electron and only
Pfund series: v = Rc  −  : n = 6, 7, 8,...
2
5 n2  resonant standing waves can persist.
• De Broglie’s hypothesis is that electrons have a
Bohr Model of the Hydrogen Atom wavelength λ =
h
.
Bohr combined classical and early quantum concepts, mv
explained the spectrum of hydrogen atom based on
quantum ideas and gave his theory in the form of three
postulates. These are:
Limitations of Bohr’s model: Bohr’s
• Bohr’s first postulate was that an electron in an model however has many limitations.
atom could revolve in certain stable orbits with- • It is applicable only to hydrogenic (single electron)
out the emission of radiant energy, contrary to the atoms.
predictions of electromagnetic theory. According • It cannot be extended to even two electron atoms
to this postulate, each atom has certain definite such as helium.
stable states in which it can exist, and each possi- • While the Bohr’s model correctly predicts the fre-
ble state has definite total energy. These are called quencies of the light emitted by hydrogenic atoms,
the stationary states of the atom. the model is unable to explain the relative intensi-
ties of the frequencies in the spectrum.
6. Show that the radius of the orbit in hydrogen
PREVIOUS YEARS’ atom varies as n2, where n is the principal

quantum number of the atom. [DELHI 2015]
EXAMINATION QUESTIONS 7. When the electron orbiting in hydrogen atom in

its ground state moves to the third excited state,
TOPIC 1 show how the de Broglie wavelength associated
with it would be affected. [All India 2015]
1 Mark Questions 8. When is Hα line in the emission spectrum
1. Show graphically, the variation of the de-Broglie of hydrogen atom obtained? Calculate the
wavelength (l) with the potential (V) through frequency of the photon emitted during this
which an electron is accelerated from rest. transition.
[All India 2011] Or
2. Write the expression for Bohr’s radius in hydrogen Calculate the wavelength of radiation emitted
atom. [DELHI 2018] when electron in a hydrogen atom jumps from
n = ∞ to n = 1. [All India 2016]
2 Mark Questions
3 Mark Questions
3. Using Rutherford model of the atom, derive the
9. An electron and a photon each have a wavelength
expression for the total energy of the electron in
1.00 nm. Find
hydrogen atom. What is the significance of total
negative energy possessed by the electron? (i) Their momenta.
OR (ii) The energy of the photon and
Using Bohr’s postulates of the atomic model (iii) The kinetic energy of electron.
derive the expression for radius of nth electron [All India 2011]
orbit. Hence obtain the expression for Bohr’s 10. The energy levels of a hypothetical atom are
radius. [All India 2014] shown below. Which of the shown transitions
4. (a) The figure shows the plot of binding energy will result in the emission of a photon of
(BE) per nucleon as a function of mass number wavelength 275 nm?
A. The letters A, B, C, D and E represent Which of these transitions correspond to
the positions of typical nuclei on the curve. emission of radiation of (i) maximum and
Point out, giving reasons, the two processes (ii) minimum wavelength?
(in terms of A, B, C, D and E ), one of which 0eV
A
can occur due to nuclear fission and the other –2eV
B C
due to nuclear fusion. –4.5 eV
D
BE –10 eV
C D
A B [All India 2011]
A E
11. (i) Using Bohr’s second postulate of quantization
Mass number (A) of orbital angular momentum show that the
circumference of the electron in the nth
(b) Identify the nature of the radioactive orbital state in hydrogen atom is ‘n’ times
radiations emitted in each step of the decay the De-Broglie wavelength associated with
process given below. it.
A
 A 4
 A 4 [All India 2015] (ii) The electron in hydrogen atom is initially
ZX Z  2Y Z 1W
in the third excited state. What is the
5. Determine the distance of closest approach when maximum number of spectral lines which
an alpha particle of kinetic energy 4.5 MeV can be emitted when it finally moves to the
ground state?
projected towards a nucleus of Z = 80, stops and
reverses its direction. [All India 2015] [All India 2012]
12. Using Bohr’s postulates, obtain the expression for 4 Mark Questions
the total energy of the electron in the stationary
17. Asha’s mother read an article in the newspaper
states of the hydrogen atom. Hence draw the
about a disaster that took place at Chernobyl.
energy level diagram showing how the line
She could not understand much from the article
spectra corresponding to Balmer series occur
and asked a few questions from Asha regarding
due to transition between energy levels.
the article. Asha tried to answer her mother’s
[All India 2013] questions based on what she learnt in Class XII
13. A 12.5 eV electron beam is used to bombard Physics
gaseous hydrogen at room temperature. Up to
(a) What was the installation at Chernobyl
which energy level the hydrogen atoms would
where the disaster took place?
be excited? Calculate the wavelength of the first
member of Lyman and first member of Balmer (b) What, according to you, was the cause of this
series. [DELHI 2014] disaster?
14. In the study of Geiger-Marsden experiment on (c) What are the values shown by Asha?
scattering of α-particles by a thin foil of gold, [All India 2014]
draw the trajectory of α-particles in the Coulomb 18. Using Bohr’s postulates, derive the expression
field of target nucleus. Explain briefly how one for the frequency of radiation emitted when
gets the information on the size of the nucleus electron in hydrogen atom undergoes transition
from this study. from higher energy state (quantum number ni) to
1 the lower state, (nf). When electron in hydrogen
From the relation R = R0 A , where R 0 is
3 atom jumps from energy state ni = 4 to nf = 3,
constant and A is the mass number of the nucleus, 2, 1, identify the spectral series to which the
show that nuclear matter density is independent emission lines belong.
of A. Or
OR (a) Draw the plot of binding energy per nucleon
Distinguish between nuclear fission and fusion.  BE  as a function of mass number A. Write
Show how in both these processes energy is  
A 
released. Calculate the energy released in MeV
in the deuterium-tritium fusion reaction:
2 3 4 1 two important conclusions that can be drawn
1 H  1 H  1 H  on regarding the nature of nuclear force.
Using the data: (b) Use this graph to explain the release of energy

m  H   2.014102u
2
1
in both the processes of nuclear fusion and
fission.
m  H   3.016049u
3
1
[All India 2014]

m  H   4.002603u
4 Solutions
1
1. Since wavelength l is inversely proportional to
mn = 1.008665u potential V the graphical variation of same is
u = 931.5MeV/c2 [All India 2015]
15. (a) Derive the mathematical expression for
(Wave length)
law of radioactive decay for a sample of a

radioactive nucleus.
(b) How is the mean life of a given radioactive
nucleus related to the decay constant?
[All India 2016]
16. (a) State Bohr’s postulate to define stable orbits
in hydrogen atom. How does de Broglie’s
V (Potential) [1]
hypothesis explain the stability of these
orbits? 2
h o
(b) A hydrogen atom initially in the ground state 2. ao  where h is planks constant, eo is
absorbs a photon which excites it to the 4th  me2
level. Estimate the frequency of the photon. permittivity of free space, m is rest mass of
[All India 2018] electron and e is charge on electron. [1]
3. Electron revolves around nucleus and required nh

centripetal force is provided by attractive force  r 
2 mr [½]
between electron and nucleus. [½]
on putting value of v in equation (i)
V n2  h2 
r  2 
m 2
z  4 kme 
r

+ze –e n2 o
r 0.53 A
z
[½] 4. (a) The nuclei at A and B undergo nuclear fusion
kze2 m 2 as their binding energy per nucleon is small
 and they are less stable so they fuse with
r2 r
other nuclei to become stable. The nuclei
kze2 at E undergo nuclear fission as its binding
 m 2  energy per nucleon is less it splits into two
r
or more lighter nuclei and become stable.
1 kze2 [1]
 KE  m 2 
2 2r [½] A
 A 4
A 4
(b) ZX Z  2Y Z 1W [½]
And potential energy
kze2
An alpha   is emitted in the
particle 24 He
U
r first reaction as atomic mass of Y is reduced
by 4 and atomic number is reduced by 2. An
∴ Total energy
E = KE + U [½]
electron is emitted  e
0
1 in the second
Negative sign of total energy shows that electron reaction as atomic mass of W remains the
is bound to revolve around nucleus. same and atomic number is increased by 1.
Or [½]
Electron revolves around the nucleus and 5. At the distance of nearest approach
required centripetal force is provided by PE = KE [½]
electrostatic force of attraction between nucleus
and electron [½] k  ze 2e
 4.5 MeV  4.5  106  1.6  10 19 J
ro
kze2 m 2

r2 r [½]
[½]
k  ze 2e
kze2
2 ro 
 m  ... (1) 4.5  1.6  10 13
r [½]

  
2
9  10  80  2  1.6  10
9 19
V   [½]
4.5  1.6  10 13
r m = 51.2 × 10–15 m
6. According to the Bohr’s theory of hydrogen atom,
+ze –e the angular momentum of a revolving electron
is given by,
mvr  n …. (i)
[½] 2
By Bohr’s postulate If an electron of mass m and velocity v is moving

in a circular orbit of radius r, then the centripetal
m r  nh ... (2) mv2
2 force required is given by, F = . [½]
r
Also, if the charge on the nucleus is Ze, then

0.4558  10 19
the force of electrostatic attraction between f 
the nucleus and the electron will provide the 10 34 [1]
necessary centripetal force. [½] 15 14
 0.4558  10  4.558  10
1  Ze e KZe 2
Or
F 
4 o r 2 r2 In hydrogen atom wavelength of radiation

1 for transition n = ∞ to n = 1
Where, K  [½]
4 o 1  1 1 
 RZ 2   
 2 2
2 2  nL nH 

 mv  KZe ... (ii)
r r3 n
 Here, nL  1, nH   |For H atom Z  1
[1]
From (i), we get v 
2 mr 1 2 1 1
 R 1   
 2 
Putting this value in (ii), we get 1 
2 2 2 1 1
m n  KZe
  R 
r 2 2 2
4 m r r2  R
o
n2 2   912 A
r ⇒ r ∝ n . 2
[½]
4 2 mKZe2
7. The velocity of a electron in a hydrogen atom is

R  1.09  10 m  7 1 [1]
given by the relation 9. Wavelength of an electron l and a photon lp, le

= lp = l = 1 nm
e2 1 1 × 10–9 m
vn  so vn  ---------- (i) [½]
2n o n n Planck’s constant, h = 6.63 × 10–34 Js
(i) The momentum of an elementary particle is
  h  h and the de Broglie wavelength given by de Broglie relation:
p m
associated with it is [½] h

p
1
So,   ... (ii)
n h
p
using equation (i) and (ii) λ∝ n  [½]
So when electron jump from n = 1 to n = 4 level It is clear that momentum depends only on
the wavelength of the particle. Since the
1 n1 1
  [½] wavelengths of an electron and a photon are
2 n2 4 equal, both have an equal momentum.
l2 = 4l1 6.63  10 34
p
so wavelength increases four times. [½] 1  10 9

8. Hα is obtained when the electron makes
p = 6.063 × 10–25 kg ms–1 [½]
transition from n = 3 to n = 2, level
(ii) The energy of a photon is given by the
E3 – E2 = hf (According to Bohr)
relation:
13.6 13.6 hf
   hc
3 2
2 2
1.6  10 19 E

1 1 hf Where,
13.6   
22
3  1.6  10 19
2
[1] Speed of light, c = 3 × 108 m/s
 5  6.63  10
34
f 6.63  10 34  3  108
13.6    E 
 36  1.6  10 19
1  10 9  1.6  10 19 [1]
1243.1eV = 1.243KeV h

Therefore, the energy of the photon is m [½]
1.243keV.
Where, l = wavelength associated with electron.
(iii) The kinetic energy (K) of an electron having
v = velocity of electron
momentum p, is given by the relation:
h = Planck’s constant
1 p2
K= m = mass of electron
2 m
h
Where,   ... (2)
m
m = Mass of electron = 9.1 × 10–31 kg
Putting value of ‘v’ from eqn. (2) in eqn.(1)
p = 6.63 × 10–25 kg ms–1
h h
1 6.63  10 25 m r  n
K    2.415  10 19 J m 2 [½]
2 9.1  10 31
rh nh
= 1.51eV [1] 
 2
Hence, the kinetic energy of the electron is
2pr = ln
1.51 eV.
Now, circumference of the electron in the nth
10. For A:
orbital state of Hydrogen atom with radius ‘r’ is
Energy change: E1 – E2 = 0 – (–2) = 2eV 2pr [½]
12.3 (ii) If ‘n’ is the quantum number of the highest
A   107  618nm
2 [1] energy level involved in the transition, then
the total number of possible spectral lines
12.3
B   107  275nm emitted is:
4 . 5
n  n  1
12.3 N
C   107  500nm 2 [½]
2 . 5
Third excited state means fourth energy level
12.3
D   107  153nm [1] i.e. n = 4. Here, electron makes transition from
8 n = 4 to n = 1. So, the highest value of ‘n’ is 4.
Maximum wavelength: emission A Therefore, the maximum number of spectral lines
would be 6. [½]
Minimum wavelength: emission D [1]
12. According to Bohr’s postulates, in a hydrogen
11. (i) According to Bohr’s second postulate of
atom, a single electron revolves around a
quantization, the electron can revolve
nucleus of charge +e. For an electron moving
round the nucleus only in those circular
with a uniform speed in a circular orbit of a
orbits in which the angular momentum
given radius, the centripetal force is provided by
of the electron is integral multiple of
Coulomb force of attraction between the electron
h
w h e r e ‘ h ’ i s P l a n c k ’s c o n s t a n t and the nucleus. The gravitational attraction
2π
may be neglected as the mass of electron and
(= 6.62 × 10–34 Js) [½] proton is very small. So, [½]
So, if ‘m’ is the mass of electron and ‘v’ is the mv2 ke2
velocity of electron in permitted quantized orbit = (1)
with radius ‘r’ them,
r r2
h Where, m = mass of the electron

m r  n ... (1)
2 r = radius of electronic orbit
Where ‘n’ is the principle quantum number and v = velocity of electron.

can take integral value i.e. n = 1, 2, 3, ….. Again,
This is Bohr’s quantization condition. Now, nh
mvr 
De-Broglie wavelength is given as, 2
nh  1
v 1 1 
2 mr [½]  R  
  n2f ni2 
2
 nh  ke2
From equation (i), we get, m    Where,
 2 mr  r
 n2 h2  ke2 R → Rydberg’s constant = 1.09678 × 10–7 m–1
m  For Balmer series, nf = 2 and ni = 3, 4, 5…..
 4 2 m2 r 2  r

1  1 1 
n2 h2 ke2  R  
  2
2
ni2 
2 2 r
4 mr [½]
Where, ni = 3, 4, 5………
n2 h2
k (2) These spectral lines lie in the visible region.
4 2 mre2
Total energy, E(ev)
Using equation (2), we get n=5
n=4
ke2 4 2 kme2
Ek 
Excited
2n2 h2
states
–0.85 n = 3
–1.5

2 2 k2 me4
Ek  n=2 Paschen
n2 h2 [½]
–3.40 Series
(ii) Potential energy,
 k  e   e ke2
Ep  
r r
4 2 kme2
E p   ke2 
n2 h2 [½]
n=1 Ground state
2 2 4
4 k me –13.6
Ep  
n2 h2 13. Energy of the electron in the nth state of an atom
= – 13.6z2 n2eV
Hence, total energy of the electron in the nth orbit,
E = Ep + Ek Here, z is the atomic number of the atom. For
hydrogen atom, z is equal to 1. Energy required
4 2 k2 me4 2 2 k2 me4 to excite an atom from the initial state (ni) to the
E 
n2 h2 n2 h2 [½] final state
2 2 k2 me4 E  13.6 ln 2f  13.6ni2 eV

E f [½]
n2 h2 This energy must be equal to or less than the
energy of the incident electron beam.
13.6
E eV −13.6 13.6
n2 = Ef + 2
n 2f ni

When the electron in a hydrogen atom jumps
from higher energy level to the lower energy
13.6  13.6  12.5
level, the difference of energies of the two energy 2
nf 2
ni
levels is emitted as a radiation of particular
wavelength. It is called a spectral line. −13.6
+ 13.6 =
12.5
In H-atom, when an electron jumps from the n 2f

orbit ni to orbit nf, the wavelength of the emitted
13.6
radiation is given by, 13.6 – 12.5 = [½]
n 2f
Energy of the electron in the ground state (2) About one alpha particle in every 8000 alpha
= 13.6eV particles deflects by more than 90. [1]
13.6 ln 2f  13.6  12.5 As most of the alpha particles go undeflected
and only a few get deflected, this shows that
13.6  12.5  13.6 ln 2f most of the space in an atom is empty and at
the centre of the atom, there exists a nucleus.
13.6 By the number of the alpha particles deflected,
=n2f = 3 .5 the information regarding size of the nucleus
1.1 [½]
can be known. [½]
State cannot be a fraction number nj = 3
If m is the average mass of a nucleon and R is
Hence, hydrogen atom would be excited up to 3rd the nuclear radius, then mass of nucleus = mA,
energy level. Rydberg formula for the spectrum where A is the mass number of the element.
of the hydrogen atom is given below. [½]
Volume of the nucleus, [½]
1  1 1 
 R  4
 2 2 V  R3
 n1 n2  3

3
Here, l is the wavelength and R is the Rydberg  1
4
constant, R = 1.097 × 107 m–1.  V    Ro A 3 
3  
For the first member of the Lyman series: [½]
n1 = 1, n2 = 2 4
V   Ro3 A
1 1 1  3
 1.097  107  
  12 22  Density of nuclear matter,
[½]
o mA
  1215  A  
  V

mA
For the first member of Balmer series: 
4
n1 = 1, n22 = 23  Ro3 A
3
1  1 1 
 1.097  107    3m
 2 2
32  
4 Ro3
[½]
o
   6563  A  This shows that the nuclear density is independent
  of A.
[½]
14. OR
Trajectory of a Particles in Coulomb Field Nuclear fission: Nuclear fission is a disintegration
of Target Nucleus process, in which a heavier nucleus gets split up
into two lighter nuclei, with the release of a large

Incident Particles
< 90° amount of energy. [1]

3 235
> 90° 92U  o n1  56 Ba
141
 36 Kr
92
 3 o n1  Q
2
1 Atomic
Here, the energy released per fission of 92U235 is
nucleus
2 200.4 MeV
3 Nuclear fusion: When two or more light nuclei

combine to form a heavy stable nuclide, part of
mass disappears and is converted into energy.
r0
This phenomenon is called nuclear fusion. [1]
From this experiment, the following is observed: 1H
1
 1 H 1  1 H 2  e  V  0.42 MeV

(1) Most of the alpha particles pass straight
1H
1
 1 H 1  1 H 2  e  V  0.42 MeV
through the gold foil. It means that they do
not suffer any collision with gold atoms.
2 element decreases exponentially with time (i.e.,

1H  1 H 2  1 H 3  1 H1  4.03 MeV
more rapidly at first and slowly afterwards.)
2
 13 H  24 H  n
No. of undercayed nuclei

1H
\ Dm = (2.104102 + 3.016049) – (4.002603 + N0

1.008665) = 0.01888 3u
Energy released, Q = 0.018883 × 931.5 MeV/e2 N0
Q = 17.589 MeV [1] 2
N0
15. (a) Radioactive decay Law: The rate of decay of
4
radioactive nuclei is directly proportional to the P
N0
number of undecayed nuclei at that time.
8 N T1 2T1 3T1 4T1
Derivation of formula: Suppose initially the 2
2 2 2 2
number of atoms in radioactive element is N0 16
and N the number of atoms after time t. After
[½]
time t, let dN be the number of atoms which
disintegrate in a short interval dt, then rate of (b) Relation between mean life and half life
disintegration will be
dN
, this is also called the 1
Mean life   where l → decay constant
dt 
activity of the substance/element. According to [½]
Rutherford-Soddy law
16. (a) Bohr’spostulate
dN
∝N 1. All the electrons revolve in circular orbit.
dt [½] Necessary centripetal force is provided by
dN electrostatic force between electron and
Or   N proton.
dt
Where λ is a constant, called decay constant
or disintegration constant of the element. Its
unit is s–1. Negative sign shows that the rate of r
disintegration decreases with increase of time. e¯
For a given element/substance λ is a constant p
and is different for different elements. Equation
(i) may be rewritten as [½]
dN
  dt
dt 2. Electron revolve only in those orbits for
which angular momentum is integral
Integrating loge N – lt + C -------- (ii) where C is
multiple of
a constant of integration.
At t = 0, N = N0  h 
 
2 
\ loge N = 0 + C
⇒ C = loge No nh
L  mvr 
∴ Equation (ii) gives loge N – lt + loge No 2 [1]
Or loge N – loge No = – lt 3. When electron jumps from n higher orbit
th
to pth lower orbit it emits energy in form of

N
Or log e   t proton.
No
N
Or  e t [1] r
No

\ N = No e–lt ... (iii)
According to this equation, the number of
undecayed atoms/nuclei of a given radioactive
En  E p  h of the emitted radiation or photon is given by

h  Eni  En f [1]
According to de Broglie hypothesis
h h
 
p mv  me4
We know that En 
nl = 2pr
8 h2 o2 n2
h
n  2 r me4 me4
mv  h   i.e.
8 h2 o2 ni2 8 h2 o2 n2f
nh = 2pmvr
nh
L [1] me4  1 1 
2 h    
8 h2 o2  n2f ni2 
(b)
me4  1 1 
n=4
or      [1]
8 h3 o2  n2f ni2 
n=1
hv Name of series
E4  E1  n = 4 to n = 3 (Paschan)
e
n = 4 to n = 2 (Balmer)
13.6 6.63  1034  
 13.6  n = 4 to n = 1 (Lyman)
4 1.6  1019
Or
34
1  6.63  10  8.8 Ni 62
Mo 100
13.6   1  I127
28
9 12 16 W184
4  19 C
1.6  10 8 6 O U238
nuclear (in Me V)
Binding Energy per
74 O18 235
2He U
1.6  1019  3
14
6 N

6
5 Li
6.63  1034  4 4
3 H3
  2.4615  1015 Hz
1
2
[1] H2
1 1
17. (a) “Oh April 1986, the world’s worst nuclear 0
25 50 75 100 125 150 175 200 225 250
accident happened at the Chernobyl. Plant Mass Number (A) [1]
near pripyat Ukraine in the soviet union. [1]
Conclusions :-
(b) An explosion and fire in the No. 4 reactorsent
(i) The intermediate nuclei have large value of
radioactivity into the atmosphere [1]
BE
(c) The value displayed by the Asha is that so they are more stable.
A
she is caring and having helping nature
towards her mother. The value displayed BE
by Asha’s mother is that she has no idea the (ii) has low value for both of light and heavy
A
outburst take place in Chernobyl (Ukraine)
but she has the curiosity about the incident nuclei so they are unstable nuclei.
that take place on April 26, 1986, at the (b) In nuclear fission, unstable heavy nuclei
Chernobyl plant near Priyat, Ukraine, in splits into two stable intermediate nuclei
the Soviyat union. [2] and in Nuclear fusion, 2 unstable light
18. According to Bohr, energy is radiated in the nuclei combines to form stable intermediate
form of a photon when the electron of an excited nuclei so in both processes energy liberates
hydrogen atom returns from higher energy as stability increases
state to the lower energy state. In other words, (c) n  P   01   Neutrinos are difficult to
energy is radiated in the form of a photon when
electron in hydrogen atom jumps from higher detect as they go through all object by
energy orbit (n = ni) where ni = nf. The energy penetrating them. [1]

CHAPTER 13
Nuclei

Composition and size of 1Q
nucleus, atomic masses,
isotopes, isobars; isotone (2 marks)
Radioactivity alpha, beta
and gamma particles/ 1Q
rays and their properties; (3 marks)
radioactive decay law.
Mass-energy relation,
mass defect; binding en-
1Q 1Q
ergy per nucleon and its
(2 marks) (1 marks)
variation with mass num-
ber
Nuclear fusion and Nu- 1Q 1Q 1Q
clear Fission (2 marks) (4 marks) (3 marks)
On the basis of above analysis, it can be said that from exam point of view Binding Energy,
Nuclear Force, Nuclear Fission, Nuclear Fusion and Radioactive Decay are the most import-
ant concepts of the chapter.
216 CHAPTER 13 : Nuclei
Topic 1: Radioactivity and Decay Law

Summary • Isotopes: Atoms of an element that have differ-
ent mass numbers but same atomic number are
• Nucleus: Nucleus can be defined as the central known as isotopes. e.g., 1H1, 1H2, 1H3 is an exam-
part of an atom, made up of neutrons, protons, ple of isotopes.
and other elementary particles. The nucleus has • Nuclear Force: Nuclear force can be referred
protons and neutrons inside it. They are called to as the force that acts inside the nucleus or
nucleons. between nucleons. These forces are neither elec-
• Mass Number: The total number of protons and trostatic nor gravitational in nature. They have
neutrons present inside the nucleus of an atom of a very short range and are independent of any
an element is referred to as mass number (A) of charge. They are a hundred times that of electro-
the element. static force and 1038 times that of gravitational
• Atomic Number: The number of protons present force.
in the nucleus of an atom of an element is known • Radioactivity: Radioactivity refers to the break-
as atomic number (Z) of the element. down of heavy elements into comparably lighter
1
elements by the emission of radiations. This phe-
• Nuclear Size: The radius of the nucleus R ∝ A 3
nomenon was discovered by Henry Becquerel in
1
1896.
• R = Ro A 3 where R0 = 1.2 × 10–15 m is an em- • Packing Fraction (P):

(Exact nuclear mass ) – (M ass number )
pirical constant. P =
M ass number
• Nuclear Density: Nuclear density is indepen-
dent of mass number and is therefore same for (A − M )
all nuclei. =
M
M ass of nucleus
ρ= For greater stability of the nucleus, the value of pack-
volume of nucleus
ing friction should be larger.
3m
ρ= where is the average mass of a • Radioactive Decay law
4π R 0 3
The Radioactive law states that the rate of disinte-
gration of radioactive atoms at any instance is direct-
nucleon. ly proportional to the number of radioactive atoms
• Atomic Mass Unit: Abbreviated as amu and is present in the given sample at that instant.
defined as one-twelfth of the mass of a carbon
dN 
nucleus. It is also denoted by u. Rate of disintegration  − ∝N
 dt 
Therefore,
1.992678 × 10 −26 dN
1 amu = kg – = λ N , where l is the decay constant.
12 dt
= 1.6 × 10–27 kg = 931 MeV The number of undecayed atoms present in the
• Isomers: The atoms that have the same mass sample at any instance N = N 0 e − λt where, N0 is
number, atomic number but different radioactive
properties are known isomers. number of atoms at time t = 0 and N is number of
• Isotones: Atoms of elements that have different atoms at time t.
mass numbers, atomic numbers but same number
• Activity of a radioactive element
of neutrons are known as isotones. e.g., 1H3, 2H4
and 6C14, 8O16 are isotones. The activity of a radioactive element is equal to its
• Isobars: The atoms of an element having differ- rate of disintegration.
ent atomic numbers but same mass numbers are dN 
Activity R =  −
known as isobars. e.g., 1H3, 2H3 and 10Na22, 10Ne22  dt 
are isobars.
CHAPTER 13 : Nuclei 217
Activity of the sample after time t, R = R 0 e – λt Relation between half-life and average life t= 1.44T
Relation between average life and decay constant
Its SI unit is Becquerel (Bq). Curie and Rutherford 1
τ =
are its other units. λ
1 Curie = 3.7 × 1010 decay/s and 1 rutherford = 106 • Alpha decay: In alpha decay, a nucleus gets
decay/s transformed into a different nucleus and an a
• Half-life of a radioactive element particle is emitted. The general form can be ex-
Half-life (T) of a radioactive element is the time tak- pressed as:
A −4
en for the radioactivity of an isotope to fall to half its A
Z X → Z − 2Y + 24H e and the Q value:
original value. The relation between disintegration 2
constant and half-life is given by Q = ( m X − mY − m H e )c
log e 2 0.6931 • Beta decay: When a nucleus undoes beta decay,
=T =
λ λ it emits an electron or a positron. When an elec-
tron is emitted, it is said to be beta minus decay
• Average Life or Mean Life (t) while in beta plus decay, a positron is emitted.
Average life or mean life (t) of a radioactive element • Gamma decay: In gamma decay, the photons
can be defined as the ratio of total life time of all the are emitted from the nuclei having MeV energy
atoms and total number of atoms present, initially in and thus the gamma rays are emitted. This is
the sample. called as gamma decay.
3 Marks Questions
PREVIOUS YEARS’ 5. State the law of radioactive decay.
Plot a graph showing the number (N) of
EXAMINATION QUESTIONS undecayed nuclei as a function of time (t) for a
given radioactive sample having half life.
TOPIC 1
1 Mark Questions Depict in the plot the number of undecayed nuclei
at (i) t = 3T1 / 2 and (ii) t = 5T1 / 2
1. Define the activity of a given radioactive
substance. Write its S.I. units. [ALL INDIA 2011]
[DELHI 2011] 6. Show that the density of nucleus over a wide
2. Why is it found experimentally difficult to detect range of nuclei is constant independent of mass
neutrinos in nuclear b-decay? number A.
[ALL INDIA 2014] [ALL INDIA 2012]
3. Two nuclei have mass numbers in the ratio 1: 2. 7. Draw a plot of potential energy between a pair
What is the ratio of their nuclear densities? of nucleons as a function of their separation.
[DELHI 2017] Mark the regions where potential energy is (i)
positive and (ii) negative
2 Marks Questions [DELHI 2013]
4. A radioactive nucleus ‘A’ undergoes a series of
8. (a) In a typical nuclear reaction, e.g.
decays according to the following scheme:
2
 1H  12 H  23 He  01 n  3.27 MeV ,

A  A1 

 A2  A3  A4


although number of nucleons is conserved,
The mass number and atomic number of A are yet energy is released. How? Explain.
180 and 72 respectively.
(b) Show that nuclear density in a given nucleus
What are these numbers for A4? is independent of mass number A.
9. Identify the nature of the radioactive radiations 

A  170176
1 A2
emitted in each step of the decay process given

below. A1  170176
1 A2
[½]
A A 4 A 4
ZX  Z  2Y  Z 1W Formation of A3:

[ALL INDIA 2015] 172 
10. (a) Derive the mathematical expression for A2  169 A3

law of radioactive decay for a sample of a Formation of A4: [½]
radioactive nucleus.
(b) How is the mean life of a given radioactive In γ-decay, mass number and atomic number
nucleus related to the decay constant? remain the same. Thus,
[DELHI 2016] 172 
A3  169 A4
11. A radioactive isotope has a half-life of 10years.
Mass number of A4 = 172
How long will it take for the activity to reduce
to 3.125% Atomic number of A4 = 169 [½]
[ALL INDIA 2018] 5. According to radioactive decay law the rate of
12. (i) Define ‘activity’ of a radioactive material and disintegration of a radioactive substance at an
write its S.I. units. instant is directly proportional to the number of
(ii) Plot a graph showing a variation of activity nuclei in the radioactive substance at that time
of a given radioactive sample with time. i.e, N = No e–lt, where symbols have their usual
(iii) The sequence of stepwise decay of a meanings. [1]
radioactive nucleus is
Number of undecayed nuclei

  1
D  D1  D2 N0
N = N0–t
If the atomic number and mass number of D2
are 71 and 176 respectively, what are their
corresponding values of D? N0/2
[DELHI 2018]
Solutions
N0/4
N0/8
N0/16
1. Activity: Rate of disintegration of radioactive
substance. i.e. number of radioactive nuclei T1/2 2T1/2 3T1/2 4T1/2 Time t
disintegrating in unit time is called activity. SI
unit: [1] From graph: [½]
1 disintegration per second (dps) or 1 Bq No

Number of un-decayed nuclei at t = 3T1 / 2 is
2. Neutrinos are difficult to detect experimentally 8
No
in β decay because they are uncharged particles Number of un-decayed nuclei at t = 5T1 / 2 is
32
with almost no mass. Also, neutrinos interact
very weakly with matter, so they are very [1½]
difficult to detect. [1]
6. To find the density of nucleus of an atom, we have
3. Nuclear density is independent of mass number. an atom with mass number suppose A. (Here,
Hence, both the atoms have the same nuclear we are neglecting mass of the orbital electrons)
density. [1]
Mass of the nucleus of the atom of the mass
4. A has mass number as 180 and atomic number number A = A a.m.u = A × 1.660565 × 10–27 kg
172.
Let the radius of nucleus is ‘R’. [1]
Formation of A1 by α decay:
4
 180  4
Then, volume of nucleus   R3
A  172  2 A1
3
[½] 3
 1
 4
176
A  170 A1    Ro A 3 
3  

Formation of A2 by β decay:
4 Mass of the nucleus

  Ro3 A
3 = A amu = A × 1.666 × 10–27 kg
Now, we know Ro = 1.1 × 10–15 m Volume of the nucleus,
3
Volume of the nucleus [1]  1
4 3 4
V   R    Ro A 3 
4
 
3
  1.1  1015  A m3 3 3  
3
Density of the nucleus, 1
Where, R = Ro A 3
Mass of nucleus A  1.660565  1027
   A  1.66  1027 1.66  1027
 
Volume of nucleus 4 3 Thus, density   ,
 1.1  1015  A 4 3 4 3
3   R0  A   R0 
3 3
d = 2.97 × 1017 m–3
shows that the density is independent of mass
Thus, we can see the density of nucleus is
number A. [1]
independent of the mass number and is constant
A A 4 A 4
for all nuclei. 9. ZX  Z  2Y  Z 1W [1]
7. The potential energy is minimum at ro: For
distance larger than ro the negative potential
An alpha particle  He is emitted in the first
4
2
energy goes on decreasing and for the distances reaction as atomic mass of Y is reduced by 4 and
less than ro the negative potential energy atomic number is reduced by 2. An electron is
decrease to zero and then becomes positive
and increases abruptly. Thus, A to B is the emitted  e in the second reaction as atomic
0
1
positive potential energy region and B to C is
the negative potential energy region. [1] mass of W remains the same and atomic number
is increased by 1. [1]
10. (a) Radioactive decay Law: The rate of decay
Potential energy (MeV)
A of radioactive nuclei is directly proportional

to the number of un decayed nuclei at that
time. [1]
100 Derivation of formula: Suppose initially the
number of atoms in radioactive element is
B C N0 and N the number of atoms after time t.
0 After time t, let dN be the number of atoms
which disintegrate in a short interval dt,
–100 dN
r0 1 2 3 then rate of disintegration will be , this
r (fm) dt
[1]
is also called the activity of the substance/
8. (a) In a nuclear reaction, the aggregate of the
element. According to Rutherford-Soddy law
masses of the target nucleus 21H and the
bombarding particle may be greater or less dN
∝N
than the aggregate of the masses of the dt
product nucleus (23He) and the outgoing
dN
particle (10n) So from the law of conservation Or   N [1]
dt
of mass-energy some energy (3.27 MeV) is
evolved or involved in a nuclear reaction. Where λ is a constant, called decay constant
This energy is called Q-value of the nuclear or disintegration constant of the element. Its
reaction. [1] unit is s–1. Negative sign shows that the rate of
(b) Density of the nucleus disintegration decreases with increase of time.
For a given element/substance λ is a constant
mass of nucleus and is different for different elements. Equation
=
volume of nucleus [1]
(i) may be rewritten as 3.125

 e0.0693t
dN 100
  dt
dt
100
Integrating loge N – lt + C -------- (ii) where C is
= e0.0693t [½]
3.125
a constant of integration.
At t = 0, N = N0 100
log e = 0.0693t
3.125
\ loge N = 0 + C
32
⇒ C = loge No 2.303 log10
t2 =
∴ Equation (ii) gives loge N – lt + loge No 0.0693
Or loge N – loge No = – lt t = 50.1 years [½]
12. (i) The activity of a radioactive material
N
Or log e   t is defined as the decay rate of a sample
No containing one or more nuclides. The S.I
N unit of radioactivity is Becquerel (B). [1]
Or  e t
No R0
\ N = No e–lt -------- (iii)
According to this equation, the number of
undecayed atoms/nuclei of a given radioactive Activity
element decreases exponentially with time (i.e.,
more rapidly at first and slowly afterwards.)
Time
N0 (ii)
N0 1
(iii) D  D1 

 176
No. of under
cayed nuclei
2 71 D2 [½]
N0
4
P
N0 Therefore,
8 N2 T1 2T1 3T 1 4T 1
  1
D  176  176
2 2 2 2
16
72 D1  71 D2 [½]
(b) Relation between mean life and half life
Therefore,
1
Mean life   where l → decay constant 180  176 
 74 D   72 D1  176
 71 D2
[½]
11. Ao = 100 [½] So, the corresponding values of atomic number
and mass number for D are74 and 180. [½]
At = 3.125
At = Aoe–µ.
0.693
t
3.125 = 100 e 10
Topic 2: Mass Defect and Binding Energy

Summary When a slow moving neutron strikes with a uranium
• Mass Defect: Mass defect can be mentioned as

nucleus ( 92U 235 ) , it splits into 56
Ba141 and 36Kr92 along
the difference between the sum of masses of all
with three neutrons and a lot of energy.
nucleons (M) and the mass of the nucleus (m).

Mass Defect (Dm) = M – m
235
92U + 0 n 1 → 56 Ba141 + 36 K r 92 → 3 0 n 1 + ener gy
= [Zmp + (A – z)mn – mn]
• Nuclear Binding Energy: Nuclear binding energy • Nuclear Chain Reaction
can be referred to as the minimum energy that is Nuclear chain reactions are defined as a chain of nu-
required to separate the nucleons up to an infinite clear fission reactions (splitting of atomic nuclei), and
distance from the nucleus. each one of them is initiated by a neutron produced in
Nuclear binding energy per nucleon = Nuclear binding the previous fission reaction. Nuclear chain reactions
energy / Total number of nucleons Binding energy, Eb are of two types:
= [Zmp + (A – Z)mn – mn]c2  Controlled chain reaction
• Nuclear Fission  Uncontrolled chain reaction
The process of the splitting a heavy nucleus into two
or more lighter nuclei is known as nuclear fission.
• Nuclear Reactor
Control Superheated
Rods Coolant Steam
Shielding
Electric
Steam Generator
Turbine
Heat
Exchanger
Water
Condenser
Fission Fuel User
Chamber Rods Steam
Moderator
Water
Pump
Cold Water
Fig.: Setup of a Nuclear Reactor
The vital parts of a nuclear reactor are the following:
1 1 2 +
 Fuel: Fissionable materials like 92U235, 92U238, 94U239 1 H + 1 H → 1 H + e + v + 0.42 MeV, where a deu-
are used as fuel. teron and a positron are formed by the combination
 Moderator: Graphite, heavy water and beryllium of two protons and 0.42 MeV energy is released.
oxide are used to slower down fast moving neutrons. The source of energy of sun and all the stars is a nu-
 Coolant: Liquid oxygen, cold water, etc. are used clear fusion reaction in which hydrogen nuclei com-
to remove heat generated in the fission process. bine to form helium nuclei.
 Control rods: Cadmium or boron rods are considered 4 11H + 2e − → 24H e + 2v + 6γ + 26.7 MeV
as good absorber of neutrons and are therefore used • Advantages of Nuclear fusion:
to control the fission reaction.  Nuclear fusion does not cause any waste as
• Nuclear Fusion: The process of combining two the only by product is helium.
light nuclei in order to form a single large nucleus  Nuclear fusion is very simple to control as
is called nuclear fusion. A large amount of energy there is no change of chain reaction.
is released in this process. The example of nuclear  There is unlimited supply of fuel for nuclear
fusion is: fusion.
in the deuterium-tritium fusion reaction:
PREVIOUS YEARS’ 2
1
H  13 H  24 He  01 n
Using the data:
EXAMINATION QUESTIONS m  H   2.014102u
2
1
TOPIC 2
m  H   3.016049u
3
1 Mark Questions
1
m  He  4.002603u
1. Four nuclei of an element undergo fusion to form 4
a heavier nucleus, with release of energy. Which 2

of the two: the parent or the daughter nucleus mn = 1.008665u
would have higher binding energy per nucleon? lu = 931.5MeV/c2
2. What characteristic property of nuclear force 7. The figure shows the plot of binding energy (BE)
explains the constancy of binding energy per per nucleon as a function of mass number A. The
BE 
nucleon  in the range of mass number ‘A’ letters A, B, C, D and E represent the positions
 A  of typical nuclei on the curve. Point out, giving
lying 30 < A < 170? [ALL INDIA 2012] reasons, the two processes (in terms of A, B, C,
D and E ), one of which can occur due to nuclear
fission and the other due to nuclear fusion.
2 Marks Questions
BE
3. A nucleus with mass number A = 240 and
A
BE C D
= 7.6 MeV breaks into two fragments each of A B
BE E
A = 120 with = 8.5 MeV Calculate the A
A
released energy. Mass number A
Or, calculate the energy in fusion reaction: [ALL INDIA 2015]
3 8. Explain the processes of nuclear fission and
12 H 12 H 2 He+n, Where BE of
nuclear fusion by using the plot of binding
2
1 He = 2.23MeV,  BE 
energy per nucleon  versus the mass
3  A 
And of 2 He=7.73MeV.,
[DELHI 2016] number A.
4. A heavy nucleus X of mass number 240 and [All India 2018]
binding energy per nucleon 7.6 MeV is split into
two fragments Y and Z of mass numbers 110 and
Solutions
130 respectively. The binding energy of nucleons 1. In fusion, parent nuclei fuse to form daughter
in Y and Z is 8.5 MeV per nucleon. Calculate atom and release some amount of energy. It
the energy Q released per fission in MeV. means daughter nuclei are more stable than
[DELHI 2018] parent nuclei hence Daughter nuclei B.E. >
5. In a nuclear reaction Parent nuclei B.E. [1]
3 3 BE
2 H  2 H  24 He  11 H  11 H  12.86 MeV 2. The approximate constancy of over the
, A
though the number of nucleons is conserved most of the range is saturation property of
on both sides of the reaction, yet the energy is nuclear force. In heavy nuclei: nuclear size >
released. Explain. [DELHI 2013] range of nuclear force. So, a nuclear sense
3 Marks Questions approximately a constant number of neighbours
BE
and thus, the nuclear levels off at high A.
6. Distinguish between nuclear fission and fusion. A
Show how in both these processes energy is This is saturation of the nuclear force. [1]
released. Calculate the energy release in MeV
3. Gain in binding energy for nucleon is about 0.9 6. Nuclear fission: Nuclear fission is a disintegration
MeV. [½] process, in which a heavier nucleus gets split up
Binding energy of the nucleus, into two lighter nuclei, with the release of a large
amount of energy. [1]
B1 = 7.6 × 240 = 1824 MeV
235
Binding energy of each product nucleus, 92U  o n1  56 Ba141  36 Kr 92  3 o n1  Q

B2 = 8.5 × 120 = 1020 MeV Here, the energy released per fission of 92U235 is
200.4MeV
Then, the energy released as the nucleus breaks.
Nuclear fusion: When two or more light nuclei
[½] combine to form a heavy stable nuclide, part of
E = 2B2 – B1 = × 120 = 1020 – 1824 = 216 MeV mass disappears and is converted into energy.
Or, calculate the energy in fusion reaction: This phenomenon is called nuclear fusion. [1]
2 2
3
1H
1
 1 H 1  1 H 2  e  V  0.42 MeV
1 H 1 H  2 He+n, where BE of
2
2 1H  1 H 2  2 H 3  o n1  3.27 MeV
1 H = 2.23MeV, [½]
2
And of 3 1H  1 H 2  1 H 3  1 H 1  4.03 MeV
2 He=7.73MeV.
DE = (7.73) – 2(2.23) 2 3 4
1H  1H  2H  n
= 7.73 – 4.46 \ Dm = (2.104102 + 3.016049) – (4.002603 +
= 3.27 MeV [½] 1.008665) = 0.01888 3u
4. Total energy of nucleus X = 240 × 7.6 = 1824 MeV Energy released, Q = 0.018883 × 931.5 MeV/e2
Total energy of nucleus V = 110 × 8.5 = 935 MeV Q = 17.589 MeV [1]
Total energy of nucleus Z = 130 × 8.5 = 1105 MeV 7. The nuclei at A and B undergo nuclear fusion
Therefore, energy released from fission, Q = 935 as their binding energy per nucleon is small
+ 1105 – 1824 [½ + ½ + ½ + ½] and they are less stable so they fuse with other
Q = 216 MeV nuclei to become stable. The nuclei at E undergo
nuclear fission as its binding energy per nucleon
5. ‘In a nuclear reaction, the sum of the masses is less it splits into two or more lighter nuclei
3
of the target nucleus 2 He may be greater or and become stable.
less the sum of the masses of tiny product 8. Nuclear Fission: When heavy nucleus bombarded
3
nucleus 4 He and the 11 He . So from the law of with neutron and it splits into smaller nucleus
conservation of mass energy some energy (12.86 and energy released. [½]
MeV) is evolved in nuclear reaction. This energy 235
 0 n1  144
 89
 3 0 n1 
is called Q-value of the nuclear reaction. The Ex : 92U 56 Ba 36 Kr
binding energy of the nuclear reaction. The Energy

binding energy of the nucleus on the left side is Nuclear Fusion: When 2 smaller nucleus fuse to
not equal to the right side. The difference in the form ea heavy nucleus again energy released.
binding energies on two sides appears as energy Ex : H 2  H 2  He4  approx. 26 MeV
released or absorbed in the nuclear reaction. [1] 1 1 2
BE Graph: [½]
10 56 100
32
S Fe Mo 122
I
per nucleon (MeV)
16
Potential energy (MeV)
CO
12 184 197 238
A W An U
Building energy
8
4 16
He14 O
N
6 6
Li
100 4
3
H
2
B C 2
H
0
0
0
100 150 200 50
250
Mass number (A)
–100
r0 1 2 3 Fig.: Graph Showing Plot of Binding Energy per
r (fm) [1] nucleon(MeV) Vs Mass Number(A)
[½]
CHAPTER 14
Semiconductor
Electronics

Energy bands in conduc-
tors; semiconductors and
insulators
Semiconductor diode: I-V
characteristics in forward 1Q
and reverse bias; diode as (1 mark)
a rectifier
Special purpose p-n junc-
tion diodes: LED, pho- 1Q
todiode, solar cell and 1Q 1Q 1Q (1 mark),
Zener diode and their (4 marks) (3 marks) (3 marks) 1Q
characteristics; Zener di- (1 mark)
ode as a voltage regulator
Junction transistor; tran-
sistor action; character-
istics of a transistor and 1Q 1Q 1Q 1Q
transistor as an amplifier (3 marks) (3 marks) (3 marks) (3 marks)
(common emitter configu-
ration);
Basic idea of analog and
digital signals; Logic 1Q
gates (OR, AND, NOT, (3 marks)
NAND and NOR)
On the basis of above analysis, it can be said that from exam point of view, Logic Gates, p-n
Junction and circuit based questions are most important concepts of the chapter.
226 CHAPTER 14 : Semiconductor Electronics
[Topic 1] Semiconductor, diode and its applications
Summary are shared by different number of atoms in the

crystal which causes splitting of energy levels.
• The materials which are present in solid state These energy levels are called energy bands. The
and their conductivity lies between insulator energy band which contains valence electrons
and conductor are called as semiconductors. is called as valence band. It always has some
Semiconductors are either pure substance like electrons and can never be empty.
silicon, germanium or they can also be formed
 The energy band which contains conduction
by addition of impurities which form a compound
electrons is called as conduction band. It can be
like gallium arsenide, cadmium selenide, etc.
empty or have some electrons which take part in
• Semiconductors have resistivity in the range flow of current.
of metals and insulators. Insulators have  The band which lies between conduction band
resistivity in the range of 1011 – 1019 Wm and and valence band is called as forbidden band. The
metals have resistivity in the range of 10–2 to minimum amount of energy required to transfer
10–8 Wm while semiconductors have resistivity in electrons from valence band to conduction band
the range of 10–5 – 106 Wm is called as band gap.
• Semiconductors can be elemental (without  Metals do not have any band gap and E g ≈ 0
doping) as well as compound (by doping).
• Intrinsic semiconductors: These are pure while band gap in insulators is greater than 3 eV
semiconductors where the conductivity is due to and the band gap for semiconductors lies between
electrons moving from valence band to conduction 0.2 eV and 3 eV.
band. Their conductivity is called intrinsic Empty 4 N states
conductivity. In intrinsic conductors, ne = nh EC
• Extrinsic semiconductors: When Impurity
is added to pure semiconductor to increase its
conductivity, is called as extrinsic semiconductor.
Eg
It can be divided into two types, i.e. p-type
semiconductors and n-type semiconductors.
In p-type semiconductors, number of holes are
EV
greater than number of electrons.
Filled 4 N states
nh >> ne
Infinitely large number of
In n-type semiconductors, number of electrons
states each occupied by
are greater than number of holes. two electrons
at 0 K
ne >> nh
Fig.: The energy band positions in a semiconductor
Trivalent atoms (B, Al, etc.) called acceptor atoms at 0 K. The upper band is conduction band and
are used for doping p-type semiconductors while the lower band is called valence bond.
pentavalent atoms (As, Sb, etc.) called donor
• p-n junction: p-type semiconductor when
atoms are used for doping n-type semiconductors.
brought in contact with n-type semiconductor
• Energy bands: Valence electrons of an atom forms a p-n junction.
CHAPTER 14 : Semiconductor Electronics 227
p-n junction
p n
+ + + +
+ + + +
p n
+ + + + Electron Symbol
Hole Depletion layer
Fig.: p–n junction depicting Depletion layer

When there are no charge carriers, a region is junction diodes are zener diode, light-emitting
created at the p-n junction called as depletion diode, photo-diode, etc.
layer.  In zener diodes, when it is reversed biased the
 Forward Biasing: When the p-side is connected current increases after a certain voltage and
to the positive terminal and n-side is connected the voltage is called breakdown voltage. This
to the negative terminal of a battery, it is called property of zener diodes is used in regulating
forward biasing. Majority charge carriers cause voltage.
forward current flow in this biasing and the
width of the depletion layer decreases.
 Reverse Biasing: When the n-side is connected
to the positive terminal and p-side is connected
Fig.: Zener diodeZener diode
to the negative terminal of the battery then it is  In photodiodes, photons are excited which
called reverse biasing. Minority charge carriers result in change of reverse saturation current to
cause reverse current flow in this biasing and measure light intensity.
width of the depletion layer increases. hv
• Junction diode as rectifier: By applying
alternating voltage across a diode the current
flows in only that part if the diode is forward
biased and by using this property diode could be
used to design a circuit which can be used as a A
rectifier.
p-side n-side
Transformer A X

Primary Fig.: An illuminated photodiode
Secondary RL
~  In light emitting diodes, electrons are excited
by a biasd voltage resulting in generation of light.
B Y
 In solar cells, emf is generated when solar
Fig.: Half-wave rectifier circuit using diode
Half wave rectifier circuit using diode
Centre-Tap radiation falls on the p-n junction. It works on
Transformer the principle of photovoltaic effect.
Diode 1(D1)
I1
Centre A
X
Tap
B
Diode 2(D2) RL Output
Y
Fig.: Full-wave rectifier circuit using diode
Depletion
• Diodes: Ac voltage can be restricted to one layer
direction using diodes. Some examples of p-n Fig.: Typical p-n junction solar cell
9. With the help of a circuit diagram, explain

the working of a junction diode as a full wave
rectifier. Draw its input and output waveforms.
PREVIOUS YEARS’ Which characteristic property makes the
junction diode suitable for rectification?
EXAMINATION QUESTIONS [ALL INDIA 2015]
10. (i) Distinguish between a conductor and a
TOPIC 1 semiconductor on the basis of energy band
2 Mark Questions diagram.
1. Name the semiconductor device that can be used (ii) The following figure shows the input
to regulate an unregulated dc power supply. With waveforms (A, B) and the output waveform
the help of I-V characteristics of this device, (Y) of a gate. Identify the gate, write its truth
explain its working principle. table and draw its logic symbol.
[DELHI 2011]
2. The current in the forward bias is known to be A
more (∼ mA) than the current in the reverse bias
(∼ μA). What is the reason, then, to operate the
photodiode in reverse bias? B
[ALL INDIA 2012]
3. Distinguish between ‘intrinsic’ and ‘extrinsic’ Y
semiconductors.
[DELHI 2015]
0 1 2 3 4 5 6 7
4. Explain, with the help of a circuit diagram, the
working of a p-n junction diode as a half-wave [DELHI 2016]
rectifier. 11. (a) In the following diagram, is the junction
[ALL INDIA 2014] diode forward biased or reverse biased ?
5. Draw the circuit diagram of an illuminated
photodiode in reverse bias. How is photodiode +5V
used to measure light intensity?
[DELHI 2018]
3 Mark Questions (b) Draw the circuit diagram of a full wave
6. Draw V–I characteristics of a p-n junction diode. rectifier and state how it works.
Answer the following questions, giving reasons: [ALL INDIA 2017]
(i) Why is the current under reverse bias almost 12. (a) A student wants to use two p-n junction
independent of the applied potential up to a diodes to convert alternating current into
critical voltage ? direct current. Draw the labelled circuit
(ii) Why does the reverse current show a sudden diagrams he would use and explain how it
increase at the critical voltage Name any works.
semiconductor device which operates under (b) Give the truth table and circuit symbol for
the reverse bias in the breakdown region NAND gate.
[ALL INDIA 2013] [ALL INDIA 2018]
7. Write any two distinguishing features between
conductors, semiconductors and insulators on
the basis of energy band diagrams. 4 Mark Questions
[DELHI 2014] 13. Meeta’s father was driving her to the school. At
8. With what considerations in view, a photo diode the traffic signal she noticed that each traffic
is fabricated? State it’s working with the help of light was made of many tiny lights instead of a
a suitable diagram. Even though the current in single bulb. When Meeta asked this question to
the forward bias is known to be more than in the her father, he explained the reason for this.
reverse bias, yet the photo diode works in reverse Answer the following questions based on above
bias. What is the reason? information:
[DELHI 2015]
(i) What were the values displayed by Meeta

and her father? Solutions
(ii) What answer did Meeta’s father give? 1. A Zener diode is a specially designed diode
which is operated in reverse breakdown region
(iii) What are the tiny lights in traffic signals
continuously with any damage. When Zener
called and how do these operate?
diode is operated in the reverse break down
[DELHI 2016]
region, the voltage across it remains practically
5 Mark Questions constant (V z) for a large change in reverse
14. (a) Draw the circuit diagram of a full wave current. Therefore, for any increase/decrease of
rectifier using p-n junction diode. Explain the input voltage there is a increase/decrease
its working and show the output, input of the voltage drop across series resistance (Rs)
waveforms. without any change in the voltage across Zener
diode [½]
(b) Show the output waveforms (Y) for the
Current
following inputs A and B of
(i) OR gate
(ii) NAND gate
Vz
t1 t2 t3 t4 t5 t6 t7 t8
Voltage
A
[½]
2. The photodiode always work under reverse
B biasing conditions although the current produced
is less. This is because in reverse bias, the width
of the depletion layer increases which reduces
[ALL INDIA 2012] the capacitance across the junction, there by
15. (a) State briefly the processes involved in the increasing response time. The sensitivity of a
formation of p-n junction explaining clearly photodiode is thus very high, a property that is
how the depletion region is formed. certainly desired. [1]
(b) Using the necessary circuit diagram, show 3.
how the V-I characteristics of a p-n junction Intrinsic Extrinsic
are obtained in semiconductor semiconductor
(i) Forward biasing It is pure semi- It is prepared by doping
(ii) Reverse biasing conducting material a small quantity of
How these characteristics are made use of in with no impurity atoms impurity atoms to the
added to it. pure semiconductor.
rectification? [DELHI 2014]
The number of free electrons The number of free
16. How is a zener diode fabricated so as to make it a
in the conduction band electrons and holes is
special purpose diode? Draw I-V characteristics
and the number of holes never equal. There is
of zener diode and explain the significance of
in valence band is exactly an excess of electrons in
breakdown voltage. Explain briefly, with the
equal. n-type semiconductors
help of a circuit diagram, how a p-n junction
and an excess of holes in
diode works as a half wave rectifier. p-type semiconductors.
[DELHI 2017] Its electrical conductivity is Its electrical conductivity
17. (a) With the help of circuit diagram explain the a function of temperature depends upon the
working principle of a transistor amplifier alone. temperature and the
as an oscillator. amount of impurity added
(b) Distinguish between a conductor, a in them.
semiconductor, a semiconductor and an [1]
insulator on the basis of energy band
diagrams. [DELHI 2018]
4. Half-wave rectifier
A D
S
To mains dc output
P Vi = Vm sin t iL RL
power (ac) (V0)
B
V1 V0
Vm
Vm
Output dc
Input ac
Vdc = 0.318 Vm
T
0 t 0 t
T
T 3T T 3T
2 2 2 2
Fig.: Circuit & Voltage characteristics of half wave rectifier. [½]

During the positive half cycle of the input area. Thus, the greater the intensity of light,
voltage, end A of the secondary is positive and the greater is the number of electron-hole pairs
end B is negative. This polarity makes the diode produced at the junction. The photocurrent
D to be forward biased. The diode conducts and is, thus, directly proportional to the intensity
a current iL flows through the load resistance of light. This can be used for measuring the
RL. Since a forward biased diode offers a very intensity of incident light.
low resistance, the voltage drop across the diode 6. [1]
is also very small. Therefore, the voltage across Ir
RL is almost equal to the voltage viat every

instant of time. During the negative half cycle
Vr
of the input voltage end A of the secondary is
negative and end B is positive. Thus, the diode Vr
Vk
is in reverse bias. The diode does not conduct.
No current flows through RL. Hence, no voltage
is developed across RL. All the input voltage Ir
appears across the diode itself. This explains Fig.: V-I Characteristics of p–n junction diode.
the output waveform. [½] (i) When p-n junction is reverse biased, the
5. The circuit diagram of an illuminated photodiode majority carriers in p and n region are
in reverse bias can be represented as below: repelled away from the junction. [¹]
There is small current due to the minority
hv carriers. This current attains its maximum
or saturation value immediately and is
independent of the applied reverse voltage.
(ii) As the reverse voltage is increased to a
A certain value, called break down voltage,
large amount of covalent bonds in p and
p-side n-side n regions are broken. As a result of this,
large electron-hole pairs are produced which
diffuse through the junction and hence there
[½] is a sudden rise in the reverse current. Once
Fig.: An illuminated photodiode break down voltage is reached, the high
Greater the intensity of light, the greater is the reverse current may damage the ordinary
number of photons falling per second per unit junction diode. Device is zener diode. [½]
7. The band model
Conduction band
Electron
Energy
Energy
Energy
Conduction band
Band gap Conduction band
Band gap
Valence band Valence band Valence band
Insulator Semiconductor Conductor

Fig.: Energy band gaps in insulators, Semiconductor & conductor [1]

p n
Conductors Semicon- Insulators
µA
ductors
(i) Conduction Conduction Conduc-
band is com- and valance tion band Fig.: An illuminated photodiode
pletely filled bands are is empty & I
partially filled valance band I4 > I3 > I2 > I1
(mA)
is completely
filled
R. B.
(ii) Conduction Forbidden Forbidden
and valance energy gap is energy gap I1 V
bands over- of the order of is very large I2
lap 1 eV i.e more than I3
5 eV. I4
I (µA) [½ + ½]
[1]
Fig.:I-VI–V character
character of photodiode for different
of photodiode
8. A photo diode is used to observe the change in for different illumination intensities
illumination intensities
current with change in the light intensity under A photo diode is preferably operated in reverse bias
reverse bias condition. condition. Consider an n- type semiconductor. Its
majority carrier (electron) density is much larger
In fabrication of photo diode, material chosen than the minority hole density i.e. n << p. When
should have band gap —1.5 eV or lower so that illuminated with light, both types of carriers
solar conversion efficiency is better. This is the increase equally in number.
reason to choose Si or GaAs material. n’ = n + Dn; p’ p + Dp
Now, n >> p and Dn = Dp
Working: It is a p-n junction fabricated with a
n p
transparent window to allow light photons to 
n p
fall on it. These photons generate electron hole
That is, the fractional increase in majority carries
pairs upon absorption. If the junction is reverse
is much less than the fractional increase in
biased using an electrical circuit, these electron
minority carriers. Consequently, the fractional
hole pair moves in opposite directions so as to
change due to the photo-effects on the minority
produce current in the circuit. This current is
carrier dominated reverse bias current is more
very small and is detected by the micro ammeter
easily measurable than the fractional change
placed in the circuit. [½]
in the majority carrier dominated forward bias
current. Hence, photo diodes are preferablly
used in the reverse bias condition for measuring
light intensity. [½]
9.
Output waveform Waveform at B Waveform at A

Centre-Tap
Transformer
Diode 1(D1)
t
Centre A
X (i)
Tap
B
Y
(a) t
(ii)
(b)
Fig.: Fill wave rectifier circuit
Due to Due to Due to Due to
(across RL)
D1 D2 D3 D4
t
(c)
Fig.: Circuit diagram, input wave form & output wave form of junction diode as a full wave rectifier. [1]
(a) A Full-wave rectifier circuit; enter (b) Input diode D1 would not conduct but diode D2 would,
wave forms given to the diode D1 at A and to giving an output current and output voltage
the diode D2 at B; (c) Output waveform across (across RL) during the negative half cycle of the
the load RL connected in the full-wave rectifier input ac. Thus, we get output voltage during
circuit. both the positive as well as the negative half
The circuit using two diodes, shown in Fig (a), of the cycle. The diode under forward biased
gives output rectified voltage corresponding to offers negligible resistance so it will conduct
both the positive as well as negative half of the while under reverse biased it offers very high
ac cycle. Hence, it is known as full-wave rectifier. resistance so it will not conduct. Therefore it is a
Here the p-side of the two diodes are connected unidirectional device which conducts only in one
to the ends of the secondary of the transformer. direction. This characteristic property makes the
The n-side of the diodes are connected together junction diode suitable for rectification. [1]
and the output is taken between this common 10.
Electron Energy
point of diodes and the midpoint of the secondary

Conduction band
of the transformer. So for a full-wave rectifier
the secondary of the transformer is provided EV
with a centre tapping and so it is called centre- Eg ≅ 0
tap transformer. As can be seen from Fig.(c) the Valence band
voltage rectified by each diode is only half the
total secondary voltage. Each diode rectifies only
[½]
for half the cycle, but the two do so for alternate Conductor
cycles. Thus, the output between their common
Conduction band
terminals and the centre tap of the transformer
Electron Energy
EC
becomes a full-wave rectifier output. Suppose
the input voltage to A with respect to the centre Eg < 3eV
tap at any instant is positive. It is clear that,
at that instant, voltage at B being out of phase
EV
will be negative as shown in Fig.(b). So, diode
D1 gets forward biased and conducts (while D2 Valence band
being reverse biased is not conducting). Hence,
during this positive half cycle we get an output Semiconductor
current (and a output voltage across the load
Distinguishing features
resistor RL) as shown in Fig.(c). In the course
of the ac cycle when the voltage at A becomes Fig.: Distinguish between conducter, semiconductive
negative with respect to centre tap, the voltage & Insulator.
at B would be positive. In this part of the cycle
(a) In conductors: Valence band and conduction Working: During its positive half cycle of the
band overlap each other. In semiconductors: input AC and diode D1 is forward biased and
Valence band and conduction band are is reverse biased. The forward current flows
separated by a small energy gap. [½] through diode D1.
(b) In conductors: Large number of free During the negative half cycle of the input AC
electrons are available in conduction band. the diode D1 is reverse biased and diode D2 is
In semiconductors: A very small number forward biased. Thus current flows through
of electrons are available for electrical diode Thus we find that during both the valves,
conduction. [½] current flows in the same direction. [½]
(ii) Gate From the given output waveform, it 12. (a) Full wave rectifier:
is clear that output is zero only when both Centre-Tap
inputs are 1, so the gate is NAND gate Transformer
Diode 1(D1)
A B Y
0 0 1 Centre A X
Tap B
0 1 1 Output
1 0 1 Diode 2(D2) RL
1 1 0
Y
Logic Symbol: Y  A  B [½] [½]
A
Output waveform Waveform at B Waveform at A
Y
B t
11. (a)
(i)
+5V

t
Voltage at P side is less than voltage at N side (ii)
of the diode so it is in “Reverse bias”. [½]
(b)
(b) Full wave rectifier: Due to Due to Due to Due to
(across RL)
D1 D2 D1 D2
+ D1
S1
B A t
RL + (c)
S2 Wave: During positive half cycle diode D1 is
–
D2 forward bias and during negative half cycle diode
For positive half cycle D2 is forward bias so due to conduction of diode
D1 D positive half will appear and due to conduction
–
S1 of D2 negative half cycle will appear. [½]
B A
(b) NAND gate:
RL +
S2
+ A
D2
[½] Y = A.B
B
The input and output wavefront have been given
in diagram. In general full wave rectifier is used
to convert AC into DC. [½]
Input Output During the positive half of the input signal,

suppose P1 and P2 are negative and positive
A B Y  A B respectively. This would mean that S1 and S2 are
0 0 1 positive and negative respectively. Therefore,
the diode D1 is forward biased and D2 is reverse
0 1 1 biased. The flow of current in the load resistance
1 0 1 RL is from A to B. [1]
1 1 0 Output waveforms (Y)
[½] t1 t2 t3 t4 t5 t6 t7 t8
13. (i) Awareness for energy conservation, power
saving and knowledge about traffic rules. [1]
(ii) Meeta’s father said that these are LED A
lights which consume less power and high
reliability. [1]
(iii) The tiny lights in traffic signals are Light B
Emitting Diode. These are operated by
connecting the P-N junction diode in forward
OR Output
biased condition. [2]
14. (a)
D1 D1 NAND Output
+
S1 + S1
P1 B A P1 B A
+ – [1]
P2 RL P2 RL
+ S2 15. (a) As we know that n-type semi-conductor
S2
+ has more concentration of electrons than that
D2 D2 of a hole and p-type semi-conductor has more
concentration of holes than an electron. Due
Input voltage
to the difference in concentration of charge

+ + carriers in the two regions of p-n junction, the
T 2T holes diffuse from p-side to n-side and electrons
T 3T diffuse from n-side to p-side.
2 2 When an electron diffuses from n to p, it leaves
behind an ionized donor on n-side. The ionised
Time donor (+ve charge) is immobile as it is bonded
by the surrounding atoms. Therefore, a layer of
positive charge is developed on the n-side of the
Output voltage
junction. Similarly, a layer of negative charge is

+ + + + developed on the p-side. [1]
Junction
T T 3T 2T Hole Electron
2 2
Time p n
[2]
Full Wave Rectifier: p region Depletion n region
When the diode rectifies the whole of the AC region
wave, it is called full wave rectifier. Hence, a space-charge region is formed on both
The figure shows the arrangement for using side of the junction, which has immobile ions
diode as full wave rectifier. The alternating and is devoid of any charge carrier, called as
input signal is fed to the primary P1P2 of a depletion layer or depletion region. [1]
transformer. The output signal appears across
the load resistance RL. [1]
(b) (i) p-n junction diode under forward bias (ii) The p-n junction under reverse bias Positive
terminal of battery is connected to n-side and
negative terminal to p-side.
Reverse bias supports the potential barrier.
mA + V Therefore, the barrier height increases and the
+ width of depletion region also increases. Due
to the majority carriers, there is no conduction
across the junction. A few minority carriers cross
the junction after being accelerated by high
reverse bias voltage.
+
+ µA V +
A p-n junction diode is said to be forward biased

if the positive terminal of the external battery B
is connected to p-side and the negative terminal
to the n-side of p-n junction.
The applied voltage of battery mostly drops
across the depletion region and the voltage drop
across the p-side and n-side of the p-n junction
in negligible small. The resistance of depletion
region is very high as it has no free change
carriers. [1]
Electron in n-region moves towards the p-n Reverse Biasing
junction and holes in the p-region move towards This constitutes a current that flows in opposite
the junction. The width of the depletion layer direction, which is called reverse current.
decreases and hence, it offers less resistance.
The V-I characteristics of p-n junction diode in
Diffusion of majority carriers takes place across
reverse bias is shown on previous page:
the junction. This leads to the forward current.
Reverse Bias (V)
The V-I characteristics of p-n junction is forward
bias is shown below: current (µA)
Reverse
current (mA)
Forward
p-n junction diode is used as a half-wave

rectifier. Its working is based on the fact that
the resistance of p-n junction becomes low when
forward biased and becomes high when reverse
biased. These characteristics of diode are used
Forward
in rectification. [1]
voltage (V) [1]
16. In an oscillator, a sustained A.C. output is

I(mA) obtained without any input oscillation. For
this to happen, the output of a transistor
amplifier is fed back into its input. This
is achieved by coupling the winding T1 to
winding T2.
Reverse bias When key S1 is closed, the collector current
VZ Forward bias begins to increase, which supports the
forward bias of the emitter-base circuit.
V(V) Collector current increases until it reaches
saturation. When the saturation is reached,
the magnetic flux linked to winding T 1
becomes steady. Hence, the forward bias
of the emitter-base circuit is no longer
supported. The transistor is now driven
I(µA) into cut-off. This cycle repeats itself and an
oscillating output is obtained. [2]
I–V Characteristics [1] (b) The energy-band diagram of a conductor is
Zener has a sharp breakdown voltage and this shown as below:
property of zener is used for voltage regulation. Overlapping
Electron energies conduction band
An ac current has a positive half cycle and a
negative half cycle. A p-n junction allows current
to pass only in one direction and that is when it (Eg = 0)
is forward biased. EV
When a positive half-cycle occurs, the p-side has
EC Valence
a lower potential. Therefore, the diode is now
forward biased and therefore conducts and this band
positive cycle is available for the load. [1] [2]
When a negative half cycle occurs, the n-side has The energy-band diagram of a semiconductor
a higher potential than the p-side. Hence, the is as below:
diode is now reverse biased and thus, does not
Electron energies
conduct. As a result, this positive half cycle also

does not conduct. Therefore, it does not appear EC
at the load and is cut-off. [1]
Eg < 3eV
We obtain a waveform, which has only positive
half cycles and therefore it is called half-wave EV
rectifier.
17. (a) The circuit diagram for a transistor amplifier

as an oscillator is represented as:

T1
1 The energy-band diagram of an insulator is
2 Mutual inductance as below:
(Coupling through Empty
magnetic field) conduction
3
C
band
Electron energies
n-p-n T2 EC
T2¢ Output
L
4 Eg > 3eV
S1(Switch) EV
[1] Valence
band
[1]

[Topic 2] Transistors, its application and logic gates
Summary  ∆V BE 
ri = 
• A thin layer of one type of semiconductor is added  ∆I B V
CE
between two thick layers of other semiconductor
 Output resistance is the ratio of change in
of same type and this forms a transistor.
collector emitter voltage to the change in collector
It can be done in two ways, i.e. adding a
current at a constant base current and is given by
p-type semiconductor between two n-type
semiconductors forming n-p-n transistor or by  ∆V 
r o =  CE 
adding an n-type semiconductor between two  ∆I C  I
B
p-type semiconductors forming p-n-p transistor.
Emitter Base Collector Emitter Base Collector
• Current gain: There are two low current gains
defined as follows:
E n p n C E p n p C  Common base current amplification factor
(a) : Ratio of the small change in collector current
to the small change in emitter current at constant
B
n-p-n transistor
B
p-n-p transistor collector-base voltage.
δ I C 
 Any transistor has 3 parts: Base (central block), α = 
Emitter and Collector (two electrodes). Therefore  δ I E VCB = const ant

the three parts of the transistor can be connected
 Common emitter current amplification
in three ways: Common Emitter (CE), Common
factor (b): Ratio of the small change in collector
Collector (CC) and Common Base (CB).
current to the small change in base current at
 For fixed IB, the plot between IC gives output
constant collector-emitter voltage.
characteristics and for fixed VCE, the plot between
IB and VBE gives input characteristics. δ I C 
β = 
• Common emitter transistor: The input is  δ I B VCE = const ant

between the base and the emitter and output is
β
between the collector and the emitter. Terms a and b are related as: α = and
IC 1+ β
+ α
mA β =
IB R1 1−α
C +
B
R2 + • A transistor can be used as an amplifier to
µA E VCE
VCC increase voltage, current or power. Voltage gain
IE
VBB VBE of an amplifier can be defined as the ratio of small
change in output voltage to small change in input
voltage. Ratio of the small change in collector
current to the small change in base current

n-p-n transistor in CE configuration at constant collector-emitter voltage is called
 Input resistance is the ratio of change in base current gain.
emitter voltage to the resulting change in base Voltage gain of amplifier is given by,
current at constant collector emitter voltage and β R 
Av = −  ac L 
is given by  r 
described as OR, AND, NOT, NAND, and NOR.

IC Different logic gates are integrated in a single
RB C RC
B chip called Integrated circuits (IC).
~ V0
IB E VCC  Boolean expression for OR gate: Y = A + B
IE
V1 ~ VBB
A
Y
Fig.: C-E transistor amplifier B

• A transistor can be used as a switch by analyzing The truth table for OR gate is shown below:
the behavior of the base-biased transistor in CE A B Output Y
configuration. When transistor works as a switch
0 0 0
a low input to the transistor gives high output
0 1 1
and a high input gives a low output. In this case
1 0 1
the transistor does not remain in active state.
1 1 1
IC
RB RC  Boolean expression for AND gate: Y = A.B
C
B +
A
~ v0
IB E VCC Y
IE
v1 ~ VBB
B
The truth table for AND gate is shown below:

Fig.: Base-biased transistor in CE configuration A B Output Y
to work as a switch 0 0 0
• Transistor oscillator: When we get ac output 0 1 0
without any external input signal then the 1 0 0
transistor works as an oscillator. 1 1 1
T1  Boolean expression for NOT gate: Y = A
1
2 Mutual inductance
(Coupling through
magnetic field) A Y
3
C T2
n-p-n T2¢ Output
L
4 The truth table for NOT gate is shown below:
A Output Y
S1(Switch)
0 1
Fig.: Tuned collector oscillator
1 0
Frequency at which the oscillator will work is
• Integrated circuits: When a entire circuit
1
given by, v = (including all passive components and active
2π L C
devices) is fabricated on a single chip or block of
• Logic gates are digital circuits which perform a semiconductor then it is known as integrated
special; logic operations. These logic gates can be circuit(IC). The most widely used technology for
making IC’s is monolithic Integrated circuit.
8. Write the truth table for the combination of the

gates shown. Name the gates used.
PREVIOUS YEARS’
EXAMINATION QUESTIONS A
TOPIC 2 A
R
S Y
1 Mark Questions B Y′
1. In a transistor, doping level in base is increased Identify the logic gates marked ‘P’ and ‘Q’ in
slightly. How will it affect (i) collector current the given circuit. Write the truth table for the
and (ii) base current? combination.
[ALL INDIA 2011] A
P
2 Mark Questions B Q X
2. Draw the output waveform at X, using the given
inputs A and B for the logic circuit shown below.
Also, identify the logic operation performed by
this circuit.
[DELHI 2014]
A 3 Mark Questions
9. The following figure shows the input waveforms
Y X (A. B) and the output waveform (Y) of a gate.
B Identify the gate, write its truth table and draw
its logic symbol.
t1 t 2 t 3 t 4 t 5 t 6 t 7 A
[ALL INDIA 2011]
B
3. Draw the transfer characteristic curve of a
transistor in CE configuration. Explain clearly Y
how the active region of the V0 versus Vi– curve
in a transistor is used as an amplifier.
0 1 2 3 4 5 6 7
[ALL INDIA 2011]
4. Draw the transfer characteristic curve of a base [DELHI 2017]
biased transistor in CE configuration. Explain 10. Output characteristics of a n-p-n transistor
clearly how the active region of the V0 versus Vi in CE configuration is shown in the figure.
curve in a transistor is used as an amplifier. Determine:
[ALL INDIA 2011] (i) Dynamic output resistance
5. Draw typical output characteristics of an n-p-n (ii) DC current gain and
transistor in CE configuration. Show how these (iii) AC current gain at an operating point VCE =
characteristics can be used to determine output 10 V, when IB = 30 μA.
resistance. [ALL INDIA 2013]
6. In the circuit shown in the figure, identify the 6 50 µA
equivalent gate of the circuit and make its truth table 5
40 µA
A A′
30 µA
Y′ 4
Y 3.5
3 20 µA
IC(mA)
B B′ 2 10 µA
[ALL INDIA 2013] 1 IB = 0
7. Draw a circuit diagram of n-p-n transistor
amplifier in CE configuration. Under what 0
0 2 4 6 8 10 12 14 16 18
condition does the transistor act as an amplifier?
VCE(V)
[DELHI 2014]
[DELHI 2013]
11. Draw a circuit diagram of a transistor amplifier 15. Draw the typical input and output characteristics
in CE configuration. Define the terms: of an n-p-n transistor in CE configuration.
(i) Input resistance and (ii) Current amplification Show how these characteristics can be used to
factor. How are these determined using typical determine (a) the input resistance (ri), and (b)
input and output characteristics? current amplification factor.
[DELHI 2015] [ALL INDIA 2018]
12. The outputs of two NOT gates are fed to a NOR 5 Mark Questions
gate. Draw the logic circuit of the combination
16. Draw a simple circuit of a CE transistor amplifier.
of gates. Write its truth table. Identify the gate
Explain its working. Show that the voltage gain,
equivalent to this circuit.
b R
Or Av, of the amplifier is given by Av = − ac L ,
ri
You are given circuits (a) and (b) as shown in the where bac is the current gain, RL is the load
figures, which consists of NAND gates. Identify resistance and r i is the input resistance of
the logic operation carried out by the two. Write the transistor. What is the significance of the
the truth tables for each. Identify the gates negative sign in the expression for the voltage
equivalent to the two circuits. gain?
[ALL INDIA 2012]
A
(a) Y 17. (a) Differentiate between three segments of a
B
transistor on the bias of their size and level
of doping.
(b) How is a transistor biased to be in active
(b) Y state?
(c) With the help of necessary circuit diagram,
[ALL INDIA 2015] describe briefly how n-p-n transistor in CE
13. (i) Write the functions of three segments of a configuration amplifies a small sinusoidal
transistor. input voltage. Write the expression for the
(ii) Draw the circuit diagram for studying the ac current gain.
input and output characteristics of n-p-n [DELHI 2014]
transistor in common emitter configuration. 18. (i) Draw a circuit diagram to study the input and
Using the circuit, explain how input, output output characteristics of an n-p-n transistor
characteristics are obtained. in its common emitter configuration. Draw
[DELHI 2016] the typical input and output characteristics.
14. (a) Write the functions of the three segments of (ii) Explain, with the help of a circuit diagram,
a transistor. the working of n-p-n transistor as a common
(b) The figure shows the input waveforms A and emitter amplifier. [DELHI 2017]
B for ‘AND’ gate. Draw the output waveform 19. (a) Explain the formation of depletion layer and
and write the truth table for this logic gate. potential barrier in p-n junction.
t1 t2 t3 t4 t5 t6 t7 t8 (b) In the given figure below the input waveform
is converted into the output waveform by
a device ‘X’. Name the device and draw its
circuit diagram.
A
(Input)
Device ‘X’
B Input Output
(c) Identify the logic gate represented by the
circuit as shown and write its truth table.
A
[ALL INDIA 2017] Y
B

[DELHI 2018]
If we plot Vo, Vs, Vi, we get a graph as shown in

Solutions figure; this characteristic curve is also called
1. If the doping level in the base of a transistor is transfer characteristic curve of a base biased
increased: transistor in CE configuration.
(i) Collector current will decrease. [½] The curve shows that there are three non-linear
(ii) Increased base doping lowers its resistance regions:
hence its base current should increase. [½] (i) between cut-off stage and active stage
2. The truth table will be (ii) between active stage and saturation
stage; [½]
A B Y X
For using the transistor as an amplifier we
0 0 1 0 will use the active region of the Vo versus Vi
1 0 0 1 curve. The slope of the linear part of the curve
1 1 0 1 represents the rate of change of the output with
the input. It is negative because the output is
0 1 0 1
Vcc − I c Rc and c not Ic R. That is why as input
0 0 1 0 voltage of the CE amplifier increases its output
1 0 0 1 voltage decreases and the output is said to
[½] be out of phase with the input. If we consider
DVo and DVi as small changes in the output
The output waveform will be
DVo
and input voltages then DV is called the
i
small signal voltage gain AV of the amplifier. If

the VBB voltage has a fixed value corresponding
[½] to the midpoint of the active region, the circuit
will behave as a CE amplifier with voltage gain
3.
DVo
DVi . We can express the voltage gain Av in
Cut of Active
regionregion
Vi
terms of the resistors in the circuit and the
Saturation current gain of the transistor as follows.
region We have, Vo  Vcc  I c Rc
Therefore, Vo  0  Rc I c
Ar Similarly, from Vi = I B RB + VBE
∆Vi = RB∆I B + ∆VBE

Vi But DVBE is negligibly small in comparison to
[½] DIB RB in this circuit
In the active region, a (small) increase of Vi So, the voltage gain of this CE amplifier is given
results in a (large, almost linear) increase in Ic. by
This results in an increase in the voltage drop RC ∆IC
across Rc. [½] AV = −
RB∆I B
4. Cutoff Active
region region R 
= −bac  c (14.14)
 RB 

Vo Saturation DIC
region Where bac is equal to DI [½]
B
Thus the linear portion of the active region of
the transistor can be exploited for the use in
amplifiers.
0.6V 1V Vf
5. OR
I0 = (IC) P-NAND GATE
Q-OR GATE
A B AB AB + B
∆I0
0 0 1 1
0 1 1 1
1 0 1 1
∆V0 V0 (= VCE)
[½] 1 1 0 1
Vo [1]
output resistance  [½]
I o 9. The gate is the NAND gate.
A
Y
6.
Input Output
1 0 0 B [1]
0 1 0 Input Output
0 0 0 0 0 1
1 1 1 0 1 1
[1] 1 0 1
7. 1 1 0
IC C2
P [1]
C1 C 10. (i) Dynamic output resistance is given as:
IBE B
VCE  ∆VCE 
Ro =  
VBE RC  ∆lC 
RB E Ibconstant
V0 12 − 8 4
= =
−3
V1 ~ (3.6 − 3.4)×10 0.2 ×10−3
VBB [1]
VCC Ro = 20 KW
(ii) DC current gain,
Q [½]
The condition necessary for the amplifier to IC 3.5 ×10−3 350
bdc = = =
work is that the base emitter junction should
IB 30 ×10−6 3
[½]
be forward biased and collector base junction
(iii) AC current gain,
should be reverse biased. [½]
8. ∆IC (4.7 − 3.5)×10−3 1.2 ×10−3
bac = = =
R-OR GATE
∆I B (40 − 30)×10−6 10 ×10−6 [½]
S-AND GATE bac = 120

A B Y’ = A+B Y = Y’.A 11.
C iC O
0 0 0 0 iB B
0 1 1 0 n-p-n
E
1 0 1 1 VCE RL
iC
1 1 1 1
Input VCC Output
[1] a.c. Signal + a.c. Signal
iB iC [½]
(i) Input resistance may be defined as the ratio 12.

of small change in the base-emitter voltage Y1
(DVBE) to the resulting change in the base A
current (IB) at constant collector-emitter Y
voltage (VCE) [½] B
 VBE  Y2
Input resistance, Ri  
 I B  V
CE
Y1 = A
800
Y2 = B
700
Y  Y1  Y2  A  B  A  B  A  B
600
The equivalent gate is AND gate.
500
Truth table
400
A B Y1 Y2 Y
300 P 0 0 1 1 0
200 0 1 1 0 0
C 1 0 0 1 0
100 B
1 1 0 0 1
0 0.20 0.40 0.60 0.80 1.00 1.20 Or
VBE (V) (b) Y1  A  A  A

To find the input resistance mark a point Y2  B  B  B
P on the input characteristic. Now, draw a
tangent at point P. The reciprocal of slope Y  Y1 Y2  A  B  A  B  A  B
of AB will give the input resistance.
The equivalent gate is OR gate
 I  Truth table
(ii) Current amplification factor r   C  is
 I  B VCE
A B Y1 Y2 Y
defined as the ratio of change in collector 0 0 1 1 0
current (DI C ) to the change in base
current. [½] 0 1 1 0 1
1 0 0 1 1
1 1 0 0 1
13. (i) Three segments of transistor are:
Base current (IB)
10 (1) Emitter (2) Base
60µA (3) Collector
8
50µA
Emitter: It is of moderate size and heavily
6 40µA doped, it supplies a large number of majority
30µA carriers which flow through the transistor.
4
20µA
Base: It is very thin and lightly doped and
2 10µA it separates emitter and collector region of
transistor and controls the flow of charge
0 2 4 6 8 10 12 14 16
carriers.
Collector to emitter voltage
(VCE) in volts Collector: This segment is moderately doped
[½] and larger in size as compared to emitter. It
collects a major portion of majority carriers
supplied by the emitter. [½]
(ii) Base: Base is the control segment and it is very

thin and lightly doped. [½]
C +
IB mA Collector: It is the segment that collects major
R1 portion of the majority carries supply by the
+
R0
+ emitter. It is moderately doped and large in size
mA B VCE VCC as compared to the emitter. Input of AND gate is
E
VBB VBE Y = A.B in this case output, will be 1 only when
IB IC both inputs are 1 [½]
(b)

Circuit to study characteristics of a High
transistor
A low
For input characteristics, base current Input
IB VS base emitter voltage VBE is plotted
High
while collector base voltage V CE is kept
B low
constant. VCE is kept large 3V to 20V. Input
characteristics for various values of VCE give
High
almost same curves.
X low
Input Characteristics (output)
100 0 1 2 3 4 5 6 7
IB [½]
(mA) 80
Input Output
60 VCE = 10V A B Y = A.B
0 0 0
40
0 1 0
20 1 0 0
VBE(V) 1 1 1
0
0.2 0.4 0.6 0.8 1.0
VBE~= 0.7V [½]
[½]
15.
Output characteristics are obtained by varying Ia/µA
IC with VCE keeping IB constant. Different curves
are obtained for different values of IB.
A
100 VCE = 10.0 V
B
80
60
Y
40
7 [½]
20
0 1 2 3 4 5 6
14. (a) These segments of a transmitter are called 0.2 0.4 0.6 0.8 1.0 VBE / V
emitter (E), Base (B) and collector (C) Emitter: (a) [½]
It is of moderate size and heavily doped. It
supplies a large number of majority carrier for
the current flow through the transistor
Where, bac = ac current gain

RL = Load resistance
r = RB + ri
10 Base current (IB) ri = input resistance
60µA RB = Base resistance
8 Current gain of the transistor will decrease if
50µA
the base is made thicker because current gain,
6 40µA Ic
30µA b=
4 Ib
20µA [1]
2 If the base of an n-p-n transistor is made thicker,
10µA
then more and more electrons will recombine
with the p-type material of the base. This
0 2 4 6 8 10 12 14 16
results in a decrease in collector current Ic.
Collector to emitter voltage (VCE) in volts Furthermore, Ib also increases.
(b) Ic
[½] Hence, ac current gain b = I decreases.
b
The above plots show the typical input output Finding expression for voltage gain of the
characteristics of an n-p-n transistor in CE amplifier:
combination. Applying Kirchhoff’s law to the output loop,
 ∆VBE  VCC = VCE + IC RC
Ri =  
 ∆I 
B VCC = VCE + IC RC
VCE
[½] VCC = VCE + IC RC
here VBE is voltage across base emitter, Ib is base VBB = VBE + IB RB
current vi ≠ 0
 V 
Ro   CE  Then, VBB  vi  VBE  I B RB  I B  RB  ri 
 IC  I  V 
B
[½]  ri   BE 
here VCE is voltage across collector emitter and  I B  V
[1]
CE
Ic is collector current.
 vi  I B  RB  ri   r I B
 ∆I 
bac =  C  ∆Ic i
 ∆I B  bac = = c
VCE
[½] ∆I B ib
16. Circuit diagram of CE transistor amplifier: It is a current gain denoted by Ai
Change Ic due to change in IB causes a change in
IC VCE and the voltage drop across resistance Rc,
RB C RC because VCC is fixed.
B
VCC  VCE  RC IC  0
+
IB E
V0
VCC
V1 VBB
IE VCE   RC IC
Change in VCC is the o/p voltage Vo.
Voltage gain of amplifier,
V VCE
AV  o 
[1] Vi r I B
R
Working: If a small sinusoidal voltage is applied C
= −bac =
r
to the input of a CE configuration, the base (Negative sign represents that o/p voltage is in
current and collector current will also have opposite direction to i/p voltage. [1]
sinusoidal variations. Because the collector 17. (a) Emitter (E): It is the left hand side thick layer
current drives the load, a large sinusoidal of the transistor, which is heavily doped.
voltage vo will be observed at the output. Base (B): It is the central thin layer of the
The expression for voltage gain of the transistor transistor, which is lightly doped. [1]
in CE configuration is:
ν −βac RL
Aν = o =
ν2 ri
[1]
Applying Kirchhoff’s law to the output loop:

IE = IB + IC
E P n P C If VCE is the collector voltage then,
VCE = VCC – IC RL [1]
When the input signal voltage is fed to the
B
emitter base circuit, it will change the emitter
base voltage and hence the emitter current,
which in turn will change the collector current.
Due to this the collector voltage VCE will vary in
E C
accordance with relation (A). This variation in
collector voltage appears as amplified output.
B Let Ri be the input resistance of emitter-base
p-n-p transistor
circuit. When an input voltage Vi is applied to the
emitter base circuit, let DIB, DIC be the change
and collector current respectively.
E n P n C Then, Vi = DIB × Ri ∆IC
The a.c. current gain of transistor is, bac = ∆I
B
B 18. (i)
IC
RB C RC
B
IB
~ V0
E VCC
E C
V1
~ IE
VBB
B
n-p-n transistor [1]
Collector (C): It is the right hand side thick
IB/µA
layer of the transistor, which is moderately
doped.
(b) There are two conditions for a transistor to
100
be into an active region. VCE = 10.0 V
(1) The input circuit should be forward biased
80
by using a low voltage battery.
(2) The output circuit should be reverse biased
60
by using a high voltage battery.
(c) n-p-n transistor as an amplifier:
40
The operating point is fixed in the middle of
its active region. [1] 20
C
IB B IC
VBE / V
0.2 0.4 0.6 0.8 1.0 [1]
npn E RL
VBE
IC
VCE
Vi ~ IE
10 Base current (IB)
VCE 60µA
8
50µA
VBB 6 40µA
30µA
4
20µA
Input a.c. voltage Amplified out put 2 10µA
An a.c. signal v1 is superimposed on bias VBB(dc).
0 2 4 6 8 10 12 14 16
The o/p is taken between the collector and the
Collector to emitter voltage (VCE) in volts
ground.
(c) n-p-n transistor as an amplifier: Similarly, a positive space-charge region is

The operating point is fixed in the middle of formed on the n-side. These two space-charge
its active region. [1] regions on either sides of the junction constitute
C what is called a depletion layer. Since the n-side
IB B IC loses electrons and p-side gains electrons, a
potential difference is developed across the
npn E RL
junction of the two regions. This potential
VBE difference tends to oppose further motions of
IC
VCE electron from the n-region into the p-region. The
Vi ~ IE same happens for holes too. The reverse polarity
VCE of this potential opposes further flow of carriers
and is thus called the barrier potential. [2]
VBB (a) The device is a full-wave rectifier. The
circuit diagram of a full-wave rectifier is
represented as:
Centre-Tap
Input a.c. voltage Amplified out put Transformer
Diode 1(D1)
An a.c. signal v1 is superimposed on bias VBB(dc).
The o/p is taken between the collector and the Centre A
X
ground. Tap
B
Applying Kirchhoff’s law to the output loop:
IE = IB + IC
If VCE is the collector voltage then, Y
(a)
VCE = VCC – IC RL [1]
When the input signal voltage is fed to the [1]
emitter base circuit, it will change the emitter (b) The logic gate represented by the circuit is
base voltage and hence the emitter current, an AND gate. The truth table of the AND
which in turn will change the collector current. gate is represented as:
Due to this the collector voltage VCE will vary in A B A.B
accordance with relation (A). This variation in 0 0 0
collector voltage appears as amplified output.
1 1 1
Let Ri be the input resistance of emitter-base
circuit. When an input voltage Vi is applied to the 1 0 0
emitter base circuit, let DIB, DIC be the change 0 1 0
and collector current respectively. 0 [1]
Then, Vi = DIB × Ri (c) A Y1
∆IC Y
The a.c. current gain of transistor is, bac = ∆I B
B
19. In a p-n junction, a p-type and an n-type material Y1 = Error! Objects cannot be
are joined together. The concentration of holes created from editing field codes.
is higher in p-type material as compared to   
Y  A  B  A  B   A  B   A  B   A  B   A  B
that in n-type material. Therefore, there is
Y = A. B
a concentration gradient between the p-type
The equivalent gate is AND Gate.
and n-type materials. As a result of this
Truth table
concentration gradient, holes move from p-side
to n-side (p → n) by the process of diffusion. A B Y1 Y
Similarly, electrons move from n-side to p-side 0 0 1 0
(n → p). [1]
0 1 1 0
As the holes diffuse from p-side, they leave
ionized spaces (negatively charged) on p-side 1 0 1 0
near the junction. These ionized spaces are 1 1 0 1
immobile. Hence, a negative space-charge
region is formed on the p-side near the junction.
CHAPTER 15
Communication System

Elements of a communi- 1Q
cation system (3 marks)
Bandwidth of signals
1Q
(speech, TV and digital
(2 marks)
data)
Bandwidth of transmis-
sion medium
Propagation of electro-
1Q
magnetic waves in the
(2 marks),
atmosphere, sky & space
1Q
wave propagation, satel-
(3 marks)
lite communication
Need for modulation, am-
plitude modulation and 1Q 1Q
frequency modulation, (1 mark), 1Q 1Q 1Q (2 marks),
advantages of frequency 1Q (1 mark) (3 marks) (3 marks) 1Q
modulation over ampli- (3 marks) (3 marks)
tude modulation
On the basis of above analysis, it can be said that from exam point of view Mode of Communi-
cation for Telephonic Communication, Attenuation and Demodulation, Component of Commu-
nication System, Mode of Communication in Satellite Communication System, Sight Commu-
nication and TV Signals are most important concepts of the chapter.
250 CHAPTER 15 : Communication System
[Topic 1] Communication
Summary
• Communication is a two way process in which exchange of information takes place either in verbal or
written form.
• Elements of communication system: There are three essential elements of communication transmitter,
medium/channel and receiver.
Communication System
Information Message Transmitter Channel Receiver Message user of

Source Signal Signal information
Noise
Transmitter transmits the signals through • Amplification: The process of increasing the
channel which is a physical medium and the amplitude and the strength of a signal using an
receiver receives the signals. electronic circuit is called amplification.
The two basic types of communication modes are • Range: The largest distance between a source
point-to-point and broadcast. and a destination is called range up to which the
signal is received with sufficient strength.
• Transducer: A device which transforms the
energy from one form into another. Example: • Bandwidth: The range of frequency over
Loudspeaker. which an equipment operates or the portion of
the spectrum occupied by the signal is called
• Signal: An information transformed into
bandwidth.
electrical form for suitable transmission is termed
as signal. Signals can be of two types: analog or • Repeater: A combination of transmitter and
digital. receiver is the repeater which amplifies the
signals picked up from the transmitter and then
• Noise: The unwanted signals which have
retransmits those signals to the receiver. In order
a tendency to create the disturbancein the
to extend the range of the communication system,
transmission and processing of message is called
the repeaters are used.
noise.
• Bandwidth of signals: The difference between
• Transmitter: The device that processes the
the upper and lower frequencies of the signals
incoming message signal in order to make it
is termed as bandwidth of signals. The different
suitable for transmission through a channel and
bandwidths of the different kinds of signals is
subsequent reception is known as transmitter.
shown in the following table:
• Receiver: In order to extract the appropriate
Types of Signals Bandwidth
message signals from the received signals at the
channel output, receiver is used. Speech signal 2800 Hz
Music signal 20 KHz
• Attenuation: When signals are propagated
through a medium, some of their strength is lost Video signal 4.2 MHz
which is known as attenuation. TV signal 6 MHz
CHAPTER 15 : Communication System 251
• Bandwidth of transmission medium: Free from few MHz to 40MHz. It uses the phenomenon
space, wire, fibre optic cable and optical fibre are of bending of EM waves so that they are diverted
the common transmission media. The bandwidths towards the earth is similar to total internal
are different for various transmission media. reflection in optics.
• Propagation of Electromagnetic Waves: In • Space wave propagation: For long distance
radio waves communication, the EM waves are transmission, antennas are used to radiatesignals
radiated at the transmitter by antenna. into space. In order to travel from transmitting
• Ground wave propagation: The ground wave antenna to the receiving antenna, space wave
propagation is also termed as surface wave takes the straight line path.They are useful for
propagation.The radio waves are travelled along line-of-sight (LOS) communication and satellite
the earth surface in this type of propagation. It is communication.
necessary for the antenna to be of a size which is • The range dT of an antenna of height hT that
comparable to the wavelength of the signal so that radiates electromagnetic waves is given by
the signals can be radiated with high efficiency. 2RhT ; R = radius of the earth.
As the frequency increases, the attenuation also
increases.
• To find out the maximum distance of line of sight
• Sky wave propagation: It is used for long (dM) between antennas with heights hT and hR:
distance communication in the frequency range =
dM 2RhT + 2RhR

Communication satellite
Space wave
Ionosphere
Los
Los Sky wave

Ground wave
Earth
(i) Receiver
PREVIOUS YEARS’ (ii) Demodulator [CBSE 2014]

10. Explain the terms (i) Attenuation and (ii)
Demodulation used in Communication System.
[CBSE 2016]
TOPIC 1 11. Why is base band signal not transmitted
directly? Give any example [CBSE 2016]
1 Mark Questions 12. Explain the function of a repeater in a
1. The figure given below-shows the block diagram
communication system. [CBSE 2018]
of a generalized communication system. Identify
13. What range of frequencies used in satellite
the element labeled ‘X’ and write its function.
communication? What is common between these
waves and light waves? [CBSE 2018]
X User
Information Transmitter Receiver 3 Mark Questions
Source 14. Draw a schematic diagram showing the (i)
Message Message ground wave (ii) sky wave and (iii) space wave
Signal Signal propagation modes for em waves.
[CBSE 2014] Write the frequency range for each of the
2. How are side bands produced? following:
[CBSE 2015] (i) Standard AM broadcast
3. What is the function of a band pass filter used (ii) Television
in a modulator for obtaining AM signal? (iii) Satellite communication [CBSE 2011]
[CBSE 2015] 15. What is space wave propagation? State the
4. A signal of 5 kHz frequency is amplitude factors which limit its range of propagation.
modulated on a carrier wave of frequency 2 Derive an expression for the maximum line of
MHz what are the frequencies of the side bands sight distance between two antennas for space
produced? wave propagation. [CBSE 2011]
[CBSE 2016] 16. Draw a schematic diagram showing the (i)
5. What is sky wave propagation? ground wave (ii) sky wave and (iii) space wave
[CBSE 2017] propagation modes for em waves.
6. Which mode of propagation is used by short wave Write the frequency range for each of the
broad cast services? following:
[CBSE 2018] (i) Standard AM broadcast
2 Mark Questions (ii) Television
7. Distinguish between ‘Analog and Digital (iii) Satellite communication [CBSE 2011]
signals’. 17. Distinguish between ‘sky waves’ and ‘space
Or waves’ modes of propagation in communication
Mention the function of any two of the following system.
used (a) Why is sky wave mode propagation restricted
i. Transducer to frequencies up to 40 MHz?
ii. Repeater (b) Give two examples where space wave mode
iii. Transmitter of propagation is used.
iv. Bandpass Filter [CBSE 2011] [CBSE 2013]
8. Block diagram of a receiver is shown in the 18. Name the type of waves which are used for
figure: line of sight (LOS) communication. What is
(a) Identify ‘X’ and ‘Y’ the range of their frequencies? A transmitting
(b) Write their functions. antenna at the top of a tower has a height of 45
Receiving Antenna m and the receiving antenna is on the ground.
Calculate the maximum distance between them
Receiving Output for satisfactory communication in LOS mode.
Signal Amplifier X Detector Y (Radius of the Earth = 6.4 ×106m)
[CBSE 2014] [CBSE 20136]
9. Write the functions of the following in 19. Draw a block diagram of a detector for AM signal
communication systems: and show, using necessary processes and the
wave forms, how the original message signal is passed to band pass filter which rejects the dc
detected from the input AM wave. and the sinusoids of frequencies wm, 2wm and 2wc
[CBSE 2015] and retains the frequencies wc, wc – wm, and wc
20. Name the three different modes of propagation + wm. The output of band pass filter is an AM
in a communication system. State briefly why do wave. [1]
the electromagnetic waves with frequency range 4. Given frequency of carrier wave
from a few MHz up to 30 MHz can reflect back
fc = 2 MHz = 2 × 103 KHz
to the earth. What happens when the frequency
range exceeds this limit? [CBSE 2015] Frequency of modulating signal fm = 5 kHz
21. Draw a block diagram of a generalized Frequency of lower side band (LSB) = fc – fm
communication system. Write the functions of LSB = (2 × 103 – 5)kHz [½]
each of the following (a) Transmitter (b) Channel
LSB = (1995) kHz
(c) Receiver.
[CBSE 2017] Frequency of upper side band (USB) fc + fm
22. Why are high frequency carrier waves used for USB = (2 × 103 + 5)kHz
transmission? [CBSE 2017] USB = 2005 kHz [½]
23. By what percentage will the transmission ranges 5. The type of propagation in which radio waves are
of TV tower be affected when the height of the transmitted towards the sky and are reflected
tower is increased by 21%? by the ionosphere towards the desired location
[CBSE 2017] on earth is called sky wave propagation. [1]
24. What is space wave propagation? Give two 6. Sky wave propagation is used by short wave
examples of communication system which use broadcast services having frequency range from
space wave mode. A TV tower is 80 m tall. a few MHz upto 30 MHz. Sky wave can travel
Calculate the maximum distance up to which very long distances and can even travel round
the signal transmitted from the tower can be the earth. [1]
received. [CBSE 2018]
7.
Solutions Analog Signal Digital Signal
1. The element labeled ‘X’ is called ‘channel’. 1. It is continuous It is a type of sig-
The function of the channel is to connect the signal, which varies nal which has only
transmitter and the receiver. A channel may continuously with two values high or
either be wireless or in the form of wires variable may be low. In digital high
connecting the transmitter and the receiver. [1] time or distance mean 1 and low
etc. means 0 (zero)
2. Side bands are produced during the process
of modulation. During modulation the audio 2. Example: Sound of Example: Tem-
frequency modulating signal wave is super human perature of day
imposed on a high frequency wave called [1 + 1]
carrier wave. Any form of modulation produces Or
frequencies that are the sum and difference of
1. Transducer: It is an electric device which
the carrier and modulating frequencies. These
converts energy from one form to another form.
frequencies are called as side bands. [½]
e.g. microphone, which converts sound energy
into electric energy and vice — versa. [½]
fc – fm fc + fm 2. Repeater: It is an electronic device used in
transmission system to regenerate the signal.
It picks up a signal amplifies it and transmits
fc it to receiver. [½]
Lower side band frequency = fc – fm 3. Transmitter: Transmitter is an electronic device
Upper side band frequency = fc + fm which is used to radiate electromagnetic waves.
The purpose of the transmitter is to boost up the
Where, fc → Carrier wave frequency
signal to be radiated to the required power level,
fm → Modulating signal frequency [½] so that it can travel long distances. The most
3. The output produced by square law device is
familiar transmitters are mobile transmitter signal. Sometimes, it also changes the carrier
antennas, radio and T.V. broadcasting antennas frequency of the pick-up signal before transmitting
etc. [½] it to the receiver. [1 + 1]
4. Band pass filter: It is an electronic filter, which 13. The waves used for satellite communication lie
pass the certain band (range) of frequency and in the following two frequency ranges:
reject rest of all. [½] (i) 3.7 – 4.2 GHz for downlink
8. From the given block diagram of demodulator of (ii) 5.9 – 6.4 GHz for uplink [1]
a typical receiver, we can conclude the following
These waves and light waves both are
things:
electromagnetic waves. They both travel in a
(a) X represents Intermediate Frequency (IF) straight line. [1]
stage while Y represents an amplifier. [½]
14. (i) Standard AM Broadcast:
(b) At IF stage, the carrier frequency is
540 – 1600 kHz
transformed to a lower frequency then
in this process, the modulated signal is (ii) Television: 54 – 890 MHz
detected. The function of amplifier is to (iii) Satellite communication: 5.925 – 6.425 GHz
amplify the detected signal which may not uplink and 3.7 – 4.2 GHz downlink [2]
be strong enough to be made use of and Communication satellite
hence is essential. [½]

9. Receiver: Receiver separates the message signal Space wave
from the carrier signal. It reconstructs actual Ionosphere
signal using output transducers. [1]
Demodulator: The process of retrieval of Los
information from the carrier wave at the receiver Los
is called demodulation and Electronic Circuit Ground wave Sky wave
Transmitting Receiving
used for it is called demodulator. [1] Antenna Antenna
Earth
10. (i) Attenuation: The loss of strength of a signal
while propagating through a medium is 0 [1]
known as attenuation. [1] 15. When a wave propagates in a straight line,
(ii) Demodulation: The process of retrieval of from the transmitting antenna to the receiving
information from the carrier wave at the antenna, its mode of propagation is called space
receiver is termed demodulation. This is the wave communication. Frequency range: Above
reverse process of modulation. [1] 40 MHz. Space waves are used for the Line of
Sight (LOS) communication.
11. The needs of modulation for transmission of a
signal are given below Space wave communication involves the
transmission from transmitter, traveling along
(i) The transmission of low frequency signal
a straight line in space, reaches to receiving
needs antenna of height 4-5 km which is
antenna. The range of their frequencies is 40
impossible to construct. So, there is need to
MHz and above.
modulate the wave in order to reduce the
height of antenna to a reasonable height. [1] The range of space wave propagation is limited
by line of sight distance between transmissions
(ii) Effective power radiated by antenna for low
to receiver /repeater antenna. [1]
wavelength or high frequency wave as
P
1 Line of Sight
P [1] h z

C A
So, for effective radiation by antenna, there is
need to modulate the wave. R
R
12. A repeater is used for extending the range of a
communication system. It consists of a receiver
and a transmitter. The receiver of a repeater
collects the signal from the transmitter of another Earth
repeater and after amplifying, it retransmits the
h<R Frequency range:

h = Height of antenna (i) Standard AM broadcast: 540 – 1600kHz
R = Radius of Earth (ii) Television: 54 – 72kHz [1]
Range 76-88 MHz: VHF (very high frequencies)
As h is very small, CA ≈ PA 174-216 MHz
Using Pythagoras theorem 420-890 MHz: UHF (ultra high frequencies)
PA2 + R2 = (R + h)2 (iii) Satellite communication: 5.925-6.425 GHz
(uplink)
PA2 + R2 = R2 + 2Rh + h2
3.7 – 4.2 GHz (downlink) [1]
PA = 2 Rh (neglecting h2)
17. Sky wave: Sky waves are the AM radio waves,
Range = 2Rh on ground surface
which are received after being reflected from the
Transmitter ionosphere. The propagation of radio wave signals
Receiver
T from one point to another via reflection from
A
R ionosphere is known as sky wave propagation.
The sky wave propagation is an important
hr hR consequence of the total internal reflection of
radio waves. As we go higher in the ionosphere,
D there is an increase in the free electron density.
Consequently, there is a decrease of refractive
Total range on ground
index. Thus, as a radio wave travels up in the
 CA  AD  2 Rhr  2 RhR ionosphere, it finds itself travelling from denser
[1] to rarer medium. It continuously bends away
from its path till it suffers total internal reflection
C B C
to reach back the Earth. [1½]
E
Output Space waves: Space waves are the waves which
nB Amcoswmt
Accosct Input (modulating modulated are used for satellite communication and line of
VCC
(carrier wave) signal) wave sight path. The waves have frequencies up to
40 MHz provides essential communication and
VBE limited the line of sight paths. [½]
(a) The e.m. waves of frequencies greater than
Circuit for an amplitude modulator 40 MHz penetrate the ionosphere and escape
[1] so, the sky wave propagation is restricted to
16. the frequencies up to 40 MHz.
Communication satellite (b) In television broadcast and satellite

communication, the space wave mode of
propagation is used. [1]
Space wave 18. High frequency waves (above 40 MHz) called space
Ionosphere waves can be transmitted from transmitting to
receiving antenna and the mode for travelling
of these waves through space is known as space
Los wave propagation. [2]
Los d = 2hRe
Ground wave Sky wave
6
d  2  45  6.4  10
Earth
d = 24000m = 24km [1]

[1]
19.
Received antenna
Received Output
Signal Amplifier X Detector Amplifier
[Block diagram for detection of AM signal] [1]

When a message is received, it gets attenuated through the channel therefore,receiving antenna is to be
followed by an amplifier and a detector. The camera frequency is usually changed to a lower frequency
in an Intermediate Frequency (IF) stage. The detected signal may not be strong enough to be made use
of and hence is required to be amplified.
In order to obtain the original message signal m(t) of angular frequency a simple method is used which
is shown below in the form of a block diagram. [½]
AM wave m(f)
Envelope
Rectifier Output
detector
(a)
(b) (c)
time time time
AM input wave
Output (without RF Rectified wave
component)
[1]
When the received modulated signal is passed occurs due to the absorption of the ultraviolet
through a rectifier, an envelope signal is produced. and other high-energy radiation coming from the
This envelope signal is the message signal. In sun by air molecules. The ionosphere is further
order to retrieve the message, the signal is passed subdivided into several layers. The degree of
through an envelope detector. [½] ionization varies with the height. The density
20. The three different modes of propagation in a of atmosphere decreases with height. At great
communication system are heights the solar radiation is intense but there
are few molecules to be ionized. Close to the
(1) Ground wave earth, even though the molecular concentration
(2) Sky wave is very high, the radiation intensity is low so that
(3) Space wave the ionization is again low. However, at some
intermediate heights, there occurs a peak of
In the frequency range from a few MHz up to 30 ionization density. The ionospheric layer acts as
to 40 MHz, long distance communication can a reflector for a certain range of frequencies (3 to
be achieved by ionospheric reflection of radio 30 MHz). Electromagnetic waves of frequencies
waves back towards the earth. This mode of higher than 30 MHz penetrate the ionosphere
propagation is called sky wave propagation and and escape. The phenomenon of bending of em
is used by short wave broadcast services. The waves so that they are diverted towards the
ionosphere is so called because of the presence earth is similar to total internal reflection in
of a large number of ions or charged particles. optics.
It extends from a height of ~ 65 Km to about
400 Km above the earth’s surface. Ionization
21. Block diagram of communication system:
Information Message Transmitted Received Message user of

Transmitter Channel Receiver
Source Signal Signal Signal Signal information
Noise
[1]
Transmitter: A transmitter is an arrangement

that converts the message signal to a form Thus, 121 h = h [1]
100 t1 t2
suitable for transmission and then transmits
it through some suitable communication
dT2 121ht2
channel. = = 1.1
[1] dT1 100 ht1

Channel: Channel is the medium through
d : d = 11 : 10
which the signal is transmitted for transmitter T2 T1 [1]
to receiver. Receiver: A receiver extracts the 24. Space wave propagation is the propagation of
desired message signals from the received waves whose frequencies lie above 40 MHz.
signals at the channel output. Examples of communication systems which use
[1] space wave mode are:
22. For transmitting a signal, the antenna should (i) Television broadcast
have a size comparable to the wavelength of
(ii) Microwave links
the signal (at least—A in dimension), where A

is the wavelength. If the frequency of the signal
[1]
is small, then its wavelength becomes very large
and it is impractical to make that large antennas The maximum distance up to which signals can
for the corresponding large wavelengths. For be received,
higher frequencies, Wavelength is smaller, d = 2 RE hT
which is the reason why high frequency carrier
waves are used for transmission.  2  6400000  80
[1+ 1]
= 32000 m
23. dt = 2 RhT = 32 km
Let the transmission of the tower before
transmission be = dT [2]
1
Range after increase in height = dT2
dT1 2 Rht2 ht2
=
Ratio = =
dT2 2 Rht1 ht1
Height increase = 21% [1]

[Topic 2] Modulation
Summary  The size of antenna is given by

λ
and low
4
• Modulation is the process by which a low frequency implies larger wavelength so the size
frequency is superimposed on a high frequency of antenna is not achievable.
carrier signal so that the low frequency can be  There are 4 types of modulation: Amplitude
transmitted to long distance. modulation, Frequency modulation, Pulse
• Demodulation: The reverse process of modulation and Phase modulation
modulation is called as demodulation in which • Amplitude Modulation: The alteration of the
the information from the carrier wave is retrieved amplitude of the carrier in accordance with the
at the receiver. information signal is amplitude modulation.
• Need of Modulation: The following expression represent the AM of a
 As there is a need of a very large antenna for low carrier wave having amplitude and frequency fc:
frequency signals, signals from different stations µ Ac µ Ac
C m ( t ) = Ac A sin ωct + cos( ωc − ω m )t − cos( ωc + ω m )t
mixes up and the attenuation is large, so the 2 2
modulation is needed.
The amplitude of the modulating wave is Am and
the frequency is fm.
Am
=
Modulation index µ ; µ ≤ 1.
Ac
Ac
Ac
2

(c – m) e (c + m) in radians
• Production of AM wave: The following block diagram shows the production of AM wave:
m(t) BANDPASS AM wave

x(t) y(t)
+ SQUARE FILTER
LAW DEVICE CENTRED
Am sin mt AT c
(Modulating 2
Bx(t) + Cx(t)
Signal) c(t)
Ac sin ct
(carrier)
The block of transmitter is as follows:
TRANSMITTING
ANTENNA
m(t)
AMPLITUDE POWER
MODULATOR AMPLIFIER
Message signal
Carrier
• Detection of AM wave: Detected signals need modification as they may not be strong enough to use. The
block diagram of receiver is given below:
Receiving Antenna
Receiving Output
Amplifier IF Stage Detector Amplifier

Signal
3. The carrier wave is given by C(t) = 2 sin (wt)
PREVIOUS YEARS’ volt. The modulating signal is a square wave as

shown. Find modulation index.
TOPIC 2
1 Mark Questions 2
1. The carrier wave is represented by C (t) = 5 sin
in (t) in volt
(10pt)V . A modulating signal is a square wave

as shown.
1 2
m(t) in vol
1 t in second
1 2 t in second
Determine modulation index. [CBSE 2014] [CBSE 2014]

2. The carrier, wave of a signal is given by C(t) = 3
sin (8πt) volt. The modulating signal is a square
wave as shown. Find its modulation index. 2 Marks Questions
4. (i) Define modulation index.
(ii) Why is the amplitude of modulating signal
kept less than the amplitude of carrier
wave? [CBSE 2011]
1.5
5. In the block diagram of a simple modulator for
m(t) in volt
obtaining an AM signal, shown in the figure,

identify the boxes A and B. Write their functions.
1 2 x(t) y(t) AM
+ A B
t in second modulating Wave
signal
carrier
wave
[CBSE 2014] [CBSE 2012]
6. Differentiate between amplitude modulated Solutions

(AM) and frequency modulated (FM) waves by
drawing suitable diagrams. Why is FM signal 1. Modulation index ‘m’ is the ratio of the amplitude
preferred over AM signal? of the modulating signal to the amplitude of the
[CBSE 2015] carrier wave
7. A carrier wave of peak voltage 15V is used The generalized equation of a carrier wave is
to transmit a message signal. Find the peak given below:
voltage of the modulating signal in order to have C(t) = AC sin wCt
a modulation index of 60%.
The generalized equation of a modulating wave
[CBSE 2018]
is given below:
CM(t) = AC sin wCt + µAC sin wmt sin wCt [½]
3 Marks Questions
Am
8. Write any two factors which justify the need for Here, µ is defined as .
Ac
modulating a signal. Draw a diagram showing
an amplitude modulated wave by superposing On comparing this with the equations of carrier
a modulating signal over a sinusoidal carrier wave and modulating wave, we get:
wave. Amplitude of modulating signal, Am = 2V
[CBSE 2012] Amplitude of carrier wave, AC = 5V
9. Write two basic modes of communication. Am 2
Explain the process of amplitude modulation. Hence, µ = [½]
Ac 5
Draw a schematic sketch showing how amplitude
modulated signal is obtained by superposing a 2. The generalized equation of a carrier wave is
modulating signal over a sinusoidal carrier given by:
wave. c(t) = AC sin wCt
[CBSE 2014] The generalized equation of a modulating wave
10. (a) Explain any two factors which justify the is given by:
need of modulating a low frequency signal. Cm(t) = AC sin wCt + µ AC sin wCt sin wCt
(b) Write two advantages of frequency Am
Here, µ is given as [½]
modulation over amplitude modulation. Ac
[CBSE 2017] On comparing this with the equations of
11. What is meant by term ‘modulation’? Draw modulation wave and carrier:
a block diagram of a simple modulator for Amplitude of a modulating signal, Am = 1.5 V
obtaining an AM signal. Amplitude of a carrier wave, Ac = 3V
[CBSE 2017] A m 1.5 1
  
12. (a) How is amplitude modulation achieved? Ac 3 2

(b) The frequencies of two side bands in an AM µ = 0.5 [½]
wave are 640 kHz and 660 kHz respectively.
Find the frequencies of carrier and 3. Modulation index (µ) is the ratio of the amplitude
modulating signal. What is the bandwidth of the modulating signal to the amplitude of the
required for amplitude modulation? carrier wave. The generated equation of a carrier
wave is given below:
[CBSE 2017]
C(t) = 2sin wct
13. (a) Give three reasons why modulation of a
The generalized equation of a modulating wave
message signal is necessary for long distance
is given below:
transmission.
CM(t) = 2sin wCt + µAC sin wmt sin wCt
(b) Show graphically an audio signal, a carrier
Am
wave and an amplitude modulated wave. Here, (µ) is defined as . [½]
Ac
[CBSE 2018]
On comparing this with the equations of carrier programs. There is a need to eliminate
wave and modulating wave, we get, amplitude amplitude-sensitive noise. This is possible if
of modulating signal, Am = 1V we eliminate amplitude variation. In other
Amplitude of carrier wave, Ac = 2V words, there is a need to keep the amplitude
of the carrier constant. This is precisely
Am 1
  what we do in frequency modulation. [½]
Ac 2 A

7. Modulation index m
µ = 0.5 [½] Ac
15
60% =
Ac
4. (i)Modulation index is the ratio of amplitude [1]
15
of modulating signal and amplitude of carrier Ac   100
60
wave.

Ac = 25V [1]
Am 8. Factors needed for modulating a signal:

Ac [1] 1. To send the signal over large distance for
communication.
Am
(ii)    1 Modulation index is Am < Ac kept 2. Practical size of antenna. [1]
Ac
less than 1, that is in order to avoid
distortion. [1]
5. In the given block diagram, block (A) is
modulator and Block (B) is power amplifier [½] time
(A) = modulator changes the amplitude of carrier
wave according to modulating signals Modulating Signal
(B) = Power amplifier enhances the voltage and
enhances power of modulated signals [½]
6. When the amplitude of carrier wave is changed
in accordance with the intensity of the signal,
it is called amplitude modulation. [½] Carrier signal
When the frequency of carrier wave is changed
in accordance with the intensity of the signal,
it is called frequency modulation.
1
c(t) 0 (a)
–1
10
0.5 1 1.5 2 2.5 3 t
m(t) 0 (b)
–1 0.5 1 1.5 2 2.5 3
20
cm(t) for AM 0 (c)
–2
Modulated Wave
10 0.5 1 1.5 2 2.5 3
cm(t) for FM 0 (d)

9. There are two basic modes of communication:
point-to-point and broadcast. In point-to-point
–1
0 0.5 1 1.5 2 2.5 3 [½]
communication mode, communication takes
FM signal is preferred over AM signal because place over a link between a single transmitter
(i) Various electrical machines and noises cause and a receiver. Telephony is an example of
amplitude disturbance in the transmission such a mode of communication. In broadcast
of amplitude modulated wave. This makes mode, there are a large number of receivers for
the reception noisy. So, there is a need for a single transmitter. Radio and television are
Frequency Modulation which can reduce the examples of broadcast mode of communication.
noise factor. [½] If 8the amplitude of the carrier wave is varied
(ii) Fidelity or audio quality of amplitude in accordance with the amplitude of the signal,
modulated transmission is poor. This type it is called amplitude modulation. Frequency
of transmission is also not good for musical and phase are kept constant. [2]
1
c (t) 0
–10 0.5 1 1.5 2 2.5 3
1
m (t) 0
–1
0 0.5 1 1.5 2 2.5 3
2
cm (t) for AM
–2
0 0.5 1 1.5 2 2.5 3
A conceptually simple method of production of amplitude modulated wave is shown in the block diagram
below.
Band pass AM
ym(t) x(t) Square y(t)
+
ym(t) = Amsinmt Law device = Bx(t) + Cx2(t) Filter (Central Wave
at wc)
Modulating signal
yc(t) = Acsinwct
Carrier wave [1]
10. (a) (i) Size of Antenna: The size of antenna increased then the power radiated will be
λ more. [1]
required will be of order of . When
4 (b) Advantage of frequency modulation over
frequency is small, the height of antenna amplitude modulation.
will be large, so audio frequency signal (i) Noise can be reduced.
should be modulated over a high frequency (ii) Transmission efficiency is more because the
carrier wave. [1] amplitude of an Fm wave is constant. [1]
(ii) Effective power radiated by an Antenna: As 11. The process of superimposing information
1 contained in a low frequency signal on a high
power radiated  2 , hence when frequency is
 frequency signal is called modulation. [1]

Bandpass
m(t) x(t) Square Filter AM Wave
+
Lqw device Centred
At C
Amsinmt
Modulating
signal
c(t)
ACsinCt
0 (carrier) [1]
12. (a) Amplitude modulation
In AM, the modulating wave is superimposed on a carrier wave in such a manner that the frequency of
the modulated wave is the same as that of the carrier wave but its amplitude varies in accordance with
the instantaneous amplitude of the modulating wave.
Fig. Block diagram for a simple modulator for So,

obtaining an AM signal. A C
x(t) y(t) AM s x  t   A C sin  c t  cos   C   m  t
+ A B 2
wave A C
modulating
signal
 cos   C   m  t
2
carrier (b) Given:
wave wc + wm = 660 kHz-------- (1)
[1] And wc – wm = 660 kHz-------- (2)
add this block diagram Adding equations (1) and (2)
Let the modulating signal be represent by 2wc = 660 + 640k Hz
m(t) = Am sinwm t 2wc = 1300 kHz
and carrier wave 1300
c   650kHz
c(t) = Am sinwm t 2
when they added the resultant wave Then wc = 650 – 600 kHz
x(t) = Am sinwm t + Ac snwc t 2wc = 10 kHz
This can be further written as wc = 2pfc = 650 kHz (wc = 2pf)
650
x (t) = AC sinwc t µ Am sinwm t × sinwc t  fc  kHz
2
Am
 is the modulation index [1] ⇒ wm = 2pfm = 10 kHz
Ac
10
In practice, µ ≤ 1 to avoid distortion and it is  fm  kHz
2
represented in percent.
Band width required for amplitude modulation
Using trigonometric relation
= upper side band – lower side band
1
sin A sin B  cos  A  B  cos  A  B (fc + fm) – (fc – fm) = 2 fm [1]
2 13. (a) (i) Height of antenna
(ii) Utilization of frequency
(iii) Power of signal [1]
(b)

1
c(t) 0 (a)
–1
0 0.5 1 1.5 2 2.5 3
1
m(t) 0 (b)
–10 0.5 1 1.5 2 2.5 3
2
cm(t) for AM 0 (c)
–20 0.5 1 1.5 2 2.5 3
1
cm(t) for FM 0 (d)
–1
0 0.5 1 1.5 2 2.5 3
1
cm(t) for PM 0 (e)
–1
0 0.5 1 1.5 2 2.5 3
time
[1]
CBSE
Sample Question Paper 1
Physics
Class XII
Time : 3 hrs Maximum Marks : 70
General Instructions
General guidelines given in the paper.
Please check that this question paper contains 5 printed pages
Code number given on the right hand side of the question paper should be written on title
page of the answer-book by the candidate.
Please check that this question paper contains 26 questions.
Please write down the Serial Number of the question paper before attempting it.
15 minutes time has been allotted to read this paper. The question paper will be distributed at
10:15 a.m. to 10:30 a.m., the student will read the question paper only and will not write any
answer on the answer-book during this period.
All questions are compulsory. There are 26 questions in all.
This question paper has five sections - Section A, Section B, Section C, Section D and
Section E
Section A contains five questions of one mark each,
Section B contains five questions of two marks each,
Section C contains twelve questions of three marks each,
Section D contains one value based question of four marks and
Section E contains three questions of five marks each.
You may use the following values of physical constants wherever necessary
c = 3 × 108 m/s
h = 6.34 × 10–34 Js
e = 1.6 × 10–19 C
0 = 4 × 10–7 T m A–1
0 = 8.854 × 10–12 C2 N –1 m–2
{1/(40)} = 9 × 109 N m2 C–2
Mass of electron = 9.1 × 10–31 kg
Mass of neutron = 1.675 × 10–27 kg
Mass of proton = 1.673 × 10–27 kg
Avogadro’s number = 6.023 × 1023 per gram mole
Boltzmann constant = 1.38 ×10–23 JK–1.
2 CBSE Sample Question Paper 1
Section A (1 × 5 = 5)
1. The speed of a charged particle moving in a magnetic field does not change. Why?
2. Which electromagnetic radiation plays an important role in maintaining the earth’s warmth
or average temperature through the green house effect?
3. For a fixed frequency of incident radiation how does photoelectric current vary with
intensity of incident light?
4. Why is it experimentally difficult to detect neutrinos in nuclear decay?
5. What is meant by the term Attenuation in communication systems?
Section B (2 × 5 = 10)
6. In the given circuit the 12V source is resistance free; If the galvanometer in the given
circuit reads zero, then find the value of the resistor.
+
12v 10k
– G
R 2v
7. When a high power heater is connected to electric mains, the bulbs lightening in the house
become dim, Why?
OR
A 10 m long potentiometer wire carries a steady current. A 1.018V standard cell is balanced
at a length of 850 cm. Then, what is the maximum emf that can be measured?
8. The optical effect in electromagnetic wave is produced by which vector?
9. From photoelectric effect the equation for maximum energy is given by Emax = hν – Φ. This
equation is stated for one photon. If an electron absorbs two photons each of frequency ν, what
will be the maximum energy of emitted electron?
10. By what percentage will the transmission range of a TV tower be affected, when the height
of tower is increased by 21%?
CBSE Sample Question Paper 1 3
Section C (3 × 12 = 36)
11. Two identical spheres having unequal, opposite charges are placed at a distance of 0.90m
apart. After touching them, they are again placed at the same distance apart. Now they
repel each other with a force of 0.025N. Answer the following questions:
(a) After touching what will be the ratio of charges on both the spheres?
(b) Final charge on each sphere.
OR
Three charges are arranged as shown. What is the electric potential energy of the system?
(Given q = 1 x 10–7C and a = 0.10 m)
–4q
a a
+q a +2q
12. (a) Define the term electrical resistance. What is its SI unit?
(b) A cylindrical wire is stretched to increase its length by 10%. Find the percentage
increase in resistance.
13. When a coil of magnetic moment 2.5 × 10–8Am2 is placed in a magnetic field such that its
plane is parallel to the field, then the moment of the couple acting on the coil is
7.5 × 10–9Nm. If the area of the coil be 1.5 cm2 and the number of turns in it be 12, find the
following :
(a) Magnitude of magnetic field.
(b) Current in it.

14. Let a magnetic dipole of moment M be rotated in a uniform magnetic field of magnitude B
through an angle θ from the field direction
(a) Derive an expression for the work done in rotating the dipole through an angle θ from
the field direction.
(b) What will be the work done if the dipole is rotated through 900 from the direction of
field?
15. (a) State law of Malus.
(b) Draw a graph showing the variation of intensity (I) of polarized light transmitted by an
analyser with angle θ between the polarizer and the analyser.
(c) What is the value of refractive index of a medium of polarizing angle 600 ?
16. (a) What are the basic conditions for interference to occur?
(b) How does the fringe width of interference fringe change when whole apparatus is
dipped in a liquid of refractive index 1.3?
17. A ray of light incident at an angle of  on the refracting face of the prism emerges from the
other face normally. If the angle of the prism is 50 and the prism is made of a material of
refractive index 1.5, what is the angle of incidence?
18. In a hydrogen atom, a transition takes place from n = 3 to n = 2 orbit. (R = 1.097 × 107
meter-1)
(a) Find the wavelength of the emitted photon.
(b) Will the photon be visible?
(c) To which spectral series will this photon belong?
19. (a) What is meant by the term half life in radioactivity? Derive a relation between half life
and decay constant.
(b) The half life of radium is 1600 years. How long will a sample of radium take for 75% of
its initial mass to disintegrate?
20. (a) How is a transistor used as an amplifier?
(b) The input and output resistances in a common base amplifier circuit are 400 h and
400 kh respectively. If the current gain α is 9.8, then find the voltage gain.
21. Give the circuit symbol and truth table of NOR gate.
22. What is the function of emitter, base and collector in the transistor?
Section D (4 × 1 = 4)
23. Sanjeev belongs to a rural area of UP. One day a storm came and the high power lines
came too close to each other; also their height got decreased. This immediately caught the
attention of sanjeev, who, keeping the safety aspect in mind, immediately informed the
electricity department. It being a Sunday, officials were not available, hence he and his
friends made a large warning sign and placed it near the wires.
(1) What are the values possessed by Sanjeev?
(2) (a) Who would have been the biggest victim of the lines coming close to each other?
(b) What solution will the authorities give for this problem?
Section E (5 × 3 = 15)
24. (a) What is meant by the term restoring couple?
(b) In what direction should a dipole be placed to electric field so that the torque acting on
it is maximum?
25. (a) Derive expression for electric field just outside a charged conductor.
(b) Does the field depend upon the shape of the conductor?
OR
(a) What is a transformer; Write a short note on step down and step up transformer.
(b) Derive an expression for efficiency of a transformer.
26 . (a) What is the advantage of reflecting telescope over refracting telescope?
(b) Draw a ray diagram to illustrate the refraction of light at convex spherical surface.
OR
(a) When light passes from one medium to another, how does its frequency vary?
(b) What is the effect on wavelength of light in going from one medium to another?
CBSE
Sample Question Paper 2
Physics
Class XII
Time : 3 hrs Maximum Marks : 70
General Instructions
General guidelines given in the paper.
Please check that this question paper contains 5 printed pages
Code number given on the right hand side of the question paper should be written on title
page of the answer-book by the candidate.
Please check that this question paper contains 26 questions.
Please write down the Serial Number of the question paper before attempting it.
15 minutes time has been allotted to read this paper. The question paper will be distributed at
10:15 a.m. to 10:30 a.m., the student will read the question paper only and will not write any
answer on the answer-book during this period.
All questions are compulsory. There are 26 questions in all.
This question paper has five sections - Section A, Section B, Section C, Section D and
Section E
Section A contains five questions of one mark each,
Section B contains five questions of two marks each,
Section C contains twelve questions of three marks each,
Section D contains one value based question of four marks and
Section E contains three questions of five marks each.
You may use the following values of physical constants wherever necessary
c = 3 × 108 m/s
h = 6.34 × 10–34 Js
e = 1.6 × 10–19 C
0 = 4 × 10–7 T m A–1
0 = 8.854 × 10–12 C2 N –1 m–2
{1/(40)} = 9 × 109 N m2 C–2
Mass of electron = 9.1 × 10–31 kg
Mass of neutron = 1.675 × 10–27 kg
Mass of proton = 1.673 × 10–27 kg
Avogadro’s number = 6.023 × 1023 per gram mole
Boltzmann constant = 1.38 ×10–23 JK–1.
Section A (1 × 5 = 5)
1. A charged particle enters a uniform magnetic field obliquely. What will be the trajectory
of the particle?
2. Which radiations are used in medicine to destroy cancer cells?
3. For a fixed frequency and intensity of incident light, how does photoelectric current vary
with increase in potential applied to collector?
4. What is the atomic number of nucleus produced when U92238 produces  decay?
5. In order to extend the range of a communication system what is used?
Section B (2 × 5 = 10)
6. Six lead accumulators each of emf 2V and internal resistance 0.015Ω are joined in series
to an external resistance of 8.5Ω. Find the current drawn from the supply.
7. In order to drive a current of 3A for 5 minutes in an electric circuit, 1350J of work is to be

done. The emf of source in the circuit is how much?
OR
Two cells of emf E1 and E2 (E1 > E2) are connected as shown. When a potentiometer is
connected between A and B, the balancing length of the potentiometer wire is 300 cm. On
connecting the same potentiometer wire between A and C , the balancing length is 100 cm.
Compute E1/E2.
E1 E2
A B C
8. What is the basic source of electromagnetic wave?

9. When monochromatic radiation of wavelength 200 A falls upon a nickel plate, the latter

acquires a positive charge. When the wavelength is increased, at 3400 A the effect is found
to cease, however intense the incident radiation may be. Explain it.
10. A TV tower has a height of 300 m, what is the maximum distance upto which this TV
transmission can be received. (R = 6400 km) ?
Section C (3 × 12 = 36)
11. A small ball of mass m is suspended by an inextensible , insulated , light thread of length
l from a hook. Each of the ball and the hook is given a charge q.
Answer the following:
(a) Draw a diagram showing the various forces acting on the ball.
(b) Find an expression for time period of the ball for small oscillations.
OR
Calculate the value of VA – VB in the given arrangement.
x y x
+q A B –q
12. (a) What do you mean by the term specific resistance. Give its SI unit.
(b) Let a wire of area of cross section A, length l and resistance R be taken. Find an
expression for specific resistance.
13. A bar magnet is suspended by a thin wire in a uniform magnetic field. On twisting the
upper end of the wire by 150°, the magnet is displaced from its initial position by 30 °. How
much should the upper end of the wire be twisted so that the magnet is displaced by 90 cm
from its initial position?
14. What do you mean by :
(a) What do you mean by Magnetic susceptibility and relative permeability?
(b) Derive a relation between relative permeability and magnetic susceptibility.
15. (a) State Brewster’s law.
(b) Show that reflected and refracted rays are perpendicular rays.
16. Two waves of amplitude a1 and a2 interfere at a point where phase difference is Φ.
(a) Write the expression of resultant amplitude.
(b) What will be the values for constructive interference and destructive interference?
17. The optical density of turpentine is higher than that of water, while its mass density is
lower. A layer of turpentine is made to flow over water in a container. Trace the path of
incident ray.
18. For Lyman series of hydrogen spectrum, find the following :
(a) Wavelength of the first line.
(b) Wavelength of the limit of this series.
19. (a) State Rutherford and Soddy law for radioactive decay.
(b) Find the half life period of a radioactive material if its activity drops to 1/16 th of its
initial value.
OR
(a) If the number of atoms in a radioactive substance be N0 and N at time t = 0 and after
time t, and be the decay constant, then derive the relation
N = N0e–t
20. Draw the circuit diagram of common base amplifier; What is the primary use of this
amplifier?
21. Give the circuit symbol and truth table of NOT gate.
22. (a) What is linearly polarized light? Describe briefly using a diagram how sunlight is
polarized.
(b) Unpolarized light is incident on a Polaroid. How the intensity of would transmitted
light change when the Polaroid is rotated?
Section D (1 × 4 = 15)
23. Last night a thunderstorm occurred in a small town of Uttar Pradesh. Because of it many
trees fell on the road and at many places water got filled. Rohan’s neighbor has a son aged
three years; He was busy playing with his ball and all of a sudden his ball went into water
in which two broken electric poles were lying. The child rushed to pick his ball from there.
Rohan saw the scene and ran to catch hold of the child before he reaches near the water.
(a) What qualities of Rohan are depicted by this act?
(b) Why did Rohan not allow the child to go near the water?
Section E (5 × 3 = 15)
24. (a) Define the term electric dipole . Is it a vector quantity?
(b) An electric dipole is situated in a uniform electric field. Is a net force acting on the
dipole? Explain.
OR
(a) Derive an expression for electric field E at any point inside a spherically symmetric
uniformly charged sphere, applying Gauss’s Law.
(b) Draw a graph to show the variation of E with distance.
25. (a) What is a choke coil. Discuss the construction of the same.
(b) Prove that average power dissipated in a choke coil is nearly zero.
OR
(a) Derive an expression for power in a circuit containing both inductance and capacitance.
(b) An alternating current of 1.5 mA rms and angular frequency = 100rad/s flows through
a 10k  resistor and 0.50µ F capacitor in series. Find the rms voltage across the
capacitor.
26. (a) Draw a ray diagram for the formation of image by a compound microscope.
(b) Derive an expression for total magnification when image is formed at infinity.
OR
(a) State Huygen’s principle and also explain their propagation of waves.
(b) What type of wavefront will emerge from a (i) point source (ii) distant light source?

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What are the important topics from the exam point of view according to the analysis given?

What are the important topics from the exam point of view according to the analysis given?

What are the three basic properties of electric charge?

What are the three basic properties of electric charge?

CHAPTER 1

Electric Charges and Fields

List of Topics 2016 2017 2018

[Topic 1] Coulomb’s law, electrostatic field and

Summary Three basic properties of electric

• If a plastic comb is rubbed with wool, it becomes o

(b) Two charged spherical conductors of radii R1

PREVIOUS YEARS’ and R2 when connected by a conducting wire

Since, rA < rB ⇒ VA > VB The volume between the plates is Ad metre3.

q q q q As the total charge q is uniformly distributed,

σ1 q1R 22 (∵ Only the axial components contribute towards

9. –q O +q EA EP (b) Graph of E versus r for r >> a

Electric field intensity due to as electric dipole r [1]

(a) Dipole at a point on the axial wire: we Fig.: E versus r

1 4 px t = (q × 2l) E sin q or t = pE sin q [½]

Topic 2: Electric Flux

Df = E.DS Fig. Infinite plane sheet (thin)

spherical shell of radius R at a point outside the

PREVIOUS YEARS’ shell. Draw a graph showing the variation of

TOPIC 2 5 Marks Questions

3. How does the electric flux due to a point charge

= fs Eo ds fL = (0 )(a2) cos 180° = 0

∴Total electric flux over the centre surface of

–+ –+ Internal electric (b) Let V0 be the potential on the surface at

List of Topics 2016 2017 2018

Topic 1: Electrostatic Potential and Electrostatic Potential

Summary q from a point R to a point P is VP – VR , which is

6. An electric dipole of length 2 cm, when placed with

PREVIOUS YEARS’ its axis making an angle of 60° with a uniform

1 Mark Questions 7. (i) Can two equipotential surfaces intersect each

2 Mark Questions (ii) Deduce the expression for the potential

A 12. (a) Explain why, for any charge configuration,

Direction will be along –Z-axis.

Pres = p Therefore, work done will be zero. [1]

equatorial plane of the dipole

(ii) The work done in bringing charge q1 from

Or 12. (a) If the field were not normal to the

Van De Graaff Generator: plastic

How will the (i) charge and (ii) potential

PREVIOUS YEARS’ difference between the plates of the capacitors

3 Mark Questions [All INDIA 2017]

5 Mark Questions 19. A parallel-plate capacitor is charged to a

Since, V is constant, dV = 0 [½]

4. According to Lenz law the polarity of induced emf q1 360  10 6

9. (i) As the dielectric slab is introduced between

Let the new voltage be V’ Q

1 shell by a wire, the charge will flow to the shell

List of Topics 2016 2017 2018

Topic 1: Electricity conduction, Ohm’s law and resistance

7. Define the term ‘electrical conductivity’ of a

PREVIOUS YEARS’ metallic wire. Write its S.I. unit.

EXAMINATION QUESTIONS 8. Show variation of resistivity of copper as a

1 Mark Questions 9. Graph showing the variation of current versus

[ALL INDIA 2012] C D

13. (a) State the working principle of a potentiometer. Solutions

Where N2 is total number of turns wound