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Wave Optics 1
Light Propagation.
Light is a form of energy which generally gives the sensation of sight.
(1) Different theories
(2) Optical phenomena explained () or not explained () by the different theories of light
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2 Wave Optics
(iv) Types of wave front
Plane WF
Spherical WF
Cylindrical WF
Point source
Line source
(v) Every point on the given wave front acts as a source of new disturbance called secondary wavelets.
Which travel in all directions with the velocity of light in the medium.
(ii) Resultant
1
y = y1 – y2
Waves are meeting at a point
y1 + y2 = with out of phase
superimpose
S
Two waves
L M2
S2
T
(iv) Time difference (T.D.) : Time difference between the waves meeting at a point is T.D.
2
(3) Resultant amplitude and intensity
If suppose we have two waves y 1 a1 sin t and y 2 a 2 sin ( t ) ; where a1 , a 2 Individual
amplitudes, = Phase difference between the waves at an instant when they are meeting a point. I1, I2 =
Intensities of individual waves
Resultant amplitude : After superimposition of the given waves resultant amplitude (or the
amplitude of resultant wave) is given by A a 12 a 22 2 a 1 a 2 cos
For the interfering waves y1 = a1 sin t and y2 = a2 cos t, Phase difference between them is 90o. So
resultant amplitude A a12 a 22
Note : The term 2 I1 I 2 cos is called interference term. For incoherent interference this term is
zero so resultant intensity I I1 I 2
(4) Coherent sources
The sources of light which emits continuous light waves of the same wavelength, same frequency and
in same phase or having a constant phase difference are called coherent sources.
Two coherent sources are produced from a single source of light by adopting any one of the following two
methods
Note : Laser light is highly coherent and monochromatic.
Two sources of light, whose frequencies are not same and phase difference between the
waves emitted by them does not remain constant w.r.t. time are called non-coherent.
The light emitted by two independent sources (candles, bulbs etc.) is non-coherent and
interference phenomenon cannot be produced by such two sources.
The average time interval in which a photon or a wave packet is emitted from an atom is
L Distance of coherence
defined as the time of coherence. It is c , it's value is of the
c Velocity of light
order of 10–10 sec.
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4 Wave Optics
Interference of Light.
When two waves of exactly same frequency (coming from two coherent sources) travels in a medium,
in the same direction simultaneously then due to their superposition, at some points intensity of light is
maximum while at some other points intensity is minimum. This phenomenon is called Interference of
light.
(1) Types : It is of following two types
(v) Resultant intensity at the point of observation will be (v) Resultant intensity at the point of observation will
maximum be minimum
I m ax I1 I 2 2 I1 I 2 I m in I1 I 2 2 I1 I 2
I max I1 I 2
2
I min I1 I 2
2
For identical source I1 I 2 I0 I I 0 I 0 2 I 0 I 0 cos 4 I0 cos 2 [1 + cos
2
2 cos 2 ]
2
Note : In interference redistribution of energy takes place in the form of maxima and minima.
I max I min
Average intensity : Iav I1 I 2 a12 a 22
2
Ratio of maximum and minimum intensities :
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Wave Optics 5
2 2
I1 I 2 2 2
I max
I1 / I 2 1 a1 a 2 a /a 1
1 2 also
I I I / I 1 a a
I min 1 2 1 2 1 2 a1 / a 2 1
I max
1
I1 a1 I min
I2 a2 I max
1
I min
If two waves having equal intensity (I1 = I2 = I0) meets at two locations P and Q with path
difference 1 and 2 respectively then the ratio of resultant intensity at point P and Q will be
1
cos 2 cos 2 1
IP
2
IQ
cos 2 2 cos 2 2
2
Young’s Double Slit Experiment (YDSE)
Monochromatic light (single wavelength) falls on two narrow slits S1 and S2 which are very close
together acts as two coherent sources, when waves coming from two coherent sources (S 1 , S 2 )
superimposes on each other, an interference pattern is obtained on the screen. In YDSE alternate bright
and dark bands obtained on the screen. These bands are called Fringes.
Screen
4 Dark
3 Bright
3 Dark
2 Bright
d = Distance between slits S1
2 Dark
D = Distance between slits and screen 1 Bright
1 Dark
= Wavelength of monochromatic S d Central bright
1 Dark fringe
light emitted from source S2 1 Bright
2 Dark (or Central
2 Bright maxima)
3 Dark
3 Bright
4 Dark
D
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6 Wave Optics
(2n 1)
and For minima at P : ; where n = 1, 2, …….
2
Note : If the slits are vertical, the path difference () is d sin , so as increases, also increases.
But if slits are horizontal path difference is d cos , so as increases, decreases.
P P
S1
d
C C
S1 S2
S2 d
D
(7) More about fringe
(i) All fringes are of
D β
equal width. Width of each fringe is and angular fringe width
d d D
(ii) If the whole YDSE set up is taken in another medium then changes so changes
3
e.g. in water w a w a a
w w 4
1
(iii) Fringe width i.e. with increase in separation between the sources, decreases.
d
nD
(iv) Position of nth bright fringe from central maxima x n n ; n 0, 1, 2 ....
d
(2 n 1) D (2 n 1)
(v) Position of nth dark fringe from central maxima x n ; n 1, 2,3 ....
2d 2
(vi) In YDSE, if n 1 fringes are visible in a field of view with light of wavelength 1 , while n 2 with light
of wavelength 2 in the same field, then n1 1 n 2 2 .
(vii) Separation (x ) between fringes
Between nth bright and mth bright fringes (n m) Between nth bright and mth dark fringe
1
(a) If n m then x n m
2
x (n m)
1
(b) If n m then x m n
2
1
(v) The sources must be close to each other : Otherwise due to small fringe width the eye can
d
not resolve fringes resulting in uniform illumination.
(10) Shifting of fringe pattern in YDSE
If a transparent thin film of mica or glass is put in the path of one of the waves, then the whole fringe
pattern gets shifted.
t
If film is put in the path of upper wave, fringe pattern shifts upward and if
film is placed in the path of lower wave, pattern shift downward.
S1
D
( 1) t ( 1) t
Fringe shift d C
d
Additional path difference ( 1)t S2
( 1) t n Screen
If shift is equivalent to n fringes then n or t
( 1) D
Shift is independent of the order of fringe (i.e. shift of zero order maxima = shift of nth order
maxima.
Shift is independent of wavelength.
Illustrations of Interference
Interference effects are commonly observed in thin films when their thickness is comparable to
wavelength of incident light (If it is too thin as compared to wavelength of light it appears dark and if it is
too thick, this will result in uniform illumination of film). Thin layer of oil on water surface and soap
bubbles shows various colours in white light due to interference of waves reflected from the two surfaces of
the film.
Air
Oil Air
Air
Water
Soap bubble in air
Oil film on water surface
(1) Thin films : In thin films interference takes place between the waves reflected from it’s two
surfaces and waves refracted through it.
Reflected rays
t r
r
Refracted rays
so 2 t (2n 1) 2 t n
2
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8 Wave Optics
Source of light moves towards the stationary Source of light moves away from the
observer (v << c) stationary observer (v << c)
v v
(i) Apparent frequency 1 and (i) Apparent frequency 1 and
c c
v v
Apparent wavelength 1 Apparent wavelength 1
c c
(ii) Doppler’s shift : Apparent wavelength < actual (ii) Doppler’s shift : Apparent wavelength > actual
wavelength, wavelength,
So spectrum of the radiation from the source of light So spectrum of the radiation from the source of light
shifts towards the red end of spectrum. This is called Red shifts towards the violet end of spectrum. This is called
shift Violet shift
v v
Doppler’s shift Δ . Doppler’s shift Δ .
c c
2r
Note : Doppler’s shift () and time period of rotation (T) of a star relates as ; r = radius
c T
of star.
Applications of Doppler effect
(i) Determination of speed of moving bodies (aeroplane, submarine etc) in RADAR and SONAR.
(ii) Determination of the velocities of stars and galaxies by spectral shift.
(iii) Determination of rotational motion of sun.
(iv) Explanation of width of spectral lines.
(v) Tracking of satellites. (vi) In medical sciences in echo cardiogram, sonography etc.
Concepts
The angular thickness of fringe width is defined as , which is independent of the screen distance D.
D d
Central maxima means the maxima formed with zero optical path difference. It may be formed anywhere on the screen.
All the wavelengths produce their central maxima at the same position.
The wave with smaller wavelength from its maxima before the wave with longer wavelength.
The first maxima of violet colour is closest and that for the red colour is farthest.
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Wave Optics 9
Fringes with blue light are thicker than those for red light.
In an interference pattern, whatever energy disappears at the minimum, appears at the maximum.
In YDSE, the nth maxima always comes before the nth minima.
Imax
In YDSE, the ratio is maximum when both the sources have same intensity.
Imin
For two interfering waves if initial phase difference between them is 0 and phase difference due to path difference
2
between them is '. Then total phase difference will be ' 0 .
Sometimes maximm number of maximas or minimas are asked in the question which can be obtained on the screen. For
this we use the fact that value of sin (or cos ) can't be greater than 1. For example in the first case when the slits are
vertical
n
sin (for maximum intensity)
d
n d
sin ≯1 ≯1 or n≯
d
Suppose in some question d/ comes out say 4.6, then total number of maximuas on the screen will be 9. Corresponding to
n 0, 1, 2, 3 and 4 .
Example
s
Example: 1 If two light waves having same frequency have intensity ratio 4 : 1 and they interfere, the ratio of
maximum to minimum intensity in the pattern will be
(a) 9 : 1 (b) 3 : 1 (c) 25 : 9 (d) 16 : 25
2
I1 4
2
1 1
I m ax I2 1 9
Solution: (a) By using 1.
I m in I1 4 1
1
I2 1
Example: 2 In Young’s double slit experiment using sodium light ( = 5898Å), 92 fringes are seen. If given
colour ( = 5461Å) is used, how many fringes will be seen
(a) 62 (b) 67 (c) 85 (d) 99
Solution: (d) By using n1 1 n 2 2 92 5898 n 2 5461 n 2 99
Example: 3 Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The
phase difference between the beams is at point A and at point B. Then the difference between the
2
resultant intensities at A and B is
(a) 2I (b) 4I (c) 5I (d) 7I
At point A : Resultant intensity I A I 4 I 2 I 4 I cos 5I
2
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10 Wave Optics
Example: 4 If two waves represented by y1 4 sin t and y 2 3 sin t interfere at a point, the amplitude of the
3
resulting wave will be about [MP PMT 2000]
(a) 7 (b) 6 (c) 5 (d) 3.
Solution: (b) By using A a12 a 22 2a1 a 2 cos A (4 ) 2 (3) 2 2 4 3 cos 37 6 .
3
Example: 5 Two waves being produced by two sources S 1 and S 2 . Both sources have zero phase difference and
have wavelength . The destructive interference of both the waves will occur of point P if (S1 P S 2 P)
has the value
[MP PET 1987]
3 11
(a) 5 (b) (c) 2 (d)
4 2
Solution: (d) For destructive interference, path difference the waves meeting at P (i.e. S 1 P S 2 P) must be odd
multiple of /2. Hence option (d) is correct.
Example: 6 Two interfering wave (having intensities are 9I and 4I) path difference between them is 11 . The
resultant intensity at this point will be
(a) I (b) 9 I (c) 4 I (d) 25 I
2
Solution: (d) Path difference 11 22 i.e. constructive interference obtained at the same
2
point
I max 144
Example: 7 In interference if then what will be the ratio of amplitudes of the interfering wave
I min 81
144 7 1 12
(a) (b) (c) (d)
81 1 7 9
I max
1 144 1 12 1
a1 I min 81 9 7
Solution: (b) By using 12
a2 I max 144 1
1 1 1
I min 81 5
Example: 8 Two interfering waves having intensities x and y meets a point with time difference 3T/2. What will
be the resultant intensity at that point
x y
(a) ( x y ) (b) ( x y xy ) (c) x y 2 xy (d)
2 xy
T 3T T
Solution: (c) Time difference T.D. 3 ; This is the condition of constructive
2 2 2
interference.
So resultant intensity I R ( I1 I 2 ) 2 ( x y ) 2 x y 2 xy .
6 10 7
So (2 3 1) 15 10 7 m 1.5 microns.
2
Example: 10 In a Young’s double slit experiment, the slit separation is 1 mm and the screen is 1 m from the slit. For a
monochromatic light of wavelength 500 nm, the distance of 3rd minima from the central maxima is
(a) 0.50 mm (b) 1.25 mm (c) 1.50 mm (d) 1.75 mm
(2n 1) D
Solution: (b) Distance of nth minima from central maxima is given as x
2d
(2 3 1) 500 10 9 1
So here x 1 .25 mm
2 10 3
Example: 11 The two slits at a distance of 1 mm are illuminated by the light of wavelength 6.5 10 7 m. The interference
fringes are observed on a screen placed at a distance of 1 m. The distance between third dark fringe and fifth
bright fringe will be
[NCERT 1982; MP PET 1995; BVP 2003]
(a) 0.65 mm (b) 1.63 mm (c) 3.25 mm (d) 4.88 mm
Solution: (b) Distance between nth bright and mth dark fringe (n > m) is given as
1 1D
x n m n m
2 2 d
1 6 .5 10 7 1
x 5 3 1 .63 mm .
2 1 10 3
Example: 12 The slits in a Young’s double slit experiment have equal widths and the source is placed symmetrically relative to
the slits. The intensity at the central fringes is I0. If one of the slits is closed, the intensity at this point will be[MP PMT 1999
(a) I0 (b) I0 / 4 (c) I0 / 2 (d) 4 I0
Solution: (b) By using I R 4 I cos 2 {where I = Intensity of each wave}
2
At central position = 0o, hence initially I0 = 4I.
If one slit is closed, no interference takes place so intensity at the same location will be I only i.e.
1 I
intensity become s th or 0 .
4 4
Example: 13 In double slit experiment, the angular width of the fringes is 0.20° for the sodium light ( = 5890 Å). In order to
increase the angular width of the fringes by 10%, the necessary change in the wavelength is
(a) Increase of 589 Å (b) Decrease of 589 Å (c) Increase of 6479 Å (d) Zero
0 .20 5890 o
0 .20 5890
Solution: (a) By using 1 1 2 6479
d 2 2 (0 .20 10 % of 0 .20 )
o 2 0 .22 2
So increase in wavelength = 6479 – 5890 = 589 Å.
Example: 14 In Young’s experiment, light of wavelength 4000 Å is used, and fringes are formed at 2 metre distance and has a
fringe width of 0.6 mm. If whole of the experiment is performed in a liquid of refractive index 1.5, then width of
fringe will be
[MP PMT 1994, 97]
(a) 0.2 mm (b) 0.3 mm (c) 0.4 mm (d) 1.2 mm
air 0 .6
Solution: (c) medium medium 0 . 4 mm .
1 .5
Example: 15 Two identical sources emitted waves which produces intensity of k unit at a point on screen where
path difference is . What will be intensity at a point on screen at which path difference is /4 [RPET 1996]
k k
(a) (b) (c) k (d) Zero
4 2
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12 Wave Optics
2
Solution: (b) By using phase difference ()
For path difference , phase difference 1 2 and for path difference /4, phase difference 2 = /2.
p
Solution: (a) By using shift x ( 1) t x (1.5 1) 2 10 6 2
5000 10 10
Since the sheet is placed in the path of the first wave, so shift will be 2 fringes upward.
Example: 17 In a YDSE fringes are observed by using light of wavelength 4800 Å, if a glass plate ( = 1.5) is
introduced in the path of one of the wave and another plates is introduced in the path of the ( = 1.8)
other wave. The central fringe takes the position of fifth bright fringe. The thickness of plate will be
(a) 8 micron (b) 80 micron (c) 0.8 micron (d) None of these
Solution: (a) Shift due to the first plate x 1 (1 1) t (Upward)
1
and shift due to the second x 2 ( 2 1) t (Downward) S1
2
d C
Hence net shift = x2 – x1 ( 2 1 ) t S2
Screen
5 5 4800 10 10
5p (1.8 1.5) t t 8 10 6 m 8 micron . D
0.3 0 .3
d
Example: 18 In young double slit experiment 10 4 (d = distance between slits, D = distance of screen from
D
the slits). At a point P on the screen resulting intensity is equal to the intensity due to individual slit
I0. Then the distance of point P from the central maxima is ( = 6000 Å)
(a) 2 mm (b) 1 mm (c) 0.5 mm (d) 4 mm
1 2
Solution: (a) By using shift I 4 I 0 cos 2 ( / 2) I 0 4 I 0 cos 2 ( / 2) cos( / 2) or
2 2 3 3
10
xd d 6000 10 2
Also path difference x x 2 10 3 m 2mm.
D 2 D 2 3
Example: 19 Two identical radiators have a separation of d = /4, where is the wavelength of the waves emitted
by either source. The initial phase difference between the sources is /4. Then the intensity on the
screen at a distance point situated at an angle = 30o from the radiators is (here I0 is the intensity at
that point due to one radiator)
(a) I0 (b) 2I0 (c) 3I0 (d) 4I0
2
Solution: (a) Initial phase difference 0 ; Phase difference due to path difference ' ()
4
2 2
where d sin ' (d sin ) (sin 30 o )
4 4
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Wave Optics 13
/ 2
Hence total phase difference 0 ' . By using I 4 I 0 cos 2 ( / 2) 4 I 0 cos 2 2 I0 .
4 2
Example: 20 In YDSE a source of wavelength 6000 Å is used. The screen is placed 1 m from the slits. Fringes
formed on the screen, are observed by a student sitting close to the slits. The student's eye can
distinguish two neighbouring fringes. If they subtend an angle more than 1 minute of arc. What will
be the maximum distance between the slits so that the fringes are clearly visible
(a) 2.06 mm (b) 2.06 cm (c) 2.06 10–3 mm (d) None of these
Solution: (a) According to given problem angular fringe width [ As 1' rad ]
d 180 60 180 60
6 10 7 180 60
i.e. d i.e. d 2.06 10 3 m d max 2.06 mm
Example: 21 the maximum intensity in case of interference of n identical waves, each of intensity I0, if the
interference is (i) coherent and (ii) incoherent respectively are
(a) n 2 I 0 , nI 0 (b) nI 0 , n 2 I 0 (c) nI 0 , I 0 (d) n 2 I 0 , (n 1)I 0
(i) In case of coherent interference does not vary with time and so I will be maximum when
cos max 1
i.e. (I m ax )co I1 I 2 2 I1 I 2 ( I1 I 2 ) 2
(ii)In case of incoherent interference at a given point, varies randomly with time, so (cos )av 0
and hence (I R ) Inco I1 I 2
Example: 23 A star is moving towards the earth with a speed of 4.5 10 6 m / s . If the true wavelength of a certain line
in the spectrum received from the star is 5890 Å, its apparent wavelength will be about
[c 3 10 8 m / s]
[MP PMT 1999]
(a) 5890 Å (b) 5978 Å (c) 5802 Å (d) 5896 Å
v 4 . 5 10 6
Solution: (c) By using ' 1 ' 5890 1 5802 Å .
c 3 10 8
Example: 24 Light coming from a star is observed to have a wavelength of 3737 Å, while its real wavelength is 3700
Å. The speed of the star relative to the earth is [Speed of light 3 10 8 m / s ] [MP PET 1997]
v v
Solution: (b) By using (3737-3700)= 3700 v 3 10 6 m / s .
c 3 10 8
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14 Wave Optics
Example: 25 Light from the constellation Virgo is observed to increase in wavelength by 0.4%. With respect to
Earth the constellation is [MP PMT 1994, 97; MP PET 2003]
(a) Moving away with velocity 1.2 10 6 m / s (b) Coming closer with velocity 1.2 10 6 m / s
(c) Moving away with velocity 4 10 6 m / s (d) Coming closer with velocity 4 10 6 m / s
v 0.4 0 .4 v
Solution: (a) By using ; where and c = 3 108 m/s v = 1.2 106 m/s
c 100 100 3 10 8
Since wavelength is increasing i.e. it is moving away.
Tricky example: 1
In YDSE, distance between the slits is 2 10–3 m, slits are illuminated by a light of wavelength
2000Å –9000 Å. In the field of view at a distance of 10 –3 m from the central position which
wavelength will be observe. Given distance between slits and screen is 2.5 m
(a) 40000 Å (b) 4500 Å (c) 5000 Å (d) 5500 Å
n D xd 10 3 2 10 3 8 10 7 8000
Solution : (b) x m Å
d nD n 2 .5 n n
8000
For n = 1, 2, 3....... = 8000 Å, 4000 Å, Å .........
3
Tricky example: 2
I is the intensity due to a source of light at any point P on the screen. If light reaches the point P via
two different paths (a) direct (b) after reflection from a plane mirror then path difference between
two paths is 3/2, the intensity at P is
(a) I (b) Zero (c) 2I (d) 4I
Solution : (d) Reflection of light from plane mirror gives additional path difference of /2 between two waves
3
Total path difference 2
2 2
Tricky example: 3
I 9I I 9
Solution : (d) From figure I1 and I 2 2
4 64 I1 16
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B Wave Optics
B 15
I I /4 9I /64
I2
1 9 1 A A 3I /64
I I1 16 49
By using max 1
I min I2 9 3I /4 3I /16
1 1
I1 16
Fresnel's Biprism.
(1) It is an optical device of producing interference of light Fresnel's biprism is made by joining base to
base two thin prism (A1BC and A2BC as shown in the figure) of very small angle or by grinding a thick glass
plate.
(2) Acute angle of prism is about 1/2o and obtuse angle of prism is about 179o.
(3) When a monochromatic light source is kept in front of biprism two coherent virtual source S1 and S2 are
produced.
(4) Interference fringes are found on the screen (in the MN region) placed behind the biprism
interference fringes are formed in the limited region which can be observed with the help eye piece.
(5) Fringe width is measured by a micrometer attached to the eye piece. Fringes are of equal width and
D d
its value is A1
d D S1
M
d S C B
N
S2
A2
a b
D
Let the separation between S1 and S2 be d and the distance of slits and the screen from the biprism be
a and b respectively i.e. D = (a + b). If angle of prism is and refractive index is then d 2a( 1)
Diffraction of Light.
It is the phenomenon of bending of light around the corners of an obstacle/aperture of the size of the
wavelength of light.
Shadow
Light
Light Shadow
Aperture Obstacle
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16 Wave Optics
The minimum distance at which the observer should be from the obstacle to observe the
d2
diffraction of light of wavelength around the obstacle of size d is given by x .
4
(1) Types of diffraction : The diffraction phenomenon is divided into two types
Source Source
Screen Screen
at
Slit Slit
(2) Diffraction of light at a single slit : In case of diffraction at a single slit, we get a central bright
band with alternate bright (maxima) and dark (minima) bands of decreasing intensity as shown
P I
x
x + x
S1
d O
S2 d sin 3 2 2 3
D Screen d d d d d d
Slit
2 D 2
(i) Width of central maxima 0 ; and angular width
d d
(ii) Minima occurs at a point on either side of the central maxima, such that the path difference
between the waves from the two ends of the aperture is given by n ; where n 1, 2, 3 ....
n
i.e. d sin n sin
d
(iii) The secondary maxima occurs, where the path difference between the waves from the two ends of
the aperture is given by (2n 1) ; where n 1, 2, 3 ....
2
(2n 1)
i.e. d sin (2n 1) sin
2 2d
(3) Comparison between interference and diffraction
Interference Diffraction
Results due to the superposition of waves from two Results due to the superposition of wavelets from
coherrent sources. different parts of same wave front. (single coherent
source)
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Wave Optics 17
(4) Diffraction and optical instruments : The objective lens of optical instrument like telescope
or microscope etc. acts like a circular aperture. Due to diffraction of light at a circular
aperture, a converging lens cannot form a point image of an object rather it produces a
brighter disc known as Airy disc surrounded by alternate dark and bright concentric
rings.
1 .22
The angular half width of Airy disc (where D = aperture of lens)
D
The lateral width of the image f (where f = focal length of the lens)
Note : Diffraction of light limits the ability of optical instruments to form clear images of objects
when they are close to each other.
(5) Diffraction grating : Consists of large number of equally spaced parallel slits. If light is incident
normally on a transmission grating, the diffraction of principle maxima (PM) is given by d sin n ;
where d = distance between two consecutive slits and is called grating element.
Light
II I PMCentral I PM II
PM maxim PM
a
Polarisation of Light R2 V2 R1 V1 V1 R1 V2 R2
Light propagates as transverse EM waves. The magnitude of electric field is much larger as compared
to magnitude of magnetic field. We generally prefer to describe light as electric field oscillations.
(1) Unpolarised light
The light having electric field oscillations in all directions in the plane perpendicular to the direction of
propagation is called Unpolarised light. The oscillation may be resolved into horizontal and vertical
component.
Direction of
propagation
(2) Polarised light Direction of Vertical oscillation Horizontal
propagation
The light having oscillations only in one plane is called Polarised or plane polarised light. oscillation
(i) The plane in which oscillation occurs in the polarised light is called plane of oscillation.
(ii) The plane perpendicular to the plane of oscillation is called plane of polarisation.
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(iii) Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.
(3) Polaroids
It is a device used to produce the plane polarised light. It is based on the principle of selective
absorption and is more effective than the tourmaline crystal. or
It is a thin film of ultramicroscopic crystals of quinine idosulphate with their optic axis parallel to
each other.
Polaroid
(i) Polaroids allow the light oscillations parallel to the transmission axis pass through them.
(ii) The crystal or polaroid on which unpolarised light is incident is called polariser. Crystal or polaroid
on which polarised light is incident is called analyser.
P A P A
No light
I0 A
If 0 o , I I 0 , A A 0 , If 45 o , I , A 0 , If 90 o , I 0 , A 0
2 2
(ii) If Ii Intensity of unpolarised light.
Ii
So I 0 i.e. if an unpolarised light is converted into plane polarised light (say by passing it through
2
I
a plaroid or a Nicol-prism), its intensity becomes half. and I i cos 2
2
(I max I min )
Note : Percentage of polarisation 100
(I max I min )
(5) Brewster’s law : Brewster discovered that when a beam of unpolarised light is reflected from a
transparent medium (refractive index =), the reflected light is completely plane polarised at a certain
angle of incidence (called the angle of polarisation p ). Unpolaris Plane
ed light P P polarised
light
Partial
polarised
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 19
Substance
The optical activity of a substance is related to the asymmetry of the molecule or crystal as a whole,
e.g., a solution of cane-sugar is dextro-rotatory due to asymmetrical molecular structure while crystals of
quartz are dextro or laevo-rotatory due to structural asymmetry which vanishes when quartz is fused.
Optical activity of a substance is measured with help of polarimeter in terms of 'specific rotation'
which is defined as the rotation produced by a solution of length 10 cm (1 dm) and of unit concentration
(i.e. 1 g/cc) for a given wavelength of light at a given temperature. i.e. [ ]t o C where is the
LC
rotation in length L at concentration C.
(7) Applications and uses of polarisation
(i) By determining the polarising angle and using Brewster's law, i.e. = tanP, refractive index of dark
transparent substance can be determined.
(ii) It is used to reduce glare.
(iii) In calculators and watches, numbers and letters are formed by liquid crystals through polarisation
of light called liquid crystal display (LCD).
(iv) In CD player polarised laser beam acts as needle for producing sound from compact disc which is
an encoded digital format.
(v) It has also been used in recording and reproducing three-dimensional pictures.
(vi) Polarisation of scattered sunlight is used for navigation in solar-compass in polar regions.
(vii) Polarised light is used in optical stress analysis known as 'photoelasticity'.
(viii) Polarisation is also used to study asymmetries in molecules and crystals through the
phenomenon of 'optical activity'.
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20 Wave Optics
Assignment
Nature of light and interference of light
1. The dual nature of light is exhibited by [KCET 1999; AIIMS 2001; BHU 2001; Bihar CEE 2004]
(a) Diffraction and photoelectric effect (b) Diffraction and reflection
(c) Refraction and interference (d) Photoelectric effect
2. Huygen wave theory allows us to know [AFMC 2004]
(a) The wavelength of the wave (b) The velocity of the wave
(c) The amplitude of the wave (d) The propagation of wave fronts
3. When a beam of light is used to determine the position of an object, the maximum accuracy is achieved if the light is [AIIMS 2003]
(a) Polarised (b) Of longer wavelength (c) Of shorter wavelength (d) Of high intensity
4. Which of the following phenomenon does not show the wave nature of light [RPET 2003; MP PMT 2003]
(a) Diffraction (b) Interference (c) Refraction (d) Photoelectric effect
5. As a result of interference of two coherent sources of light, energy is [MP PMT 2002; KCET 2003]
(a) Increased
(b) Redistributed and the distribution does not vary with time
(c) Decreased
(d) Redistributed and the distribution changes with time
6. To demonstrate the phenomenon of interference, we require two sources which emit radiation [AIEEE 2003]
(a) Of the same frequency and having a definite phase relationship
(b) Of nearly the same frequency
(c) Of the same frequency
(d) Of different wavelengths
7. Consider the following statements
Assertion (A): Thin films such as soap bubble or a thin layer of oil on water show beautiful colours, when illuminated by
white light.
Reason (R) : It happens due to the interference of light reflected from the upper surface of the thin film.
Of these statements [AIIMS 2002]
(a) Both A and R are true but R is a correct explanation of A (b) Both A and R are true but R is not a correct
explanation of A
(c) A is true but R is false (d) A is false but R is true
(e) Both A and R are false
8. When light passes from one medium into another medium, then the physical property which does not change is
[CPMT 1990; MNR 1995; AMU 1995; UPSEAT 1999, 2000; MP PET 2002; RPET 1996, 2003; AFMC 1993, 98, 2003]
(a) Velocity (b) Wavelength (c) Frequency (d) Refractive index
9. The frequency of light ray having the wavelength 3000Å is [DPMT 2002]
(a) 9 10 13
cycles/sec (b) 10 15
cycles/sec (c) 90 cycles/sec (d) 3000 cycles/sec
10. Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum
intensity is 25. The intensities of the sources are in the ratio [RPMT 1989; UPSEAT 2002]
(a) 25 : 1 (b) 5 : 1 (c) 9 : 4 (d) 25 : 16
11. What is the path difference of destructive interference [AIIMS 2002]
(n 1) (2n 1)
(a) n (b) n( + 1) (c) (d)
2 2
12. Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum
possible intensities in the resulting beam are
[IIT-JEE 1988; AIIMS 1997; MP PMT 1997; MP PET 1999; KCET (Engg./Med.) 2000; MP PET 2002]
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 21
(a) 5I and I (b) 5I and 3I (c) 9I and I (d) 9I and 3I
13. Laser beams are used to measure long distance because [DCE 2001]
(a) They are monochromatic (b) They are highly polarised
(c) They are coherent (d) They have high degree of parallelism
14. Wave nature of light is verified by [RPET 2001]
(a) Interference (b) Photoelectric effect (c) Reflection (d) Refraction
15. If the wavelength of light in vacuum be , the wavelength in a medium of refractive index n will be [UPSEAT 2001; MP PET 2001]
(a) n (b) (c) (d) n 2
n n2
16. Newton postulated his corpuscular theory on the basis of [UPSEAT 2001; KCET 2001]
(a) Newton’s rings (b) Colours of thin films
(c) Rectilinear propagation of light (d) Dispersion of white light
17. Two coherent sources of intensities. I1 and I2 produce an interference pattern. The maximum intensity in the interference
pattern will be [UPSEAT 2001; MP PET 2001]
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31. An oil flowing on water seems coloured due to interference. For observing this effect, the approximate thickness of the oil
film should be [DPMT 1987; JIPMER 1997]
(a) 100 Å (b) 10000 Å (c) 1 mm (d) 1 cm
32. If L is the coherence length and c the velocity of light, the coherent time is [MP PMT 1996]
L c 1
(a) cL (b) (c) (d)
c L Lc
33. By a monochromatic wave, we mean [AFMC 1995]
(a) A single ray (b) A single ray of a single colour
(c) Wave having a single wavelength (d) Many rays of a single colour
34. Two coherent sources of light produce destructive interference when phase difference between them is [MP PMT 1994; CPMT 1995]
(a) 2 (b) (c) /2 (d) 0
35. Which one of the following statements is correct [KCET 1994]
(a) In vacuum, the speed of light depends upon frequency
(b) In vacuum, the speed of light does not depend upon frequency
(c) In vacuum, the speed of light is independent of frequency and wavelength
(d) In vacuum, the speed of light depends upon wavelength
36. Figure here shows P and Q as two equally intense coherent sources emitting radiations of wavelength 20 m. The separation
PQ is 5.0 m and phase of P is ahead of the phase of Q by 90°. A, B and C are three distant points of observation equidistant
from the mid-point of PQ. The intensity of radiations at A, B, C will bear the ratio [NSEP 1994]
B
(a) 0 : 1 : 4
(b) 4 : 1 : 0
(c) 0 : 1 : 2 P Q
(d) 2 : 1 : 0 C A
37. In Huygen’s wave theory, the locus of all points in the same state of vibration is called [CBSE PMT 1993]
(a) A half period zone (b) Vibrator (c) A wavefront (d) A ray
38. The idea of the quantum nature of light has emerged in an attempt to explain [CPMT 1990]
(a) Interference (b) Diffraction
(c) Radiation spectrum of a black body (d) Polarisation
39. The necessary condition for an interference by two source of light is that the [RPMT 1988; CPMT 1989]
(a) Two monochromatic sources should be of same amplitude but with a constant phase
(b) Two sources should be of same amplitude
(c) Two point sources should have phase difference varying with time
(d) Two sources should be of same wavelength
40. If the intensity of the waves observed by two coherent sources is I. Then the intensity of resultant waves in constructive
interference will be [RPET 1988]
(a) 2I (b) 4I (c) I (d) None of these
41. In figure, a wavefront AB moving in air is incident on a plane glass surface xy. Its position CD after refraction through a
glass slab is shown also along with normals drawn at A and D. the refractive index of glass with respect to air will be equal to [CPMT 19
sin B
(a)
sin
sin x D
(b) A y
sin
C
(c) (BD/AC)
(d) (AB/CD)
42. Four independent waves are expressed as
(i) y1 a1 sin t (ii) y 2 a2 sin 2t (iii) y3 a3 cos t (iv) y4 a4 sin(t / 3)
The interference is possible between [CPMT 1986]
(a) (i) and (ii) (b) (i) and (iv) (c) (iii) and (iv) (d) Not possible at all
43. Colour of light is known by its [MP PMT 1984]
(a) Velocity (b) Amplitude (c) Frequency (d) Polarisation
44. Laser light is considered to be coherent because it consists of [CPMT 1972]
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 23
(a) Many wavelengths (b) Uncoordinated wavelengths
(c) Coordinated waves of exactly the same wavelength (d) Divergent beams
45. A laser beam may be used to measure very large distances because [CPMT 1972]
(a) It is unidirectional (b) It is coherent (c) It is monochromatic (d) It is not absorbed
46. Interference patterns are not observed in thick films, because
(a) Most of the incident light intensity is observed within the film
(b) A thick film has a high coefficient of reflection
(c) The maxima of interference patterns are far from the minima
(d) There is too much overlapping of colours washing out the interference pattern
47. Phenomenon of interference is not observed by two sodium lamps of same power. It is because both waves have
(a) Not constant phase difference (b) Zero phase difference
(c) Different intensity (d) Different frequencies
Basic Level
48. In a Young’s double slit experiment, the separation between the two slits is 0.9 mm and the fringes are observed one metre
away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic
source of light used is
[KCET 2004]
(a) 500 nm (b) 600 nm (c) 450 nm (d) 400 nm
49. A monochromatic beams of light is used for the formation of fringes on the screen by illuminating the two slits in the
Young’s double slit mica is interposed in the path of one of the interfering beams then [AIIMS 2004]
(c) The fringe width remains the same but the pattern shifts
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56. In a Young’s double slit experiment, the source illuminating the slits is changed from blue to violet. The width of the fringes
[Kerala CET (Med.) 2002]
(a) Increases (b) Decreases (c) Becomes unequal (d) Remains constant
57. In Young’s double slit experiment, the intensity of light coming from the first slit is double the intensity from the second slit.
The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is
[KCET (Med.) 2002]
(a) 34 (b) 40 (c) 25 (d) 38
58. In Young’s double slit experiment the wavelength of light was changed from 7000Å to 3500Å. While doubling the separation
between the slits which of the following is not true for this experiment [Orissa JEE 2002]
(a) The width of the fringes changes
(b) The colour of bright fringes changes
(c) The separation between successive bright fringes changes
(d) The separation between successive dark fringes remains unchanged
59. In Young’s double slit experiment, the central bright fringe can be identified [KCET (Engg.) 2002]
(a) By using white light instead of monochromatic light (b) As it is narrower than other bright fringes
(c) As it is wider than other bright fringes (d) As it has a greater intensity than the other bright
fringes
60. Interference was observed in interference chamber when air was present, now the chamber is evacuated and if the same light
is used, a careful observer will see [CBSE PMT 1993; DPMT 2000; BHU 2002]
(a) No interference
(b) Interference with bright bands
(c) Interference with dark bands
(d) Interference in which width of the fringe will be slightly increased
61. A slit of width a is illuminated by white light. For red light ( 6500 Å) . The first minima is obtained at 30 o . Then the
value of a will be [MP PMT 1987; CPMT 2002]
(a) 3250 Å (b) 6.5 10 4 mm (c) 1.24 microns (d) 2.6 10 4 cm
62. In the Young’s double slit experiment with sodium light. The slits are 0.589 m apart. The angular separation of the third
maximum from the central maximum will be (given = 589 mm) [Pb. PMT 2002]
(a) sin1 (0.33 10 8 ) (b) sin1 (0.33 10 6 ) (c) sin1 (3 10 8 ) (d) sin1 (3 10 6 )
63. In the Young’s double slit experiment for which colour the fringe width is least [MP PMT 1994; UPSEAT 2001; MP PET 2001]
(a) Red (b) Green (c) Blue (d) Yellow
64. In a Young’s double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the
distance D of the screen from the slits should be made [AMU (Engg.) 2001]
D D
(a) (b) (c) 2D (d) 4D
2 2
65. Consider the following statements
Assertion (A): In Young’s experiment, the fringe width for dark fringes is different from that for bright fringes.
Reason (R) : In Young’s double slit experiment performed with a source of white light, only black and bright fringes are
observed
Of these statements [AIIMS 2001]
(a) Both A and R are true and R is a correct explanation of A (b) Both A and R are true but R is not a correct
explanation of A
(c) Both A and R are false (d) A is false but R is true
(e) A is true but R is false
66. In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of
wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same
segment of the screen is given by [IIT-JEE (Screening) 2001]
(a) 12 (b) 18 (c) 24 (d) 30
67. In Young’s double slit experiment, a mica slit of thickness t and refractive index is introduced in the ray from the first
source S 1 . By how much distance the fringes pattern will be displaced [RPMT 1996, 97; JIPMER 2000]
d D d D
(a) ( 1)t (b) ( 1)t (c) (d) ( 1)
D d ( 1)D d
68. Young’s double slit experiment is performed with light of wavelength 550 nm. The separation between the slits is 1.10 mm
and screen is placed at distance of 1 m. What is the distance between the consecutive bright or dark fringes [Pb. PMT 2000]
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 25
(a) 1.5 mm (b) 1.0 m (c) 0.5 mm (d) None of these
69. In interference obtained by two coherent sources, the fringe width () has the following relation with wavelength ()
[CPMT 1997; MP PMT 2000]
(a) 2
(b) (c) 1/ (d) 2
70. In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the
interference pattern [IIT-JEE (Screening) 2000]
(a) The intensities of both the maxima and the minima increase
(b) The intensity of maxima increases and the minima has zero intensity
(c) The intensity of maxima decreases and that of the minima increases
(d) The intensity of maxima decreases and the minima has zero intensity
71. In Young’s double slit experiment with a source of light of wavelength 6320Å, the first maxima will occur when
[Roorkee 1999]
(a) Path difference is 9480 Å (b) Phase difference is 2 radian
(c) Path difference is 6320 Å (d) Phase difference is radian
72. If a transparent medium of refractive index = 1.5 and thickness t 2.5 10 5 m is inserted in front of one of the slits of
Young’s double slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5
mm and that between slits and screen is 100 cm [AIIMS 1999]
(a) 5 cm (b) 2.5 cm (c) 0.25 cm (d) 0.1 cm
73. If a torch is used in place of monochromatic light in Young’s experiment what will happens
[MH CET (Med.) 1999; KCET (Med.) 1999]
(a) Fringe will appear for a moment then it will disappear (b) Fringes will occur as from monochromatic light
(c) Only bright fringes will appear (d) No fringes will appear
74. When a thin metal plate is placed in the path of one of the interfering beams of light [KCET (Engg./Med.) 1999]
(a) Fringe width increases (b) Fringes disappear (c) Fringes become brighter (d) Fringes become
blurred
75. What happens by the use of white light in Young’s double slit experiment
[Similar to (AIIMS 2001; Kerala 2000); IIT-JEE 1987; RPMT 1993; MP PMT 1996; RPET 1998; UPSEAT 1999]
(a) Bright fringes are obtained
(b) Only bright and dark fringes are obtained
(c) Central fringe is bright and two or three coloured and dark fringes are observed
(d) None of these
76. Young’s experiment is performed in air and then performed in water, the fringe width [CPMT 1990; MP PMT 1994; RPMT 1997]
(a) Will remain same (b) Will decrease (c) Will increase (d) Will be infinite
77. In Young’s experiment, one slit is covered with a blue filter and the other (slit) with a yellow filter. Then the interference
pattern
[MP PET 1997]
(a) Will be blue (b) Will be yellow (c) Will be green (d) Will not be formed
78. Two sources give interference pattern which is observed on a screen. D distance apart from the sources. The fringe width is
2w. If the distance D is now doubled, the fringe width will [MP PET 1997]
(a) Become w/2 (b) Remain the same (c) Become w (d) Become 4w
79. In Young’s double slit experiment, angular width of fringes is 0.20° for sodium light of wavelength 5890 Å. If complete
system is dipped in water, then angular width of fringes becomes [RPET 1997]
(a) 0.11° (b) 0.15° (c) 0.22° (d) 0.30°
80. In two separate set-ups of the Young’s double slit experiment, fringes of equal width are observed when lights of wavelengths
in the ratio 1 : 2 are used. If the ratio of the slit separation in the two cases is 2 : 1, the ratio of the distances between the
plane of the slits and the screen in the two set-ups is [Kurukshetra CEE 1996]
(a) 4 : 1 (b) 1 : 1 (c) 1 : 4 (d) 2 : 1
81. In a Young’s double slit experiment, the central point on the screen is [MP PMT 1996]
(a) Bright (b) Dark (c) First bright and then dark (d) First dark and then
bright
82. In Young’s double slit experiment, the distance between sources is 1 mm and distance between the screen and source is 1m.
If the fringe width on the screen is 0.06 cm, then = [CPMT 1996]
(a) 6000 Å (b) 4000 Å (c) 1200 Å (d) 2400 Å
83. In a Young’s double slit experiment, the distance between two coherent sources is 0.1 mm and the distance between the slits
and the screen is 20 cm. If the wavelength of light is 5460 Å then the distance between two consecutive maxima is [RPMT 1995]
(a) 0.5 mm (b) 1.1 mm (c) 1.5 mm (d) 2.2 mm
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84. If a thin mica sheet of thickness t and refractive index (5 / 3) is placed in the path of one of the interfering beams as
shown in figure, then the displacement of the fringe system is [CPMT 1995]
Dt
(a)
3d P
t
Dt
(b) S1
5d
2d
Dt
(c) S2
4d D
2 Dt
(d)
5d
85. In a double slit experiment, the first minimum on either side of the central maximum occurs where the path difference
between the two paths is [CPMT 1995]
(a) (b) (c) (d) 2
4 2
86. In Young’s double slit experiment, the phase difference between the light waves reaching third bright fringe from the central
fringe will be ( = 6000 Å) [MP PMT 1994]
(a) Zero (b) 2 (c) 4 (d) 6
7
87. Sodium light ( 6 10 m) is used to produce interference pattern. The observed fringe width is 0.12 mm. The angle
between the two interfering wave trains is [CPMT 1993]
(a) 5 10 1 rad (b) 5 10 3 rad (c) 1 10 2 rad (d) 1 10 3 rad
88. The contrast in the fringes in any interference pattern depends on [Roorkee 1992]
(a) Fringe width (b) Intensity ratio of the sources
(c) Distance between the slits (d) Wavelength
89. In Young’s double slit experiment, carried out with light of wavelength = 5000 Å, the distance between the slits is 0.2 mm
and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central
maximum as zeroth maximum) will be at x equal to [CBSE PMT 1992]
(a) 1.67 cm (b) 1.5 cm (c) 0.5 cm (d) 5.0 cm
90. In a Young’s experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one metre away. If it
produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used
would be
[CBSE PMT 1992]
4 4 5 5
(a) 60 10 cm (b) 10 10 cm (c) 10 10 cm (d) 60 10 cm
91. In Fresnel’s biprism, coherent sources are obtained by [RPET 1991]
(a) Division of wavefront (b) Division of amplitude (c) Division of wavelength (d) None of these
92. In Young’s experiment, the ratio of maximum and minimum intensities in the fringe system is 9 : 1. The ratio of amplitudes
of coherent sources is [NCERT 1990]
(a) 9 : 1 (b) 3 : 1 (c) 2 : 1 (d) 1 : 1
93. In a certain double slit experimental arrangement interference fringes of width 1.0 mm each are observed when light of
wavelength 5000 Å is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength 6000 Å,
the fringe width will be [CPMT 1988]
(a) 0.5 mm (b) 1.0 mm (c) 1.2 mm (d) 1.5 mm
94. In Young’s double slit experiment, if the slit widths are in the ratio 1 : 9, then the ratio of the intensity at minima to that at
maxima will be [MP PET 1987]
(a) 1 (b) 1/9 (c) 1/4 (d) 1/3
95. The Young’s experiment is performed with the lights of blue ( = 4360 Å) and green colour ( = 5460 Å). If the distance of
the 4th fringe from the centre is x, then [CPMT 1987]
x (Blue) 5460
(a) x(Blue) x(Green ) (b) x(Blue) x(Green ) (c) x(Blue) x(Green ) (d)
x (Green ) 4360
96. In Young’s experiment, keeping the distance of the slit from screen constant if the slit width is reduced to half, then [CPMT 1986]
(a) The fringe width will be doubled (b) The fringe width will
reduce to half
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 27
(c) The fringe width will not change (d) The fringe width will
become 2 times
97. In Young’s experiment, if the distance between screen and the slit aperture is increased the fringe width will [RPET 1986]
(a) Decrease (b) Increases but intensity will decrease
(c) Increase but intensity remains unchanged (d) Remains unchanged but intensity decreases
98. In Fresnel’s biprism experiment, the two coherent sources are [RPET 1985]
(a) Real (b) Imaginary
(c) One is real and the other is imaginary (d) None of these
99. In Fresnel’s experiment, the width of the fringe depends upon the distance [RPET 1985]
(a) Between the prism and the slit aperture
(b) Of the prism from the screen
(c) Of screen from the imaginary light sources
(d) Of the screen from the prism and the distance from the imaginary sources
100. In the Young’s double slit experiment, the ratio of intensities of bright and dark fringes is 9. This means that [IIT-JEE 1982]
(a) The intensities of individual sources are 5 and 4 units respectively
(b) The intensities of individual sources are 4 and 1 units respectively
(c) The ratio of their amplitudes is 3
(d) The ratio of their amplitudes is 2
101. The figure below shows a double slit experiment. P and Q are the slits. The path lengths PX and QX are n and (n + 2)
respectively where n is a whole number and is the wavelength. Taking the central bright fringe as zero, what is formed at X
n X
(a) First bright
P
(n+2)
(b) First dark
(c) Second bright Q
Screen
(d) Second dark
102. A plate of thickness t made of a material of refractive index is placed in front of one of the slits in a double slit experiment.
What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero
(a) ( 1) (b) ( 1) (c) (d)
2 2( 1) ( 1)
3
103. The thickness of a plate (refractive index for light of wavelength ) which will introduce a path difference of is
4
3 3 3
(a) (b) (c) (d)
4 ( 1) 2( 1) 2( 1) 4
Advance Level
104. In the Young’s double slit experiment, if the phase difference between the two waves interfering at a point is , the intensity
at that point can be expressed by the expression (where A + B depends upon the amplitude of the two waves)
[MP PMT/PET 1998; MP PMT 2003]
A
(a) I A 2 B 2 cos 2 (b) I cos (c) I A B cos / 2 (d) I A B cos
B
105. In the adjacent diagram CP represents wavefronts and AO and BP the corresponding two rays. Find the condition on for
constructive interference at P between the ray BP and reflected ray OP [IIT-JEE (Screening) 2003]
O
Q R
(a) cos 3 / 2d
C
d
(b) cos / 4d
A
(c) sec cos / d
P
B
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(d) sec cos 4 / d
106. When one of the slits of Young’s experiment is covered with a transparent sheet of thickness 4.8 mm, the central fringe shifts
to a position originally occupied by the 30th bright fringe. What should be the thickness of the sheet if the central fringe has
to shift to the position occupied by 20th bright fringe [KCET (Engg.) 2002]
(a) 3.8 mm (b) 1.6 mm (c) 7.6 mm (d) 3.2 mm
107. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of
the interfering beams (wavelength ), the intensity at the position where the central maximum occurred previously remains
unchanged. The minimum thickness of the glass-plate is [IIT-JEE (Screening) 2002)]
2
(a) 2 (b) (c) (d)
3 3
108. In an interference arrangement similar to Young’s double slit experiment, the slits S 1 and S 2 are illuminated with coherent
microwave sources each of frequency 106 Hz. The sources are synchronized to have zero phase difference. The slits are
separated by distance d = 150 m. The intensity I() is measured as a function of , where is defined as shown. If I0 is
maximum intensity, then I() for 0 90 is given by [IIT-JEE 1995]
b2 2b 2 b2 2b 2
(a) (b) (c) (d)
d d 3d 3d
110. In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If
the screen is moved by 5 10 2 m towards the slits, the change in fringe width is 3 10 5 m . If separation between the slits is
10 3 m , the wavelength of light used is [Roorkee 1992]
(a) 6000 Å (b) 5000 Å (c) 3000 Å (d) 4500 Å
111. In the figure is shown Young’s double slit experiment. Q is the position of the first bright fringe on the right side of O. P is the
11th fringe on the other side, as measured from Q. If the wavelength of the light used is 6000 10 10 m , then S1 B will be
equal to
[CPMT 1986, 92]
6
(a) 6 10 m Q
S1 B
6
(b) 6.6 10 m O
S2
(c) 3.138 10 7 m
Slit P
(d) 3.144 10 7 m
112. In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude A and wavelength . In another
experiment with the same set up the two slits are of equal amplitude A and wavelength but are incoherent. The ratio of the
intensity of light at the mid-point of the screen in the first case to that in the second case is [IIT-JJE 1986]
(a) 1 : 2 (b) 2 : 1 (c) 4 : 1 (d) 1 : 1
113. When light of wavelength falls on a thin film of thickness t and refractive index n, the essential condition for the
production of constructive interference fringes by the rays A and B are (m = 1, 2, 3, ……)
1 A
(a) 2nt cos r m
2 B
S1
(a) At O O
S2
(b) Above O
(c) Below O
(d) Anywhere depending on angle , thickness of plate t and refractive index of glass
118. In Young's double slit experiment how many maximas can be obtained on a screen (including the central maximum) on both
sides of the central fringe if 2000 Å and d 7000 Å
(a) 12 (b) 7 (c) 18 (d) 4
119. Young's double slit experiment is made in a liquid. The 10 th bright fringe in liquid lies where 6th dark fringe lies in vacuum.
The refractive index of the liquid is approximately
(a) 1.8 (b) 1.54 (c) 1.67 (d) 1.2
120. Light of wavelength 0 in air enters a medium of refractive index n. If two points A and B in this medium lie along the path
of this light at a distance x, then phase difference 0 between these two points is
1 2
2 2 1 2
(a) 0 x (b) 0 n x
(c) 0 (n 1) x
(d) 0 x
n 0
0
0 (n 1) 0
121. In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength
0 750 nm and 900nm . The minimum distance from the common central bright fringe on a screen 2m from the slits
where a bright fringe from one interference pattern coincides with a bright fringe from the other is
(a) 1.5 mm (b) 3 mm (c) 4.5 mm (d) 6 mm
122. In the ideal double slit experiment, when a glass plate (refractive index 1.5) of thickness t is introduced in the path of one of
the interfering beams (wavelength ), the intensity at the position where the central maximum occurred previously remains
unchanged. The minimum thickness of the glass plate is
2
(a) 2 (b) (c) (d)
3 3
123. Two wavelengths of light 1 and 2 are sent through a Young's double slit apparatus simultaneously. If the third order 1
bright fringe coincides with the fourth order 2 bright fringe then
1 4 1 3 1 5 1 4
(a) (b) (c) (d)
2 3 2 4 2 4 2 5
124. A flake of glass (refractive index 1.5) is placed over one of the openings of a double slit apparatus. The interference pattern
displaces itself through seven successive maxima towards the side where the flake is placed. if wavelength of the diffracted
light is 600nm , then the thickness of the flake is
(a) 2100 nm (b) 4200 nm (c) 8400 nm (d) None of these
125. In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the
interference pattern
(a) The intensitites of both the maxima and the minima increase
(b) The intensity of the maxima increases and minima has zero intensity
(c) The intensity of the maxima decreases and that of minima increases
(d) The intensity of the maxima decreases and the minima has zero intensity
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30 Wave Optics
126. In Young's experiment the wavelength of red light is 7800 Å and that of blue light is 5200 Å. The value of n for which the
(n 1)th blue bright band coincides with the nth red band is
(a) 4 (b) 3 (c) 2 (d) 1
127. In a double slit experiment if 5th dark fringe is formed opposite to one of the slits, the wavelength of light is
2
d d2 d2 d2
(a) (b) (c) (d)
6D 5D 15 D 9D
128. In a Young's double slit experiment one of the slits is advanced towards the screen by a distance d / 2 and d n where n is
an odd integer and d is the initial distance between the slits. If I0 is the intensity of each wave from the slits, the intensity at
O is
(a) I0 S2
S
I0 O
(b)
4 S1
(c) 0
(d) 2I0
129. Two ideal slits S 1 and S 2 are at a distance d apart, and illuminated by light of wavelength passing through an ideal
source slit S placed on the line through S 2 as shown. The distance between the planes of slits and the source slit is D. A
screen is held at a distance D from the plane of the slits. The minimum value of d for which there is darkness at O is
3 D
(a)
2 S1
(b) D S O
S2
D
(c)
2
D D
(d) 3D
130. In a double slit experiment interference is obtained from electron waves produced in an electron gun supplied with voltage
V. if is the wavelength of the beam, D is the distance of screen, d is the spacing between coherent source, h is Planck’s
constant, e is charge on electron and m is mass of electron then fringe width is given as
hD 2 hD hd 2 hd
(a) (b) (c) (d)
2meV d meV d 2 meV D meV D
131. In a double slit arrangement fringes are produced using light of wavelength 4800 Å. One slit is covered by a thin plate of
glass of refractive index 1.4 and the other with another glass plate of same thickness but of refractive index 1.7. By doing so
the central bright shifts to original fifth bright fringe from centre. Thickness of glass plate is
(a) 8 m (b) 6 m (c) 4 m (d) 10 m
132. Two point sources X and Y emit waves of same frequency and speed but Y lags in phase behind X by 2l radian. If there is a
maximum in direction D the distance XO using n as an integer is given by
D
(a) (n l)
2 O
(b) (n l)
X
(c) (n l)
2
Y
(d) (n l)
133. A student is asked to measure the wavelength of monochromatic light. He sets up the apparatus sketched below. S1 , S 2 , S 3
are narrow parallel slits, L is a sodium lamp and M is a micrometer eye-piece. The student fails to observe interference
fringes. You would advise him to
(a) Increase the width of S 1 S2
L S1 5 cm M
(b) Decrease the distance between S 2 and S 3
S3
(c) Replace L with a white light source
(d) Replace M with a telescope 10 cm 60
cm
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 31
134. A beam with wavelength falls on a stack of partially reflecting planes with separation d. The angle that the beam should
make with the planes so that the beams reflected from successive planes may interfere constructively is (where n =1, 2, ……)
n
(a) sin1
d
n
(b) tan 1
d
n d
(c) sin1
2d
n
(d) cos 1
2d
135. In a double slit experiment the source slit S is at a distance D1 and the screen at a distance D2 from the plane of ideal slit
cuts S 1 and S 2 as shown. If the source slit is shifted to by parallel to S 1 S 2 ,
the central bright fringe will be shifted by S
S1
D D y
(a) y (b) – y (c) 2 y (d) 2y O
D1 D1 S2
136. A parallel beam of monochromatic light is used in a Young’s double slit experiment. The D1 slits are separated
D2 by a distance d
and the screen is placed parallel to the plane of the slits. The angle which the incident beam must make with the normal to
the plane of the slits to produce darkness at the position of central brightness is
(a) cos 1
d
2 S1
(b) cos 1
d
(c) sin 1 S2
d
(d) sin 1
2d
137. In a Young’s double slit experiment, let be the fringe width, and let I0 be the intensity at the central bright fringe. At a
distance x from the central bright fringe, the intensity will be
x x x I x
(a) I0 cos (b) I0 cos 2 (c) I0 cos 2 (d) 0 cos 2
4
138. In Young’s double slit experiment the distance d between the slits S 1 and S 2 is 1 mm. What should be the width of each slit
be so as to obtain 10 maxima of the two slit interference pattern with in the central maximum of the single slit diffraction
pattern
(a) 0.1 mm (b) 0.2 mm (c) 0.3 mm (d) 0.4 mm
Diffraction of light
139. When light is incident on a diffraction grating the zero order principal maximum will be [KCET 2004]
(a) One of the component colours (b) Absent
(c) Spectrum of the colours (d) White
140. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction
pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright
fringe is
[IIT-JEE 1994; KCET 2004]
(a) 1.2 mm (b) 1.2 cm (c) 2.4 cm (d) 2.4 mm
141. Consider the following statements
Assertion (A): When a tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the
centre of the shadow of the obstacle.
Reason (R) : Destructive interference occurs at the centre of the shadow.
Of these statements [AIIMS 2002]
(a) Both A and R are true and R is a correct explanation of A (b) Both A and R are true but R is not a correct
explanation of A
(c) A is true but R is false (d) A is false but R is true
(e) Both A and R are false
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genius PHYSICS by Pradeep Kshetrapal
32 Wave Optics
142. The light of wavelength 6328 Å is incident on a slit of width 0.2 mm perpendicularly situated at a distance of 9 m and the
central maxima between two minima, the angular is approximately [MP PMT 1987; Pb. PMT 2002]
(a) 0.36° (b) 0.18° (c) 0.72° (d) 0.08°
143. A diffraction pattern is obtained using a beam of red light. What happens if the red light is replaced by blue light
[KCET (Eng./Med.) 2000; BHU 2001]
(a) No change (b) diffraction bands become narrower and crowded
together
(c) Bands become broader and farther apart (d) Bands disappear
144. Angular width () of central maximum of a diffraction pattern on a single slit does not depend upon [DCE 2000, 2001]
(a) Distance between slit and source (b) Wavelength of light used
(c) Width of the slit (d) Frequency of light used
145. In order to see diffraction the thickness of the film is [J&K CEE 2001]
(a) 100 Å (b) 10,000 Å (c) 1 mm (d) 1 cm
146. What will be the angle of diffracting for the first minimum due to Fraunhoffer diffraction with sources of light of wave length
550 nm and slit of width 0.55 mm [Pb. PMT 2001]
(a) 0.001 rad (b) 0.01 rad (c) 1 rad (d) 0.1 rad
147. The bending of beam of light around corners of obstacles is called
[NCERT 1990; AFMC 1995; RPET 1997; CPMT 1999; JIPMER 2000]
(a) Reflection (b) Diffraction (c) Refraction (d) Interference
148. Diffraction effects are easier to notice in the case of sound waves than in the case of light waves because [RPET 1978; KCET 2000]
(a) Sound waves are longitudinal (b) Sound is perceived by the ear
(c) Sound waves are mechanical waves (d) Sound waves are of longer wavelength
149. Direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of
the slit)
[KCET 1999]
3 3
(a) a sin (b) a cos (c) a sin (d) a sin
2 2 2
150. A slit of size 0.15 cm is placed at 2.1 m from a screen. On illuminated it by a light of wavelength 5 10 5 cm . The width of
diffraction pattern will be [RPET 1999]
(a) 70 mm (b) 0.14 mm (c) 1.4 cm (d) 0.14 cm
151. Yellow light is used in a single slit diffraction experiment with a slit of 0.6 mm. If yellow light is replaced by x-rays, than the
observed pattern will reveal [IIT-JEE 1999]
(a) That the central maxima is narrower (b) More number of fringes
(c) Less number of fringes (d) No diffraction pattern
152. A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen
placed perpendicular to the direction of incident beam. At the first maximum of the diffraction pattern the phase difference
between the rays coming from the edges of the slit is [IIT–JEE 1995, 98]
(a) 0 (b) (c) (d) 2
2
153. Diffraction and interference of light suggest [CPMT 1995; RPMT 1998]
(a) Nature of light is electro-magnetic (b) Wave nature
(c) Nature is quantum (d) Nature of light is transverse
154. A light wave is incident normally over a slit of width 24 10 5 cm . The angular position of second dark fringe from the
central maxima is 30o. What is the wavelength of light [RPET 1995]
(a) 6000 Å (b) 5000 Å (c) 3000 Å (d) 1500 Å
155. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 nm wide and the resulting diffraction
pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright
fringe is [IIT-JEE 1994]
(a) 1.2 cm (b) 1.2 mm (c) 2.4 cm (d) 2.4 mm
156. A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of width 0.001
mm. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the
angle of diffraction equal to [CBSE PMT 1993]
(a) 0o (b) 15o (c) 30o (d) 60o
157. Light appears to travel in straight lines since [RPMT 1997; AIIMS 1998; CPMT 1987, 89, 90, 2001; KCET (Engg.) 2002; BHU 2002]
(a) It is not absorbed by the atmosphere (b) It is reflected by the atmosphere
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 33
(c) It's wavelength is very small (d) It's velocity is very large
158. The condition for observing Fraunhofer diffraction from a single slit is that the light wavefront incident on the slit should be
[MP PMT 1987]
(a) Spherical (b) Cylindrical (c) Plane (d) Elliptical
159. The position of the direct image obtained at O, when a monochromatic beam of light is passed through a plane transmission
grating at normal incidence is shown in fig.
O A B C
The diffracted images A, B and C correspond to the first, second and third order diffraction when the source is replaced by
an another source of shorter wavelength [CPMT 1986]
(a) All the four shift in the direction C to O (b) All the four will shift in the direction O to C
(c) The images C, B and A will shift toward O (d) The images C, B and A will shift away from O
160. To observe diffraction the size of an obstacle [CPMT 1982]
(a) Should be of the same order as wavelength (b) Should be much larger than the wavelength
(c) Have no relation to wavelength (d) Should be exactly
2
161. The first diffraction minima due to a single slit diffraction is at 30 o for a light of wavelength 5000 Å. The width of the
slit is
[CPMT 1985]
(a) 5 10 5 cm (b) 1.0 10 4 cm (c) 2.5 10 5 cm (d) 1.25 10 5 cm
162. Radio waves diffract pronoucedly around buildings while light waves which are also electromagnetic waves do not because [PPE 1978]
(a) Wavelength of the radio waves is not comparable with the size of the obstacle
(b) Wavelength of radio waves is of the order of 200-500 m hence they bend more than the light waves whose wavelength is
very small
(c) Light waves are transverse whereas radio waves are longitudinal
(d) None of the above
163. One cannot obtain diffraction from a wide slit illuminated by a monochromatic light because [PPE 1978]
(a) The half period elements contained in a wide slit are very large so the resultant effect is general illumination
(b) The half period elements contained in a wide slit are small so the resultant effect is general illumination
(c) Diffraction patterns are superimposed by interference pattern and hence the result is general illumination
(d) None of these
164. In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength
1 is found to be coincident with the third maximum at 2 . So
(a) 31 0.32 (b) 31 2 (c) 1 3.5 2 (d) 0.31 32
165. In case of Fresnel diffraction
(a) Both source and screen are at finite distance from diffracting device
(b) Source is at finite distance while screen at infinity from diffraction device
(c) Screen is at finite distance while source at infinity from diffracting device
(d) Both source and screen are effectively at infinity from diffracting device
166. Light of wavelength = 5000 Å falls normally on a narrow slit. A screen placed at a distance of 1 m from the slit and
perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 mm from the centre of central
maximum. The width of the slit is
(a) 0.1 mm (b) 1.0 mm (c) 0.5 mm (d) 0.2 mm
167. Light falls normally on a slit of width 0.3 mm. A lens of focal length 40 cm collects the rays at its focal plane. The distance of
the first dark band from the direct one is 0.8 mm. The wavelength of light is
(a) 4800 Å (b) 5000 Å (c) 6000 Å (d) 5896 Å
168. A parallel monochromatic beam of light is incident at an angle to the normal of a slit of width e. The central point O of the
screen will be dark if
(a) e sin n where n = 1, 3, 5 ...
S1
(b) e sin n where n = 1, 2, 3 ...
O
S2
(c) e sin (2n 1) / 2 where n = 1, 2, 3 ......
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34 Wave Optics
(d) e cos n where n = 1, 2, 3, 4 ........
Polarization of Light
169. The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refraction index n) is [AIEEE 2004]
1 1
(a) sin 1 (n) (b) sin 1 (c) tan 1 (d) tan 1 (n)
n n
170. Through which character we can distinguish the light waves from sound waves [CBSE PMT 1990; RPET 2002]
(a) Interference (b) Refraction (c) Polarisation (d) Reflection
171. Which of following can not be polarised [Kerala PMT 2001]
(a) Radio waves (b) Ultraviolet rays (c) Infrared rays (d) Ultrasonic waves
172. A polaroid is placed at 45o to an incoming light of intensity I0 . Now the intensity of light passing through polaroid after
polarisation would be [CPMT 1995]
(a) I0 (b) I0 / 2 (c) I0 / 4 (d) Zero
173. Plane polarised light is passed through a polaroid. On viewing through the polaroid we find that when the polariod is given
one complete rotation about the direction of the light, one of the following is observed [MNR 1993]
(a) The intensity of light gradually decreases to zero and remains at zero
(b) The intensity of light gradually increases to a maximum and remains at maximum
(c) There is no change in intensity
(d) The intensity of light is twice maximum and twice zero
174. Out of the following statements which is not correct [CPMT 1991]
(a) When unpolarised light passes through a Nicol's prism, the emergent light is elliptically polarised
(b) Nicol's prism works on the principle of double refraction and total internal reflection
(c) Nicol's prism can be used to produce and analyse polarised light
(d) Calcite and Quartz are both doubly refracting crystals
175. A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster's angle . If represents
the refractive index of glass with respect to air, then the angle between reflected and refracted rays is [CPMT 1989]
A B
(a) The intensity is reduced down to zero and remains zero N
(b) The intensity reduces down some what and rises again 33° 33°
(c) There is no change in intensity O
(d) The intensity gradually reduces to zero and then again increases
178. Polarised glass is used in sun glasses because [CPMT 1981]
(a) It reduces the light intensity to half an account of polarisation (b) It is fashionable
(c) It has good colour (d) It is cheaper
179. In the propagation of electromagnetic waves the angle between the direction of propagation and plane of polarisation is [CPMT 1978]
(a) 0o (b) 45o (c) 90o (d) 180o
180. The transverse nature of light is shown by
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 35
[CPMT 1972, 74, 78; RPMT 1999; MP PMT 2000; AFMC 2001; AIEEE 2002; MP PET 2004]
(a) Interference of light (b) Refraction of light (c) Polarisation of light (d) Dispersion of light
181. A calcite crystal is placed over a dot on a piece of paper and rotated, on seeing through the calcite one will be see [CPMT 1971]
(a) One dot (b) Two stationary dots
(c) Two rotating dots (d) One dot rotating about the other
182. In a doubly refracting crystal, optic axis is a direction along which
(a) A plane polarised beam does not suffer deviation
(b) Any beam of light does not suffer any deviation
(c) Double refraction does not take place
(d) Ordinary and extraordinary rays undergo maximum deviation
183. Which is incorrect with reference to polarisation by reflection
(a) The degree of polarisation varies with the angle of incidence
(b) Percentage of the polarising light in the reflected beam is the greatest at the angle of polarisation
(c) Reflected light is plane polarised in the plane of incidence
(d) Reflected light is plane polarised in the plane perpendicular to plane of incidence
184. Two polarising plates have polarising directions parallel so as to transmit maximum intensity of light. Through what angle
must either plate be turned if the intensities of the transmitted beam is to drop by one-third
(a) 55o18' (b) 144o22' (c) Both of these (d) None of these
185. The polaroid is
(a) Celluloid film (b) Big crystal
(c) Cluster of small crystals arranged in a regular way (d) Cluster of small crystals arranged in a haphazard way
186. Light from the cloudless sky is
(a) Fully polarised (b) Partially polarised (c) Unpolarised (d) Can not be said
187. The observed wavelength of light coming from a distant galaxy is found to be increased by 0.5% as compared with that
comparing from a terrestrial source. The galaxy is [MP PMT 1993, 2003]
(a) Stationary with respect to the earth
(b) Approaching the earth with velocity of light
(c) Receding from the earth with the velocity of light
(d) Receding from the earth with a velocity equal to 1.5 10 6 m / s
188. In hydrogen spectrum the wavelength of H line is 656 nm whereas in the spectrum of a distant galaxy. H line wavelength
is 706nm. Estimated speed of the galaxy with respect to earth is [IIT-JEE 1999; UPSEAT 2003]
(a) 2 10 m / s
8
(b) 2 10 m / s
7
(c) 2 10 m / s
6
(d) 2 10 5 m / s
189. A star emits light of 5500 Å wavelength. Its appears blue to an observer on the earth, it means [DPMT 2002]
(a) Star is going away from the earth (b) Star is stationary
(c) Star is coming towards earth (d) None of the above
190. The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5 Å. The speed with which the star is
receding from the earth is [Pb. PMT 2002]
192. A star emitting light of wavelength 5896 Å is moving away from the earth with a speed of 3600 km/sec. The wavelength of
light observed on earth will ( c 3 10 8 m / sec is the speed of light) [MP PET 1995, 2002]
(a) Decrease by 5825.25 Å (b) Increase by 5966.75 Å (c) Decrease by 70.75 Å (d) Increase by 70.75 Å
193. The periodic time of rotation of a certain star is 22 days and its radius is 7 10 m . If the wavelength of light emitted by its
8
surface be 4320 Å, the Doppler shift will be (1 day = 86400 sec) [MP PET 2001]
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genius PHYSICS by Pradeep Kshetrapal
36 Wave Optics
(a) 0.033 Å (b) 0.33 Å (c) 3.3 Å (d) 33 Å
194. A heavenly body is receding from earth such that the fractional change in is 1, then its velocity is [DCE 2000]
3C C 2C
(a) C (b) (c) (d)
5 5 5
195. A star is going away from the earth. An observer on the earth will see the wavelength of light coming from the star [MP PMT 1999]
(a) Decreased
(b) Increased
(c) Neither decreased nor increased
(d) Decreased or increased depending upon the velocity of the star
196. If the shift of wavelength of light emitted by a star is towards violet, then this shows that star is [RPET 1996; RPMT 1999]
(a) Stationary (b) Moving towards earth (c) Moving away from earth (d) Information is
incomplete
197. When the wavelength of light coming from a distant star is measured it is found shifted towards red. Then the conclusion is
[JIPMER 1999]
(a) The star is approaching the observer (b) The star recedes away from earth
(c) There is gravitational effect on the light (d) The star remains stationary
198. In the spectrum of light of a luminous heavenly body the wavelength of a spectral line is measured to be 4747 Å while actual
wavelength of the line is 4700 Å. The relative velocity of the heavenly body with respect to earth will be (velocity of light is
3 10 8 m / s ) [MP PET 1997; MP PMT/PET 1998]
(a) 3 10 5 m / s moving towards the earth (b) 3 10 5 m / s moving away from the earth
(c) 3 10 6 m / s moving towards the earth (d) 3 10 6 m / s moving away from the earth
199. The wavelength of light observed on the earth, from a moving star is found to decrease by 0.05%. Relative to the earth the
star is
[MP PMT/PET 1998]
(c) Moving away with a velocity of 1.5 10 4 m / s (d) Coming closer with a velocity of 1.5 10 4 m / s
200. Due to Doppler's effect, the shift in wavelength observed is 0.1 Å for a star producing wavelength 6000 Å. Velocity of
recession of the star will be [KCET 1998]
(a) 2.5 km/s (b) 10 km/s (c) 5 km/s (d) 20 km/s
201. A rocket is going away from the earth at a speed of 10 m / s . If the wavelength of the light wave emitted by it be 5700 Å,
6
what will be its Doppler's shift [MP PMT 1990, 94; RPMT 1996]
(a) 200 Å (b) 19 Å (c) 20 Å (d) 0.2 Å
202. A rocket is going away from the earth at a speed 0.2 c, where c = speed of light, it emits a signal of frequency 4 10 7 Hz .
What will be the frequency observed by an observer on the earth [RPMT 1996]
203. A star moves away from earth at speed 0.8 c while emitting light of frequency 6 10 14 Hz . What frequency will be observed
on the earth (in units of 10 14 Hz ) (c = speed of light) [MP PMT 1995]
(a) 0.24 (b) 1.2 (c) 30 (d) 3.3
204. The sun is rotating about its own axis. The spectral lines emitted from the two ends of its equator, for an observer on the
earth, will show [MP PMT 1994]
(a) Shift towards red end
(b) Shift towards violet end
(c) Shift towards red end by one line and towards violet end by other
(d) No shift
205. The time period of rotation of the sun is 25 days and its radius is 7 10 8 m . The Doppler shift for the light of wavelength
6000 Å emitted from the surface of the sun will be [MP PMT 1994]
(a) 0.04 Å (b) 0.40 Å (c) 4.00 Å (d) 40.0 Å
genius PHYSICS by Pradeep Kshetrapal
Wave Optics 37
206. The apparent wavelength of the light from a star moving away from the earth is 0.01 % more than its real wavelength. Then
the velocity of star is [CPMT 1979]
(a) 60 km/sec (b) 15 km/sec (c) 150 km/sec (d) 30 km/sec
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
a d c d b a c c b c d c d a b c d a c c
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
c b b c a d c d a d b b c b c d c c a b
41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
c d c c a d a b c a d d c a b b a d a d
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
c d c c c b b c b a b, c b d b c b d d b a
81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
a a b a b d b b b d a c c c c a b b d b,d
101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
c c a d b d a a,b a,c a a b a a,d c a d b a b
121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
c a a c a c d c c a a b b c d d c b d d
141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160
c a b a b a b d d b a c b a d c c c c a
161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180
b b a c a a c b d c d b d a c a d a a c
181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
d c c c c d d b c d c d a a b b b d b c
201 202 203 204 205 206
b b b c a d
Only for students of genius academy Jamanipali, Maxwell classes Korba, and Gupta classes Kusmunda.
genius PHYSICS
Reflection of Light 1
O I
When a ray of light after incidenting on a boundary separating two media comes back into the
same media, then this phenomenon, is called reflection of light.
Normal
∠i = ∠r
Incident Reflected ray After reflection, velocity, wave length and
ray frequency of light remains same but intensity
i r decreases
Boundary There is a phase change of if reflection takes
place from denser medium
Note : After reflection velocity, wavelength and frequency of light remains same but
intensity decreases.
If light ray incident normally on a surface, after reflection it retraces the path.
(Real image)
I O
O Real image
I
(Virtual object)
(Real image) (Virtual object)
(Real object)
O I
(Virtual image)
(Real object) (Virtual image) (Virtual image)
Plane Mirror.
The image formed by a plane mirror is virtual, erect, laterally inverted, equal in size that of
the object and at a distance equal to the distance of the object in front of the mirror.
x x
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2 Reflection of Light
(1) Deviation : Deviation produced by a plane mirror and by two inclined plane mirrors.
Final path
2
i r Original
path
Note : If two plane mirrors are inclined to each other at 90o, the emergent ray is anti-
parallel to incident ray, if it suffers one reflection from each. Whatever be the angle
to incidence.
(2) Rotation : If a plane mirror is rotated in the plane of incidence through angle , by
keeping the incident ray fixed, the reflected ray turned through an angle 2.
IR
IR RR
RR
(3) Images by two inclined plane mirrors : When two plane mirrors are inclined to each
other at an angle , then number of images (n) formed of an object which is kept between them.
360 360
(i) n 1 ; If even integer
360
(ii) If odd integer then there are two possibilities
(a) Object is placed symmetrically (b) Object is placed asymmetrically
360 360
n 1 Object n
Object
/2
/2
Note : If θ = 0o i.e. mirrors are parallel to each other so n i.e. infinite images will be
formed.
360
If θ = 90o, n 1 3
90
360
If θ = 72o, n 1 4 (If nothing is said object is supposed to be symmetrically
72
placed).
(4) Other important informations
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Reflection of Light 3
(i) When the object moves with speed u towards (or away) from the plane mirror then image
also moves toward (or away) with speed u. But relative speed of image w.r.t. object is 2u.
(ii) When mirror moves towards the stationary object with speed u, the image will move with
speed 2u.
O I O I
u u Rest 2u
u
Mirror at rest Mirror is
moving
(iii) A man of height h requires a mirror of length at least equal to h/2, to see his own
complete image.
(iv) To see complete wall behind himself a person requires a plane mirror of at least one third
the height of wall. It should be noted that person is standing in the middle of the room.
H
H E
E M' M'
h E h
h 3
h
2
M' M'
L B
d d
Example Concepts
s
The reflection from a denser medium causes an additional phase change of or path change of /2 while reflection from
rarer medium doesn't cause any phase change.
Incident light
We observe number of images in a thick plane mirror, out of them only second is brightest.
(100%)
10%
9%
0.9%
Correct Wrong
O OM = MI
OM = MI
Example: 1 A plane mirror makes an angle of 30o with horizontal. If a vertical ray strikes the mirror,
find the angle between mirror and reflected ray
(a) 30o (b) 45o (c) 60o (d) 90o
Solution : (c) Since angle between mirror and normal is 90o and reflected ray (RR) makes IR
30o
an angle of 30o with the normal so required angle will be 60 . o
30o
RR
= 60o
30o
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4 Reflection of Light
Example: 2 Two vertical plane mirrors are inclined at an angle of 60 o with each other. A ray of light
travelling horizontally is reflected first from one mirror and then from the other. The
resultant deviation is
(a) 60o (b) 120o (c) 180o (d) 240o
Solution : (d) By using (360 2 ) 360 2 60 240 o
Example: 3 A person is in a room whose ceiling and two adjacent walls are mirrors. How many images
are formed
[AFMC 2002]
I1 O
I1 O
I2 I3
Four images by
I2 I3 ceiling
Three images by walls
Note : The person will see only six images of himself (I1 , I 2 , I 3 , I1' , I 2' , I 3' )
Example: 4 A ray of light makes an angle of 10 o with the horizontal above it and strikes a plane mirror
which is inclined at an angle to the horizontal. The angle for which the reflected ray
becomes vertical is
(a) 40o (b) 50o (c) 80o (d) 100o
Solution : (a) From figure Vertical RR
10 90
IR
40 o
10o Horizontal line
Plane mirror
Example: 5 A ray of light incident on the first mirror parallel to the second and is reflected from the
second mirror parallel to first mirror. The angle between two mirrors is
(a) 30o (b) 60o (c) 75o (d) 90o
Solution : (b) From geometry of figure
180 o
60 o
Example: 6 A point object is placed mid-way between two plane mirrors distance 'a' apart. The plane
mirror forms an infinite number of images due to multiple reflection. The distance between
the nth order image formed in the two mirrors is
(a) na (b) 2na (c) na/2 (d) n2 a
Solution : (b)
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genius PHYSICS
M' M Reflection of Light 5
III II order I order I order II order III
order image image image image order
image O image
From above figure it can be proved that seperation between nth order image formed in the
two mirrors = 2na
Example: 7 Two plane mirrors P and Q are aligned parallel to each other, as shown in the figure. A light
ray is incident at an angle of at a point just inside one end of A. The plane of incidence
coincides with the plane of the figure. The maximum number of times the ray undergoes
reflections (including the first one) before it emerges out is
l
(a)
d tan l
d
(b) d
l tan
(c) ld tan
(d) None of these
Solution : (a) Suppose n = Total number of reflection light ray undergoes before exist out.
x = Horizontal distance travelled by light ray in one reflection.
l
x x
So nx = l also tan
d
l d
n
d tan
Example: 8 A plane mirror and a person are moving towards each other with same velocity v. Then the
velocity of the image is
(a) v (b) 2v (c) 3v (d) 4v
Solution : (c) If mirror would be at rest, then velocity of image should be 2v. but due to the motion of
mirror, velocity of image will be 2v + v = 3v.
Example: 9 A ray reflected successively from two plane mirrors inclined at a certain angle undergoes a
deviation of 300o. The number of images observable are
(a) 10 (b) 11 (c) 12 (d) 13
Solution : (b) By using (360 2 ) 300 360 2
360
30 o . Hence number of images 1 11
30
Tricky example: 1
A small plane mirror placed at the centre of a spherical screen of radius R. A beam of light is falling
on the mirror. If the mirror makes n revolution. per second, the speed of light on the screen after
reflection from the mirror will be
nR nR
(a) 4nR (b) 2nR (c) (d)
2 4
Solution : (a) When plane mirror rotates through an angle , the reflected ray rotates through an angle 2. So
spot on the screen will make 2n revolution per second
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genius PHYSICS
6 Reflection of Light
Tricky example: 2
A watch shows time as 3 : 25 when seen through a mirror, time appeared will be
[RPMT 1997; JIPMER 2001, 2002]
Tricky example: 3
When a plane mirror is placed horizontally on a level ground at a distance of 60 m from the foot of
a tower, the top of the tower and its image in the mirror subtend an angle of 90 o at the eye. The
height of the tower will be [CPMT 1984]
h
Solution : (b) Form the figure it is clear that tan 45 o
60
h = 60 m
Tower
45o
45o 60 m
Image
Curved Mirror.
It is a part of a transparent hollow sphere whose one surface is polished.
C P P C
F F
Principle axis
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genius PHYSICS
Reflection of Light 7
R
(vii) Relation between f and R : f (fconcare = –ve , fconvex = + ve , fplane = )
2
(viii) Power : The converging or diverging ability of mirror
(ix) Aperture : Effective diameter of light reflecting area.
Intensity of image Area (Aperture)2
(x) Focal plane : A plane passing from focus and perpendicular to principle
axis.
(2) Rules of image formation and sign convention :
Rule (i) Rule (ii) Rule (iii)
F F F F C C
Distance of object u – u – u –
Distance of image v – v + v +
Focal length f – f – f +
Height of object O + O+ O +
Height of image I – I + I +
Radius of curvature R – R – R +
Magnification m – m+ m +
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8 Reflection of Light
(4) Position, size and nature of image formed by the spherical mirror
Note : In case of convex mirrors, as the object moves away from the mirror, the image
becomes smaller and moves closer to the focus.
Images formed by mirrors do not show chromatic aberration.
For convex mirror maximum image distance is it’s focal length.
In concave mirror, minimum distance between a real object and it's real image is
zero.
(i.e. when u = v = 2f)
1 1 1
(1) Mirror formula : ; (use sign convention while solving the problems).
f v u
Note : Newton’s formula : If object distance (x1) and image distance (x2) are measured
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Reflection of Light 9
Size of object
(2) Magnification : m =
Size of image
Linear magnification
Areal magnification
Transverse Longitudinal
f
2
Ai
(L i ) .L o ms m2
u f Ao
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10 Reflection of Light
Different graphs
1 1
Graph between and
v u
(a) Real image formed by (b) Virtual image formed by (c) Virtual image formed by
concave mirror concave mirror convex mirror
1 1 1
v v v
1
1 1
u
u u
Graph between u and v Graph between u and m Graph between u and m
for real image of concave for virtual image by for virtual image by
mirror concave mirror convex mirror.
Hyperbola
m m
2f
f 1 1
f 2f f u u
Concepts
Focal length of a mirror is independent of material of mirror, medium in which it is placed, wavelength of
incident light
Divergence or Convergence power of a mirror does not change with the change in medium.
If an object is moving at a speed vo towards a spherical mirror along it’s axis then speed of image away from
2
f
mirror is vi .vo (use sign convention)
u f O
C P
When object is moved from focus to infinity at constant speed, the image will move F
faster in the beginning and slower later on, towards the mirror.
I
As every part of mirror forms a complete image, if a part of the mirror is obstructed,
full image will be formed but intensity will be reduced.
Example
s
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Reflection of Light 11
Example: 10 A convex mirror of focal length f forms an image which is 1/n times the object. The distance
of the object from the mirror is
n 1 n 1
(a) (n – 1) f (b) f (c) f (d) (n + 1) f
n n
f
Solution : (a) By using m
f u
1 1 f
Here m , f f So, u (n 1) f
n n f u
Example: 11 An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of
curvature of 20 cm. The size of the image is
(a) 0.11 cm (b) 0.50 cm (c) 0.55 cm (d) 0.60 cm
I f
Solution : (c) By using
O f u
R
Here O 5 cm , f 10 cm , u 1 m 100 cm
2
I 10
So, I = – 0.55 cm.
5 10 (100 )
Example: 12 An object of length 2.5 cm is placed at a distance of 1.5 f from a concave mirror where f is the
magnitude of the focal length of the mirror. The length of the object is perpendicular to the
principle axis. The length of the image is
(a) 5 cm, erect (b) 10 cm, erect (c) 15 cm, erect (d) 5 cm,
inverted
I f
Solution : (d) By using ; where I = ? , O = + 2.5 cm. f f , u = – 1.5 f
O f u
I f
I 5 cm. (Negative sign indicates that image is
2.5 f (1.5 f )
inverted.)
Example: 13 A convex mirror has a focal length f. A real object is placed at a distance f in front of it from
the pole produces an image at
(a) Infinity (b) f (c) f / 2 (d) 2f
1 1 1 1 1 1 f
Solution : (c) By using v
f v u f v f 2
Example: 14 Two objects A and B when placed one after another infront of a concave mirror of focal
length 10 cm from images of same size. Size of object A is four times that of B. If object A is
placed at a distance of 50 cm from the mirror, what should be the distance of B from the
mirror
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
I f I O f uB 1 1 10 u B
Solution : (b) By using A B u B 20 cm .
O f u IB O A f uA 1 4 10 50
Example: 15 A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length
10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the
axis. The area enclosed by the image of the wire is
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genius PHYSICS
12 Reflection of Light
(a) 4 cm2 (b) 6 cm2 (c) 16 cm2 (d) 36 cm2
Ai f
Solution : (a) By using m 2 ; where m
Ao f u
10 2
Hence from given values m and A o 9 cm 2
10 25 3
2
2
Ai 9 4 cm
2
3
Example: 16 A convex mirror of focal length 10 cm is placed in water. The refractive index of water is 4/3.
What will be the focal length of the mirror in water
(a) 10 cm (b) 40/3 cm (c) 30/4 cm (d) None of
these
Solution : (a) No change in focal length, because f depends only upon radius of curvature R.
Example: 17 A candle flame 3 cm is placed at distance of 3 m from a wall. How far from wall must a
concave mirror be placed in order that it may form an image of flame 9 cm high on the wall
(a) 225 cm (b) 300 cm (c) 450 cm (d) 650 cm
Solution : (c) Let the mirror be placed at a distance x from wall
3cm
By using
I v 9 x (x–
x 4.5m 450 cm. 3m
O u 3 x 3 3)m
x
Example: 18 A concave mirror of focal length 100 cm is used to obtain the image of the sun which
subtends an angle of 30'. The diameter of the image of the sun will be
(a) 1.74 cm (b) 0.87 cm (c) 0.435 cm (d) 100 cm
Solution : (b) Diameter of image of sun d f Image of
sun
30
o
30 d 30 '
d 100 60
60 180 F
d 0.87 cm .
Example: 19 A thin rod of length f / 3 lies along the axis of a concave mirror of focal length f. One end of
its magnified image touches an end of the rod. The length of the image is [MP PET 1995]
1 1
(a) f (b) f (c) 2 f (d) f
2 4
f 5f
Solution : (b) If end A of rod acts an object for mirror then it's image will be A' and if u 2 f
3 3
1 1 1 1 1 1 5 2f
So by using v f
f v u f v 5 f 2 f/3 u = 2f – (f/3)
3 A
F
A' C
5 f
Length of image f 2 f
2 2 v
Example: 20 A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let
O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has
a real image, also located at C. If the mirror is now filled with water, the image will be
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Reflection of Light 13
(a) Real, and will remain at C (b) Real, and located at a point between
C and
(c) Virtual and located at a point between C and O (d) Real, and located at a point between
C and O
C Object C Object
Solution : (d) image
Image
O O
Initiall Finally
y
Tricky example: 4
An object is placed infront of a convex mirror at a distance of 50 cm. A plane mirror is introduced
covering the lower half of the convex mirror. If the distance between the object and plane mirror is
30 cm, it is found that there is no parallel between the images formed by two mirrors. Radius of
curvature of mirror will be
50
(a) 12.5 cm (b) 25 cm (c) cm (d) 18 cm
3
Solution : (b) Since there is no parallel, it means that both images (By plane mirror
and convex mirror) coinciding each other.
According to property of plane mirror it will form image at a distance of 30 cm behind it. Hence for
convex mirror u = – 50 cm, v = + 10 cm
Object
1 1 1 1 1 1 4
By using A
f v u f 10 50 50
30 cm 20 cm
25
f cm R 2 f 25 cm.
2 50 cm
10 cm
Tricky example: 5
A convergent beam of light is incident on a convex mirror so as to converge to a
distance 12 cm from the pole of the mirror. An inverted image of the same size
is formed coincident with the virtual object. What is the focal length of the mirror
(a) 24 cm (b) 12 cm (c) 6 cm (d) 3 cm
Solution : (c) Here object and image are at the same position so this position must
C
be centre of curvature
R
R = 12 cm f
2
1. A light bulb is placed between two mirrors (plane) inclined at an angle of 60o. Number of images formed are
[NCERT 1980; CPMT 1996, 97; SCRA 1994; AIIMS 1997; RPMT 1999; AIEEE 2002; Orissa JEE 2003; MP PET 2004]
(a) 2 (b) 4 (c) 5 (d) 6
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14 Reflection of Light
2. Two plane mirrors are inclined at an angle of 72 o . The number of images of a point object placed between them
will be
[KCET (Engg. & Med.)1999; BCECE 2003]
(a) 2 (b) 3 (c) 4 (d) 5
3. To get three images of a single object, one should have two plane mirrors at an angle of [AIEEE 2003]
(a) 30 o
(b) 60 o
(c) 90 o
(d) 120 o
4. A man of length h requires a mirror of length at least equal to, to see his own complete image [MP PET 2003]
h h h
(a) (b) (c) (d) h
4 3 2
5. Two plane mirrors are at 45o to each other. If an object is placed between them then the number of images will be [MP PMT 2003
(a) 5 (b) 9 (c) 7 (d) 8
6. An object is at a distance of 0.5 m in front of a plane mirror. Distance between the object and image is [CPMT 2002]
(a) 0.5 m (b) 1 m (c) 0.25 m (d) 1.5 m
7. A man runs towards a mirror at a speed 15 m/s. The speed of the image relative to the man is [RPMT 1999;
Kerala PET 2002]
(a) 15 ms 1 (b) 30 ms 1 (c) 35 ms 1 (d) 20 ms 1
8. The light reflected by a plane mirror may form a real image [KCET (Engg. & Med.) 2002]
(a) If the rays incident on the mirror are diverging (b) If the rays incident on the mirror are
converging
(c) If the object is placed very close to the mirror (d) Under no circumstances
9. A man is 180 cm tall and his eyes are 10 cm below the top of his head. In order to see his entire height right from
toe to head, he uses a plane mirror kept at a distance of 1 m from him. The minimum length of the plane mirror
required is [MP PMT 1993; DPMT 2001]
(a) 180 cm (b) 90 cm (c) 85 cm (d) 170 cm
10. A small object is placed 10 cm infront of a plane mirror. If you stand behind the object 30 cm from the object and
look at its image, the distance focused for your eye will be
(a) 60 cm (b) 20 cm (c) 40 cm (d) 80 cm
11. Two plane mirrors are at right angles to each other. A man stands between them and combs his hair with his right
hand. In how many of the images will he be seen using his right hand
(a) None (b) 1 (c) 2 (d) 3
12. A man runs towards mirror at a speed of 15 m/s. What is the speed of his image [CBSE PMT 2000]
(a) 7.5 m/s (b) 15 m/s (c) 30 m/s (d) 45 m/s
13. A ray of light is incidenting normally on a plane mirror. The angle of reflection will be [MP PET 2000]
(a) 0o (b) 90o (c) Will not be reflected (d) None of
these
14. A plane mirror produces a magnification of [MP PMT/PET 1997]
(a) – 1 (b) + 1 (c) Zero (d) Between 0
and +
15. When a plane mirror is rotated through an angle , then the reflected ray turns through the angle 2, then the size
of the image [MP PAT 1996]
(a) Is doubled (b) Is halved (c) Remains the same (d) Becomes
infinite
16. What should be the angle between two plane mirrors so that whatever be the angle of incidence, the incident ray
and the reflected ray from the two mirrors be parallel to each other
(a) 60o (b) 90o (c) 120o (d) 175o
17. Ray optics is valid, when characteristic dimensions are [CBSE PMT 1994]
(a) Of the same order as the wavelength of light (b) Much smaller than the wavelength of light
(c) Of the order of one millimeter (d) Much larger than the wavelength of light
18. It is desired to photograph the image of an object placed at a distance of 3 m from the plane mirror. The camera
which is at a distance of 4.5 m from the mirror should be focussed for a distance of
(a) 3 m (b) 4.5 m (c) 6 m (d) 7.5 m
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Reflection of Light 15
19. Two plane mirrors are parallel to each other an spaced 20 cm apart. An object is kept in between them at 15 cm
from A. Out of the following at which point an image is not formed in mirror A (distance measured from mirror A)
(a) 15 cm (b) 25 cm (c) 45 cm (d) 55 cm
Advance Level
20. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an
angle of 30o at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The
maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
2 3m
(a) 28
(b) 30 0.2m B
30o
(c) 32
(d) 34
A
21. A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a
wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. The
greatest distance over which he can see the image of the light source in the mirror is
(a) d/2
(b) d d
(c) 2d L B
2L A
(d) 3d
22. The figure shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror is
23. An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of
5 cm/s, then after 6 s the distance between the object and its image will be
X X
1 2 3 4
26. A point object O is placed between two plan mirrors as shown is fig. The distance of the first three images formed
by mirror M 2 from it are
M1 M2
(a) 2 mm, 8 mm, 18 mm
O
(b) 2 mm, 18 mm, 28 mm
(c) 2 mm, 18 mm, 22 mm 10mm 2mm
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16 Reflection of Light
(d) 2 mm, 18 mm, 58 mm
27. A plane mirror is placed at the bottom of the tank containing a liquid of refractive index . P is a small object at a
height h above the mirror. An observer O-vertically above P outside the liquid see P and its image in the mirror.
The apparent distance between these two will be
2h O
(a) 2h (b)
P
2h 1 h
(c) (d) h 1
1
28. One side of a glass slab is silvered as shown. A ray of light is incident on the other side at angle of incidence
i 45 o . Refractive index of glass is given as 1.5. The deviation of the ray of light from its initial path when it
comes out of the slab is
(a) 90o 45o
(b) 180o
= 1.5
(c) 120o
(d) 45o
29. If an object moves towards a plane mirror with a speed v at an angle to the perpendicular to the plane of the
mirror, find the relative velocity between the object and the image
y
(a) v O I
(b) 2v
vO vI
(c) 2v cos x
(d) 2v sin
30. Figure shows a cubical room ABCD will the wall CD as a plane mirror. Each side of the room is 3m. We place a
camera at the midpoint of the wall AB. At what distance should the camera be focussed to photograph an object
placed at A
(a) 1.5 m (b) 3 m A B
D C
Basic Level
31. A man having height 6 m, want to see full height in mirror. They observe image of 2m height erect, then used mirror
is [J & K CET 2004]
(a) Concave (b) Convex (c) Plane (d) None of
these
32. An object of length 6cm is placed on the principal axis of a concave mirror of focal length f at a distance of 4 f. The
length of the image will be [MP PET 2003]
(a) 2 cm (b) 12 cm (c) 4 cm (d) 1.2 cm
33. Convergence of concave mirror can be decreased by dipping in [AFMC 2003]
(a) Water (b) Oil (c) Both (d) None of these
34. In an experiment of find the focal length of a concave mirror a graph is drawn between the magnitudes of u and v.
The graph looks like
v v v v
(a) (b) (c) (d)
u u u u
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Reflection of Light 17
35. An object 2.5 cm high is placed at a distance of 10 cm from a concave mirror of radius of curvature 30 cm The size
of the image is
[BVP 2003]
(a) 9.2 cm (b) 10.5 cm (c) 5.6 cm (d) 7.5 cm
36. A diminished virtual image can be formed only in [MP PMT 2002]
(a) Plane mirror (b) A concave mirror (c) A convex mirror (d) Concave-
parabolic mirror
37. A point object is placed at a distance of 30 cm from a convex mirror of focal length 30cm. The image will form at [JIPMER 2002]
(a) Infinity (b) Focus (c) Pole (d) 15 cm
behind the mirror
38. The focal length of a convex mirror is 20 cm its radius of curvature will be [MP PMT 2001]
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
39. A concave mirror of focal length 15 cm forms an image having twice the linear dimensions of the object. The
position of the object when the image is virtual will be
(a) 22.5 cm (b) 7.5 cm (c) 30 cm (d) 45 cm
40. Under which of the following conditions will a convex mirror of focal length f produce an image that is erect,
diminished and virtual
[AMU (Engg.) 2001]
(a) Only when 2f > u > f (b) Only when u = f (c) Only when u < f (d) Always
41. A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the
image to be real, the focal length should be [SCRA 1998; JIPMER 2
(a) 10 cm (b) 15 cm (c) 20 cm (d) 30 cm
42. A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20cm from a concave
mirror. If the object is moved by 0.1cm towards the mirror, the image will shift by about
(a) 0.4 cm away from the mirror (b) 0.4 cm towards the mirror
(c) 0.8 cm away from the mirror (d) 0.8 cm towards the mirror
43. The minimum distance between the object and its real image for concave mirror is [RPMT 1999]
(a) f (b) 2f (c) 4f (d) Zero
44. An object is placed at 20 cm from a convex mirror of focal length 10 cm. The image formed by the mirror is [JIPMER 1999]
(a) Real and at 20 cm from the mirror (b) Virtual and at 20 cm from the mirror
(c) Virtual and at 20/3 cm from the mirror (d) Real and at 20/3 cm from the mirror
45. An object is placed 40 cm from a concave mirror of focal length 20 cm. The image formed is [MP PET 1986; MP PMT/PET 199
(a) Real, inverted and same in size (b) Real, inverted and smaller
(c) Virtual, erect and larger (d) Virtual, erect and smaller
46. Match List I with List II and select the correct answer using the codes given below the lists [SCRA 1998]
List I List II
(Position of the object) (Magnification)
(I) An object is placed at focus before a convex mirror (A) Magnification is –
(II) An object is placed at centre of curvature before a concave mirror (B) Magnification is 0.5
(III) An object is placed at focus before a concave mirror (C) Magnification is + 1
(IV) An object is placed at centre of curvature before a convex mirror (D) Magnification is – 1
(E) Magnification is 0.33
Codes :
(a) I-B, II-D, III-A, IV-E (b) I-A, II-D, III-C, IV-B (c) I-C, II-B, III-A, IV-E (d) I-B, II-E,
III-D, IV-C
47. In a concave mirror experiment, an object is placed at a distance x 1 from the focus and the image is formed at a
distance x 2 from the focus. The focal length of the mirror would be
x1 x 2 x1
(a) x1 x 2 (b) x1 x 2 (c) (d)
2 x2
48. Which of the following forms a virtual and erect image for all positions of the object [IIT-JEE 1996]
(a) Convex lens (b) Concave lens (c) Convex mirror (d) Concave
mirror
49. A convex mirror has a focal length f. A real object is placed at a distance f in front of it from the pole produces an
image at
[MP PAT 1996]
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18 Reflection of Light
(a) Infinity (b) f (c) f / 2 (d) 2f
50. Radius of curvature of concave mirror is 40 cm and the size of image is twice as that of object, then the object
distance is [AFMC 1995]
(a) 60 cm (b) 20 cm (c) 40 cm (d) 30 cm
51. All of the following statements are correct except [Manipal MEE 1995]
(a) The magnification produced by a convex mirror is always less than one
(b) A virtual, erect, same-sized image can be obtained using a plane mirror
(c) A virtual, erect, magnified image can be formed using a concave mirror
(d) A real, inverted, same-sized image can be formed using a convex mirror
52. Radius of curvature of convex mirror is 40 cm and the size of object is twice as that of image, then the image
distance is [AFMC 1995]
(a) 10 cm (b) 20 cm (c) 40 cm (d) 30 cm
53. If an object is placed 10 cm in front of a concave mirror of focal length 20 cm, the image will be [MP PMT 1995]
(a) Diminished, upright, virtual (b) Enlarged, upright, virtual (c)
Diminished, inverted, real (d) Enlarged, upright, real
54. An object 1 cm tall is placed 4 cm in front of a mirror. In order to produce an upright image of 3 cm height one needs
a [SCRA 1994]
(a) Convex mirror of radius of curvature 12 cm (b) Concave mirror of radius of curvature 12
cm
(c) Concave mirror of radius of curvature 4 cm (d) Plane mirror of height 12 cm
55. The image formed by a convex mirror of a real object is larger than the object [CPMT 1994]
(a) When u < 2f (b) When u > 2f (c) For all values of u (d) For no
value of u
56. An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm. The
size of the image is
[MP PET 1993]
(a) 0.11 cm (b) 0.50 cm (c) 0.55 cm (d) 0.60 cm
57. A virtual image three times the size of the object is obtained with a concave mirror of radius of curvature 36 cm.
The distance of the object from the mirror is
(a) 5 cm (b) 12 cm (c) 10 cm (d) 20 cm
58. Given a point source of light, which of the following can produce a parallel beam of light [CPMT 1974]
(a) Convex mirror (b) Concave mirror
(c) Concave lens (d) Two plane mirrors inclined at an angle of
90o
59. A convex mirror is used to form the image of an object. Then which of the following statements is wrong
(a) The images lies between the pole and the focus (b) The image is diminished in size
(c) The images is erect (d) The image is real
60. A boy stands straight infront of a mirror at a distance of 30 cm away from it. He sees his erect image whose height
1 th
is of his real height. The mirror he is using is
5
(a) Plane mirror (b) Convex mirror (c) Concave mirror (d) Plano-
convex mirror
61. For the largest distance of the image from a concave mirror of focal length 10cm, the object should be kept at
(a) 10 cm (b) Infinite (c) 40 cm (d) 60 cm
62. A dentist uses a small mirror that gives a magnification of 4 when it is held 0.60 cm from a tooth. The radius of
curvature of the mirror is
(a) 1.60 cm (convex) (b) 0.8 cm (concave) (c) 1.60 cm (concave) (d) 0.8 cm
(convex)
63. A dice is placed with its one edge parallel to the principal axis between the principal focus and the centre of the
curvature of a concave mirror. Then the image has the shape of
(a) Cube (b) Cuboid (c) Barrel shaped (d) Spherical
Advance Level
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Reflection of Light 19
64. A short linear object of length l lies along the axis of a concave mirror of focal length f at a distance u form the pole
of the mirror. The size of the image is approximately equal to [IIT 1988; BHU 2003]
1/2 2 1/2 2
u f u f f f
(a) l (b) l (c) l (d) l
f f u f u f
65. A point object is moving on the principal axis of a concave mirror of focal length 24 cm towards the mirror. When
it is at a distance of 60 cm from the mirror, its velocity is 9 cm/sec. What is the velocity of the image at that instant
(a) 5 cm/sec towards the mirror (b) 4 cm/sec
towards the mirror
(c) 4 cm/sec away from the mirror (d) 9 cm/sec away from the mirror
66. A convex mirror of focal length 10 cm forms an image which is half of the size of the object. The distance of the
object from the mirror is
(a) 10 cm (b) 20 cm (c) 5 cm (d) 15 cm
67. A concave mirror is used to focus the image of a flower on a nearby well 120 cm from the flower. If a lateral
magnification of 16 is desired, the distance of the flower from the mirror should be
(a) 8 cm (b) 12 cm (c) 80 cm (d) 120 cm
68. A thin rod of 5 cm length is kept along the axis of a concave mirror of 10 cm focal length such that its image is real
and magnified and one end touches the rod. Its magnification will be
(a) 1 (b) 2 (c) 3 (d) 4
69. A luminous object is placed 20 cm from surface of a convex mirror and a plane mirror is set so that virtual images
formed in two mirrors coincide. If plane mirror is at a distance of 12 cm from object, then focal length of convex
mirror, is
(a) 5 cm (b) 10 cm (c) 20 cm (d) 40 cm
70. A rear mirror of a vehicle is cylindrical having radius of curvature 10 cm. The length of arc of curved surface is also
10 cm. If the eye of driver is assumed to be at large distance, from the mirror, then the field of view in radian is
(a) 0.5 (b) 1 (c) 2 (d) 4
71. A vehicle has a driving mirror of focal length 30 cm. Another vehicle of dimension 2 4 1 .75 m 3 is 9 m away
from the mirror of first vehicle. Position of the second vehicle as seen in the mirror of first vehicle is
(a) 30 cm
(b) 60 cm
9m
(c) 90 cm
(d) 9 cm
72. A cube of side 2 m is placed in front of a concave mirror focal length 1m with its face P at a distance of 3 m and
face Q at a distance of 5 m from the mirror. The distance between the images of face P and Q and height of images
of P and Q are
(a) 1 m, 0.5 m, 0.25 m
Q 2m P
(b) 0.5 m, 1 m, 0.25 m 2m
(c) 0.5 m, 0.25 m, 1m 3m
(d) 0.25 m, 1m, 0.5 m
73. A concave mirror of radius of curvature 60 cm is placed at the bottom of tank containing water upto a height of 20
cm. The mirror faces upwards with its axis vertical. Solar light falls normally on the surface of water and the image
4
of the sun is formed. If a w then with the observer in air, the distance of the image from the surface of water
3
is
(a) 30 cm (b) 10 cm (c) 7.5 cm above (d) 7.5 cm
below
74. A concave mirror forms an image of the sun at a distance of 12 cm from it
(a) The radius of curvature of this mirror is 6 cm
(b) To use it as a shaving mirror, it must be held at a distance of 8-10 cm from the face
(c) If an object is kept at a distance of 12 cm from it, the image formed will be of the same size as the object
(d) All the above a alternatives are correct
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20 Reflection of Light
75. A small piece of wire bent into an L shape with upright and horizontal portions of equal lengths, is placed with the
horizontal portion along the axis of the concave mirror whose radius of curvature is 10 cm. If the bend is 20 cm
from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of
the wire is
(a) 1 : 2 (b) 3 : 1 (c) 1 : 3 (d) 2 : 1
76. As the position of an object (u) reflected from a concave mirror is varied, the position of the image (v) also varies.
By letting the u changes from 0 to the graph between v versus u will be
77.
v v v v
(a) b c d
u u u u
78. A concave mirror has a focal length 20 cm. The distance between the two positions of the object for which the
image size is double of the object size is
(a) 20 cm (b) 40 cm (c) 30 cm (d) 60 cm
79. A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are placed facing each other 40
cm apart. A point object is placed between the mirrors, on their common axis and 15 cm from the concave mirror.
Find the position and nature of the image produced by the successive reflections, first at concave mirror and then
at convex mirror
(a) 2 cm (b) 4 cm (c) 6 cm (d) 8 cm
Answer Sheet
Assignments
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
c c c c c b b b b c b b a b c b d d c b
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
d a b b c c b a c d b a d c d c d d b d
41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
b a d c a a b b, c c d d a b b d c b b d b
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78
a c b d c a a a a b a d c b b a a c
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Reflection of Light 21
F F
The bending of the ray of light passing from one medium to the other medium is called refraction.
Incident ray
i Rarer medium
Denser medium
i
r
r
Denser medium Refracted ray
Rarer medium
Deviation = (i – r)
Deviation = (r – i )
Snell’s law
The ratio of sine of the angle of incidence to the angle of refraction (r) is a constant called
refractive index
sin i sin i
i.e. (a constant). For two media, Snell's law can be written as 1 2 2
sin r 1 sin r
1 sin i 2 sin r i.e. sin constant
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22 Reflection of Light
3 4 g 3 / 2 9
a glass 1 . 5 , a water 1 . 33 w g
2 3 w 4 / 3 8
a diamond 2.4, a Cs 2 1 .62 (b) When light enters from glass to diamond :
D 2 .4 8
a crown 1.52, vacuum 1 , air 1.0003 1 g D
g 1 .5 5
B C
Note : Cauchy’s equation : A ...... ( Red violet so Red violet) 1
2
4
2 1 v 1 1
If a light ray travels from medium (1) to medium (2), then 1 2
1 2 v 2 v
v
(2) Dependence of Refractive index
(i) Nature of the media of incidence and refraction.
(ii) Colour of light or wavelength of light.
(iii) Temperature of the media : Refractive index decreases with the increase in temperature.
(3) Principle of reversibility of light and refraction through several media :
Principle of reversibility Refraction through several media
Incident ray 1
1
i
2
r
3
2
1
1
1 2
1 2 2 3 3 1 1
2 1
Normal shift
1
Normal shift OO ' x 1 t
O O' Glass
x slab
t
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Reflection of Light 23
1 2
Light
For two medium in contact optical path = 1 x 1 2 x 2
x1 x2
Note : Since for all media 1, so optical path length (x ) is always greater than the
geometrical path length (x).
Real and Apparent Depth.
If object and observer are situated in different medium then due to refraction, object appears
to be displaced from it’s real position. There are two possible conditions.
(1) When object is in denser medium and observer is (1) Object is in rarer medium and observer is in denser
in rarer medium medium. O d
h
O h
h
h O
d
O
1 (3) d ( 1)h
(3) Shift d h h ' 1 h
4 h h
(4) For water d (4) Shift for water d w
3 4 3
3 h h
For glass d Shift for glass d g
2 3 2
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24 Reflection of Light
1 d1
2 d2
d1 d2 d3
Apparent depth of bottom .... 3 d3
1 2 3
d AC d d 2 ..... 2 1 2
combination = 1 (In case of two liquids if d 1 d 2 than )
d App . d1 d 2 1 2
....
1 2
Total Internal Reflection.
When a ray of light goes from denser to rarer medium it bends away from the normal and as
the angle of incidence in denser medium increases, the angle of refraction in rarer medium also
increases and at a certain angle, angle of refraction becomes 90o, this angle of incidence is called
critical angle (C).
When Angle of incidence exceeds the critical angle than light ray comes back in to the same medium
after reflection from interface. This phenomenon is called Total internal reflection (TIR).
Rarer r
90o
Denser i
i=C
>C TIR
1
Important formula μ cosec C ; where Rerer Denser
sin C
Note : When a light ray travels from denser to rarer medium, then deviation of the ray is
2 max. when min. C
i.e. max ( 2C) ; C critical angle
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Reflection of Light 25
(iii) Temperature : With temperature rise refractive index of the material decreases therefore
critical angle increases.
(2) Examples of total internal reflection (TIR)
(i) Denser Rarer
i>C Sky I
O i>C
Rarer O
I Desner
Earth Earth
(iv) Field of vision of fish (or swimmer) : A fish (diver) inside the water can see the
whole world through a cone with.
(a) Apex angle 2C 98 o r
h
(b) Radius of base r h tan C C >C
2 1 h
C C
h 2
(c) Area of base A
( 2 1)
9h 2
Note : For water 4 so r
3h
and A .
37 7
(v) Porro prism : A right angled isosceles prism, which is used in periscopes or binoculars.
It is used to deviate light rays through 90 o and 180 o and also to erect the image.
B A
45o
B
90o A
90o
45o 45o 45o
45o 45o
90o 45o 45o 45o
Example
s
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26 Reflection of Light
Example: 1 A beam of monochromatic blue light of wavelength 4200 Å in air travels in water ( 4 / 3) .
Its wavelength in water will be
(a) 2800 Å (b) 5600 Å (c) 3150 Å (d) 4000 Å
1 1 2 1 2
Solution: (c) 2 3150 Å
2 1 4 4200
3
Example: 2 On a glass plate a light wave is incident at an angle of 60 o. If the reflected and the refracted
waves are mutually perpendicular, the refractive index of material is [MP
PMT 1994; Haryana CEE 1996]
3 3 1
(a) (b) 3 (c) (d)
2 2 3
Example: 3 Velocity of light in glass whose refractive index with respect to air is 1.5 is 2 × 10 8 m / s and in
certain liquid the velocity of light found to be 2.50 10 8 m / s . The refractive index of the
liquid with respect to air is
[CPMT 1978; MP
PET/PMT 1988]
(a) 0.64 (b) 0.80 (c) 1.20 (d) 1.44
1 li v g 2 10 8
Solution: (c) l l 1.2
v g vl 1 .5 2 .5 10 8
Example: 4 A ray of light passes through four transparent media with refractive indices
1 . 2 , 3 , and 4 as shown in the figure. The surfaces of all media are parallel. If the
emergent ray CD is parallel to the incident ray AB, we must have
(a) 1 2
D
(b) 2 3 1
2 3 4
B C
(c) 3 4
A
(d) 4 1
Solution: (d) For successive refraction through difference media sin constant.
sin i sin r
Solution: (b) For glass water interface g ......(i) and For water-air interface a
sin r sin 90
.....(ii)
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Reflection of Light 27
1
g a sin i g
sin i
Example: 6 The ratio of thickness of plates of two transparent mediums A and B is 6 : 4. If light takes
equal time in passing through them, then refractive index of B with respect to A will be
(a) 1.4 (b) 1.5 (c) 1.75 (d) 1.33
x
Solution: (b) By using t
c
B x A 6 3
AB 1 .5
A xB 4 2
Example: 7 A ray of light passes from vacuum into a medium of refractive index , the angle of incidence
is found to be twice the angle of refraction. Then the angle of incidence is
(a) cos 1 / 2 (b) 2 cos 1 / 2 (c) 2 sin 1 (d)
2 sin 1 / 2
r cos 1 . So, i 2r 2 cos 1 .
2 2
Example: 8 A ray of light falls on the surface of a spherical glass paper weight making an angle with
the normal and is refracted in the medium at an angle . The angle of deviation of the
emergent ray from the direction of the incident ray is
(a) (b) 2 (c) / 2 (d)
( ) 2
Example: 9 A rectangular slab of refractive index is placed over another slab of refractive index 3, both
slabs being identical in dimensions. If a coin is placed below the lower slab, for what value of
will the coin appear to be placed at the interface between the slabs when viewed from the
top
(a) 1.8 (b) 2 (c) 1.5 (d) 2.5
x x
Solution: (c) Apparent depth of coin as seen from top x
1 2 2 = x
1 1 1 1
1 1 1.5 1 = 3 x
1 2 3
Example: 10 A coin is kept at bottom of an empty beaker. A travelling microscope is focussed on the coin
from top, now water is poured in beaker up to a height of 10 cm. By what distance and in
which direction should the microscope be moved to bring the coin again in focus
(a) 10 cm up ward (b) 10 cm down ward (c) 2.5 cm up wards (d) 2.5 cm
down wards
h 10
Solution: (c) When water is poured in the beaker. Coin appears to shift by a distance d 2 .5 cm
4 4
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28 Reflection of Light
Hence to bring the coil again in focus, the microscope should be moved by 2.5 cm in upward
direction.
4
Example: 11 Consider the situation shown in figure. Water w is filled in a breaker upto a height
3
of 10 cm. A plane mirror fixed at a height of 5 cm from the surface of water. Distance of
image from the mirror after reflection from it of an object O at the bottom of the beaker is
(a) 15 cm (b) 12.5 cm (c) 7.5 cm (d) 10 cm
cm 2 .5 cm
10
Solution: (b) From figure it is clear that object appears to be raised by
4 5 cm
Hence distance between mirror and O' 5 7.5 12 .5 cm
10 cm
10 O'
So final image will be formed at 12.5 cm behind the plane mirror. c
4
O
m
Example: 12 The wavelength of light in two liquids 'x' and 'y' is 3500 Å and 7000 Å, then the critical angle of x
relative to y will be
(a) 60o (b) 45o (c) 30o (d) 15o
2 1 3500 1
Solution: (c) sin C C 30 o
1 2 7000 2
Example: 13 A light ray from air is incident (as shown in figure) at one end of a glass fiber (refractive
index = 1.5) making an incidence angle of 60o on the lateral surface, so that it undergoes a
total internal reflection. How much time would it take to traverse the straight fiber of length
1 km [Orissa JEE 2002]
(a) sin 8 / 9 B A
(b) 2 / 3 sin 8 / 9
(c) sin 2 / 3
C
(d) cos 8 / 9
Solution: (a) From figure it is clear that
Total internal reflection takes place at AC, only if > C
B A
1
sin sin C sin
g
C
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Reflection of Light 29
1 8
sin sin
9/8 9
Example: 15 When light is incident on a medium at angle i and refracted into a second medium at an
angle r, the graph of sin i vs sin r is as shown in the graph. From this, one can conclude that
(a) Velocity of light in the second medium is 1.73 times the
velocity of light in the I medium
(b) Velocity of light in the I medium is 1.73 times the velocity in sin
the II medium r
1
(c) The critical angle for the two media is given by sin ic 30o
3 sin
i
1
(d) sin ic
2
sin r 1 2 v1
Solution: (b, c) From graph tan 30 o 1 2 3 1 .73 v1 1.75 v 2
sin i 1 2 1 v 2
1 1 1 1
Also from sin C sin C .
sin C Rarer Denser 1 2 3
Example: 16 A beam of light consisting of red, green and blue colours is incident on a right angled prism.
The refractive indices of the material of the prism for the above red, green and blue
wavelength are 1.39, 1.44 and 1.47 respectively. The prism will
(a) Separate part of red colour from the green and the blue
colours
(b) Separate part of the blue colour from the red and green
colours 45
°
(c) Separate all the colours from one another
(d) Not separate even partially any colour from the other two colours
Solution: (a) At face AB, i = 0 so r = 0, i.e., no refraction will take place. So light will be incident on face
AC at an angle of incidence of 45o. The face AC will not transmit the light for which i C ,
i.e., sin i sin C
A
or sin 45 o 1 / i.e., 2 ( 1 .41)
Now as R while G and B , so red will be transmitted
45°
through the face AC while green and blue will be reflected. So the
45°
prism will separate red colour from green and blue. B C
Example: 17 An air bubble in a glass slab 1.5 is 6 cm deep when viewed from
one face and 4 cm deep when viewed from the opposite face. The thickness of the glass plate
is
(a) 10 cm (b) 6.67 cm (c) 15 cm (d) None of
these
Solution: (c) Let thickness of slab be t and distance of air bubble from one side is x 6 cm 4 cm
x
When viewed from side (1) : 1 . 5 x 9 cm Air
6 bubble
Side 1 x Side 2
When viewed from side (2) : 1 . 5
(t x )
1 .5
t 9 t 15 cm t
4 4
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30 Reflection of Light
Tricky example: 1
One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the first
face, forms an image 12 cm behind the silvered face. The refractive index of the glass is
(a) 0.4 (b) 0.8 (c) 1.2 (d) 1.6
Solution : (c) From figure thickness of glass plate t = 6 cm. x
t 6 8 cm 12 cm
Also 1 .2
Tricky example:
x 2 5
12 +(6–x)
t
A ray of light is incident on a glass sphere of refractive index 3/2. What should be the angle of incidence so that the
ray which enters the sphere doesn't come out of the sphere
2 2 1
(a) tan 1 (b) sin 1 (c) 90o (d) cos 1
3 3 3
Solution : (c) Ray doesn't come out from the sphere means TIR takes place.
Hence from figure ABO OAB C
A B
1 1 2 i
sin C C C
sin C 3
O
sin i 3 3 3 2
Applying Snell's Law at A sin i sin C 1 i 90 o
sin C 2 2 2 3
Tricky example: 3
The image of point P when viewed from top of the slabs will be
(a) 2.0 cm above P (b) 1.5 cm above P (c) 2.0 cm below P (d) 1 cm above P
Solution: (d) The two slabs will shift the image a distance
1.5 cm
1 1
d 2 1 t 2 1 1 . 5 1 cm 1.5 cm
1 .5 1.5 cm
2.0 cm
Therefore, final image will be 1 cm above point P. P
1 2 1 2
O P I O P I
1 Refractive index of the medium from which light rays are coming (from object).
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Reflection of Light 31
Lens.
Lens is a transparent medium bounded by two refracting surfaces, such that at least one
surface is spherical.
Convex lens (Converges the light rays) Concave lens (Diverges the light rays)
Double convex Plano convex Concavo convex Double concave Plane concave Convexo concave
Thick at middle Thin at middle
It forms real and virtual images both It forms only virtual images
– R2 +R1 – R1 +R2
(i) Optical centre (O) : A point for a given lens through which light ray passes undeviated
(Light ray passes undeviated through optical centre).
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32 Reflection of Light
F1 F1 F2 F2
1 2
(iii) Focal length (f) : Distance of second principle focus from optical centre is called focal
length
fconvex positive, fconcave negative, fplane
(v) Power of lens (P) : Means the ability of a lens to converge the light rays. Unit of power
is Diopter (D).
1 100
P ; Pconvex positive, Pconcave negative, Pplane zero .
f (m ) f (cm )
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2f f f 2f
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Reflection of Light 33
Note : Minimum distance between an object and it’s real image formed by a convex lens is
4f. Maximum image distance for concave lens is it’s focal length.
Equiconvex lens Plano convex lens Equi concave lens Plano concave lens
R1 R and R 2 R R1 , R 2 R R1 R , R 2 R R1 , R 2 R
R R R R
f f f f
2( 1) ( 1) 2( 1) 2( 1)
fl ( a μ g 1)
(Lens is supposed to be made of glass).
fa ( l μ g 1)
Note : Focal length of a glass lens ( 1 .5) is f in air then inside the water it’s focal length
is 4f.
In liquids focal length of lens increases () and it’s power decreases ().
(6) Opposite behaviour of a lens
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34 Reflection of Light
L > M L < M L = M
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Reflection of Light 35
f, P
(11) Combination of lens
2f 2f f, P
(i) For a system of lenses,
f, P the net power, net focal length and magnification given as follows :
P/2 P/2
1 1 1 1
P P1 P2 P3 .......... , .......... . ,
F f1 f2 f3
m m 1 m 2 m 3 .......... ..
(ii) In case when two thin lens are in contact : Combination will behave as a lens, which have
more power or lesser focal length.
1 1 1 f1 f2
F and P P1 P2
F f1 f2 f1 f2
(iii) If two lens of equal focal length but of opposite nature are in contact then combination
will behave as a plane glass plate and Fcombinatio n
(iv) When two lenses are placed co-axially at a distance d from each other then equivalent
focal length (F).
f1 f2
1 1 1 d
and P P1 P2 dP1 P2
F f1 f2 f1 f2
d
and
and
F = f/2
F=
f
f F =f F=f
+ +
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fl Only fm
F fl fm
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36 Reflection of Light
R R R R R
fm , fl so F fm , fl so F
2 ( 1) 2 ( 1) 2 ( 1)
(ii) Double convex lens is silvered
R R
Since fl , fm
2 ( 1) 2 +
R
So F
2 (2 1) F fl fm
White
Real Violet V R so fR fV
light FV FR Mathematically chromatic aberration
= f R f V ωfy
fV = Dispersion power of lens.
fR
fy = Focal length for mean colour f R fV
Removal : To remove this defect i.e. for Achromatism we use two or more lenses in contact in
place of single lens.
1 2
Mathematically condition of Achromatism is : 0 or 1 f2 2 f1
f1 f2
Note : Component lenses of an achromatic doublet cemented by canada blasam because it
is transparent and has a refractive index almost equal to the refractive of the glass.
(ii) Spherical aberration : Inability of a lens to form the point image of a point object on
the axis is called Spherical aberration.
In this defect all the rays passing through a lens are not focussed at a single point and the
image of a point object on the axis is blurred.
Marginal rays
Paraxial ray F F F F
Removal : A simple method to reduce spherical aberration is to use a stop before and infront
of the lens. (but this method reduces the intensity of the image as most of the light is cut off). Also
by using plano-convex lens, using two lenses separated by distance d = F – F ', using crossed lens.
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Reflection of Light 37
Note : Marginal rays : The rays farthest from the principal axis.
Paraxial rays : The rays close to the principal axis.
Spherical aberration can be reduced by either stopping paraxial rays or marginal
rays, which can be done by using a circular annular mask over the lens.
Parabolic mirrors are free from spherical aberration.
(iii) Coma : When the point object is placed away from the principle axis and the image is
received on a screen perpendicular to the axis, the shape of the image is like a comet. This defect is
called Coma.
It refers to spreading of a point object in a plane to principle axis.
Image of P
P
Axis
P
Removal : It can be reduced by properly designing radii of curvature of the lens surfaces. It
can also be reduced by appropriate stops placed at appropriate distances from the lens.
(iv) Curvature : For a point object placed off the axis, the image is spread both along and
perpendicular to the principal axis. The best image is, in general, obtained not on a plane but on a
curved surface. This defect is known as Curvature.
Removal : Astigmatism or the curvature may be reduced by using proper stops placed at
proper locations along the axis.
(v) Distortion : When extended objects are imaged, different portions of the object are in
general at different distances from the axis. The magnification is not the same for all portions of
the extended object. As a result a line object is not imaged into a line but into a curve.
(vi) Astigmatism : The spreading of image (of a point object placed away from the principal
axis) along the principal axis is called Astigmatism.
Concepts
If a sphere of radius R made of material of refractive index 2 is placed in a medium of refractive index 1 , Then if the
1
object is placed at a distance R from the pole, the real image formed is equidistant from the sphere.
2 1
1 2
O P1 P2 I
2
The lens doublets used in telescope are achromatic for blue and red colours, while these used in camera are achromatic for
x 2x x
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38 Reflection of Light
violet and green colours. The reason for this is that our eye is most sensitive between blue and red colours, while the
photographic plates are most sensitive between violet and green colours.
Position of optical centre
Equiconvex and equiconcave Exactly at centre of lens
Convexo-concave and concavo-convex Outside the glass position
Plano convex and plano concave On the pole of curved surface
Composite lens : If a lens is made of several materials then
Number of images formed = Number of materials used
Here no. of images = 5
Example
s
Example: 18 A thin lens focal length f1 and its aperture has diameter d. It forms an image of intensity I.
Now the central part of the aperture upto diameter d/2 is blocked by an opaque paper. The
focal length and image intensity will change to
f I I 3f I 3I
(a) and (b) f and (c) and (d) f and
2 2 4 4 2 4
d 1
Solution: (d) Centre part of the aperture up to diameter is blocked i.e. th area is blocked
2 4
2
A d . Hence remaining area A 3 A . Also, we know that intensity Area
4 4
I A 3 3
I I .
I A 4 4
Focal length doesn't depend upon aperture.
Example: 19 The power of a thin convex lens (a g 1.5) is + 5.0 D. When it is placed in a liquid of
refractive index a l , then it behaves as a concave lens of local length 100 cm. The refractive
index of the liquid a l will be
(c) 36 cm
(d) 48 cm
Solution: (d) From the figure shown it is clear that
For lens : u = 12 cm and v = x = ?
P' P
1 1 1 1 1 1
By using x = 48 cm.
f v u 16 x 12 x
12c
Example: 22 m 25 cm.
A convex lens of focal length 40 cm is an contact with a concave lens of focal length
The power of combination is
(a) – 1.5 D (b) – 6.5 D (c) + 6.5 D (d) + 6.67 D
1 1 1 1 1 1
Solution: (a) By using
F f1 f2 F 40 25
1 1 1 1 1 1 f1 f2
Hence by using ......(i)
F f1 f2 f1 f2 60 f1 f2
1 1 1 d
Finally by using where F 30 cm and d = 10 cm
F f1 f2 f1 f2
1 1 1 10
......(ii)
30 f1 f2 f1 f2
Example: 24 A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of
glass with refractive index 1.65. Its focal length is nearly
(a) 20 cm (b) 31 cm (c) 35 cm (d) 50 cm
R 40
Solution: (b) By using f f 30 .7 cm 31 cm.
2 1 21.65 1
Example: 25 A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass
(refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is
found to have a real image Q in the glass. The line PQ cuts the surface at a point O and
PO OQ . The distance PO is equal to
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40 Reflection of Light
[MP PMT 1994; Haryana CEE 1996]
OP = 5 R
Example: 26 The distance between an object and the screen is 100 cm. A lens produces an image on the
screen when placed at either of the positions 40 cm apart. The power of the lens is
(a) 3 D (b) 5 D (c) 7 D (d) 9 D
D x2 2
100 40
2 2
Solution: (b) By using f f 21 cm
4D 4 100
100 100
Hence power P 5D
Fcm 21
Example: 27 Shown in figure here is a convergent lens placed inside a cell filled with a liquid. The lens has
focal length +20 cm when in air and its material has refractive index 1.50. If the liquid has
refractive index 1.60, the focal length of the system is
Liquid
(a) + 80 cm (b) – 80 cm
(c) – 24 cm (d) – 100 cm Lens
1 1 3
1 .6 1
1
Solution: (d) Here .......(i)
f1 20 100
1 1
1 .5 1
1 1
.......(ii) + +
f2 20 20 20
1 1 3
1 .6 1
1
.......(iii)
f3 20 100 f1 f2 f3
F
1 1 1 1 1 3 1 3
By using F 100 cm
F f1 f2 f3 F 100 20 100
Example: 28 A concave lens of focal length 20 cm placed in contact with a plane mirror acts as a RA 1998]
(a) Convex mirror of focal length 10 cm (b) Concave mirror of focal length 40 cm
(c) Concave mirror of focal length 60 cm (d) Concave mirror of focal length 10 cm
1 2 1
Solution: (a) By using
F fl fm
+
f 20
Since fm F l 10 cm
2 2
F Fe Fm
(After silvering concave lens behave as convex mirror)
Example: 29 A candle placed 25 cm from a lens, forms an image on a screen placed 75 cm on the other
end of the lens. The focal length and type of the lens should be
(a) + 18.75 cm and convex lens (b) – 18.75 cm and concave lens
(c) + 20.25 cm and convex lens (d) – 20.25 cm and concave lens
Solution: (a) In concave lens, image is always formed on the same side of the object. Hence the given lens
is a convex lens for which u = – 25 cm, v = 75 cm.
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Reflection of Light 41
1 1 1 1 1 1
By using f = + 18.75 cm.
f v u f 75 25
Example: 30 A convex lens forms a real image of an object for its two different positions on a screen. If
height of the image in both the cases be 8 cm and 2 cm, then height of the object is [KCET (Engg./Med.) 20
(a) 16 cm (b) 8 cm (c) 4 cm (d) 2 cm
Solution: (c) By using O I1 I 2 O 8 2 4 cm
Example: 31 A convex lens produces a real image m times the size of the object. What will be the distance
of the object from the lens [JIPMER 2002]
m 1 m 1 m 1
(a) f (b) (m 1) f (c) f (d)
m m f
By using m
f
here m
f
1
f u
1
u m 1
u
Solution: (a) . f
f u f u m f f m
Example: 32 An air bubble in a glass sphere having 4 cm diameter appears 1 cm from surface nearest to
eye when looked along diameter. If a g 1.5 , the distance of bubble from refracting surface
is [CPMT 2002]
(a) 1.2 cm (b) 3.2 cm (c) 2.8 cm (d) 1.6 cm
Solution: (a) By using
2 1 2 1 2=1
v u R 1=1.5
Example: 33 The sun's diameter is 1.4 10 9 m and its distance from the earth is 10 11 m . The diameter of its
image, formed by a convex lens of focal length 2m will be
(a) 0.7 cm (b) 1.4 cm (c) 2.8 cm (d) Zero (i.e.
point image)
f
Solution: (c) From figure Sun
(D)
D 10 11 2 1 .4 10 9
d 2 .8 cm. (d)
d 2 10 11 Image
1011 m
Example: 34 Two point light sources are 24 cm apart. Where should a convex lens of focal length 9 cm be
put in between them from one source so that the images of both the sources are formed at
the same place
(a) 6 cm (b) 9 cm (c) 12 cm (d) 15 cm
Solution: (a) The given condition will be satisfied only if one source (S1) placed on one side such that u < f
(i.e. it lies under the focus). The other source (S2) is placed on the other side of the lens such
that u > f (i.e. it lies beyond the focus).
1 1 1 1 1 1
If S 1 is the object for lens then ........(i)
f y x y x f
1 1 1 1 1 1
If S 2 is the object for lens then ........(ii)
f y (24 x ) y f (24 x )
I1 S1 S2
I2
x (24 –
24 cm 4)
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42 Reflection of Light
From equation (i) and (ii)
1 1 1 1 1 1 2 2
x 2 24 x 108 0
x f f (24 x ) x (24 x ) f 9
On solving the equation x 18 cm , 6 cm
Example: 35 There is an equiconvex glass lens with radius of each face as R and a g 3 / 2 and
a w 4 / 3 . If there is water in object space and air in image space, then the focal length is
3 3
1
2 2 1 v 3 R
2
R 9R v2 2
3
The image will be formed at a distance do R . This is equal to the focal length of the lens.
2
Tricky example: 4
A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On
the other side of the lens. At what distance from the lens a convex mirror of radius of curvature 10
cm be placed in order to have an upright image of the object coincident with it
[CBSE PMT 1998; JIPMER 2001, 2002]
A convex lens of local length 30 cm and a concave lens of 10 cm focal length are placed so as to have
the same axis. If a parallel beam of light falling on convex lens leaves concave lens as a parallel
beam, then the distance between two lenses will be
(a) 40 cm (b) 30 cm (c) 20 cm (d) 10 cm
Solution : (c) According to figure the combination behaves as plane glass plate (i.e., F= )
1 1 1 d
Hence by using
F f1 f2 f1 f2
1 1 1 d
d = 20 cm
30 10 30 10
d
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Prism
Prism is a transparent medium bounded by refracting surfaces, such that the incident surface (on which
light ray is incidenting) and emergent surface (from which light rays emerges) are plane and non
parallel.
i r1 r2
sin r1 sin e
C B
ma It is observed if
x e
i = 90o
r2 i e and r1 r2 r
r1 = C i e
r r
then :
In this condition of maximum deviation (i) Refracted ray inside the prism is parallel to the base
i 90 ,o
r1 C, r2 A C and from of the prism
Snell’s law on emergent surface
m Only
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44 Reflection of Light
sin( A C) A A m
e sin 1 (ii) r and i
sin C 2 2
A m
sin
sin i 2
(iii) or
sin A / 2 sin A / 2
i = 0o e = 0o
r1 = 0o and r2 = 0o
r2 e i r1 or
sin i
In any of the above case use and i A
sin A
(4) Grazing emergence and TIR through a prism
When a light ray falls on one surface of prism, it is not necessary that it will exit out from the
prism. It may or may not be exit out as shown below
Normal incidence Grazing incidence
Ray –1 : General emergence Ray –1 : General emergence
A < C and A < 2C and
< cosec A A < cosec (A/2)
A
Screen
Incident R
white light Y
V
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Reflection of Light 45
R R
V A A
V
Crown Crown
A' ( y 1) A' ( R )
(i) (i) V
A ( ' y 1) A ( ' V ' R )
'
(ii) net 1 ( ' ' ) (ii) net 1
'
Scattering of Light
Molecules of a medium after absorbing incoming light radiations, emits them in all direction.
This phenomenon is called Scattering.
1
(1) According to scientist Rayleigh : Intensity of scattered light 4
(2) Some phenomenon based on scattering : (i) Sky looks blue due to scattering.
(ii) At the time of sunrise or sunset it looks reddish. (iii) Danger signals are made from red.
(3) Elastic scattering : When the wavelength of radiation remains unchanged, the
scattering is called elastic.
(4) Inelastic scattering (Raman’s effect) : Under specific condition, light can also suffer
inelastic scattering from molecules in which it’s wavelength changes.
Rainbow
Rainbow is formed due to the dispersion of light suffering
Re
d Violet
42o40o
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(1) Primary rainbow : (i) Two refraction and one TIR. (ii) Innermost arc is violet and
outermost is red. (iii) Subtends an angle of 42 o at the eye of the observer. (iv) More bright
(2) Secondary rainbow : (i) Two refraction and two TIR. (ii) Innermost arc is red and
outermost is violet.
(iii) It subtends an angle of 52 .5 o at the eye. (iv) Comparatively less bright.
Colours
Colour is defined as the sensation received by the eye (cone cells of the eye) due to light coming from
object.
(1) Types of colours
Spectral colours Colours of pigment and dyes
Green (P) Yellow (P)
(2) Colours of object : The perception of a colour by eye depends on the nature of object
and the light incident on it.
Colours of opaque object Colours of transparent object
(i) Due to selective reflection. (i) Due to selective transmission.
(ii) A rose appears red in white light because it reflects (ii) A red glass appears red because it absorbs all
red colour and absorbs all remaining colours. colours, except red which it transmits.
(iii) When yellow light falls on a bunch of flowers, then (iii) When we look on objects through a green glass or
yellow and white flowers looks yellow. Other flowers green filter then green and white objects will appear
looks black. green while other black.
Note : A hot object will emit light of that colour only which it has observed when it was
heated.
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Spectrum.
The ordered arrangements of radiations according to wavelengths or frequencies is called
Spectrum. Spectrum can be divided in two parts (I) Emission spectrum and (II) Absorption
spectrum.
(1) Emission spectrum : When light emitted by a self luminous object is dispersed by
a prism to get the spectrum, the spectrum is called emission spectra.
(2) Absorption spectrum : When white light passes through a semi-transparent solid, or
liquid or gas, it’s spectrum contains certain dark lines or bands, such spectrum is called absorption
spectrum (of the substance through which light is passed).
(i) Substances in atomic state produces line absorption spectra. Polyatomic substances such
as H 2 , CO 2 and KMnO 4 produces band absorption spectrum.
(ii) Absorption spectra of sodium vapour have two (yellow lines) wavelengths D1 (5890 Å) and
D2 (5896 Å)
Note : If a substance emits spectral lines at high temperature then it absorbs the same lines
at low temperature. This is Kirchoff’s law.
(3) Fraunhoffer’s lines : The central part (photosphere) of the sun is very hot and emits all
possible wavelengths of the visible light. However, the outer part (chromosphere) consists of
vapours of different elements. When the light emitted from the photosphere passes through the
chromosphere, certain wavelengths are absorbed. Hence, in the spectrum of sunlight a large
number of dark lines are seen called Fraunhoffer lines.
(i) The prominent lines in the yellow part of the visible spectrum were labelled as D-lines,
those in blue part as F-lines and in red part as C-line.
(ii) From the study of Fraunhoffer’s lines the presence of various elements in the sun’s
atmosphere can be identified e.g. abundance of hydrogen and helium.
(4) Spectrometer : A spectrometer is used for obtaining pure spectrum of a source in
laboratory and calculation of of material of prism and of a transparent liquid.
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It consists of three parts : Collimator which provides a parallel beam of light; Prism Table for
holding the prism and Telescope for observing the spectrum and making measurements on it.
The telescope is first set for parallel rays and then collimator is set for parallel rays. When
prism is set in minimum deviation position, the spectrum seen is pure spectrum. Angle of prism
(A) and angle of minimum deviation ( m ) are measured and of material of prism is calculated
using prism formula. For of a transparent liquid, we take a hollow prism with thin glass sides.
Fill it with the liquid and measure ( m ) and A of liquid prism. of liquid is calculated using prism
formula.
(5) Direct vision spectroscope : It is an instrument used to observe pure spectrum. It
produces dispersion without deviation with the help of n crown prisms and (n 1) flint prisms
alternately arranged in a tabular structure.
For no deviation n ( 1)A (n 1) ( '1)A' .
Concepts
When a ray of white light passes through a glass prism red light is deviated less than blue light.
For a hollow prism A 0 but 0
If an opaque coloured object or crystal is crushed to fine powder it will appear white (in sun light) as it will lose it's
property of selective reflection.
Our eye is most sensitive to that part at the spectrum which lies between the F line (sky green) one the C-line (red)
of hydrogen equal to the refractive index for the D line (yellow) of sodium. Hence for the dispersive power, the
C
following formula is internationally accepted F
D 1
Sometimes a part of prism is given and we keep on thinking whether how should we proceed ? To solve such
problems first complete the prism then solve as the problems of prism are solved A
60o 70o 60o 70o
B C
Example: 36 When light rays are incident on a prism at an angle of 45 o, the minimum deviation is
obtained. If refractive index of the material of prism is 2 , then the angle of prism will be
(a) 30o (b) 40o (c) 50o (d) 60o
1
sin i sin 45 A 2 1 A
Solution: (d) 2 sin 30 o A 60 o
A A 2 2 2 2
sin sin
2 2
Example: 37 Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism.
The angle of prism is (cos 41 o 0.75)
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Example: 40 Angle of a prism is 30o and its refractive index is 2 and one of the surface is silvered. At
what angle of incidence, a ray should be incident on one surface so that after reflection from
the silvered surface, it retraces its path
(a) 30o (b) 60o (c) 45o (d) sin 1 1.5
Solution: (c) This is the case when light ray is falling normally an second surface.
sin i sin i 1
Hence by using 2 o
sin i 2 i 45 o
sin A sin 30 2
A
Example: 41 The refracting angle of prism is A and refractive index of material of prism is cot . The
2
angle of minimum deviation is
(a) 180 o 3 A (b) 180 o 2 A (c) 90 o A (d) 180 o 2 A
A m A m A A m
sin sin cos sin
2 A 2 2 2
Solution: (d) By using cot
A 2 A A A
sin sin sin sin
2 2 2 2
A A m A A m
sin 90 sin 90
m 180 2 A
2 2 2 2
Example: 42 A ray of light passes through an equilateral glass prism in such a manner that the angle of
incidence is equal to the angle of emergence and each of these angles is equal to 3/4 of the
angle of the prism. The angle of deviation is
(a) 45o (b) 39o (c) 20o (d) 30o
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50 Reflection of Light
3 3
Solution: (d) Given that A 60 o and i e A 60 45 o
4 4
By using i e A 45 45 60 30 o
Example: 43 PQR is a right angled prism with other angles as 60 o and 30o. Refractive index of prism is
1.5. PQ has a thin layer of liquid. Light falls normally on the face PR. For total internal
reflection, maximum refractive index of liquid is
(a) 1.4
P Q
(b) 1.3 60 30°
°
(c) 1.2
(d) 1.6
R
Solution: (c) For TIR at PQ C
From geometry of figure 60 i.e. 60 C sin 60 sin C
3 Liquid 3 3
Liquid Pr ism Liquid 1.5 Liquid 1.3 .
2 Pr ism 2 2
Example: 44 Two identical prisms 1 and 2, each will angles of 30 o, 60o and 90o are placed in contact as
shown in figure. A ray of light passed through the combination in the position of minimum
deviation and suffers a deviation of 30o. If the prism 2 is removed, then the angle of
deviation of the same ray is [PMT (Andhra) 1995]
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(a) 0.034 and 0.064 (b) 0.064 and 0.034 (c) 1.00 and 0.064 (d) 0.034 and 1.0
v r 1 .5318 1 .5140
Solution: (a) Crown 0 .034 and
y 1 (1 .5170 1)
v' r' 1 . 6852 1 . 6434
Flint 0 . 064
y' 1 1 . 6499 1
Example: 48 Flint glass prism is joined by a crown glass prism to produce dispersion without deviation.
The refractive indices of these for mean rays are 1.602 and 1.500 respectively. Angle of
prism of flint prism is 10o, then the angle of prism for crown prism will be
(a) 12 o 2.4 ' (b) 12 o 4 ' (c) 1 .24 o (d) 12o
Solution: (a) For dispersion without deviation
AC ( 1) A (1 .602 1)
F A 12.04 o 12 o 2.4 '
AF ( C 1) 10 (1 .500 1)
Tricky example: 6
An achromatic prism is made by crown glass prism ( A C 19 o ) and flint glass prism
( A F 6 o ) . If C v 1 .5 and F v 1.66 , then resultant deviation for red coloured ray will be
(a) 1.04o (b) 5o (c) 0.96o (d) 13.5o
Solution : (d) For achromatic combination w C w F [( v r )A]C [( v r )A] F
[ r A]C [ r A] F [ v A]C [ v A] F 1.5 19 6 1.66 38.5
Resultant deviation [( r 1)A]C [( r 1)A] F
[ r A]C [ r A] F ( AC A F ) 38 .5 (19 6) 13 .5 o
Tricky example: 7
The light is incident at an angle of 60o on a prism of which the refracting angle of prism is
30o. The refractive index of material of prism will be
(a) 2 (b) 2 3 (c) 2 (d) 3
Solution : (d) By using i e A 60 e 30 30 e 0 .
Hence ray will emerge out normally so by using the formula
sin i sin 60
3
sin A sin 30
60°
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52 Reflection of Light
Human Eye.
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Reflection of Light 53
Specific Example
A person wishes to distinguish between two pillars located at a distances of 11 Km. What should be the minimum
distance between the pillars.
d
o
1
Solution : As the limit of resolution of eye is
60
o
1 d 1
So d 3 .2 m 11 km
60 11 10 3
60 180
(iii) Far point comes closer. (iii) Near point moves away
Presbyopia : In this defect both near and far objects are not clearly visible. It is an old age
disease and it is due to the loosing power of accommodation. It can be removed by using bifocal
lens.
Concave
Convex
Astigmatism : In this defect eye cannot see horizontal and vertical lines clearly,
simultaneously. It is due to imperfect spherical nature of eye lens. This defect can be removed by
using cylindrical lens (Torric lenses).
Microscope.
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54 Reflection of Light
It is an optical instrument used to see very small objects. It’s magnifying power is given by
Visual angle with instrument ( )
m
Visual angle when object is placed at least distance of distinct vision ( )
D D
m D 1 and m
f max f min
Da Da
If lens is kept at a distance a from the eye then m D 1 and m
f f
(2) Compound microscope
(i) Consist of two converging lenses called
objective and eye lens.
(ii) fey e lens fobj ective and
(diameter) ey e lens (diameter )objective
v0 D f0 D (v f0 ) D
m . 0 .
u 0 Fe (u 0 f0 ) fe f0 Fe
u 0 f0
When final images is formed at ; L v 0 fe fe
u 0 f0
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(L f0 fe )D
Note : m
f0 fe
m m objective m ey e lens
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56 Reflection of Light
fe D
(v) Length : L D f0 4 f u e f0 4 f and L f0 4 f fe
fe D
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Concepts
As magnifying power is negative, the image seen in astronomical telescope is truly inverted, i.e., left is turned right
with upside down simultaneously. However, as most of the astronomical objects are symmetrical this inversion does
not affect the observations.
Objective and eye lens of a telescope are interchanged, it will not behave as a microscope but object appears very small.
In a telescope, if field and eye lenses are interchanged magnification will change from (f o / fe) to (fe / fo), i.e., it will
change from m to (1/m), i.e., will become (1/m2) times of its initial value.
As magnification for normal setting as (fo / fe), so to have large magnification, fo must be as large as practically
possible and fe small. This is why in a telescope, objective is of large focal length while eye piece of small.
In a telescope, aperture of the field lens is made as large as practically possible to increase its resolving power as
resolving power of a telescope (D/)*. Large aperture of objective also helps in improving the brightness of
image by gathering more light from distant object. However, it increases aberrations particularly spherical.
For a telescope with increase in length of the tube, magnification decreases.
In case of a telescope if object and final image are at infinity then :
f D
m o
fe d
If we are given four convex lenses having focal lengths f1 f2 f3 f4 . For making a good telescope and
microscope. We choose the following lenses respectively. Telescope f1 (o), f4 (e) Microscope f4 (o), f3 (e )
If a parrot is sitting on the objective of a large telescope and we look towards (or take a photograph)of distant
astronomical object (say moon) through it, the parrot will not be seen but the intensity of the image will be slightly
reduced as the parrot will act as obstruction to light and will reduce the aperture of the objective.
Example
s
Example: 1 A man can see the objects upto a distance of one metre from his eyes. For correcting his eye
sight so that he can see an object at infinity, he requires a lens whose power is
or
A man can see upto 100 cm of the distant object. The power of the lens required to see far
objects will be
[MP PMT 1993, 2003]
(a) +0.5 D (b) +1.0 D (c) +2.0 D (d) –1.0 D
100 100
Solution: (d) f = –(defected far point) = – 100 cm. So power of the lens P 1 D
f 100
Example: 2 A man can see clearly up to 3 metres. Prescribe a lens for his spectacles so that he can see clearly
up to 12 metres
[DPMT 2002]
(a) – 3/4 D (b) 3 D (c) – 1/4 D (d) – 4 D
xy 3 12 1 1
Solution: (c) By using f f 4 m . Hence power P D
x y 3 12 f 4
Example: 3 The diameter of the eye-ball of a normal eye is about 2.5 cm. The power of the eye lens varies
from
(a) 2 D to 10 D (b) 40 D to 32 D (c) 9 D to 8 D (d) 44 D to 40
D
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58 Reflection of Light
1 1 1
Solution: (d) An eye sees distant objects with full relaxation so 2
or
2.5 10 f
1 1
P 40 D
f 25 10 2
1 1 1
An eye sees an object at 25 cm with strain so or
2.5 10 2 25 10 2 f
1
P 40 4 44 D
f
Example: 4 The resolution limit of eye is 1 minute. At a distance of r from the eye, two persons stand
with a lateral separation of 3 metre. For the two persons to be just resolved by the naked eye,
r should be
(a) 10 km (b) 15 km (c) 20 km (d) 30 km
o d = 3m
d 1 1
Solution: (a) From figure ; where 1' rad
r
60 60 180
1 3
1 r = 10 km
60 180 r r
Example: 5 Two points separated by a distance of 0.1 mm can just be resolved in a microscope when a
light of wavelength 6000 Å is used. If the light of wavelength 4800 Å is used this limit of
resolution becomes
[UPSEAT 2002]
(a) 0.08 mm (b) 0.10 mm (c) 0.12 mm (d) 0.06 mm
(R.L.)1 0 .1 6000
Solution: (a) By using resolving limit (R.L.) 1 (R.L.) 2 0.08 mm .
(R.L.) 2 2 (R.L.) 2 4800
Example: 6 In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm an object
is placed at 2 cm form objective and the final image is formed at 25 cm from eye lens. The
distance between the two lenses is
[EAMCET (Med.) 2000]
(a) 6.00 cm (b) 7.75 cm (c) 9.25 cm (d) 11.00 cm
Solution: (d) It is given that fo = 1.5 cm, fe = 6.25 cm, uo = 2 cm
When final image is formed at least distance of distinct vision, length of the tube
uo fo fe D
LD
uo fo fe D
2 1 .5 6.25 25
LD 11 cm .
(2 1.5) (6.25 25)
Example: 7 The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm
and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm.
The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances
in cm of the object and the image produced by the objective measured from the objective
lens are respectively [IIT-JEE 1995]
(a) 2.4 and 12.0 (b) 2.4 and 15.0 (c) 2.3 and 12.0 (d) 2.3 and
3.0
Solution: (a) Given that fo 2 cm , fe 3 cm , L 15 cm
vo vo fo 12 12 2
By using L vo fe 15 vo 3 vo 12 cm . Also
uo fo uo 2
uo 2.4 cm .
Example: 8 The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm
respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is
(a) 30 cm (b) 25 cm (c) 15 cm (d) 12 cm
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60 Reflection of Light
Example: 15 A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5
cm. Assuming the final image to be at the least distance of distinct vision. The magnification
produced by the objective will be
(a) +5 (b) – 5 (c) +6 (d) – 6
Solution (b) Magnification produced by compound microscope m mo me
D 25
where mo = ? and m e 1 1 6 30 mo 6 m o 5 .
fe 5
Tricky Example 1 : A man is looking at a small object placed at his least distance of distinct vision. Without
changing his position and that of the object he puts a simple microscope of magnifying power 10 X and just sees the
clear image again. The angular magnification obtained is
(a) 2.5 (b) 10.0 (c) 5.0 (d) 1.0
tan I/D I
Solution : (d) Angular magnification
tan O / D O
I v
Since image and object are at the same position, 1 Angular magnification = 1
O u
Tricky Example 2: A compound microscope is used to enlarge an object kept at a distance 0.03m from it’s
objective which consists of several convex lenses in contact and has focal length 0.02m. If a lens of focal length 0.1m
is removed from the objective, then by what distance the eye-piece of the microscope must be moved to refocus the
image
(a) 2.5 cm (b) 6 cm (c) 15 cm (d) 9 cm
Solution : (d) If initially the objective (focal length Fo ) forms the image at distance vo then
u f 32
vo o o 6 cm
uo fo 3 2
1 1 1 1 1 1 1 1 1
Now as in case of lenses in contact ..... where .....
Fo f1 f2 f3 f1 Fo Fo f2 f3
So if one of the lens is removed, the focal length of the remaining lens system
1 1 1 1 1
Fo 2.5 cm
Fo F0 f1 2 10
u o Fo 3 2.5
This lens will form the image of same object at a distance v o such that v o 15 cm
u o Fo (3 2.5)
So to refocus the image, eye-piece must be moved by the same distance through which the image
formed by the objective has shifted i.e. 15 – 6 = 9 cm.
Assignment
Human eye
80. Near and far points of human eye are [EAMCET (Med.) 1995; MP PET 2001; Bihar CECE 2004]
(a) 25 cm and infinite (b) 50 cm and 100 cm (c) 25 cm and 50 cm (d) 0 cm and
25 cm
81. A defective eye cannot see close objects clearly because their image is formed [MP PET 2003]
(a) On the eye lens (b) Between eye lens and retina
(c) On the retina (d) Beyond retina
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84. An astronaut is looking down on earth's surface from a space shuttle at an altitude of 400km. Assuming
that the astronaut's pupil diameter is 5 mm and the wavelength of visible light is 500nm. The astronaut
will be able to resolve linear object of the size of about [AIIMS 2003]
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62 Reflection of Light
97. A person wears glasses of power –2.0 D. The defect of the eye and the far point of the person without the glasses
will be
[MP PMT 1999]
(a) Nearsighted, 50 cm (b) Farsighted, 50 cm (c) Nearsighted, 250 cm (d)
Astigmatism, 50 cm
98. A person is suffering from the defect astigmatism. Its main reason is [MP PMT 1997]
(a) Distance of the eye lens from retina is increased (b) Distance of the eye lens from retina is
decreased
(c) The cornea is not spherical (d) Power of accommodation of the eye is
decreased
99. Myopia is due to [AFMC 1996]
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Reflection of Light 63
110. If the distance of the far point for a myopia patient is doubled, the focal length of the lens required to cure it will
become [MP PET 1989]
(a) Half (b) Double
(c) The same but a convex lens (d) The same but a
concave lens
111. Image is formed for the short sighted person at [AFMC 1988]
(a) Retina (b) Before retina (c) Behind the retina (d) Image is
not formed at all
112. A man who cannot see clearly beyond 5 m wants to see stars clearly. He should use a lens of focal length
[MP PET/PMT 1988]
(a) – 100 metre (b) + 5 metre (c) – 5 metre (d) Very large
113. Far point of myopic eye is 250 cm, then the focal length of the lens to be used will be [CPMT 1986; DPMT 2002]
(a) + 250 cm (b) – 250 cm (c) + 250/9 cm (d) – 250/9 cm
114. One can take pictures of objects which are completely invisible to the eye using camera film which are invisible to [MNR 1985]
(a) Ultra-violet rays (b) Sodium light (c) Visible light (d) Infra-red
rays
115. In human eye the focussing is done by [CPMT 1983]
(a) To and fro movement of eye lens (b) To and fro movement of the retina
(c) Change in the convexity of the lens surface (d) Change in the refractive index of the eye
fluids
116. The minimum light intensity that can be perceived by the eye is about 10 10 watt / metre 2 . The number of photons
of wavelength 5.6 10 7 metre that must enter per second the pupil of area 10 4 metre 2 for vision, is
approximately equal to (h 6.6 10 34 joule sec) [NCERT 1982]
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64 Reflection of Light
123. A man, wearing glasses of power +2D can read clearly a book placed at a distance of 40 cm from the eye. The
power of the lens required so that he can read at 25 cm from the eye is
(a) +4.5 D (b) +4.0 D (c) +3.5 D (d) +3.0 D
124. A person can see clearly between 1 m and 2m. His corrective lenses should be
(a) Bifocals with power –0.5D and additional +3.5D (b) Bifocals with power –1.0D and additional
+3.0 D
(c) Concave with power 1.0 D (d) Convex with power 0.5 D
125. While reading the book a man keeps the page at a distance of 2.5 cm from his eye. He wants to read the book by
holding the page at 25 cm. What is the nature of spectacles one should advice him to use to completely cure his
eye sight
(a) Convex lens of focal length 25 cm (b) Concave lens of focal length 25 cm
(c) Convex lens of focal length 2.5 cm (d) Concave lens of focal length 2.5 cm
126. The blades of a rotating fan can not be distinguished from each other due to
(a) Parallex (b) Power of accommodation (c) Persistence of vision (d) Binocular
vision
127. Aperture of the human eye is 2 mm. Assuming the mean wavelength of light to be 5000 Å, the angular resolution
limit of the eye is nearly
(a) 2 minutes (b) 1 minute (c) 0.5 minute (d) 1.5 minutes
128. If there had been one eye of the man, then
(a) Image of the object would have been inverted (b) Visible region would have decreased
(c) Image would have not been seen three dimensional (d) (b) and (c) both
129. A man can see the object between 15cm and 30cm. He uses the lens to see the far objects. Then due to the lens
used, the near point will be at
10 100
(a) cm (b) 30 cm (c) 15 cm (d) cm
3 3
130. A presbyopic patient has near point as 30 cm and far point as 40 cm. The dioptric power for the corrective lens for
seeing distant objects is
(a) 40 D (b) 4 D (c) 2.5 D (d) 0.25 D
131. A man swimming under clear water is unable to see clearly because
(a) The size of the aperture decreases (b) The size of the aperture increases
(c) The focal length of eye lens increases (d) The focal length of eye lens decreases
132. The distance between retina and eye-lens in a normal eye is 2.0 cm. The accommodated power of eye lens range
from
(a) 45 D to 50 D (b) 50 D to 54 D (c) 10 D to 16 D (d) 5 D to 8 D
133. If the eye is taken as a spherical ball of radius 1 cm, the range of accommodated focal length of eye-lens is
(a) 1.85 cm to 2.0 cm (b) 1.0 cm to 2.8 cm (c) 1.56 cm to 2.5 cm (d) 1.6 cm to
2.0 cm
134. A person cannot read printed matter within 100 cm from his eye. The power of the correcting lens required to
read at 20 cm from his eye if the distance between the eye lens and the correcting lens is 2 cm is
(a) 4.8 D (b) 1.25 D (c) 4.25 D (d) 4.55 D
135. A student having –1.5 D spectacles uses a lens of focal length 5 cm as a simple microscope to read minute scale
divisions in the laboratory. The least distance of distinct vision without glasses is 20 cm for the student. The
maximum magnifying power he gets with spectacles on is
(a) 6 (b) 9 (c) 5 (d) 4
Microscope
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136. In a compound microscope the object of fo and eyepiece of fe are placed at distance L such that L equals [Kerala PMT 2004]
(a) fo fe (b) fo fe
(c) Much greater than fo or fe (d) Need not depend either value of focal
lengths
137. In a simple microscope, if the final image is located at infinity then its magnifying power is [CPMT 1985; MP PMT 2004]
25 D f f
(a) (b) (c) (d)
f 25 25 D 1
138. In a simple microscope, if the final image is located at 25 cm from the eye placed close to the lens, then the
magnifying power is
[BVP 2003]
25 25 f f
(a) (b) 1 (c) (d) 1
f f 25 25
139. The maximum magnification that can be obtained with a convex lens of focal length 2.5 cm is (the least
distance of distinct vision is 25 cm) [MP PET 2003]
(a) 10 (b) 0.1 (c) 62.5 (d) 11
140. In a compound microscope, the intermediate image is [IIT-JEE (Screening) 2000; AIEEE 2003]
(a) Virtual, erect and magnified (b) Real, erect and
magnified
(c) Real, inverted and magnified (d) Virtual, erect and reduced
141. A compound microscope has two lenses. The magnifying power of one is 5 and the combined magnifying power is
100. The magnifying power of the other lens is [Kerala PMT 2002]
(a) 10 (b) 20 (c) 50 (d) 25
142. Wavelength of light used in an optical instrument are 1 4000 Å and 2 5000 Å , then ratio of their respective
resolving power (corresponding to 1 and 2 ) is [AIEEE 2002]
149. If the red light is replaced by blue light illuminating the object in a microscope the resolving power of the
microscope
[DCE 2001]
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66 Reflection of Light
(a) Decreases (b) Increases (c) Gets halved (d) Remains
unchanged
150. In case of a simple microscope, the object is placed at [UPSEAT 2000]
(a) Focus f of the convex lens (b) A position between f and 2f (c) Beyond 2f (d) Between
the lens and f
151. In a compound microscope cross-wires are fixed at the point [EAMCET (Engg.) 2000]
(a) Where the image is formed by the objective (b) Where the image is formed by the eye-piece
(c) Where the focal point of the objective lies (d) Where the focal point of the eye-piece lies
152. The length of the tube of a microscope is 10 cm. The focal lengths of the objective and eye lenses are 0.5 cm and
1.0 cm. The magnifying power of the microscope is about [MP PMT 2000]
(a) A concave lens of small focal length and small aperture (b) Convex lens of small focal length and large
aperture
(c) Convex lens of large focal length and large aperture (d) Convex lens of small focal length and small
aperture
155. For relaxed eye, the magnifying power of a microscope is [CBSE PMT 1998]
vo D v o fe uo D uo D
(a) (b) (c) (d)
u o fe uo D v o fe vo fe
(a) Equal to the focal length of its eye piece (b) Less than the focal length of eye piece
(c) Greater than the focal length of eye piece (d) Any of the above three
158. To produce magnified erect image of a far object, we will be required along with a convex lens, is
[MNR 1983; MP PAT 1996]
(a) Another convex lens (b) Concave lens (c) A plane mirror (d) A concave
mirror
159. An object placed 10 cm in front of a lens has an image 20 cm behind the lens. What is the power of the lens (in
dioptres)
[MP PMT 1995]
(a) 1.5 (b) 3.0 (c) – 15.0 (d) +15.0
160. Resolving power of a microscope depends upon [MP PET 1995]
(a) The focal length and aperture of the eye lens (b) The focal lengths of the objective and the
eye lens
(c) The apertures of the objective and the eye lens (d) The wavelength of light illuminating the
object
161. If the focal length of the objective lens is increased then [MP PMT 1994]
(a) Magnifying power of microscope will increase but that of telescope will decrease
(b) Magnifying power of microscope and telescope both will increase
(c) Magnifying power of microscope and telescope both will decrease
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(d) Magnifying power of microscope will decrease but that of telescope will increase
162. If in compound microscope m 1 and m 2 be the linear magnification of the objective lens and eye lens respectively,
then magnifying power of the compound microscope will be [CPMT 1985; KCET 1994]
163. The magnifying power of a microscope with an objective of 5 mm focal length is 400. The length of its tube is 20
cm. Then the focal length of the eye-piece is [MP PMT 1991]
(a) Decreases (b) Increases (c) Does not change (d) May
decrease or increase
167. An electron microscope is superior to an optical microscope in [CPMT 1984]
(a) Large (b) Smaller (c) Equal to that of objective (d) Less than
that of objective
169. An electron microscope gives better resolution than optical microscope because [CPMT 1982]
172. Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective, a 5
diopter lens as the eyepiece and the separation 30 cm between the two lenses. The least distance for clear vision is
25 cm
(a) 8.4 (b) 7.4 (c) 9.4 (d) 10.4
173. The focal length of the objective and the eye-piece of a microscope are 2 cm and 5 cm respectively and the distance
between them is 30 cm. If the image seen by the eye is 25 cm from the eye-piece, the distance of the object from
the objective is
(a) 0.8 cm (b) 2.3 cm (c) 0.4 cm (d) 1.2 cm
174. The focal length of objective and eye-piece of a microscope are 1 cm and 5 cm respectively. If the magnifying
power for relaxed eye is 45, then length of the tube is
(a) 6 cm (b) 9 cm (c) 12 cm (d) 15 cm
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175. A microscope has an objective of focal length 1.5 cm and an eye-piece of focal length 2.5 cm. If the distance
between objective and eye-piece is 25 cm. What is the approximate value of magnification produced for relaxed
eye is
(a) 75 (b) 110 (c) 140 (d) 25
176. The magnifying power of a microscope is generally marked as 10X, 100 X, etc. These markings are for a normal
relaxed eye. A microscope marked 10X is used by an old man having his near point at 40 cm. The magnifying
power of the microscope for the old man with his eyes completely relaxed is
(a) 10 (b) 18 (c) 12 (d) 16
177. If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away
from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is
(a) 150 (b) 200 (c) 250 (d) 400
178. A compound microscope is adjusted for viewing the distant image of an object, the distance of the object from the
object glass is now slightly increased, what re-adjustment of the instrument would be necessary for obtaining a
distant image again
(a) Objective should be moved away from the eye-piece (b) Eye-piece should be moved towards the
objective
(c) Both should be moved towards each other (d) Both should be moved away from each
other
179. When the object is self-luminous, the resolving power of a microscope is given by the expression
182. The focal length of the objective and eyepiece of an astronomical telescope for normal adjustments are 50 cm and
5 cm. The length of the telescope should be [MP PMT 2004]
(a) 50 cm (b) 55 cm (c) 60 cm (d) 45 cm
183. The resolving power of an astronomical telescope is 0.2 seconds. If the central half portion of the objective lens is
covered, the resolving power will be [MP PMT 2004]
(a) 0.1 sec (b) 0.2 sec (c) 1.0 sec (d) 0.6 sec
184. If Fo and Fe are the focal length of the objective and eye-piece respectively of a telescope, then its magnifying
power will be
[CPMT 1977, 82, 97, 99, 2003; SCRA 1994; KCET (Engg./Med.) 1999; Pb. PMT 2000; BHU 2001; BCECE 2003, 2004]
1
(a) Fo Fe (b) Fo Fe (c) Fo / Fe (d) (Fo Fe )
2
185. The length of an astronomical telescope for normal vision (relaxed eye) (fo = focal length of objective lens and fe =
focal length of eye lens) is [EAMCET (Med.) 1995; MP PAT 1996; CPMT 1999; BVP 2003]
A collection by Pradeep Kshetrapal for Physics students at genius, Maxwell and Gupta classes Only
genius PHYSICS
Reflection of Light 69
fo
(a) fo fe (b) (c) fo fe (d) fo fe
fe
186. A telescope of diameter 2m uses light of wavelength 5000 Å for viewing stars. The minimum angular separation
between two stars whose image is just resolved by this telescope is [MP PET 2003]
(a) 4 10 4 rad (b) 0.25 10 6 rad (c) 0.31 10 6 rad (d) 5.0 10 3
rad
187. The aperture of the objective lens of a telescope is made large so as to [AIEEE 2003; KCET 2003]
(a) Increase the magnifying power of the telescope (b) Increase the resolving power of the
telescope
(c) Make image aberration less (d) Focus on distant
objects
188. The distance of the moon from earth is 3.8 10 5 km. The eye is most sensitive to light of wavelength 5500 Å. The
separation of two points on the moon that can be resolved by a 500 cm telescope will be [AMU (Med.) 2002]
(a) 51 m (b) 60 m (c) 70 m (d) All of the
above
189. To increase both the resolving power and magnifying power of a telescope [Kerala PET 2002; KCET (Engg.) 2002]
(a) Both the focal length and aperture of the objective has to be increased
(b) The focal length of the objective has to be increased
(c) The aperture of the objective has to be increased
(d) The wavelength of light has to be decreased
190. The focal lengths of the objective and eye lenses of a telescope are respectively 200cm and 5cm. The maximum
magnifying power of the telescope will be [MP PMT/PET 1998; JIPMER 2001, 2002]
(a) – 40 (b) – 48 (c) – 60 (d) – 100
191. A telescope has an objective of focal length 50 cm and an eye piece of focal length 5 cm. The least distance of
distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away. The separation
between the objective and the eye-piece is
[Kerala PET 2002]
(a) 75 cm (b) 60 cm (c) 71 cm (d) 74 cm
192. In a laboratory four convex lenses L1 , L2 , L3 and L4 of focal lengths 2, 4, 6 and 8cm respectively are available.
Two of these lenses form a telescope of length 10cm and magnifying power 4. The objective and eye lenses are [MP PMT 2001]
193. Four lenses of focal length + 15 cm, + 20 cm, + 150 cm and + 250 cm are available for making an
astronomical telescope. To produce the largest magnification, the focal length of the eye-piece should
be [CPMT 2001; AIIMS 2001]
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genius PHYSICS
70 Reflection of Light
(c) The image of the planet is inverted
(d) All of the above
198. The astronomical telescope consists of objective and eye-piece. The focal length of the objective is [AIIMS 1998; BHU 2000]
(a) Equal to that of the eye-piece (b) Greater than that of the eye-piece
(c) Shorter than that of the eye-piece (d) Five times shorter than that of the eye-
piece
199. The diameter of the objective of a telescope is a, the magnifying power is m and wavelength of light is . The
resolving power of the telescope is [MP PMT 2000]
(a) (1.22)/a (b) (1.22a)/ (c) m/(1.22a) (d) a/(1.22m)
200. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation
between the objective and the eyepiece is 36 cm and final image is formed at infinity. The focal lengths of the
objective and eyepiece are respectively
[IIT-JEE 1989; MP PET 1995; JIPMER 2000]
(a) 20 cm, 16 cm (b) 50 cm, 10 cm (c) 30 cm, 6 cm (d) 45 cm, –9
cm
201. A photograph of the moon was taken with telescope. Later on, it was found that a housefly was sitting on the
objective lens of the telescope. In photograph [NCERT 1970; MP PET 1999]
(a) The image of housefly will be reduced (b) There is a reduction in the intensity of the
image
(c) There is an increase in the intensity of the image (d) The image of the housefly will be enlarged
202. The magnifying power of a telescope is M. If the focal length of eye piece is doubled, then the magnifying power
will become
[Haryana CEET 1998]
203. The minimum magnifying power of a telescope is M. If the focal length of its eyelens is halved, the magnifying
power will become
[MP PMT/PET 1998]
(a) M/2 (b) 2 M (c) 3 M (d) 4 M
204. The final image in an astronomical telescope is [EAMCET (Engg.) 1998]
(a) Real and errect (b) Virtual and inverted (c) Real and inverted (d) Virtual and
errect
205. The astronomical telescope has two lenses of focal powers 0.5 D and 20 D. Its magnifying power will be [CPMT 1997]
(a) 40 (b) 10 (c) 100 (d) 35
206. An astronomical telescope of ten-fold angular magnification has a length of 44 cm. The focal length of the objective is[CBSE PMT 199
(a) 4 cm (b) 40 cm (c) 44 cm (d) 440 cm
207. A telescope consisting of an objective of focal length 100 cm and a single eyes lens of focal length 10 cm is focussed
on a distant object in such a way that parallel rays emerge from the eye lens. If the object subtends an angle of 2°
at the objective, the angular width of the image is [JIPMER 1997]
(a) 20° (b) 1/6° (c) 10° (d) 24°
208. When diameter of the aperture of the objective of an astronomical telescope is increased, its [MP PMT 1997]
(a) Magnifying power is increased and resolving power is decreased
(b) Magnifying power and resolving power both are increased
(c) Magnifying power remains the same but resolving power is increased
(d) Magnifying power and resolving power both are decreased
209. The focal length of objective and eye-piece of a telescope are 100 cm and 5 cm respectively. Final image is formed
at least distance of distinct vision. The magnification of telescope is [RPET 1997]
A collection by Pradeep Kshetrapal for Physics students at genius, Maxwell and Gupta classes Only
genius PHYSICS
Reflection of Light 71
211. The diameter of the objective of the telescope is 0.1 metre and wavelength of light is 6000 Å. Its resolving power
would be approximately [MP PET 1997]
(a) 7.32 10 6 radian (b) 1.36 10 6 radian (c) 7.32 10 5 radian (d)
1.36 10 radian
5
212. A Gallilean telescope has objective and eye-piece of focal lengths 200 cm and 2 cm respectively. The magnifying
power of the telescope for normal vision is [MP PMT 1996]
(a) The total focal length of an astronomical telescope is the sum of the focal lengths of its two lenses
(b) The image formed by the astronomical telescope is always erect because the effect of the combination of the
two lenses its divergent
(c) The magnification of an astronomical telescope can be increased by decreasing the focal length of the eye-
piece
(d) The magnifying power of the refracting type of astronomical telescope is the ratio of the focal length of the
objective to that of the eye-piece
214. The length of a telescope is 36 cm. The focal length of its lenses can be [Bihar MEE 1995]
220. To increase the magnifying power of telescope (fo = focal length of the objective and fe = focal length of the eye
lens)
[MP PET/PMT 1988; MP PMT 1992, 94]
(a) fo should be large and fe should be small (b) fo should be small and fe should be large
(c) fo and fe both should be large (d) fo and fe both should be small
221. The limit of resolution of a 100 cm telescope ( 5.5 10 7 m ) is [BHU 1993]
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genius PHYSICS
72 Reflection of Light
223. A planet is observed by an astronomical refracting telescope having an objective of focal length 16 m and an
eyepiece of focal length 2 cm [IIT-JEE 1993]
(a) The distance between the objective and the eyepiece is 16.02 m
(b) The angular magnification of the planet is 800
(c) The image of the planet is inverted
(d) The objective is larger than the eyepiece
224. The average distance between the earth and moon is 38 .6 10 4 km. The minimum separation between the two
points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of 5 m
with = 6000 Å is [MP PMT 1993]
(a) 5.65 m (b) 28.25 m (c) 11.30 m (d) 56.51 m
225. The focal length of the objective and eye piece of a telescope are respectively 60 cm and 10 cm. The magnitude of
the magnifying power when the image is formed at infinity is [MP PET 1991]
(a) 50 (b) 6 (c) 70 (d) 5
226. The focal length of an objective of a telescope is 3 metre and diameter 15 cm. Assuming for a normal eye, the
diameter of the pupil is 3 mm for its complete use, the focal length of eye piece must be [MP PET 1989]
(a) 6 cm (b) 6.3 cm (c) 20 cm (d) 60 cm
227. An opera glass (Gallilean telescope) measures 9 cm from the objective to the eyepiece. The focal length of the
objective is 15 cm. Its magnifying power is [DPMT 1988]
(a) 2.5 (b) 2/5 (c) 5/3 (d) 0.4
228. The focal length of objective and eye lens of a astronomical telescope are respectively 2 m and 5 cm. Final image is
formed at (i) least distance of distinct vision (ii) infinity. The magnifying power in both cases will be [MP PMT/PET 1988]
(a) – 48, – 40 (b) – 40, – 48 (c) – 40, 48 (d) – 48, 40
229. An optical device that enables an observer to see over or around opaque objects, is called [CPMT 1986]
(a) Microscope (b) Telescope (c) Periscope (d)
Hydrometer
230. The magnifying power of a telescope can be increased by [CPMT 1979]
(a) Increasing focal length of the system (b) Fitting eye piece of high power
(c) Fitting eye piece of low power (d) Increasing the distance of objects
231. An achromatic telescope objective is to be made by combining the lenses of flint and crown glasses. This proper
choice is [CPMT 1977]
(a) Convergent of crown and divergent of flint (b) Divergent of crown and convergent of flint
(c) Both divergent (d) Both convergent
232. An observer looks at a tree of height 15 m with a telescope of magnifying power 10. To him, the tree appears [CPMT 1975]
(a) 10 times taller (b) 15 times taller (c) 10 times nearer (d) 15 times
nearer
233. The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the
telescope is 1.1 m. The magnification when the image is formed at least distance of distinct vision (D = 25 cm) is
(a) 14 (b) 6 (c) 16 (d) 18
234. The objective of a telescope has a focal length of 1.2 m. it is used to view a 10.0 m tall tower 2 km away. What is
the height of the image of the tower formed by the objective
(a) 2 mm (b) 4 mm (c) 6 mm (d) 8 mm
235. A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In
normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed
by the objective? The diameter of the moon is 3.5 10 6 m and the radius of the lunar orbit round the earth is
3.8 10 8 m
(a) 10 cm (b) 12.5 cm (c) 15 cm (d) 17.5 cm
236. The aperture of the largest telescope in the world is 5 metre. If the separation between the moon and the earth is
4 10 5 km and the wavelength of the visible light is ~ 5000 Å , then the minimum separation between the
objects on the surface of the moon which can be just resolved is
(a) 1 metre approximately (b) 10 metre approximately (c) 50 metre approximately (d) 200 metre
approximately
237. In Galileo’s telescope, magnifying power for normal vision is 20 and power of eye-piece is –20 D. Distance
between the objective and eye-piece should be
A collection by Pradeep Kshetrapal for Physics students at genius, Maxwell and Gupta classes Only
genius PHYSICS
Reflection of Light 73
79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98
A collection by Pradeep Kshetrapal for Physics students at genius, Maxwell and Gupta classes Only
genius PHYSICS
74 Reflection of Light
a d b a c a b b c a b b d b a d d a c a
99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
b c c c b b c b c a b b c b d c c a a b
119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138
a d a c a d c b d b c c b a d a c a b d
139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158
c b d d c a a b c b d a d d d a c b b d
159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178
d d d c b c a a b c c b,d a b d c d b b a
179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198
b b b c c c c b a a b c d a c c a, d b d
c
199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218
c b b b b a b a c b b d b b a a a a b d
219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238
a a d a d b a a a c b a c a c d c b b a
239 240 241 242 243 244 245 246 247
b b a a c d a b b,
c
A collection by Pradeep Kshetrapal for Physics students at genius, Maxwell and Gupta classes Only
genius PHYSICS
Ge Solids and Semi-conductor 1
Ge P Ge
Ge
Energy Bands.
In isolated atom the valence electrons can exist only in one of the allowed orbitals each of a sharply
defined energy called energy levels. But when two atoms are brought nearer to each other, there are
alterations in energy levels and they spread in the form of bands.
Energy bands are of following types
(1) Valence band
The energy band formed by a series of energy levels containing valence electrons is known as valence
band. At 0 K, the electrons fills the energy levels in valence band starting from lowest one.
(i) This band is always fulfill by electron.
(ii) This is the band of maximum energy.
(iii) Electrons are not capable of gaining energy from external electric field.
(iv) No flow of current due to such electrons.
(v) The highest energy level which can be occupied by an electron in valence band at 0 K is called
fermi level.
(2) Conduction band
The higher energy level band is called the conduction band.
(i) It is also called empty band of minimum energy.
(ii) This band is partially filled by the electrons.
(iii) In this band the electrons can gain energy from external electric field.
(iv) The electrons in the conduction band are called the free electrons. They are able to move any
where within the volume of the solid.
(v) Current flows due to such electrons.
(3) Forbidden energy gap (Eg)
Energy gap between conduction band and valence band E g (C.B.) min (V .B.) max max.
C.B.
(i) No free electron present in forbidden energy gap. min.
Eg
(ii) Width of forbidden energy gap upon the nature of substance. max.
V.B.
(iii) As temperature increases (), forbidden energy gap decreases () very slightly. min.
Types of Solids.
On the basis of band structure of crystals, solids are divided in three categories.
genius PHYSICS
2 Solids and Semi-conductor
(4) Energy gap Zero or very small Very large; for For Ge Eg = 0.7 eV
diamond it is 6 eV for Si Eg = 1.1 eV
(5) Current carries Free electrons –– Free electrons and
holes
(6) Condition of V.B. and V.B. and C.B. are V.B. – completely filled V.B. – somewhat
C.B. at ordinary completely filled or C.B. is C.B. – completely empty
temperature some what empty unfilled C.B. – somewhat
filled
(7) Temperature co-efficient Positive Zero Negative
of resistance ()
(8) Effect of temperature on Decreases — Increases
conductivity
(9) Effect of temperature on Increases — Decreases
resistance
(11) Examples Cu, Ag, Au, Na, Pt, Hg etc. Wood, plastic, mica, Ge, Si, Ga, As etc.
diamond, glass etc.
(12) Electron density 1029/m3 — Ge ~ 1019 /m3
Si ~ 1016 /m3
Holes in semiconductors
At absolute zero temperature (0 K) conduction band of semiconductor is completely empty and the
semiconductor behaves as an insulator.
When temperature increases the valence electrons acquires thermal energy to jump to the conduction
band (Due to the braking of covalent bond). If they jumps to C.B. they leaves behind the deficiency of
electrons in the valence band. This deficiency of electron is known as hole or cotter. A hole is considered
as a seat of positive charge, having magnitude of charge equal to that of an electron.
(1) Holes acts as virtual charge, although there is no physical charge on it.
(2) Effective mass of hole is more than electron.
(3) Mobility of hole is less than electron.
Types of Semiconductors.
genius PHYSICS
Solids and Semi-conductor 3
v v
E = Applied electric field e e mobility of e– and n h mobility of holes
E E V
(iv) The number of atoms of impurity element is about 1 in 10 8 atoms of the semiconductor.
(v) n e n h
genius PHYSICS
4 Solids and Semi-conductor
(vi) In these fermi level shifts towards valence or conduction energy bands.
(vii) Their conductivity is high and they are practically used.
(3) Types of extrinsic semiconductor
Ge P Ge Ge B Ge
N-types P-types
Ge S.C. Ge S.C.
V.B. V.B.
P-N Junction
P N
P N
Anode Cathode
Due to diffusion, neutrality of both N and P-type semiconductor is disturbed, a layer of negative
charged ions appear near the junction in the P-crystal and a layer of positive ions appears near the
junction in N-crystal. This layer is called depletion layer
(i) The thickness of depletion layer is 1 micron = 10–6 m. – +
1 VB
(ii) Width of depletion layer
Dopping
(iii) Depletion is directly proportional to temperature.
(iv) The P-N junction diode is equivalent to capacitor in which the P N
Depletion layer
depletion layer acts as a dielectric.
(2) Potential barrier
The potential difference created across the P-N junction due to the diffusion of electron and holes is
called potential barrier.
For Ge VB 0 .3 V and for silicon VB 0 .7 V
On the average the potential barrier in P-N junction is ~ 0.5 V and the width of depletion region ~ 10–
6.
V 0 .5
So the barrier electric field E 6 5 10 5 V / m
d 10
Some important graphs
Potential Charge density
Electric
P N P N field
P N
+ve
–ve
Distance distance Distance
P N P N
+ – – +
(ii) Width of depletion layer decreases (ii) Width of depletion layer increases
(iii) RForward 10 - 25 (iii) RReverse 105
(iv) Forward bias opposes the potential barrier and (iv) Reverse bias supports the potential barrier and
for V > VB a forward current is set up across the no current flows across the junction due to the
junction. diffusion of the majority carriers.
(A very small reverse currents may exist in the circuit
due to the drifting of minority carriers across the
junction)
(v) Cut-in (Knee) voltage : The voltage at which the (v) Break down voltage : Reverse voltage at which
current starts to increase. For Ge it is 0.3 V and for Si break down of semiconductor occurs. For Ge it is 25
it is 0.7 V. V and for Si it is 35 V.
(vi) df – diffusion (vi)
Forward current in mA
Reverse voltage
dr – drift
Break down Reverse current
P N voltage
Idf
Idr
Idf
Inet Idr
Knee
voltage Forward voltage Inet
output RL
Input RL
dc
ac
O/P (dc)
D2
Input Input
ac signal + + ac signal + +
– – – –
Output Output
+ + D1 D2 D1 D2
dc signal dc signal
Fluctuating dc
During positive half cycle During positive half cycle
Diode forward biased Diode : D1 forward biased
Output signal obtained D2 reverse biased
During negative half cycle Output signal obtained due to D1 only
Diode reverse biased During negative half cycle
Output signal not obtained Diode : D1 reverse biased
D2 forward biased
Output signal obtained due to D2 only
Transistor.
genius PHYSICS
8 Solids and Semi-conductor
A junction transistor is formed by sandwiching a thin layer of P-type semiconductor between two N-
type semiconductors or by sandwiching a thin layer of n-type semiconductor between two P-type
semiconductor.
E N P N C E P N P C
Note : In normal operation base-emitter is forward biased and collector base junction is reverse
biased.
(1) Working of Transistor : In both transistor emitter - base junction is forward biased and
collector – base junction is reverse biased.
Ie Ic
Ie Ic
Ib
Ib
VEB VCB
VEB VCB
5% emitter electron combine with the holes in the base 5% emitter holes combine with the electrons in the base
region resulting in small base current. Remaining 95% region resulting in small base current. Remaining 95%
electrons enter the collector region. holes enter the collector region.
Ie > Ic , and Ic = Ib + Ic Ie > Ic , and Ic = Ib + Ic
Note : In a transistor circuit the reverse bias is high as compared to the forward bias. So
that it may exert a large attractive force on the charge carriers to enter the collector region.
(2) Characteristics of transistors : A transistor can be connected in a circuit in the following
three different configurations.
(i) Common base (CB) (ii) Common emitter (CE) (iii) Common collector (CC)
(i) CB characteristics : The graphs between voltages and currents when base of a transistor is
common to input and output circuits are known as CB characteristic of a transistor.
PNP VCB = – 10 V
Ie Ic
E C
+ Ie (mA) VCB = – 20 V VCB = 0 Ic (mA) Ie = 15 mA
B Ie = 10 mA
– RL
Ib VCB = output Ie = 5 mA
VEB = input – Ie = 0 mA
+
VEB (in volt) VCB (in volt)
genius PHYSICS
Solids and Semi-conductor 9
(ii) CE characteristics : The graphs between voltages an d currents when emitter of a transistor is
common to input and output circuits are known as CE characteristics of a transistor.
PNP
C Ic
Ie B
VCE = – 0.1 V VCE = – 0.2 V
– Ib (A) Ic (mA) Ib = 30 mA
E Ib = 20 mA
+
Ie RL VCE = output Ib = 10 mA
– Ib = 0 mA
VEB = input
+
VEB (in volt) VCE (in volt)
Knee voltage
(3) Transistor as an amplifier : A device which increases the amplitude of the input signal is
called amplifier.
Amplifier
Input signal
Output amplified signal
RL Output
signal
Input VC Input Ie RL
~ Ib ~ signal
signa – signa –
l VCB l VEB VCE
VEB + +
+ – – +
CB amplifier CE amplifier
(5) Relation between and : or
1 1
(6) Comparison between CB, CE and CC amplifier
S.No. Characteristic Amplifier
CB CE CC
(i) Input resistance (Ri) 50 to 200 low 1 to 2 k medium 150 – 800 k high
(ii) Output resistance (Ro) 1 – 2 k high 50 k medium k low
(iii) Current gain 0.8 – 0.9 low 20 – 200 high 20 – 200 high
(iv) Voltage gain Medium High Low
(v) Power gain Medium High Low
(vi) Phase difference Zero 180o Zero
between input and
output voltages
(vii) Used as amplifier for current Power Voltage
Example
s
Example: 2 A Ge specimen is doped with Al. The concentration of acceptor atoms is ~1021 atoms/m3. Given that
the intrinsic concentration of electron hole pairs is ~ 10 19 / m 3 , the concentration of electrons in the
specimen is [AIIMS 2004]
(a) 10 17 / m 3 (b) 10 15 / m 3 (c) 10 4 / m 3 (d) 10 2 / m 3
Solution : (a) n i2 n h n e (10 19 ) 2 10 21 n e n e 10 17 / m 3 .
Example: 3 A silicon specimen is made into a P-type semi-conductor by doping, on an average, one Indium atom
per 5 10 7 silicon atoms. If the number density of atoms in the silicon specimen is 5 10 28
atoms/m3, then the number of acceptor atoms in silicon will be
(a) 2.5 10 30 atoms/cm3 (b) 1.0 10 13 atoms/cm3 (c) 1.0 10 15 atoms/cm3 (d) 2.5 10 36 atoms/cm3
Solution : (c) Number density of atoms in silicon specimen = 5 1028 atom/m3 = 5 1022 atom/cm3
Since one atom of indium is doped in 5 107 Si atom. So number of indium atoms doped per cm-3 of
silicon.
5 10 22
n 1 10 15 atom / cm 3 .
5 10 7
Example: 4 A P-type semiconductor has acceptor levels 57 meV above the valence band. The maximum
wavelength of light required to create a hole is (Planck’s constant h = 6.6 10 34 J-s)
(a) 57 Å (b) 57 10 3 Å (c) 217100 Å (d) 11 .61 10 33 Å
hc hc 6 .6 10 34 3 10 8
Solution : (c) E = 217100 Å.
E 57 10 3 1 .6 10 19
Example: 5 A potential barrier of 0.50V exists across a P-N junction. If the depletion region is 5.0 10 7 m wide,
the intensity of the electric field in this region is [UPSEAT 2002]
(a) 1 .0 10 6 V / m (b) 1 .0 10 5 V / m (c) 2.0 10 5 V / m (d) 2.0 10 6 V / m
genius PHYSICS
Solids and Semi-conductor 11
V 0.50
Solution : (a) E = 1 106 V/m.
d 5 10 7
Example: 6 A 2V battery is connected across the points A and B as shown in the figure given below. Assuming
that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current
supplied by the battery when its positive terminal is connected to A is
10
(a) 0.2 A
(b) 0.4 A 10
(c) Zero
(d) 0.1 A A B
Solution : (a) Since diode in upper branch is forward biased and in lower branch is reversed biased. So current
V
through circuit i ; here rd = diode resistance in forward biasing = 0
R rd
V 2
So i 0 .2 A .
R 10
Example: 7 Current in the circuit will be
5
(a) A 20
40
5
(b) A 30
50
5 i
(c) A
10 20 5V
5
(d) A
20
Solution : (b) The diode in lower branch is forward biased and diode in upper branch is reverse biased
5 5
i A
20 30 50
Example: 8 Find the magnitude of current in the following circuit [RPMT 2001]
(a) 0 –4 3 –1
(b) 1 amp V V
Example: 11 In the following common emitter configuration an NPN transistor with current gain = 100 is used.
The output voltage of the amplifier will be [AIIMS 2003]
(a) 10 mV
10K
(b) 0.1 V 1K Vout
(c) 1.0 V 1mV
(d) 10 V
Output voltage
Solution : (c) Voltage gain Vout = Vin Voltage gain
Input voltage
RL 10
Vout = Vin Current gain Resistance gain = Vin = 10 3 100 1V .
R BE 1
Example: 12 While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2
mA when the emitter current changes by 8.3 mA. The value of forward current ratio hfe is
(a) 82 (b) 83 (c) 8.2 (d) 8.3
i 8 .2
Solution : (a) h fe c 82
ib Vce 8 . 3 8 .2
Example: 13 The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the
common-emitter configuration is 1 K. The peak value for an ac input voltage of 0.01 V peak is
(a) 100 A (b) 0.01 mA (c) 0.25 mA (d) 500 A
Vi 0.01
Solution : (d) ic ib 50 500 10 6 A 500 A
Ri 1000
Example: 14 In a common base amplifier circuit, calculate the change in base current if that in the emitter current
is 2 mA and = 0.98 [BHU 1995]
(a) 0.04 mA (b) 1.96 mA (c) 0.98 mA (d) 2 mA
Solution : (a) ic ie 0.98 2 196 mA
ib ie ic 2 1.96 0.04 mA .
EMI_AC “genius” Physics classes by Pradeep Kshetrapal
Revision Notes
Formulas
Current element
It is the product of current and length of infinitesimal segment of current carrying wire.
B
A
The current element is taken as a vector quantity. Its direction is same as the direction of current.
i
dl
Current element AB = i dl
In the figure shown below, there is a segment of current carrying wire and P is a point where magnetic
field is to be calculated. i d l is a current element and r is the distance of the point ‘P’ with respect to the current
element i d l . According to Biot-Savart Law, magnetic field at point ‘P’ due to the current element i d l is given
i dlsin θ 0 i dl sin
by the expression, dB k
r 2
also B dB
4 r 2
.
P
idl sin
In C.G.S. : k = 1 dB Gauss dl r
r2
i
0 idl sin
In S.I. : k dB 0 Tesla
4 4 r2
Wb Henry
where 0 = Absolute permeability of air or vacuum 4 10 7 . It's other units are
Amp metre metre
N Tesla metre
or 2
or
Amp Ampere
2 Magnetic Effect of Current
genius PHYSICS
(2) Similarities and differences between Biot-Savart law and Coulomb’s Law
(i) The current element produces a magnetic field, whereas a point charge produces an electric field.
(ii) The magnitude of magnetic field varies as the inverse square of the distance from the current element,
as does the electric field due to a point charge.
0 id l rˆ 1 q1 q 2
dB Biot-Savart Law F rˆ Coulomb’s Law
4 r 2 4 0 r2
(iii) The electric field created by a point charge is radial, but the magnetic field created by a current
element is perpendicular to both the length element d l and the unit vector r̂ .
dl E
i
r̂ B
+q r̂
B
(3) Right hand thumb rule of circular currents
According to this rule if the direction of current in circular i
genius PHYSICS
Magnetic Effect of Current 3
3
conducting coil is in the direction of folding fingers of right hand, then the direction of magnetic field will be in
the direction of stretched thumb.
(4) Right hand palm rule
If we stretch our right hand such that fingers point towards the point. At
which magnetic field is required while thumb is in the direction of current then
normal to the palm will show the direction of magnetic field.
B
Note : If magnetic field is directed perpendicular and into the plane of the paper it is represented by
(cross) while if magnetic field is directed perpendicular and out of the plane of the paper it is
represented by (dot)
i i
i i CW ACW
B B B B
Out In In Out
In Out
If a coil of radius r, carrying current i then magnetic field on it's axis at a distance x from its centre given
by
μ0 2 πNir 2
B axis . 2 ; where N = number of turns in coil.
4 π ( x r 2 ) 3/2 r
P
B
Different cases O x
i
Case 1 : Magnetic field at the centre of the coil
0 2Ni Ni
(i) At centre x = 0 Bcentre . = 0 B max
4 r 2r
2i 0 i 0 2i
(ii) For single turn coil N = 1 Bcentre 0 . (iii) In C.G.S. 1 B centre
4 r 2r 4 r
r
(ii) They locates at x from the centre of the coil.
2 x = – r/2 x = 0 x = r/2
a a
O A A A
O1 O2
O1 O O2 x
+ +
a a
– – x x
2 2
Note : The device whose working principle based on this arrangement and in which uniform
magnetic field is used called as "Halmholtz galvanometer".
(3) Magnetic field due to current carrying circular arc : Magnetic field at centre O
genius PHYSICS
Magnetic Effect of Current 5
5
i i
i
r O r
O r
O
0 i 0 i 0 i 0 (2 ) i
B . B . B .
4 r 4r 4 r 4 r
Special results
Note : B1 r2 r1
B 2 r2 r1
(ii) Non-coplanar and concentric : Plane of both coils are perpendicular to each other
B2
Different cases
Case 1 : When the linear conductor Case 2 : When the linear conductor Case 3 : When the linear conductor
XY is of finite length and the point P XY is of infinite length and the point is of infinite length and the point P
lies on it's perpendicular bisector as P lies near the centre of the lies near the end Y or X
shown conductor
Y
P
P
i r i i
P
X
1 2 1 90 o and 2 0 o .
0 i 1 = 2 = 90o. 0 i i
So B . (2 sin ) So, B [sin 90 o sin 0 o ] 0
4 r i 2i 4 r 4 r
So, B 0 [sin 90 o sin 90 o ] 0
4 r 4 r
Note : When point P lies on axial position of current carrying conductor then magnetic field at P
i
P
B=0
The value of magnetic field induction at a point, on the centre of separation of two linear parallel
conductors carrying equal currents in the same direction is zero.
(6) Zero magnetic field : If in a symmetrical geometry, current enters from one end and exists from the
other, then magnetic field at the centre is zero.
i i
O
O
O
O O O
Concepts
If a current carrying circular loop (n = 1) is turned into a coil having n identical turns then magnetic field at the centre of
the coil becomes n2 times the previous field i.e. B (n turn) = n2 B(single turn)
When a current carrying coil is suspended freely in earth's magnetic field, it's plane stays in East-West direction.
0 q(v r ) 0 q(v rˆ )
Magnetic field ( B ) produced by a moving charge q is given by B ; where v = velocity of charge
4 r3 4 r 2
and v << c (speed of light).
B
r
q v
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Magnetic Effect of Current 7
7
If an electron is revolving in a circular path of radius r with speed v then magnetic field produced at the centre of circular
ev
path B 0 . 2 .
4 r
Example
s
Example: 1 Current flows due north in a horizontal transmission line. Magnetic field at a point P vertically above it
directed
P
(a) North wards
(b) South wards N
i
W E
(c) Toward east S
1
(a) 2 (b) 1 (c) (d) None of these
2
Solution : (a) Current carrying coil behaves as a bar magnet as shown in figure.
Q B2
We also knows for a bar magnet, if axial and equatorial distance
are same then B a 2 B e x
P
B1 2 S N
Hence, in this equation B1
B2 1 x
Example: 3 Find the position of point from wire 'B' where net magnetic field is zero due to following current
distribution
(a) 4 cm
30 A B
(b) cm
7
5i 2i
12
(c) cm 6cm
7
(d) 2 cm
Solution : (c) Suppose P is the point between the conductors where net magnetic field is zero.
So at P |Magnetic field due to conductor 1| = |Magnetic field due to conductor 2|
0 2(5i) 0 2(2i) 5 9 5i 2i
30
i.e. . . x cm B1
4 i 4 (6 x ) x 6x 7
B2 P
6 cm
1 2
x cm (6 – x)
cm
8 Magnetic Effect of Current
genius PHYSICS
30 12
Hence position from B 6 cm
7 7
Example: 4 Find out the magnitude of the magnetic field at point P due to following current distribution
0 ia
(a)
r 2 i B r
ia 2 a
(b) 0
r M P
0 ia a
(c) r B
2r 2 i
2 0 ia
(d)
r 2
0 2i
Solution : (a) Net magnetic field at P, Bnet = 2B sin ; where B = magnetic field due to one wire at P .
4 r
a 0 2i a 0 ia
and sin Bnet 2
. .
r 4 r r r 2
Example: 5 What will be the resultant magnetic field at origin due to four infinite length wires. If each wire produces
magnetic field 'B' at origin
Y
(a) 4 B 1
i
x
(b) 2B i 4 2
x i X
(c) 2 2 B
3 i
(d) Zero
Solution : (c) Direction of magnetic field (B1, B2, B3 and B4) at origin due to wires 1, 2, 3 and 4 are shown in the following
figure. 1
i 2
0 2i i i
B1 B 2 B 3 B 4 . B . So net magnetic field at origin O B4
4 x B2
O B3 B1
Bnet (B1 B 2 ) 2 (B 2 B 4 ) 2
4 i
(2 B)2 (2 B)2 2 2 B 3
Example: 6 Two parallel, long wires carry currents i1 and i2 with i1 i2 . When the currents are in the same direction,
the magnetic field at a point midway between the wires is 10 T. If the direction of i2 is reversed, the field
becomes 30 T. The ratio i1 / i2 is
(a) 4 (b) 3 (c) 2 (d) 1
Solution : (c) Initially when wires carry currents in the same direction as shown.
Magnetic field at mid point O due to wires 1 and 2 are respectively i1 i2
2i 2i O
B1 0 . 1 and B2 0 . 2
4 x 4 x x x
0 2
Hence net magnetic field at O Bnet (i1 i2 ) 1 2
4 x
0 2
10 10 6 . (i1 i2 ) .....(i)
4 x i1 i2
If the direction of i2 is reversed then O
x x
1 2
genius PHYSICS
Magnetic Effect of Current 9
9
0 2i1 2i
B1 . and B 2 0 . 2
4 x 4 x
2 2
So Bnet 0 . (i1 i2 ) 30 10 6 0 . (i1 i2 ) ......(ii)
4 x 4 x
i1 i2 3 i 2
Dividing equation (ii) by (i) 1
i1 i2 1 i2 1
Example: 7 A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of three turns. If
in both cases same amount of current is passed, then the ratio of the intensities of magnetic field
produced at the centre of a coil will be
1 1
(a) 9 times of first case (b) times of first case (c) 3 times of first case (d) times of first case
9 3
0 2ni
Solution : (a) Magnetic field at the centre of n turn coil carrying current i B . ......(i)
4 r
0 2i
For single turn n =1 B . .....(ii)
4 r
r
If the same wire is turn again to form a coil of three turns i.e. n = 3 and radius of each turn r'
3
0 2 (3) 0 2i
So new magnetic field at centre B' . B' 9 . ......(iii)
4 r' 4 r
Comparing equation (ii) and (iii) gives B' 9 B .
Example: 8 A wire in the form of a square of side a carries a current i. Then the magnetic induction at the centre of the
square wire is (Magnetic permeability of free space = 0)
0i
(a)
2 a
i
0i 2
(b) O
a
2 20i
(c)
a a
0i
(d)
2 a
Solution : (c) Magnetic field due to one side of the square at centre O i
0 2i sin 45 o
B1 . i
4 a/2 45o
O
45o
2 2i
B1 0 .
4 a
a/2
(2 2 i) a
Hence magnetic field at centre due to all side B net 4 B1 0
a
Example: 9 The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at
the centre of a square coil made from the same length of wire will be
2 2
(a) (b) (c) (d)
4 2 8 2 2 2 4 2
i
r 45o
O
45o
i a/2
a
10 Magnetic Effect of Current
genius PHYSICS
Length L = 2 r Length L = 4a
0 2i 0 4 2 i 0 2 2 i 8 2i
Magnetic field B . . B . B 0 .
4 r 4 r 4 a 4 a
B circular 2
Hence
B square 8 2
Example: 10 Find magnetic field at centre O in each of the following figure
(i) (ii) (iii)
r1
r2
r O
i
r2 r1
O
O
0 i
While due to part (2) B2 . r
4 r
(1) O (3)
Net magnetic field at centre O,
0 i i
Bnet B1 B 2 B 3 . B net 0
4 r 4r
2
(ii) (b) B1 B 3 0
0 i 1 r1 3
B2 .
4 r1 O
0 i r2
B4 .
4 r2
4
0 1 1
So B net B 2 B 4 . i
4 r1 r2
4
(iii) (a) B1 B 3 0
r2 2
0 i
B2 .
4 r1 r1
0 i 1 O 3
B4 . As | B 2 | | B 4 |
4 r2
genius PHYSICS
Magnetic Effect of Current 11
11
0i 1 1
So B net B 2 B 4 B net
4 r1 r2
Example: 11 Find magnetic field at centre O in each of each of the following figure
(i) (ii) (iii)
i
i i
r r r
90o
O
O O
0i 0 i 0 2i
(a) (a) ( 2) (a) ( 1)
2r 2 r 2r r
0i 0 i i 0 i 2i
(b) (b) . ( 2) (b) . ( 1)
2r 4 r 4r r
30i 0i
(c) (c) (c) Zero
8r 4r
30 i 0i
(d) (d) (d) Infinite
8r 4r
0 (2 )i (2 / 2)i 3 0 i
Solution : (i) (d) By using B . B 0.
4 r 4 r 8r
(ii) (b) Magnetic field at centre O due to section 1, 2 and 3 are respectively
0 i
B1 . 1 i
4 r
i
0 i r
B2 .
4 r 2
O
0 i
B3 .
4 r 3
0 i
Bnet B1 B 2 B 3 ( 2)
4 r
(iii) (b) The given figure is equivalent to following figure, magnetic field at O due to long wire (part 1)
0 2i
B1 . i
4 r r
2
2 i
Due to circular coil B 2 0 . O
4 r
Hence net magnetic field at O
0 2i 1
Bnet B 2 B1 . ( 1)
4 r
Example: 12 The field B at the centre of a circular coil of radius r is times that due to a long straight wire at a
distance r from it, for equal currents here shows three cases; in all cases the circular part has radius r and
straight ones are infinitely long. For same current the field B is the centre P in cases 1, 2, 3 has the ratio[CPMT 198
i
r
i r
O r O
i
90o
O
12 Magnetic Effect of Current
genius PHYSICS
3 1 3 1
(a) : : (b) 1 : 1 :
2 2 4 2 2 2 4 2
3 1 3 1
(c) : : (d) 1 : :
2 2 4 2 2 2 4 2
0 i (A)
Solution : (a) Case 1 : B A .
4 r i
r
i
BB 0 . (B) O
4 r
0 i
BC .
4 r (C)
70 10 0 14 0 50
(a) (b) (c) (d)
d d d d
Solution : (d) Magnetic field at P
P (0, 0, d) Y
2 (8) B2
Due to wire 1, B1 0 .
4 d B1
6A
0 2 (16)
and due to wire 2, B 2 . X
4 d 8A
2 2
16 12 0 2 50
B net B12 B 22 0 . 0 . 10
4 d 4 d 4 d d
Example: 14 An equilateral triangle of side 'a' carries a current i then find out the magnetic field at point P which is vertex of
triangle
genius PHYSICS
Magnetic Effect of Current 13
13
0 i
(a)
2 3a P
0 i
(b) i i
2 3a
2 30i
(c) a
a
(d) Zero
Solution : (b) As shown in the following figure magnetic field at P due to side 1 and side 2 is zero.
Magnetic field at P is only due to side 3, P
1 2
0 2i sin 30 o
which is B1 . i 30o 30o i 3a
4 3a 2
2
3
0 2i 0 i
. = a
4 3a 2 3a
Example: 15 A battery is connected between two points A and B on the circumference of a uniform conducting ring of
radius r and resistance R. One of the arcs AB of the ring subtends an angle at the centre. The value of,
the magnetic induction at the centre due to the current in the ring is [IIT-JEE 1995]
H1 3
H2 = magnetic field at M due to R + due to QS + due to PQ = 0 H1 H1
2 2
H1 2
H2 3
Example: 21 Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC
is 2r. The value of magnetic field at the centre of the loop assuming uniform wire is
2 0i
(a)
3 a B
2 0i A
(b) C
3 a O
i
2 0i
(c)
a D
2 0i
(d)
a
Solution : (b) According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC
i 1 2i i
is half that of ABC i.e. 2 . Also i1 i2 i i1 and i2
i1 2 3 3
0 2i1 sin 45 o 2 2 i1
Magnetic field at centre O due to wire AB and BC (part 1 and 2) B1 . 0.
4 a/2 4 a
genius PHYSICS
Magnetic Effect of Current 15
15
0 2 2 i2
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) B 3 B 4
4 a
Also i1 = 2i2. So (B1 = B2) > (B3 = B4)
B
Hence net magnetic field at centre O (1) (2)
Bnet (B1 B 2 ) (B 3 B 4 ) A i1 C
i i2 O
2 i
2 2 i 2 2 2 (3) (4)
3 0 3 4 2i 2 0i
2 0 . . 0 . (2 1) D
4 a 4 a 4 3 a 3 a
Tricky example: 1
Figure shows a straight wire of length l current i. The magnitude of magnetic field produced by the
current at point P is l
P
i
l
Tricky example: 2
A cell is connected between the points A and C of a circular conductor ABCD of centre 'O' with
angle AOC = 60°, If B 1 and B 2 are the magnitudes of the magnetic fields at O due to the currents
B1
in ABC and ADC respectively, the ratio is i1
B
[KCET (Engg./ Med.) 1999]
B2 300o
O
(a) 0.2 60o
A C
(b) 6 i2 D
(c) 1
1A
(d) 5
0 i
Solution: (c) B .
4 r
i1
1
B i 300o
O
B1 i 60o
1 1
B 2 2 i2
i2 2
1A
16 Magnetic Effect of Current
genius PHYSICS
i1 l B1 1
Also 2 2 Hence
i2 l1 1 B2 1
Amperes Law.
Amperes law gives another method to calculate the magnetic field due to a given current distribution.
Line integral of the magnetic field B around any closed curve is equal to 0 times the net current i
threading through the area enclosed by the curve
B
i.e. Bd l 0 i 0 (i1 i3 i2 )
i5
i3
i1
Also using B 0 H (where H = magnetising field)
i2
0 H .dl 0 i H .dl i i4
Note : Total current crossing the above area is (i1 i3 i2 ) . Any current outside the area is not
O O
R R r r
R1
i i i
R2
0 i
B out
2r
0 i
In all the above cases B surface
2R
(ii) Inside the cylinder : Magnetic field inside the hollow cylinder is zero.
B0
B=0 B=0
B0
Cross sectional view Solid cylinder Thin hollow cylinder Thick hollow
cylinder
genius PHYSICS
Magnetic Effect of Current 17
17
R Loop
r r Loop R1
Loop i
R2
Current enclosed by loop (i) is lesser then the total current Current enclosed by loop (i) is lesser then the total current
(i) (i)
Baxis = 0 (min.), Bsurface = max (distance r always from axis of cylinder), Bout 1/r.
(2) Magnetic field due to an infinite sheet carrying current : The figure shows an infinite sheet of
current with linear current density j (A/m). Due to symmetry the field line pattern above and below the sheet is
uniform. Consider a square loop of side l as shown in the figure.
B
P
d c
i
l
a l b
b c d a
According to Ampere’s law, a
b
B.dl B.dl B.dl B.dl 0 i .
c d
c a
Since B dl along the path b c and d a, therefore,
b
B.dl 0 ; B.dl 0
d
b a
Also, B || dl along the path a b and c d, thus B.dl B.dl 2 Bl
a d
(3) Solenoid
18 Magnetic Effect of Current
genius PHYSICS
A cylinderical coil of many tightly wound turns of insulated wire with generally diameter of the coil
smaller than its length is called a solenoid.
One end of the solenoid behaves like the north pole and opposite end behaves like the south pole. As the
length of the solenoid increases, the interior field becomes more uniform and the external field becomes
weaker.
B=0
S N
B
i Solenoid i
A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is
uniform and parallel to the axis of solenoid.
(i) Finite length solenoid : If N = total number of turns,
r
l = length of the solenoid
P
N
n = number of turns per unit length
l
0
Magnetic field inside the solenoid at point P is given by B (2 ni)[sin sin ]
4
(ii) Infinite length solenoid : If the solenoid is of infinite length and the point is well inside the
solenoid i.e. ( / 2) .
So B in μ 0 ni
(ii) If the solenoid is of infinite length and the point is near one end i.e. 0 and ( / 2)
1
So B end (μ 0 ni )
2
Winding
Core
r P
r dl
O
i B
0 Ni N
B (2r) 0 Ni B o ni where n
2r 2r
For any point inside the empty space surrounded by toroid and outside the toroid, magnetic field B is zero
because the net current enclosed in these spaces is zero.
Concepts
The line integral of magnetising field (H ) for any closed path called magnetomotive force (MMF). It's S.I. unit is amp.
Example
s
Example: 22 A long solenoid has 200 turns per cm and carries a current of 2.5 A. The magnetic field at its centre is
[0 = 4 10–7 Wb/m2] [MP PET 2000]
(a) 3.14 10–2 Wb/m2 (b) 6.28 10–2 Wb/m2 (c) 9.42 10–2 Wb/m2 (d) 12.56 10–2 Wb/m2
200
Solution : (b) B 0 ni 4 10 7 2.5 6.28 10 2 Wb / m 2 .
10 2
Example: 23 A long solenoid is formed by winding 20 turns/cm. The current necessary to produce a magnetic field of
20 mili tesla inside the solenoid will be approximately 0 10 7 Tesla - metre / ampere [MP PMT 1994]
4
(a) 8.0 A (b) 4.0 A (c) 2.0 A (d) 1.0 A
20 turn turn
Solution : (a) B 0 ni ; where n = 2000 . So, 20 10 5 4 2000 i i 8 A.
10 cm m
Example: 24 Two solenoids having lengths L and 2L and the number of loops N and 4N, both have the same current,
then the ratio of the magnetic field will be [CPMT 1994]
Tricky example: 3
A winding wire which is used to frame a solenoid can bear a maximum 10 A current. If length of
solenoid is 80cm and it's cross sectional radius is 3 cm then required length of winding wire is
(B 0.2 T )
0 Ni
Solution : (c) B where N Total number of turns, l length of the solenoid
l
4 10 7 N 10 4 10 4
0 .2 N
0 .8
Since N turns are made from the winding wire so length of the wire
(L) 2 r N 2 r length of each turns
4 10 4
L 2 3 10 2 2 .4 10 3 m.
genius PHYSICS
Magnetic Effect of Current 21
21
Fm Fm
B
B
v v
90°
Direction of force on charged particle in magnetic field can also be find by Flemings Left Hand Rule
(FLHR).
Here, First finger (indicates) Direction of magnetic field F
× × × × × × ×
+ F +
× × × × × × ×
× × × × × × ×
v + v
× × × × × × ×
22 Magnetic Effect of Current
genius PHYSICS
= 90o, hence from F = qvB sin particle will experience a maximum magnetic force Fmax = qvB which
act's in a direction perpendicular to the motion of charged particle. (By Flemings left hand rule).
(i) Radius of the path : In this case path of charged particle is circular and magnetic force provides the
mv 2 mv
necessary centripetal force i.e. qvB radius of path r
r qB
If p = momentum of charged particle and K = kinetic energy of charged particle (gained by charged
particle after accelerating through potential difference V) then p mv 2mK 2mqV
mv p 2mK 1 2 mV
So r
qB qB qB B q
r v p K i.e. with increase in speed or kinetic energy, the radius of the orbit increases.
(ii) Direction of path : If a charge particle enters perpendicularly in a magnetic field, then direction of
path described by it will be
–q
B
Clockwise
Negative Inward
–q
B
Anticlockwise
Positive Inward
+q
B
Clockwise
Positive Outward
+q
(iii) Time period : As in uniform circular motion v = r, so the angular frequency of circular motion,
v qB 2 m
called cyclotron or gyro-frequency, will be given by and hence the time period, T 2
r m qB
i.e., time period (or frequency) is independent of speed of particle and radius of the orbit and depends
q
only on the field B and the nature, i.e., specific charge , of the particle.
m
When the charged particle is moving at an angle to the field (other than 0o, 90o, or 180o).
In this situation resolving the velocity of the particle along and perpendicular to the field, we find that the
particle moves with constant velocity v cos along the field (as no force acts on a charged particle when it
moves parallel to the field) and at the same time it is also moving with velocity v sin perpendicular to the field
m (vsin θ)
due to which it will describe a circle (in a plane perpendicular to the field) of radius. r
qB
Y
p B
B
v v sin v r
X
q, m
v cos
Z
2 m qB
Time period and frequency do not depend on velocity and so they are given by T and
qB 2 m
So the resultant path will be a helix with its axis parallel to the field B as shown in figure in this situation.
The pitch of the helix, (i.e., linear distance travelled in one rotation) will be given by
m
p T (v cos ) 2 (v cos )
qB
(ii) When E is parallel to B and both these fields are perpendicular to v then : Fe is
perpendicular to Fm and they cannot cancel each other. The path of charged particle is curved in both these
fields. E B
v
q
(iii) v , E and B are mutually perpendicular : In this situation if E and B are such that
y
E
F Fe Fm 0 i.e., a (F / m ) 0 Fe
+q +q
v x
as shown in figure, the particle will pass through the field with same velocity.
Fm
And in this situation, as Fe Fm i.e., qE qvB v E / B B
z
This principle is used in ‘velocity-selector’ to get a charged beam having a specific velocity.
S
genius PHYSICS
Magnetic Effect of Current 25
25
enormously large velocities if the particle is made to traverse the potential difference a number of times.
It consists of two hollow D-shaped metallic chambers D1 and D2 called dees. The two dees are placed
horizontally with a small gap separating them. The dees are connected to the source of high frequency electric
field. The dees are enclosed in a metal box containing a gas at a low pressure of the order of 10 –3 mm mercury.
The whole apparatus is placed between the two poles of a strong electromagnet NS as shown in fig. The
magnetic field acts perpendicular to the plane of the dees.
Note : The positive ions are produced in the gap between the two dees by the ionisation of the
gas. To produce proton, hydrogen gas is used; while for producing alpha-particles, helium gas is
used.
r m
(1) Cyclotron frequency : Time taken by ion to describe q semicircular path is given by t
v qB
2 m 1 Bq
If T = time period of oscillating electric field then T 2 t the cyclotron frequency
qB T 2m
(2) Maximum energy of position : Maximum energy gained by the charged particle
q2B2 2
E max r
2m
where r0 = maximum radius of the circular path followed by the positive ion.
force acting on electron Fm e (v B). This force acts Force acting on the hole due to magnetic field
along x-axis and hence electrons will move towards Fm e (v B) force acts along x-axis and hence holes
face (2) and it becomes negatively charged. move towards face (2) and it becomes positively
charged.
26 Magnetic Effect of Current
genius PHYSICS
Concepts
The energy of a charged particle moving in a uniform magnetic field does not change because it experiences a force in a
direction, perpendicular to it's direction of motion. Due to which the speed of charged particle remains unchanged and
hence it's K.E. remains same.
Magnetic force does no work when the charged particle is displaced while electric force does work in displacing the
charged particle.
Magnetic force is velocity dependent, while electric force is independent of the state of rest or motion of the charged
particle.
If a particle enters a magnetic field normally to the magnetic field, then it starts moving in a circular orbit. The point at
which it enters the magnetic field lies on the circumference. (Most of us confuse it with the centre of the orbit)
Deviation of charged particle in magnetic field : If a charged particle (q, m) enters a uniform magnetic field
B (extends upto a length x) at right angles with speed v as shown in figure. v
The speed of the particle in magnetic field does not change. But it gets deviated in the magnetic
field.
B
q, m v
x
Bq
Deviation in terms of time t; t t
m
v
x
Deviation in terms of length of the magnetic field ; sin 1 . This relation can be used only when x r .
r
r
For x > r, the deviation will be 180o as shown in the following figure
v
x
Example
s
Example: 28 Electrons move at right angles to a magnetic field of 1.5 10 2 Tesla with a speed of 6 10 27 m / s. If the
specific charge of the electron is 1.7 10 11 Coul/kg. The radius of the circular path will be [BHU 2003]
Example: 29 An electron (mass 9 10 31 kg . charge 1.6 10 19 coul. ) whose kinetic energy is 7 .2 10 18 joule is
moving in a circular orbit in a magnetic field of 9 10 5 weber / m 2 . The radius of the orbit is[MP PMT 2002]
(a) 1.25 cm (b) 2.5 cm (c) 12.5 cm (d) 25.0 cm
2mk
Solution : (b) By using r ; For both particles q same, B same, k same
qB
re me
Hence r m m p m e so rp re
rp mp
genius PHYSICS
Magnetic Effect of Current 27
27
Since radius of the path of proton is more, hence it's trajectory is less curved.
Example: 31 A proton and an particles enters in a uniform magnetic field with same velocity, then ratio of the radii
of path describe by them
(a) 1 : 2 (b) 1 : 1 (c) 2 : 1 (d) None of these
mv m rp mp q mp 2q p 1
Solution : (b) By using r ; v same, B same r
qB 2 r m qp 4m p qp 2
Example: 32 A proton of mass m and charge e is moving in a circular orbit of a magnetic field with energy 1MeV.
What should be the energy of -particle (mass = 4 m and charge = +2e), so that it can revolve in the path
of same radius [BHU 1997]
2mK q2
Solution : (a) By using r ; r same, B same K
qB m
2 2
K q m 2q
p p
m
p 1 K K p 1 meV .
Hence
K p q p
m qp
4m p
Example: 33 A proton and an particle enter a uniform magnetic field perpendicularly with the same speed. If
proton takes 25 sec to make 5 revolutions, then the periodic time for the particle would be [MP PET 1993]
25
Solution : (c) Time period of proton T p 5 sec
5
2 m T m qp 4m p qp
By using T T 2T p 10 sec .
qB Tp m p q mp 2q p
Example: 34 A particle with 10–11 coulomb of charge and 10–7 kg mass is moving with a velocity of 108 m/s along the y-
axis. A uniform static magnetic field B = 0.5 Tesla is acting along the x-direction. The force on the particle
is
[MP PMT 1997]
(a) 5 10–11 N along î (b) 5 103 N along k̂ (c) 5 10–11 N along ˆj (d) 5 10–4 N along k̂
Example: 35 An electron is moving along positive x-axis. To get it moving on an anticlockwise circular path in x-y
plane, a magnetic filed is applied
(a) Along positive y-axis (b) Along positive z-axis
(c) Along negative y-axis (d) Along negative z-axis
Solution : (a) The given situation can be drawn as follows
According to figure, for deflecting electron in x-y plane, force must be acting an it towards y-axis.
Hence according to Flemings left hand rule, magnetic field directed along positive y axis.
y x-y plane
e–
e–
x
z
28 Magnetic Effect of Current
genius PHYSICS
Example: 36 A particle of charge 16 10 18 coulomb moving with velocity 10 m/s along the x-axis enters a region
where a magnetic field of induction B is along the y-axis, and an electric field of magnitude 104 V/m is
along the negative z-axis. If the charged particle continuous moving along the x-axis, the magnitude of B
is [AIEEE 2003]
Solution : (b) Particles is moving undeflected in the presence of both electric field as well as magnetic field so it's speed
E E 10 4
v B 10 3 Wb / m 2 .
B v 10
Example: 37 A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It
enters a region containing a uniform magnetic field B directed along the negative z direction extending
from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b
is
[IIT-JEE (Screening) 2002]
Example: 38 At a certain place magnetic field vertically downwards. An electron approaches horizontally towards you
and enters in this magnetic fields. It's trajectory, when seen from above will be a circle which is
(a) Vertical clockwise (b) Vertical anticlockwise
(c) Horizontal clockwise (d) Horizontal anticlockwise
Solution : (c) By using Flemings left hand rule.
Example: 39 When a charged particle circulates in a normal magnetic field, then the area of it's circulation is
proportional to
(a) It's kinetic energy (b) It's momentum
(c) It's charge (d) Magnetic fields intensity
2mK (2mK )
Solution : (a) r and A Aq 2 A A K.
qB q 2b 2
Example: 40 An electron moves straight inside a charged parallel plate capacitor at uniform charge density . The
space between the plates is filled with constant magnetic field of induction B. Time of straight line
motion of the electron in the capacitor is
e
(a)
0 lB
0 lB
(b) e–
l
genius PHYSICS
Magnetic Effect of Current 29
29
e
(c)
0 B
0B
(d)
e
Solution : (b) The net force acting on the electron is zero because it moves with constant velocity, due to it's motion on
straight line.
E
Fnet F e F m 0 | F e | | F m | e E evB ve E
B 0 B o
l 0 lB
The time of motion inside the capacitor t .
v
Example: 41 A proton of mass 1.67 10 27 kg and charge 1.6 10 19 C is projected with a speed of 2 10 6 m / s at an
angle of 600 to the X-axis. If a uniform magnetic field of 0.104 Tesla is applied along Y-axis, the path of
proton is
Example: 42 A charge particle, having charge q accelerated through a potential difference V enter a perpendicular
magnetic field in which it experiences a force F. If V is increased to 5V, the particle will experience a force
F
(a) F (b) 5F (c) (d) 5F
5
1 2qV
Solution : (d) mv 2 qV v . Also F qvB
2 m
2qV
F qB hence F V which gives F' 5 F.
m
Example: 43 The magnetic field is downward perpendicular to the plane of the paper and a few charged particles are
projected in it. Which of the following is true [CPMT 1997]
Solution : (c) Both particles are deflecting in same direction so they must be of same sign.(i.e., both A and B represents
protons)
mv
By using r rv
qB
30 Magnetic Effect of Current
genius PHYSICS
From given figure radius of the path described by particle B is more than that of A. Hence v B v A .
Example: 44 Two very long straight, particle wires carry steady currents i and – i respectively. The distance between
the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in
the plane of the wires. It's instantaneous velocity v is perpendicular to this plane. The magnitude of the
force due to the magnetic field acting on the charge at this instant is
0 iqv 0 iqv 2 0 iqv
(a) (b) (c) (d) Zero
2d d d
Solution : (d) According to gives information following figure can be drawn, which shows that direction of magnetic
field is along the direction of motion of charge so net on it is zero.
v
q
Example: 45 A metallic block carrying current i is subjected to a uniform magnetic induction B as shown in the figure.
The moving charges experience a force F given by ……. which results in the lowering
d of the potential of the
face ……. Assume the speed of the carriers to be v d/2 d/2 [IIT-JEE 1996]
B Y
(a) eVB kˆ , ABCD
E G
(b) eVB kˆ , ABCD A
F H X
B
(c) eVB kˆ , ABCD C
I
D
(d) eVB kˆ , EFGH
Solution : (c) As the block is of metal, the charge carriers are electrons; so for current along positive x-axis, the electrons
are moving along negative x-axis, i.e. v vi
E G y
A
and as the magnetic field is along the y-axis, i.e. B Bˆj B e– B
x
d F v F z
i H
so F q(v B) for this case yield F (e )[vˆi Bˆj] C D
Tricky example: 4
An ionised gas contains both positive and negative ions. If it is subjected simultaneously to an
electric field along the +ve x-axis and a magnetic field along the +z direction then [IIT-JEE (Screening) 2000]
(a) Positive ions deflect towards +y direction and negative ions towards – y direction
(b) All ions deflect towards + y direction
(c) All ions deflect towards – y direction
(d) Positive ions deflect towards – y direction and negative ions towards + y direction.
Solution : (c) As the electric field is switched on, positive ion will start to move along positive x-direction and
negative ion along negative x-direction. Current associated with motion of both types of ions is
along positive x-direction. According to Flemings left hand rule force on both types of ions will be
along negative y-direction.
× × × × × × ×
× × × × × ×
dF
× × × × × × ×
B
× ×
i × × × ×
dl
× × × × × × ×
genius PHYSICS
Magnetic Effect of Current 31
31
Total magnetic force F d F i(d l B)
If magnetic field is uniform i.e., B = constant
F i d l B i(L B)
d l L vector sum of all the length elements from initial to final point. Which is in accordance with
the law of vector addition is equal to length vector L joining initial to final point.
(1) Direction of force : The direction of force is always perpendicular to the plane containing id l and
B and is same as that of cross-product of two vectors ( A B) with A i d l .
dF
i dl
P
P B
i dl B
dF
The direction of force when current element i d l and B are perpendicular to each other can also be
determined by applying either of the following rules
Magnetic
Magnetic field
field
Current Force
(2) Force on a straight wire : If a current carrying straight conductor (length l) is placed in an uniform
magnetic field (B) such that it makes an angle with the direction of field then force experienced by it is
F Bil sin
F
If 0 , F 0
o
B
If 90 o , Fmax Bil
i l
× × × × × × × × × × × × × ×
b
b F B
× × × × × × B × × × × × × × ×
× × × × × × × × × × × × × ×
× × × × × × × × × ×
× × × × i L
× × ×
L × × × × × × × × × × ×
a a
× × × × × × × × × × × × × ×
Specific Example
The force experienced by a semicircular wire of radius R when it is carrying a current i and is placed in a
uniform magnetic field of induction B as shown.
× × × × × × ×
Y Y B Y
× × × × × × ×
B
i i × × × × × × ×
× × × × × × ×
× × × × × × ×
P O Q X P O Q X × P× × O × × Q ×
X×
L 2Rˆi and B Bˆi L 2Rˆi and B Bˆj L 2Rˆi and B B (kˆ )
So by using F i(L' B) force on F i 2 BR(ˆi ˆj) F i 2 BR (ˆj)
the wire F 2 BiR (along Y-axis)
F 2 BiR kˆ i.e. F 2 BiR
F i(2 R)(B)(ˆi ˆi ) F 0
(perpendicular to paper outward)
When two long straight conductors carrying currents i1 and i2 placed parallel to each other at a distance
‘a’ from each other. A mutual force act between them when is given as
0 2i1 i2
F1 F2 F l i1 i2
4 a
where l is the length of that portion of the conductor on which force is to be calculated. a
Direction of force : If conductors carries current in same direction, then force between them will be
attractive. If conductor carries current in opposite direction, then force between them will be repulsive.
1 2× 1× ×
2×
× ×
i1 i2 i1 i2
× × × × × ×
× × × × × ×
F1 F2 F1 F2
× × × × × ×
× × × × × ×
× × × × × ×
genius PHYSICS
Magnetic Effect of Current 33
33
0 q1 q 2 v1v 2
Magnetic force between them is Fm . ..... (i)
4 r2
1 q1 q 2
and Electric force between them is Fe . ..... (ii)
4 0 r2
Fm 1
From equation (i) and (ii) 0 0 v 2 but 0 0 2 ; where c is the velocity light in vacuum. So
Fe c
2
Fm v
Fe c
T A B T
B
d
B R
O
Specific example
In the above circular loop tension in part A and B.
In balanced condition of small part AB of the loop is shown below dF
i
A B
Note : If no magnetic field is present, the loop will still open into a circle as in it’s adjacent parts
current will be in opposite direction and opposite currents repel each other.
i
Case 2 : Equilibrium of a current carrying conductor : When a finite length current carrying wire
is kept parallel to another infinite length current carrying wire, it can suspend freely in air as shown below
Fixed i1
l
Movable
X i2 Y
h
h i2
Movable
X Y
Fixed i1 l
In both the situations for equilibrium of XY it's downward weight = upward magnetic force i.e.
μ 2i i
mg 0 . 1 2 .l
4π h
Note : In the first case if wire XY is slightly displaced from its equilibrium position, it executes
h
SHM and it’s time period is given by T 2 .
g
If direction of current in movable wire is reversed then it’s instantaneous acceleration produced is
2g .
Case 3 : Current carrying wire and circular loop : If a current carrying straight wire is placed in
the magnetic field of current carrying circular loop.
i1
i1
i2
i2
l
Wire is placed in the perpendicular magnetic field wire is placed along the axis of coil so magnetic
due to coil at it's centre, so it will experience a field produced by the coil is parallel to the wire.
0 i1 Hence it will not experience any force.
maximum force F Bil i2 l
2r
Case 4 : Current carrying spring : If current is passed through a spring, then it will contract because
current will flow through all the turns in the same direction.
+
–
Spring Spring
K
m Hg
genius PHYSICS
Magnetic Effect of Current 35
35
Case 5 : Tension less strings : In the following figure the value and direction of current through the
conductor XY so that strings becomes tensionless?
Strings becomes tensionless if weight of conductor XY balanced by magnetic force (Fm ) .
String
Fm
× × × × × × ×
B
× × × × × × × i
T T X Y
× × × l× × × ×
X m Y
× × × × × × ×
mg
× × × × mg
Hence direction of current is from X Y and in balanced condition Fm mg B i l mg i
Bl
Case 6 : A current carrying conductor floating in air such that it is making an angle with the direction
of magnetic field, while magnetic field and conductor both lies in a horizontal plane.
Fm
mg
In equilibrium mg B i l sin i
B l sin
mg
Case 7 : Sliding of conducting rod on inclined rails : When a conducting rod slides on conducting
rails.
X F cos
R
B
i
v F
i
Insulated
stand
Y
i mg sin
+
mg
–
Concepts
Electric force is an absolute concept while magnetic force is a relative concept for an observer.
The nature of force between two parallel charge beams decided by electric force, as it is dominator. The nature of force
between two parallel current carrying wires decided by magnetic force.
+ + – +
i1 i2 + + – +
+ + – +
+ + – +
Example
s
Example: 46 A vertical wire carrying a current in the upward direction is placed in a horizontal magnetic field directed
towards north. The wire will experience a force directed towards
(a) North (b) South (c) East (d) West
Solution : (d) By applying Flemings left hand rule, direction of force is found towards west.
i BH
F
N
W
S E
Example: 47 3 A of current is flowing in a linear conductor having a length of 40 cm. The conductor is placed in a
magnetic field of strength 500 gauss and makes an angle of 30o with the direction of the field. It
experiences a force of magnitude
Solution : (c) By using F Bil sin F = (500 10–4) 0.4 sin 30o 3 10–2 N.
Example: 48 Wires 1 and 2 carrying currents t1 and t2 respectively are inclined at an angle to each other. What is
the force on a small element dl of wire 2 at a distance of r from 1 (as shown in figure) due to the magnetic
field of wire 1 [AIEEE 2002]
0
(a) i1 , i2 dl tan
2r
0 i1 i2
(b) i1 , i2 dl sin r
2r
dl
0
(c) i1 , i2 dl cos
2r
0
(d) i1 , i2 dl sin
4r
Solution : (c) Length of the component dl which is parallel to wire (1) is dl cos , so force on it
2i i i i dl cos
F 0 1 2 (dl cos ) 0 1 2 .
4 r 2r
Example: 49 A conductor PQRSTU, each side of length L, bent as shown in the figure, carries a current i and is placed in
a uniform magnetic induction B directed parallel to the positive Y-axis. The force experience by the wire
and its direction are
R
Z B
(a) 2iBL directed along the negative Z-axis
i
(b) 5iBL directed along the positive Z-axis S Q P
(c) iBL direction along the positive Z-axis Y
T U
(d) 2iBL directed along the positive Z-axis X
The current in TS and RQ are in mutually opposite direction. Hence, FTS FRQ 0
Therefore the force will act only on the segment SR whose value is Bil and it’s direction is +z.
Alternate method :
R
Z B
i F B
S P
Q P i
Y
T U U
X
genius PHYSICS
Magnetic Effect of Current 37
37
The given shape of the wire can be replaced by a straight wire of length l between P and U as shown below
Hence force on replaced wire PU will be F Bil
and according to FLHR it is directed towards +z-axis
Example: 50 A conductor in the form of a right angle ABC with AB = 3cm and BC = 4 cm carries a current of 10 A.
There is a uniform magnetic field of 5T perpendicular to the plane of the conductor. The force on the
conductor will be
(a) 1.5 N (b) 2.0 N (c) 2.5 N (d) 3.5 N
A A
B 10 A F
10 A
Solution : (c) According to the question figure can be drawn as shown below.
Force on the conductor ABC = Force on the conductor AC
3 cm
= 5 10 (5 10–2)
B 4 cm C B C
= 2.5 N
Example: 51 A wire of length l carries a current i along the X-axis. A magnetic field exists which is given as B B0
( ˆi ˆj kˆ ) T. Find the magnitude of the magnetic force acting on the wire
1
(a) B0 il (b) B0 i l 2 (c) 2 B0i l (d) B0 i l
2
It's magnitude F 2 B 0 il
Example: 52 A conducting loop carrying a current i is placed in a uniform magnetic field pointing into the plane of the
paper as shown. The loop will have a tendency to [IIT-JEE (Screening) 2003]]
(a) Contract (b) Expand
(c) Move towards + ve x-axis (d) Move towards – ve x-axis
Solution : (b) Net force on a current carrying loop in uniform magnetic field is zero. Hence the loop can't translate. So,
options (c) and (d) are wrong. From Flemings left hand rule we can see that if Y
i Fm
magnetic field is perpendicular to paper inwards and current in the loop is
clockwise (as shown) the magnetic force Fm on each element of the loop is X
radially outwards, or the loops will have a tendency to expand.
Example: 53 A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The centre
of the loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the
loop is B. Find the magnetic force on the wire
Y
(a) i a B i
(b) 4 i a B
a B
(c) Zero
(d) 2 i a B
Solution : (d) The direction of the magnetic force will be vertically downwards at each element of the wire.
Thus F = Bil = Bi (2a) = 2iaB.
Example: 54 A wire abc is carrying current i. It is bent as shown in fig and is placed in a uniform magnetic field of
magnetic induction B. Length ab = l and abc = 45o. The ratio of force on ab and on bc is
38 Magnetic Effect of Current
genius PHYSICS
1 2
(a) (b) 2 (c) 1 (d) b
3
2 B
Solution : (c) Force on portion ab of wire F1 = Bil sin 90o = Bil l 45o i
i c
l F
Force on portion bc of wire F2 = Bi sin 45 o Bil . So 1 1 .
a
2 F2
Example: 55 Current i flows through a long conducting wire bent at right angle as shown in figure. The magnetic field
at a point P on the right bisector of the angle XOY at a distance r from O is
0i
(a) Y
r
2 0 i P
(b) i
r r
0 i 45o
(c) ( 2 1)
4r O X
0 2i
(d) . ( 2 1)
4 r
0 i r
Solution : (d) By using B . (sin 1 sin 2 ) , from figure d r sin 45 o Y
4 r 2
d P
0 i 45o
Magnetic field due to each wire at P B . (sin 45 o sin 90 o ) i
4 (r / 2 ) r 45o
d
0 i
. ( 2 1) 45o
4 r X
O
0 i i
Hence net magnetic field at P Bnet 2 . ( 2 1) 0 . ( 2 1)
4 r 2 r
Example: 56 A long wire A carries a current of 10 amp. Another long wire B, which is parallel to A and separated by 0.1
m from A, carries a current of 5 amp. in the opposite direction to that in A. What is the magnitude and
nature of the force experienced per unit length of B [ 0 4 10 7 weber/amp – m]
Example: 57 Three long, straight and parallel wires carrying currents are arranged as shown in figure. The force
experienced by 10 cm length of wire Q is [MP PET 1997]
R Q P
(a) 1.4×10–4 N towards the right
2cm 10cm
(b) 1.4×10–4 N towards the left
(c) 2.6 × 10–4 N to the right
(d) 2.6×10–4 N to the left 20 A 10 A 30 A
2 20 10
Solution : (a) Force on wire Q due to R ; FR 10 7 (10 10 2 ) = 2 10–4 m (Repulsive)
(2 10 2 )
10 30
Force on wire Q due to P ; FP 10 7 2 2
(10 10 2 ) = 0.6 10–4 N (Repulsive)
(10 10 )
genius PHYSICS
Magnetic Effect of Current 39
39
Hence net force Fnet = FR – FP = 2 10–4 – 0.6 10–4 = 1.4 10–4 N (towards right i.e. in the direction of
FR .
Example: 58 What is the net force on the coil [DCE 2000]
7
(a) 25 10 N moving towards wire 10 cm
7 2A
(b) 25 10 N moving away from wire 1A
15 cm
(c) 35 10 7 N moving towards wire
2 cm
(d) 35 10 7 N moving away from wire
2 2 1
Force on side AB FAB 10 7 2
15 10 2 3 10 6 N B 10 cm C
2 10 2A
1A 15 cm
FAB FCD
2 2 1
Force on side CD FAB 10 7 15 10 2 0 . 5 10 6 N
12 10 2 2 cm
A D
Hence net force on loop = FAB – FCD = 25 10–7 N (towards the wire).
Example: 59 A long wire AB is placed on a table. Another wire PQ of mass 1.0 g and length 50 cm is set to slide on two
rails PS and QR. A current of 50A is passed through the wires. At what distance above AB, will the wire PQ
be in equilibrium
S R
(a) 25 mm Q
P
(b) 50 mm
(c) 75 mm B A
50 A
(d) 100 mm
Solution : (a) Suppose in equilibrium wire PQ lies at a distance r above the wire AB
0 2i 3 7 2 (50)2
Hence in equilibrium mg Bil mg il 10 10 10 0.5 r 25 mm
4 r r
Example: 60 An infinitely long, straight conductor AB is fixed and a current is passed through it. Another movable
straight wire CD of finite length and carrying current is held perpendicular to it and released. Neglect
weight of the wire
A
(a) The rod CD will move upwards parallel to itself i1
A
i1
C D i2
B
40 Magnetic Effect of Current
genius PHYSICS
Tricky example: 5
A current carrying wire LN is bent in the from shown below. If wire carries a current of 10 A and it is
placed in a magnetic field a 5T which acts perpendicular to the paper outwards then it will experience a
force N
(a) Zero 4 cm
10A
(b) 5 N
6 cm
(c) 30 N 4 cm
(d) 20 N L
N N
Solution : (b) The given wire can be replaced
N by a straight wire as shown below
4 cm
10A
10A 10 cm
6 cm
4 cm
L L
L 6 cm
Tricky example: 6
2
A wire, carrying a current i, is kept in X Y plane along the curve y A sin x . A magnetic
field B exists in the Z-direction find the magnitude of the magnetic force on the portion of the wire
between x 0 and x
i B
(a) iB (b) Zero (c) (d) 3 / 2iB
2
Solution : (a) The given curve is a sine curve as shown below.
The given portion of the curved wire may be treated as a straight wire AB of length which
experiences a magnetic force Fm Bi
Y
i
A B
X
x=0 x=
Z
Current
S N Magnetic
moment M
genius PHYSICS
Magnetic Effect of Current 41
41
Specific examples
A given length constant current carrying straight wire moulded into different shaped loops. as shown
Linear Square Equilateral Circle
i
i
i
l
i
a
l = 4a l = 3a l = 2r
3
A A = r2
2
A = a2 a
4
il 2 3 2 3 il 2 il 2
M ia 2 M i a M i (r 2 ) max.
16 4 36 4
Note : For a given perimeter circular shape have maximum area. Hence maximum magnetic
moment.
For a any loop or coil B and M are always parallel.
B,M B,M
max NBiA
The above expression is valid for coils of all shapes. Q
R
(2) Workdone
If coil is rotated through an angle from it's equilibrium position then required work. W MB(1 cos ).
It is maximum when = 180o Wmax = 2 MB
(3) Potential energy
Is given by U = – MB cos U M .B
Note : Direction of M is found by using Right hand thumb rule according to which curl the
fingers of right hand in the direction of circulation of conventional current, then the thumb gives
the direction of M .
42 Magnetic Effect of Current
genius PHYSICS
Instruments such as electric motor, moving coil galvanometer and tangent galvanometers etc. are
based on the fact that a current-carrying coil in a uniform magnetic field experiences a torque (or
couple).
5 0 1
4 2
3 3
2 4 F
1 5 Core
Spring N S
Magnet Moving
(fixed) N S
coil
F
Moving coil F=nBil
Galvanometer
The field in a moving coil galvanometer radial in nature in order to have a linear relation between the current and the
deflection.
A rectangular current loop is in an arbitrary orientation in an external magnetic field. No work required to rotate the loop
genius PHYSICS
Magnetic Effect of Current 43
43
Example
s
Example: 61 A circular coil of radius 4 cm and 20 turns carries a current of 3 ampere. It is placed in a magnetic field of
0.5 T. The magnetic dipole moment of the coil is [MP PMT 2001]
(a) 0.60 A-m2 (b) 0.45 A-m2 (c) 0.3 A-m2 (d) 0.15 A-m2
Solution : (c) M = niA M = 20 3 ( 4 10–2)2 = 0.3 A-m2.
Example: 62 A steady current i flows in a small square loop of wire of side L in a horizontal plane. The loop is now
folded about its middle such that half of it lies in a vertical plane. Let 1 and 2 respectively denote the
magnetic moments due to the current loop before and after folding. Then [IIT-JEE 1993]
(b) 0.3 N
12 cm
N S
(c) 0.45 N
D C
(d) 0.6 N 10 cm
Solution : (b) Since plane of the coil is parallel to magnetic field. So = 90o
Hence = NBiA sin 90o = NBiA = 50 0.25 2 (12 10–2 10 10–2) = 0.3 N.
Example: 64 A circular loop of area 1 cm2, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular
to the plane of the loop. The torque on the loop due to the magnetic field is
(a) Zero (b) 10–4 N-m (c) 10–2 N-m (d) 1 N-m
Solution : (a) = NBiA sin ; given = 0 so = 0.
Example: 65 A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic
field of 0.1 weber/m2. The amount of work done in rotating it through 180 o from its equilibrium position
will be
[CPMT 1977]
(a) 0.1 J (b) 0.2 J (c) 0.4 (d) 0.8 J
44 Magnetic Effect of Current
genius PHYSICS
Solution : (a) Work done in rotating a coil through an angle from it's equilibrium position is W = MB(1 – cos) where = 180o
and M = 50 2 (4 10–2) = 50.24 10–2 A-m2. Hence W = 0.1 J
Example: 66 A wire of length L is bent in the form of a circular coil and current i is passed through it. If this coil is
placed in a magnetic field then the torque acting on the coil will be maximum when the number of turns is
(a) As large as possible (b) Any number (c) 2 (d) 1
l
Solution : (d) max MB or max nia 2 B . Let number of turns in length l is n so l n(2a) or a
2n
niBl 2 l 2iB 1
max max nmin 1
4 n 2 2
4nmin nmin
Example: 67 A square coil of N turns (with length of each side equal L) carrying current i is placed in a uniform
magnetic field B B ˆj as shown in figure. What is the torque acting on the coil
0
i
(a) B 0 NiL 2 kˆ B B 0 ˆj
(b) B 0 NiL kˆ
2 k̂
ˆj
(c) B 0 NiL 2 ˆj î
L
(d) B 0 NiL 2 ˆj
iNB 0 L2 kˆ X
Example: 68 The coil of a galvanometer consists of 100 turns and effective area of 1 square cm. The restoring couple is 10– 8 N-m
rad. The magnetic field between the pole pieces is 5 T. The current sensitivity of this galvanometer will be
[MP PMT 1997]
(a) 5 104 rad/ amp (b) 5 10– 6 per amp (c) 2 10– 7 per amp (d) 5 rad./ amp
NBA 100 5 10 4
Solution : (d) Current sensitivity (Si) = 5 rad / amp .
i C i 10 8
Example: 69 The sensitivity of a moving coil galvanometer can be increased by [SCRA 2000]]
(a) Increasing the number of turns in the coil (b) Decreasing the area of the coil
(c) Increasing the current in the coil (d) Introducing a soft iron core inside the coil
NBA
Solution : (a) Sensitivity (Si) = Si N .
C
Tricky example: 7
The square loop ABCD, carrying a current i, is placed in uniform magnetic field B, as shown. The
loop can rotate about the axis XX'. The plane of the loop makes and angle ( < 90°) with the
direction of B. Through what angle will the loop rotate by itself before the torque on it becomes zero
X C
B
(a) Z
B
(b) 90°–
i
(c) 90° +
D
(d) 180°– A
X
Solution : (c) In the position shown, AB is outside and CD is inside the plane of the paper. The Ampere force on
AB acts into the paper. The torque on the loop will be clockwise, as seen from above. The loop must
rotate through an angle (90o + ) before the plane of the loop becomes normal to the direction of B
and the torque becomes zero.
genius PHYSICS
Magnetic Effect of Current 45
45
genius PHYSICS
B Nuclear Physics & Radioactivity 1
a
92U 92U
235 236
K
r
Rutherford's -scattering experiment established that the mass of atom is concentrated with small
positively charged region at the centre which is called 'nucleus'.
e–
Nuclei are made up of proton and neutron. The number of protons in a
nucleus (called the atomic number or proton number) is represented by the
symbol Z. The number of neutrons (neutron number) is represented by N. The
total number of neutrons and protons in a nucleus is called it's mass number A
so A = Z + N. e–
e–
Neutrons and proton, when described collectively are called nucleons.
Nucleus contains two types of particles : Protons and neutrons
Nuclides are represented as Z X A ; where X denotes the chemical symbol of the element.
Neutron.
Neutron is a fundamental particle which is essential constituent of all nuclei except that of hydrogen
atom. It was discovered by Chadwick.
(1) The charge of neutron : It is neutral
(2) The mass of neutron : 1.6750 10–27 kg A free neutron outside the nucleus is unstable
1 h and decays into proton and electron.
(3) It's spin angular momentum : J -s
2 2 0n
1
1H
1
1
0
Proton Electron Antinutrin o
(4) It's magnetic moment : 9.57 10–27 J/Tesla
(5) It's half life : 12 minutes
(6) Penetration power : High
(7) Types : Neutrons are of two types slow neutron and fast neutron, both are fully capable of penetrating
a nucleus and causing artificial disintegration.
Thermal neutrons
Fast neutrons can be converted into slow neutrons by certain materials called moderator's (Paraffin wax,
heavy water, graphite) when fast moving neutrons pass through a moderator, they collide with the molecules of
the moderator, as a result of this, the energy of moving neutron decreases while that of the molecules of the
moderator increases. After sometime they both attains same energy. The neutrons are then in thermal
equilibrium with the molecules of the moderator and are called thermal neutrons.
Note : Energy of thermal neutron is about 0.025 eV and speed is about 2.2 km/s.
Nucleus.
(1) Different types of nuclei
The nuclei have been classified on the basis of the number of protons (atomic number) or the total
number of nucleons (mass number) as follows
(i) Isotopes : The atoms of element having same atomic number but different mass number are called
isotopes. All isotopes have the same chemical properties. The isotopes of some elements are the following
1 H 1, 1H 2, 1H 3 8O
16
, 8 O 17 , 8 O 18 2 He 3 , 2 He
4
17 Cl
35
, 17 Cl
37
92 U
235
, 92 U
238
genius PHYSICS
2 Nuclear Physics & Radioactivity
(ii) Isobars : The nuclei which have the same mass number (A) but different atomic number (Z) are
called isobars. Isobars occupy different positions in periodic table so all isobars have different chemical
properties. Some of the examples of isobars are
1 H 3 and 2 He
3
, 6C
14
and 7N
14
, 8 O 17 and 9F
17
(iii) Isotones : The nuclei having equal number of neutrons are called isotones. For them both the atomic
number (Z) and mass number (A) are different, but the value of (A – Z) is same. Some examples are
4 Be 9 and 5B
10
, 6 C 13 and 7N
14
, 8 O 18 and 9F
19
, 3 Li 7 and 4 Be
8
, 1 H 3 and 2 He
4
(iv) Mirror nuclei : Nuclei having the same mass number A but with the proton number (Z) and
neutron number (A – Z) interchanged (or whose atomic number differ by 1) are called mirror nuclei for
example.
3
1H and 2 He 3 , 3 Li 7 and 4 Be
7
4 4
(ii) Nuclear volume : The volume of nucleus is given by V R 3 R 03 A V A
3 3
(iii) Nuclear density : Mass per unit volume of a nucleus is called nuclear density.
Massof nucleus mA
Nuclear density( )
Volume of nucleus 4
(R 0 A 1 / 3 )3
3
where m = Average of mass of a nucleon (= mass of proton + mass of neutron = 1.66 10–27 kg)
and mA = Mass of nucleus
3m
2 . 38 10 17 kg / m 3
4R 03
(iii) These are attractive force and causes stability of the nucleus.
(iv) These forces are charge independent.
At high speeds, nuclei come
(v) Nuclear forces are non-central force. close enough for the strong force
to bind them together.
Nuclear Physics & Radioactivity 3
- mesons are of three types – Positive meson (+), negative meson ( –), neutral meson (0)
The force between neutron and proton is due to exchange of charged meson between them i.e.
p n, n p
The forces between a pair of neutrons or a pair of protons are the result of the exchange of neutral meson
(o) between them i.e. p p ' 0 and n n' 0
Thus exchange of meson between nucleons keeps the nucleons bound together. It is responsible for the
nuclear forces.
Dog-Bone analogy
The above interactions can be explained with the dog bone analogy according to
which we consider the two interacting nucleons to be two dogs having a common bone
clenched in between their teeth very firmly. Each one of these dogs wants to take the
bone and hence they cannot be separated easily. They seem to be bound to each other
with a strong attractive force (which is the bone) though the dogs themselves are strong
enemies. The meson plays the same role of the common bone in between two nucleons.
(4) Atomic mass unit (amu)
The unit in which atomic and nuclear masses are measured is called atomic mass unit (amu)
1
1 amu (or 1u) = th of mass of 6 C 12 atom = 1.66 10–27 kg
12
Masses of electron, proton and neutrons
Mass of electron (me) = 9.1 10–31 kg = 0.0005486 amu, Mass of proton (mp) = 1.6726 10–27 kg = 1.007276 amu
Mass of neutron (mn) = 1.6750 10–27 kg = 1.00865 amu, Mass of hydrogen atom (me + mp) = 1.6729 10–27 kg = 1.0078 amu
Mass-energy equivalence
According to Einstein, mass and energy are inter convertible. The Einstein's mass energy relationship is given by
E mc 2
If m = 1 amu, c = 3 108 m/sec then E = 931 MeV i.e. 1 amu is equivalent to 931 MeV or 1 amu (or 1 u) = 931 MeV
(5) Pair production and pair-annihilation
When an energetic -ray photon falls on a heavy substance. It is absorbed by some nucleus of the
substance and an electron and a positron are produced. This phenomenon is called pair production and may be
represented by the following equation h 1
0
1
0
( photon) (Positron) (Electron) 0
+1
+Ze
The rest-mass energy of each of positron and electron is h
-photon Nucleus
E0 = m0c2 = (9.1 10–31 kg) (3.0 108 m/s)2
–1 0
= 8.2 10–14 J = 0.51 MeV
4 Nuclear Physics & Radioactivity
Hence, for pair-production it is essential that the energy of -photon must be at least 2 0.51 = 1.02 MeV.
If the energy of -photon is less than this, it would cause photo-electric effect or Compton effect on striking the
matter.
The converse phenomenon pair-annihilation is also possible. Whenever an electron and a positron come
very close to each other, they annihilate each other by combining together and two -photons (energy) are
produced. This phenomenon is called pair annihilation and is represented by the following equation.
1 1 h h
0 0
N
(i) Neutron-proton ratio Ratio
Z
The chemical properties of an atom are governed entirely by the number of protons (Z) in the nucleus, the
stability of an atom appears to depend on both the number of protons and the number of neutrons.
For lighter nuclei, the greatest stability is achieved when the number of protons and neutrons are
N
approximately equal (N Z) i.e. 1
Z
Heavy nuclei are stable only when they have more neutrons than protons. Thus heavy nuclei are neutron
rich compared to lighter nuclei (for heavy nuclei, more is the number of protons in the nucleus, greater is the
electrical repulsive force between them. Therefore more neutrons are added to provide the strong attractive
forces necessary to keep the nucleus stable.)
Stable nuclei
number (N)
(Line of
Neutron
stability)
N=Z
10 20 30405060708090
Proton number
(Z)
Figure shows a plot of N verses Z for the stable nuclei. For mass number upto about A = 40. For larger
value of Z the nuclear force is unable to hold the nucleus together against the electrical repulsion of the protons
unless the number of neutrons exceeds the number of protons. At Bi (Z = 83, A = 209), the neutron excess in N
– Z = 43. There are no stable nuclides with Z > 83.
A nuclide above the line of stability i.e. having excess neutrons, decay through emission
(neutron changes into proton). Thus increasing atomic number Z and decreasing neutron
N
number N. In emission, ratio decreases.
Z
Number (N)
Neutron
A nuclide below the line of stability have excess number of protons. It – N=Z
decays by emission, results in decreasing Z and increasing N. In +
N
emission, the ratio increases. Proton number (Z)
Z
Nuclear Physics & Radioactivity 5
(ii) Even or odd numbers of Z or N : The stability of a nuclide is also determined by the consideration
whether it contains an even or odd number of protons and neutrons.
It is found that an even-even nucleus (even Z and even N) is more stable (60% of stable nuclide have even
Z and even N).
An even-odd nucleus (even Z and odd N) or odd-even nuclide (odd Z and even N) is found to be lesser
sable while the odd-odd nucleus is found to be less stable.
(iii) Binding energy per nucleon : The stability of a nucleus is determined by value of it's binding energy
per nucleon. In general higher the value of binding energy per nucleon, more stable the nucleus is
Mass Defect and Binding Energy.
(1) Mass defect (m)
It is found that the mass of a nucleus is always less than the sum of masses of it's constituent nucleons in
free state. This difference in masses is called mass defect. Hence mass defect
Note : The mass of a typical nucleus is about 1% less than the sum of masses of nucleons.
If m is measured in amu then binding energy = m amu = [{mpZ + mn(A – Z)} – M] amu = m 931
MeV
(4) Binding energy per nucleon
The average energy required to release a nucleon from the nucleus is called binding energy per nucleon.
8.0 He 26Fe56
nucleon (MeV)
6.0
4.0 Li
2.0 H2
0
56 100 150 200
Mass number A
(1) Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour
nuclei. For example 2 He 4 , 4 Be 8 , 6 C 12 , 8 O 16 and 10 Ne 20 . These nuclei are more stable than their neighbours.
(2) The binding energy per nucleon is maximum for nuclei of mass number A = 56 ( 26 Fe 56 ) . It's value is
8.8 MeV per nucleon.
(3) For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the
value of binding energy per nucleon drops to 7.5 MeV.
Note : When a heavy nucleus splits up into lighter nuclei, then binding energy per nucleon of
lighter nuclei is more than that of the original heavy nucleus. Thus a large amount of energy is
liberated in this process (nuclear fission).
When two very light nuclei combines to form a relatively heavy nucleus, then binding energy per
nucleon increases. Thus, energy is released in this process (nuclear fusion).
B. E.
A +
Fusion Fission
+
A
Nuclear Reactions.
The process by which the identity of a nucleus is changed when it is bombarded by an energetic particle is
called nuclear reaction. The general expression for the nuclear reaction is as follows.
X a C Y b Q
(Parent nucleus) (Incident particle) (Com pound nucleus) (Com pound nucleus) (Product particles) (Energy )
Here X and a are known as reactants and Y and b are known as products. This reaction is known as (a, b)
reaction and can be represented as X(a, b) Y
Nuclear Physics & Radioactivity 7
2 He 4 7 N 14 8 O 17 1 H 1
2 He 4 7 N 14 9 F 18 8 O 17 1 H 1
It is called (, p) reaction. Some other nuclear reactions are given as follows.
(p, n) reaction 1 H 1 5 B 11 6 C 12 6 C 11 0 n 1
(p, ) reaction 1 H 1 3 Li 11 4 Be 8 2 He 4 2 He 4
(p, ) reaction 1H
1
6 C 12 7 N 13 7 N 13
(n, p) reaction 0n
1
7 N 14 7 N 15 6 C 14 1 H 1
(, n) reaction 1 H 2 1 H 1 0 n1
92 U
235
0 n1 92 U
236
56 Ba
141
36 Kr
92
3 0 n1 Q
(unstable nucleus)
(ii) The energy released in U235 fission is about 200 MeV or 0.8 MeV per nucleon.
(iii) By fission of 92 U
235
, on an average 2.5 neutrons are liberated. These neutrons are called fast neutrons
and their energy is about 2 MeV (for each). These fast neutrons can escape from the reaction so as to proceed
the chain reaction they are need to slow down.
(iv) Fission of U235 occurs by slow neutrons only (of energy about 1eV) or even by thermal neutrons (of
energy about 0.025 eV).
(v) 50 kg of U235 on fission will release 4 × 1015 J of energy. This is equivalence to 20,000 tones of TNT
explosion. The nuclear bomb dropped at Hiroshima had this much explosion power.
(vi) The mass of the compound nucleus must be greater than the sum of masses of fission products.
Binding energy
(vii) The of compound nucleus must be less than that of the fission products.
A
(viii) It may be pointed out that it is not necessary that in each fission of uranium, the two fragments
56 Ba and 36 Kr are formed but they may be any stable isotopes of middle weight atoms.
Same other U 235 fission reactions are
92 U
235
0 n 1 54 Xe 140 38 Sr
94
2 0 n1
57 La 148 35 Br
85
3 0 n1
Many more
(ix) The neutrons released during the fission process are called prompt neutrons.
(x) Most of energy released appears in the form of kinetic energy of fission fragments.
Ba
Energy
Energy
Slow
Neutron
92U235 92U236
Energy
Energy
Kr
(2) Chain reaction
In nuclear fission, three neutrons are produced along with the release of large energy. Under favourable
conditions, these neutrons can cause further fission of other nuclei, producing large number of neutrons. Thus
a chain of nuclear fissions is established which continues until the whole of the uranium is consumed.
Kr
U
Kr Ba
n
U
Kr
Ba U
Ba
In the chain reaction, the number of nuclei undergoing fission increases very fast. So, the energy produced
takes a tremendous magnitude very soon.
Difficulties in chain reaction
(i) Absorption of neutrons by U 238 , the major part in natural uranium is the isotope U238 (99.3%), the
isotope U 235 is very little (0.7%). It is found that U 238 is fissionable with fast neutrons, whereas U 235 is
Nuclear Physics & Radioactivity 9
fissionable with slow neutrons. Due to the large percentage of U 238 , there is more possibility of collision of
neutrons with U 238 . It is found that the neutrons get slowed on coliding with U 238 , as a result of it further
fission of U238 is not possible (Because they are slow and they are absorbed by U238). This stops the chain
reaction.
Note : The energy released in the explosion of an atom bomb is equal to the energy released by
2000 tonn of TNT and the temperature at the place of explosion is of the order of 107 oC.
Nuclear Reactor.
10 Nuclear Physics & Radioactivity
A nuclear reactor is a device in which nuclear fission can be carried out through a sustained and a
controlled chain reaction. It is also called an atomic pile. It is thus a source of controlled energy which is
utilised for many useful purposes. Cadmium rods
Core
Coolant
Coolant out
Turbine
Concrete To electric
wall generator
Condenser
Moderator Water
Heat
Coolant in exchanger
(1) Parts of nuclear reactor
Fuel elements
(i) Fissionable material (Fuel) : The fissionable material used in the reactor is called the fuel of the
reactor. Uranium isotope (U235) Thorium isotope (Th232) and Plutonium isotopes (Pu239, Pu240 and Pu241) are
the most commonly used fuels in the reactor.
(ii) Moderator : Moderator is used to slow down the fast moving neutrons. Most commonly used
moderators are graphite and heavy water (D2O).
(iii) Control Material : Control material is used to control the chain reaction and to maintain a stable
rate of reaction. This material controls the number of neutrons available for the fission. For example, cadmium
rods are inserted into the core of the reactor because they can absorb the neutrons. The neutrons available for
fission are controlled by moving the cadmium rods in or out of the core of the reactor.
(iv) Coolant : Coolant is a cooling material which removes the heat generated due to fission in the
reactor. Commonly used coolants are water, CO2 nitrogen etc.
(v) Protective shield : A protective shield in the form a concrete thick wall surrounds the core of the
reactor to save the persons working around the reactor from the hazardous radiations.
Note : It may be noted that Plutonium is the best fuel as compared to other fissionable material.
It is because fission in Plutonium can be initiated by both slow and fast neutrons. Moreover it
can be obtained from U 238 .
Nuclear reactor is firstly devised by fermi. Apsara was the first Indian nuclear reactor.
(2) Uses of nuclear reactor
(i) In electric power generation.
(ii) To produce radioactive isotopes for their use in medical science, agriculture and industry.
(iii) In manufacturing of PU 239 which is used in atom bomb.
(iv) They are used to produce neutron beam of high intensity which is used in the treatment of cancer and
nuclear research.
Note : A type of reactor that can produce more fissile fuel than it consumes is the breeder
reactor.
Nuclear fusion
In nuclear fusion two or more than two lighter nuclei combine to form a single heavy nucleus. The mass of
single nucleus so formed is less than the sum of the masses of parent nuclei. This difference in mass results in
the release of tremendous amount of energy
1 H 2 1 H 2 1 H 3 1 H 1 4 MeV
1 H 3 1 H 2 2 He 4 0 n 1 17 .6 MeV
or 1 H 2 1 H 2 2 He 4 24 MeV
For fusion high pressure ( 106 atm) and high temperature (of the order of 107 K to 108 K) is required and
so the reaction is called thermonuclear reaction.
Nuclear Physics & Radioactivity 11
Fusion energy is greater then fission energy fission of one uranium atom releases about 200 MeV of
energy. But the fusion of a deutron (1 H 2 ) and triton (1 H 3 ) releases about 17.6 MeV of energy. However the
energy released per nucleon in fission is about 0.85 MeV but that in fusion is 4.4 MeV. So for the same mass of
the fuel, the energy released in fusion is much larger than in fission.
Plasma : The temperature of the order of 108 K required for thermonuclear reactions leads to the
complete ionisation of the atom of light elements. The combination of base nuclei and electron cloud is called
plasma. The enormous gravitational field of the sun confines the plasma in the interior of the sun.
The main problem to carryout nuclear fusion in the laboratory is to contain the plasma at a temperature
of 10 K. No solid container can tolerate this much temperature. If this problem of containing plasma is solved,
8
then the large quantity of deuterium present in sea water would be able to serve as in-exhaustible source of
energy.
Note : To achieve fusion in laboratory a device is used to confine the plasma, called Tokamak.
Stellar Energy
Stellar energy is the energy obtained continuously from the sun and the stars. Sun radiates energy at the
rate of about 1026 joules per second.
Scientist Hans Bethe suggested that the fusion of hydrogen to form helium (thermo nuclear reaction) is
continuously taking place in the sun (or in the other stars) and it is the source of sun's (star's) energy.
The stellar energy is explained by two cycles
Proton-proton cycle Carbon-nitrogen cycle
1H
1
1 H 1 H 1 e Q1
1 2 0
1H
1
6 C 12 7 N 13 Q1
1H
2
1 H 1 2 He 3 Q 2 7 N 13 6 C 13 1 e
0
2 He 3 2 He 3 2 He 4 2 1 H 1 Q 3 1H
1
6 C 13 7 N 14
Q2
4 1 H 2 He 2
1 4
1 e
0
2 26 .7 MeV 1H
1
7 N 14
8O 15
Q3
8 O 15
7N 15
1e Q4
0
1H
1
7N 15
6 C 12 2 He 4
4 1 H 1 2 He 4 2 1 e 0 24 .7 MeV
About 90% of the mass of the sun consists of hydrogen and helium.
Nuclear Bomb. Based on uncontrolled nuclear reactions.
Atom bomb Hydrogen bomb
Based on fission process it involves the fission of U235 Based on fusion process. Mixture of deutron and
tritium is used in it
In this critical size is important There is no limit to critical size
Explosion is possible at normal temperature and High temperature and pressure are required
pressure
Less energy is released compared to hydrogen bomb More energy is released as compared to atom bomb so
it is more dangerous than atom bomb
Concepts
A test tube full of base nuclei will weight heavier than the earth.
The nucleus of hydrogen contains only one proton. Therefore we may say that the proton is the nucleus of hydrogen
atom.
If the relative abundance of isotopes in an element has a ratio n1 : n2 whose atomic masses are m1 and m2 then atomic
n m n2 m 2
mass of the element is M 1 1
n1 n 2
Example
s
12 Nuclear Physics & Radioactivity
Example: 1 A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 8 : 1. The ratio
of radii of the fragments is
(a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 2 : 1
Solution : (a)
v1
By conservation of momentum m1v1 = m2v2
m1 v1 8 m 2
…… (i)
M v 2 1 m1
m2
1/3
r1 A1
1/3
1 1
v2 Also from r A1 / 3
r2 A2
8 2
Example: 3 If Avogadro’s number is 6 1023 then the number of protons, neutrons and electrons in 14 g of 6C14 are
respectively
(a) 36 1023, 48 1023, 36 1023 (b) 36 1023, 36 1023, 36 1021
(c) 48 1023, 36 1023, 48 1021 (d) 48 1023, 48 1023, 36 1021
14
Solution : (a) Since the number of protons, neutrons and electrons in an atom of 6C are 6, 8 and 6 respectively. As 14
gm of 6C
14
contains 6 1023 atoms, therefore the numbers of protons, neutrons and electrons in 14 gm
of 6 C 14
are 6 6 10 23 36 10 23 , 8 6 10 23 48 10 23 , 6 6 10 23 36 10 23 .
Example: 4 Two Cu64 nuclei touch each other. The electrostatics repulsive energy of the system will be
(a) 0.788 MeV (b) 7.88 MeV (c) 126.15 MeV (d) 788 MeV
Solution : (c) Radius of each nucleus R R0 ( A) 1/3
1 .2 (64 ) 1/3
4 .8 fm
k q2 9 10 9 (1 .6 10 19 29 ) 2
So potential energy U 126 .15 MeV .
r 2 4 .8 10 15 1 .6 10 19
235
Example: 5 When 92 U undergoes fission. 0.1% of its original mass is changed into energy. How much energy is
235
released if 1 kg of 92 U undergoes fission [MP PET 1994; MP PMT/PET 1998; BHU 2001; BVP 2003]
Example: 8 The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and
protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of
the helium nucleus will be [1 amu= 931 MeV] [CPMT 1986; MP PMT 1995; CBSE 2003]
(a) 28.4 MeV (b) 20.8 MeV (c) 27.3 MeV (d) 14.2 MeV
Solution : (a) Helium nucleus consist of two neutrons and two protons.
So binding energy E = m amu = m 931 MeV
E = (2 mp + 2mn – M) 931 MeV = (2 1.0073 + 2 1.0087 – 4.0015) 931 = 28.4 MeV
Example: 9 A atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of
uranium atom U 238 is 170 MeV. The number of uranium atoms fissioned per hour will be
(a) 5 1015 (b) 10 1020 (c) 40 1021 (d) 30 1025
W n E
Solution : (c) By using P where n = Number of uranium atom fissioned and E = Energy released due to
t t
n 170 10 6 1.6 10 19
each fission so 300 10 6 n = 40 1021
3600
Example: 10 The binding energy per nucleon of O16 is 7.97 MeV and that of O17 is 7.75 MeV. The energy (in MeV)
required to remove a neutron from O17 is [IIT-JEE 1995]
(a) 3.52 (b) 3.64 (c) 4.23 (d) 7.86
Solution : (c) O17 O16 0 n1
Energy required = Binding of O17 – binding energy of O16 = 17 7.75 – 16 7.97 = 4.23 MeV
Example: 11 A gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is 0.5 MeV
and the total kinetic energy of the electron-positron pair is 0.78 MeV, then the energy of the gamma ray
photon must be [MP PMT 1991]
(a) 0.78 MeV (b) 1.78 MeV (c) 1.28 MeV (d) 0.28 MeV
Solution : (b) Energy of -rays photon = 0.5 + 0.5 +0.78 = 1.78 MeV
Example: 12 What is the mass of one Curie of U234 [MNR 1985]
(a) 3.7 1010 gm (b) 2.348 1023 gm (c) 1.48 10–11 gm (d) 6.25 10–34 gm
Solution : (c) 1 curie = 3.71 1010 disintegration/sec and mass of 6.02 1023 atoms of U 234
234 gm
234 3 .71 10 10
Mass of 3.71 1010 atoms 1 .48 10 11 gm
6 .02 10 23
Example: 13 In the nuclear fusion reaction 2
1H 13 H 42 He n, given that the repulsive potential energy between the
two nuclei is 7.7 10 14 J , the temperature at which the gases must be heated to initiate the reaction is
nearly [Boltzmann’s constant k 1 .38 10 23 J /K ] [AIEEE 2003]
Solution : (c, d) Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus
20
is always less then the sum of masses of it's constituent particles 10 Ne is made up of 10 protons plus 10
neutrons. Therefore, mass of 20
10 Ne nucleus M 1 10 (m p m n )
Also heavier the nucleus, more is he mass defect thus 20 (mn m p ) M 2 10(m p mn ) M1
or 10 (m p m n ) M 2 M 1
M 2 M1 10 (m p mn ) M 2 M 1 M 1 M 2 2M 1
Tricky example: 1
Binding energy per nucleon vs mass number curve for nuclei is shown in the figure. W, X, Y and Z
are four nuclei indicated on the curve. The process that would release energy is [IIT-JEE 1999]
(a) Y 2Z
Y
Binding energy
8.5 X
(b) W X + Z
nucleon in
8.0 W
7.5
MeV
(c) W 2Y 5.0 Z
(d) X Y + Z
30 60 90 120
Mass number of nuclei
Solution : (c) Energy is released in a process when total binding energy of the nucleus (= binding energy per
nucleon number of nucleon) is increased or we can say, when total binding energy of products
is more than the reactants. By calculation we can see that only in case of option (c) this happens.
Given W 2Y
Binding energy of reactants = 120 7.5 = 900 MeV
and binding energy of products = 2 (60 8.5) = 1020 MeV > 900 MeV
Radioactivity.
The phenomenon of spontaneous emission of radiatons by heavy elements is called radioactivity. The
elements which shows this phenomenon are called radioactive elements.
(1) Radioactivity was discovered by Henery Becquerel in uranium salt in the year 1896.
(2) After the discovery of radioactivity in uranium, Piere Curie and Madame Curie discovered a new
radioactive element called radium (which is 106 times more radioactive than uranium)
(3) Some examples of radio active substances are : Uranium, Radium, Thorium, Polonium, Neptunium
etc.
(4) Radioactivity of a sample cannot be controlled by any physical (pressure, temperature, electric or
magnetic field) or chemical changes.
(5) All the elements with atomic number (Z ) > 82 are naturally radioactive.
(6) The conversion of lighter elements into radioactive elements by the bombardment of fast moving
particles is called artificial or induced radioactivity.
(7) Radioactivity is a nuclear event and not atomic. Hence electronic configuration of atom don't have any
relationship with radioactivity.
Nuclear radiatons
According to Rutherford's experiment when a sample of radioactive substance is put in a lead box and
allow the emission of radiation through a small hole only. When the radiation enters into the external electric
field, they splits into three parts
-rays -rays
Magnetic
– - + -
– +
– rays
-rays
+
rays
-rays field
– +
– +
Nuclear Physics & Radioactivity 15
(i) Radiations which deflects towards negative plate are called -rays (stream of positively charged
particles)
(ii) Radiations which deflects towards positive plate are called particles (stream of negatively charged
particles)
(iii) Radiations which are undeflected called -rays. (E.M. waves or photons)
Note : Exactly same results were obtained when these radiations were subjected to magnetic
field.
No radioactive substance emits both and particles simultaneously. Also -rays are emitted
after the emission of or -particles.
-particles are not orbital electrons they come from nucleus. The neutron in the nucleus decays
into proton and an electron. This electron is emitted out of the nucleus in the form of -rays.
atom (2He4)
2. Charge + 2e –e Zero
3. Mass 4 mp (mp = mass of 4 mp me Massless
proton = 1.87 10–27
4. Speed 107 m/s 1% to 99% of speed of light Speed of light
5. Range of kinetic energy 4 MeV to 9 MeV All possible values between Between a minimum
a minimum certain value to value to 2.23 MeV
1.2 MeV
6. Penetration power (, , 1 100 10,000
) (Stopped by a paper) (100 times of ) (100 times of upto 30
cm of iron (or Pb) sheet
7. Ionisation power ( > > ) 10,000 100 1
8. Effect of electric or Deflected Deflected Not deflected
magnetic field
9. Energy spectrum Line and discrete Continuous Line and discrete
10. Mutual interaction with Produces heat Produces heat Produces, photo-electric
matter effect, Compton effect,
pair production
11. Equation of decay decay
X A XA A
0
X A zXa
Z Z Z 1 Y 1 e z
A 4
Z 2 Y 2 He 4 Z
n
X A
Z' X
A
n β (2 n α Z Z')
X A
n
A'
Z Z' Y
16 Nuclear Physics & Radioactivity
A' A
nα
4
Radioactive Disintegration.
(1) Law of radioactive disintegration
According to Rutherford and Soddy law for radioactive decay is as follows.
"At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at
dN dN
that instant" i.e. N N . It can be proved that N = N0e–t
dt dt
This equation can also be written in terms of mass i.e. M = M0e–t
where N = Number of atoms remains undecayed after time t, N0 = Number of atoms present initially (i.e. at t =
0), M = Mass of radioactive nuclei at time t, M0 = Mass of radioactive nuclei at time t = 0, N0 – N = Number of
disintegrated nucleus in time t
dN
= rate of decay, = Decay constant or disintegration constant or radioactivity constant or Rutherford
dt
Soddy's constant or the probability of decay per unit time of a nucleus.
Note : depends only on the nature of substance. It is independent of time and any physical or
chemical changes. N0
N N = N0e–t
0 t
(2) Activity
It is defined as the rate of disintegration (or count rate) of the substance (or the number of atoms of any
dN
material decaying per second) i.e. A N N 0 e t A0 e t
dt
where A0 = Activity of t = 0, A = Activity after time t
Note : Activity per gm of a substance is known as specific activity. The specific activity of 1 gm of
radium – 226 is 1 Curie.
1 millicurie = 37 Rutherford
The activity of a radioactive substance decreases as the number of undecayed nuclei decreases with
time.
1
Activity
Half life
Time (t) Number of undecayed atoms (N) Remaining fraction of Fraction of atoms
(N0 = Number of initial atoms) active atoms (N/N0) decayed (N0 – N) /N0
probability of survival probability of decay
t=0 N0 1 (100%) 0
Nuclear Physics & Radioactivity 17
t = T1/2 N0 1 1
(50%) (50%)
2 2 2
t = 2(T1/2) 1 N0 N 1 3
02 (25%) (75%)
2 2 (2) 4 4
t = 3(T1/2) 1 N N 1 7
0 03 (12.5%) (87.5%)
2 (2) (2) 8 8
t = 10 (T1/2) N0 10 99.9%
1
10 0 .1 %
(2) 2
t = n (N1/2) N n
1 n
1
(2) 2 1
2 2
Useful relation
n t / T1 / 2
1 1
After n half-lives, number of undecayed atoms N N 0 N 0
2 2
(4) Mean (or average) life ()
The time for which a radioactive material remains active is defined as mean (average) life of that material.
Other definitions
(i) It is defined as the sum of lives of all atoms divided by the total number of atoms
Sum of the lives of all the atoms 1
i.e.
Total number of atoms
N
ln N
t N0
(ii) From N N 0 e slope of the line shown in the graph ln
N0 Slope = –
t
N
i.e. the magnitude of inverse of slope of ln vs t curve is known as mean life ().
N0
t
(iii) From N N 0 e t
1 1
If t N N 0 e 1 N 0 0 .37 N 0 37 % of N0.
e
1
i.e. mean life is the time interval in which number of undecayed atoms (N) becomes times or 0.37 times
e
or 37% of original number of atoms. or
1
It is the time in which number of decayed atoms (N0 – N) becomes 1 times or 0.63 times or 63% of
e
original number of atoms.
0.693 1 1
(iv) From T1 / 2 . (t1 / 2 ) 1.44 (T1 / 2 )
0.693
i.e. mean life is about 44% more than that of half life. Which gives us > T(1/2)
Note : Half life and mean life of a substance doesn't change with time or with pressure,
temperature etc.
Radioactive Series.
If the isotope that results from a radioactive decay is itself radioactive then it will also decay and so on.
The sequence of decays is known as radioactive decay series. Most of the radio-nuclides found in nature
are members of four radioactive series. These are as follows
Mass number Series (Nature) Parent Stable and Integer Number of lost
product n particles
4n Thorium (natural) Th 232 Pb 208 52 = 6, = 4
90 82
4n + 1 Neptunium Np 237
(Artificial)
93
83 Bi 209 52 = 8, = 5
18 Nuclear Physics & Radioactivity
Note : The 4n + 1 series starts from 94 PU 241 but commonly known as neptunium series because
neptunium is the longest lived member of the series.
The 4n + 3 series actually starts from 92 U 235 .
dN 1
Rate of disintegration of A 1 N 1 (which is also the rate of formation of B)
dt
dN 2
Rate of disintegration of B 2 N 2
dt
Net rate of formation of B = Rate of disintegration of A – Rate of disintegration of B
= 1N1 – 2N2
Equilibrium
In radioactive equilibrium, the rate of decay of any radioactive product is just equal to it's rate of
production from the previous member.
1 N 2 2 (T )
i.e. 1N1 = 2N2 1/2
2 N 2 1 (T1 / 2 )1
(4) As tracers - (Tracer) : Very small quantity of radioisotopes present in a mixture is known as tracer
(i) Tracer technique is used for studying biochemical reaction in tracer and animals.
(5) In industries
(i) For detecting leakage in oil or water pipe lines (ii) For determining the age of planets.
Concept
If a nuclide can decay simultaneously by two different process which have decay constant 1 and 2, half life T1 and T2 and
mean lives 1 and 2 respectively then
= 1 + 2
1 1, T1, 1
T
2 2, T2, 2
Nuclear Physics & Radioactivity 19
T1T2
T
T1 T2
1 2
1 2
Example: 16 When 90 Th
228
transforms to 83 Bi
212
, then the number of the emitted –and –particles is, respectively
[MP PET 2002]
(a) 8, 7 (b) 4, 7 (c) 4, 4 (d) 4, 1
A 228 A ' 212
Solution : (d) Z 90 Th Z ' 83 Bi
A A ' 228 212
Number of -particles emitted n 4
4 4
Number of -particles emitted n 2n Z Z ' 2 4 90 83 1 .
Example: 17 A radioactive substance decays to 1/16th of its initial activity in 40 days. The half-life of the radioactive
substance expressed in days is
(a) 2.5 (b) 5 (c) 10 (d) 20
t/T 40 / T
1 1/2 N 1 1 1/2
Example: 23 nucleus of mass number A, originally at rest, emits an -particle with speed v. The daughter nucleus
recoils with a speed [DCE 2000; AIIMS 2004]
(a) 2v /( A 4) (b) 4v /( A 4) (c) 4v /( A 4) (d) 2v /( A 4)
Solution : (c) m
m
v v
A A–4 + 4
Rest 4v
According to conservation of momentum 4v ( A 4)v' v ' .
A4
Example: 24 The counting rate observed from a radioactive source at t = 0 second was 1600 co unts per second and
at t = 8 seconds it was 100 counts per second. The counting rate observed as counts per second at t =
6 seconds will be [MP PET 1996; UPSEAT 2000]
(a) 400 (b) 300 (c) 200 (d) 150
n 8 / T1 / 2 8 / T1 / 2
1 1 1 1
Solution : (c) By using A A0 100 1600 T1 / 2 2 sec
2 2 16 2
6/2
1
Again by using the same relation the count rate at t = 6 sec will be A 1600 200 .
2
Example: 25 The kinetic energy of a neutron beam is 0.0837 eV. The half-life of neutrons is 693s and the mass of
neutrons is 1.675 10 27 kg. The fraction of decay in travelling a distance of 40m will be
(a) 10 3 (b) 10 4 (c) 10 5 (d) 10 6
2E 2 0 . 0837 1 . 6 10 19
Solution : (c) v = 4 103 m/sec
m 1 . 675 10 27
40
Time taken by neutrons to travel a distance of 40 m t' 10 2 sec
4 10 3
dN dN
N dt
dt N
N 0 . 693 0 . 693
Fraction of neutrons decayed in t sec in t t 10 2 10 5
N T 693
Example: 26 The fraction of atoms of radioactive element that decays in 6 days is 7/8. The fraction that decays in 10 days
will be
(a) 77/80 (b) 71/80 (c) 31/32 (d) 15/16
N N
t / T1 / 2 T1 / 2 log e 0 log e 0
1 N N t N 1
Solution : (c) By using N N 0 t t log e 0 1
2 log e (2) N t2 N
log e 0
N 2
6 log e (8 / 1) N 10 N
Hence log e 0 log e (8 ) log e 32 0 32 .
10 log e ( N 0 / N ) N 6 N
1 31
So fraction that decays 1 .
32 32
Tricky example: 2
Half-life of a substance is 20 minutes. What is the time between 33% decay and 67% decay [AIIMS 2000]
(a) 40 minutes (b) 20 minutes (c) 30 minutes (d) 25 minutes
Solution : (b) Let N0 be the number of nuclei at beginning
Number of undecayed nuclei after 33% decay = 0.67 N0
and number of undecayed nuclei after 67% of decay = 0.33 N0
0 . 67 N 0
0.33 N0 ~ and in the half-life time the number of undecayed nuclei becomes half.
2
genius Physics……..Pradeep Kshetrapal Electrostatics 2011
ELECTROSTATICS : Study of Electricity in which Electrostatic series :If two substances are
electric charges are static i.e. not moving, is rubbed together the former in series acquires
called electrostatics the positive charge and later, the –ve.
• STATIC CLING (i) Glass (ii) Flannel (iii) Wool (iv) Silk (v) Hard Metal (vi)
• An electrical phenomenon that accompanies Hard rubber (vii) Sealing wax (viii) Resin (ix) Sulphur
dry weather, causes these pieces of papers to
stick to one another and to the plastic comb. Electron theory of Electrification
• Due to this reason our clothes stick to our
body. • Nucleus of atom is positively charged.
• The electron revolving around it is negatively
• ELECTRIC CHARGE : Electric charge is charged.
characteristic developed in particle of material • They are equal in numbers, hence atom is
due to which it exert force on other such electrically neutral.
particles. It automatically accompanies the • With friction there is transfer of electrons,
particle wherever it goes. hence net charge is developed in the particles.
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• But because free quarks do not exist and their directly proportional to the product of the charges,
sum is always an integral number, it does not inversely proportional to the square of the distance
violet the quantization rules.) between them and
acts along the straight line joining the two charges.
•
Conservation of Charges
•
Like conservation of energy, and Momentum,
the electric charges also follow the rules of
conservation.
1. Isolated (Individual) Electric charge can neither
be created nor destroyed, it can only be
transferred.
2. Charges in pair can be created or destroyed.
Example for 1.
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e0 is permittivity of free space or vacuum and its value • No, In S.I. System, the fundamental quantity is
is e0 = 8.85 x 10-12 coul2 / N x m2 Electric current and its unit is Ampere.
If point charges are immersed in a dielectric medium, Therefore coulomb is defined in it’s terms as
then e0 is replaced by e a quantity-characteristic of the under:
matter involved In such case. For vacuum e = e0 • Coulomb is that quantity of charge which
passes across any section of a conductor per
second when current of one ampere flows
through it, i.e.
• 1 coulomb=1 Ampere x 1 sec
Permittivity, Relative Permittivity and Dielectric
Constant
In cgs electrostatic system, the unit of charge is
Permittivity is a measure of the property of the
called as STATECOULUMB or esu of charge.
medium surrounding electric charge which determine
• In this system electrostatic constant c=1 for
the forces between the charges.
Its value is known as Absolute permittivity of that
Medium e vacuum or air,
More is Permittivity of medium, Less is coulombs One stat coulomb is defined that amount of charge
Force. which when placed at a distance of 1 cm in air from an
For water, permittivity is 80 times then that of vacuum, equal and similar charge repel it with a force of one
hence force between two charges in water will be 1/80 dyne.
time force in vacuum (or air.)
Relative Permittivity(er) : It is a dimension-less In cgs electromagnetic system, the unit of charge is
characteristic constant, which express absolute called ABCOULOMB or emu of charge
permittivity of a medium w.r.t. permittivity of vacuum 1 Coulomb = 3 x 109 statcoulomb
or air. It is also called = 1/10 abcoulomb
Dielectric constant (K) K= er = e/e0 \ 1 emu = 3x1010 esu of charge
q1 12 q2
• Unit of charge:- In S.I. System of units, the 12 21
position vectors be (OA) and (OB). Then If a number of Forces F11, F12,F13,……F1n are acting on a single
AB= . According to triangle law of vectors : charge q1 then charge will experience force F1 equal to
+ = = - and vector sum of all these forces .
= - F1 = F11 + F12 + F13 + …… + F1n
X
(ii) According to Coulumb’s law, the Force 12 exerted on q1
by q2 is given by : 12 = 21 where 21 is a unit
9
Ans : Fe= = 9x10 Newton
-11
Fg=G =6.67x10 Newton
Fe / Fg = 2.26 x 1039
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q X 2q
Bearing
h W h
Q Q
+q a -q (b) If one of the Cs+ ion is missing the crystal is said to have defect.
How much will be the force on chlorine ion in that case?
: if intensity of electric field at a location is E and a 2- Areal distribution: charge is evenly distributed
charge ‘q’ is placed ,then force experienced by this over a surface area,S.
charges F is
The surface charge density is ‘σ‘
given by
Direction of force F is in direction of electric field E Where Q is charge given to a surface of area ‘S’.
DISTRIBUTION OF CHARGE
Electric charge on a body may be concentrated at a
point, then it is called a ‘point charge’. If it is
distributed all over, then it is called distribution of
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E DUE TO +q
ALONG
OPPOSITE TO
NET ELECTRIC FIELD
DIPOLE
2-CO2 & CH4 are non-polar because centers of –ve & SINCE >
+ve charges co-incide and there is no distance between
them.
: IS IN THE DIRECTION OF
3-if non polar atom is placed in an elect.field a distance
is created between +ve & -ve charge: it become polar. IF R>>L THE, E=
E due to +q , E+q
E due to -q E-q
Where ‘q’ is each charge and ‘2L’ is distance between
them.(each charge is at a distance L from ‘center’ of |E+q| = |E-q| = Eq
dipole)
each Eq is resolved in two direction. One along
Dipole moment = q x 2 is a vector quantity it has
magnitude p=2qL equatorial line and other in axial directions which are
And its direction is along line from –q to +q. the Esinθ and normal direction E cosθ .
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E(net) = 2Ecosθ
E= 3/2
IF R>>L Then, E=
------------------------------------------------------------------------
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+q A F
2L
F θ
Field Lines due to some charge configurations.
1.Single positive or negative charge -q B C
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or τ =P x E Or, W = PE – = pE
NOTE :
3.If a dipole is rotated through 900 from the direction
1.Direction of torque is normal to the plane containing of the field, then work done will be
dipole moment P and electric field E and is governed
by right hand screw rule. W = pE = pE
2. If Dipole is parallel to E the torque is Zero. 4. If the dipole is rotated through 1800 from the
direction of the field, then work done will be :
3. Torque is maximum when Dipole is perpendicular to
E and that torque is PE
W = pE = 2 pE
4. This equation gives the definition of dipole moment.
If E is 1 N/C then P=T.
Potential Energy of a dipole kept in Electric field :
Therefore; Dipole Moment of a dipole is equal to the 1. dipole in Equilibrium ( P along E ):-
Torque experience by that dipole when placed in an A dipole is kept in Electric field in equilibrium
electric field of strength 1 N/C at right angle to it. condition, dipole moment P is along E
To calculate Potential Energy of dipole we calculate
5. If a dipole experiencing a torque in electric field is
work done in bringing +q from zero potential i.e. to
allowed to rotate, then it will rotate to align itself to
location B, and add to the work done in bringing –q
the Electric field. But when it reach along the direction
from to position A.
of E the torque become zero. But due to inertia it
1.The work done on –q from up to A
overshoots this equilibrium condition and then starts
= -(Work done up to B + Work done from B to A)
oscillating about this mean position.
2.Work done on +q = +(Work done up to B)
6.Dipole in Non-Uniform Electric field : Adding the two
Total work done = Work done on –q from B to A
In case Electric field is non-uniform, magnitude of force = Force x displacement
on +q and –q will be different, hence a net force will be = -qE x 2L = - 2qLE
acting on centre of mass of dipole and it will make a =- P.E
linear motion. At the same time due to couple of This work done convert into Potential Energy of dipole
forces acting, a torque will also be acting on it. U= -P. E
Work done in rotating a dipole in a uniform Electric If P and E are inclined at angle θ to each other then
field: magnitude of this Potential Energy is
U = - P E Cos θ
1.If a dipole is placed in a uniform electric field
experience a torque. If it is rotated from its equilibrium
position, work has to be done on it. If an Electric dipole
with moment P is placed in electric field E making an
angle α, then torque acting on it at that instant is
Electric – Potential
τ = PESinα
2. If it is rotated further by a small angle dα then work (1) Electric Potential is characteristic of a
done dw = (PEsinα).dα location in the electric field. If a unit
Then work done for rotating it through an angle θ from charge is placed at that location it has
equilibrium position of angle 0 is :- potential energy (due to work done on
its placement at that location). This
potential energy or work done on unit
θ
W= = PE charge in bringing it from infinity is
called potential at that point.
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Energy with Q at B is q VB = -
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genius Physics……..Pradeep Kshetrapal Electrostatics 2011
VA VB +q -q
2L
VA > vB
Q
This also show that an electric charge At P – V+q =
4 0 ( r l )
experience force from high potential towards
low potential if allowed to move, it will do so in Q
this direction only. V-q =
4 0 ( r l )
If E and are not collinear and make angle
between them, then according to relation of 1 1
Total V =– V+q + V-q =
work & force r l r l
dv = - E dr Cos
Or, - dv / dr = E Cos =
2Ql
=
P
Or, dv = - E . dr 4 0 ( r l )
2 2
4 0 (r 2 l 2 )
Or V = E . dr P
If r > > L Then V =
4 0 r 2
( work) given by dot product of two vector - q & + q are placed at A & B. Point P is on equatorial
E & dr.
V = V1 + V2 + V3
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Q r L cos r LCos
=
4 0 r 2 L2Cos 2
Q X 2 LCos
=
4 0 (r 2 L2Cos )
PCos v
Then, Or, V =
4 0 r 2
0,0 R r
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V+v ’
V2
v’ V
V1
Gauss's Law
Electric Flux
Think of air blowing in through a window. How much
air comes through the window depends upon the speed
of the air, the direction of the air, and the area of the
Proof:- Suppose E is window. We might call this air that comes through the
not at right angle to equipotential surface, and makes window the "air flux".
angle with it. Then it has two components, E Cos
along surface and E Sin normal to surface due to We will define the electric flux for an electric field
that is perpendicular to an area as
component E Cos , force q E Cos should be
created on surface and it should move the charge. But =EA
we find that charges are in equilibrium. i.e.
E Cos = 0 ;
since E = 0, therefore Cos = 0 or ∠ = 900
Hence E is always at right angle to equip. surface.
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genius Physics……..Pradeep Kshetrapal Electrostatics 2011
For a curved surface, that will not be the case. For that
case, we can apply this definition of the electric flux
over a small area A or A or An.
=E A cos or
=E A
=E A
A=An
Gauss’s Law : Total electric flux though a closed
where n is a unit vector pointing perpendicular to the surface is 1/ε₀ times the charge enclosed in the
area. In that case, we could also write the electric flux surface.
across an area as
ΦE=q / ε₀
=E nA But we know that Electrical flux through a closed
surface is
Both forms say the same thing. For this to make any
sense, we must be talking about an area where the
= q / ε₀
direction of A or n is constant.
This is Gauss’s theorem.
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genius Physics……..Pradeep Kshetrapal Electrostatics 2011
PROOF : Let’s consider an hypothetical spherical Consider a Gaussian Surface in the shape of a cylinder
surface having charge q placed at its centre. At every having axis along conductor. It has radius r so that
point of sphere the electrical field is radial, hence point P lies on the surface. Let its length be l.
making angle 0 degree with area vector. The electric field is normal to conductor, hence it is
symmetrical to the surfaces of these cylinder.
Electric field due to line charge : Let’s consider a Gaussian surface, in shape of a cylinder
Electric charge is distributed on an infinite long straight which has axis normal to the sheet of charge and
conductor with linear charge density λ. We have to containing point P at its plain surface (radius a ).
find Electric field on a point P at normal distance r.
Electric field E is normal to the surface containing
charge hence it is normal to the plain surface of
cylinder and parallel to curved surface.
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genius Physics……..Pradeep Kshetrapal Electrostatics 2011
Or E=
Now = for curved surface + for 2
plane surfaces.
= + + This expression is same as electric field due to a point
= for plain surfaces 2E ( E is uniform) charge q placed at distance r from P. i.e. In this case if
= 2Eπa2 complete charge q is placed at the centre of shell the
The charge enclosed inside Gaussian surface q = σ.A electric field is same.
Or, q = σπa2
Case 2. If P is on the surface.
Applying Gauss’s Law : = q / ε₀ In above formula when r decrease to R the electric
field increase.
Putting values 2Eπa2 =
On the surface (replace r with R) E=
Or E= Hence this is electric field on the surface of a shell and
its value is maximum compared to any other point.
Electric Field due to charge distributed over a
spherical shell :- Case 3. If P is within the surface. Or ‘r’ R
Charged Shell +
+ +
+ +
+ R +
+
r +
+ + + Gaussian Surface
+ +
+ +
+ + + +
+ R + + +
+ +
r Let’s consider +a Gaussian surface, a concentric
+ + P E spherical shell of radius r passing through P.
+ +
+ +
+ + Then charge contained inside Gaussian surface is Zero.
+
According to Gauss’s Theorem = q / ε₀
If q is zero then = 0.
As ds is not zero then E = 0
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genius Physics……..Pradeep Kshetrapal Electrostatics 2011
Gaussian Surface E
Charged Sphere
+++++++++
++++++++
++ + + + + + + +
r
+++++++++ R
E
+ + + + + + + +P +
+ + + + + +++ + +
+++++++++
+++++++++
++ + + + + + + +
+++++++++ Electric field due to two charged parallel surface
Charges of similar nature
+++++++++
1 2
+++++++++
E1 = - + E1 = + + E1 = +
+ +++ + + + + + +
+ +
+++++++++ + +
+++++++++ E2 = - + E2 = - E2 = +
+
+++++++++ + +
Case I. When+P+is+ Out
+ + +side
+ + sphere.
+ Same as in the case + +
+ + + + + + + + + E=E1+E2= - + E=0 E=E1+E2= +
of charged shell E = +
+++++++++ + +
a. Charges of + opposite nature + :-
Case 2. When + +point
+ + +P+ is+ +on+ the surface of shell: Same
+1 2+
+++++++++ +
as in case of shell . E = ++ E1 = + + E =+
+++++++++ E1 = - -+ 1
Case 3 If point++P + +is+inside
+ + + +the charged sphere. ++
-+
Consider Gaussian surface, a concentric spherical shell +
++++++++ E2 = + -
of radius r, such that point P lies on the surface. + E2 = + E2 = -
Electric field is normal to the surface. + -
+ -
E= -E1 + E2= 0 E=+E1+E2 E= +E1 - E2= 0
Now = for complete area of Gaussian surface + -
+ =+ -
= = E (E is uniform)
+ -
= E x 4πr . (for spherical shell
2
= 4πr2 )
+
-
+ :
Equipotential Surface
Charge within Gaussian surface = charge density x -
+
volume. Energyof a charged + particle in terms - of potential:-
= ρ πr3 (where ρ is the charge per unit volume.) -
Applying Gauss’s Law = q / ε₀
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genius Physics……..Pradeep Kshetrapal Electrostatics 2011
Electron-Volt : By relation Work/energy = qV, the Corona discharge : when an uncharged body is
smallest unit of work/energy is Electron Volt. brought near a charged body having sharp corners
there is large number of charges at the corners. Due to
One electron volt is the work done by/on one induction, they induce large number of opposite
electron for moving between two points having charges. This creates a very strong Electric field
potential difference of one Volt. between them. Finally the dielectric strength breaks-
down and there is fast flow of charges. This Spray of
1 eV = 1.6 x 10-19 Joules charges by spiked object is called Corona discharge.
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CAPACITOR
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It is a device to store charge and in turn store the PARALLEL PLATE CAPACITOR : -
electrical energy.
Since single conductor capacitor do not have large
Any conductor can store charge to some extent. But capacitance , parallel plate capacitors are
we cannot give infinite charge to a conductor. When constructed.
charge is given to a conductor its potential increases.
But charge cannot escape the conductor because air, Principle : Principle of a parallel plate capacitor is
or medium around conductor is di-electric. that an uncharged plate brought bear a charged
When due to increasing charge the potential increase plate decrease the potential of charged plate and
to such extent that air touching the conductor starts hence its capacitance (C = ) increase. Now it can
getting ionized and hence charge gets leaked. No take more charge. Now if uncharged conductor is
more charge can be stored and no more potential earthed, the potential of charged plate further
increase. This is limit of charging a conductor. decreases and capacitance further increases. This
arrangement of two parallel plates is called parallel
The electric field which can ionize air is 3 x 109 vm-1. plate capacitor.
CAPACITANCE OF A CONDUCTOR Expression for capacitance :
q
Term capacitance of a conductor is the ratio of charge Charge q is given to a plate
to it by rise in its Potential + -
Of area ‘A’. Another plate
C= -
+ -
is kept at a distance ‘d’.
A E
In this relation if V=1 then C= q. Therefore , -
After induction an +
-
Capacitance of a conductor is equal to the charge
Electric field E is set-up + -
which can change its potential by one volt.
-
Unit of capacitance : Unit of capacitance is Between the plates. Here + d -
farad, (symbol F ). q = σA and E= -
+ - by
One farad is capacitance of such a conductor whose The Potential difference between plates is given
potential increase by one volt when charge of one V = Ed = d -
+
coulomb is given to it. -
Now C = = = +
One coulomb is a very large unit. The practical smaller
units are
+
i. Micro farad ( μF ) = 10-6F.(used in electrical circuits) C =
Ii Pieco farad ( pF) = 10-12 used in electronics circuits
+
If a dielectric of dielectric constant K is inserted
Expression for capacitance of a spherical conductor : between the plates, then capacitance increase by
+
If charge q is given to a spherical conductor of radius r, factor K and become
its potential rise by V = +
Therefore capacitance C = = q/ = C =
+
Or for a sphere C= Note : The capacitance depends only on its
configuration i.e. plate area and distance, and on the
The capacitor depends only on the radius or size medium between them.
of the conductor. The other examples of parallel plate capacitors is
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or p = αε₀E₀
= χeεE
E. = E - Ep = =
or ( =σ
or =σ
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Ligh genius PHYSICS
t
Electron, Photon, Photoelectric Effect and X-rays 1
A
+ –
Manometer
Vacuum pump
The discharge tube is filled with the gas through which discharge is to be studied. The pressure of the
enclosed gas can be reduced with the help of a vacuum pump and it's value is read by manometer.
Sequence of phenomenon
As the pressure inside the discharge tube is gradually reduced, the following is the sequence of
phenomenon that are observed.
Negative glow Positive
column
– Streamers + – Positive + – +
column
Gas Colour
H2 Blue
N2 Red
Cl2 Green
Na Yellow
Cathode Rays.
Cathode rays, discovered by sir Willium Crooke are the stream of electrons. They can be produced by
using a discharge tube containing gas at a low pressure of the order of 10–2 mm of Hg. At this pressure the gas
molecules ionise and the emitted electrons travel towards positive potential of anode. The positive ions hit the
cathode to cause emission of electrons from cathode. These electrons also move towards anode. Thus the
cathode rays in the discharge tube are the electrons produced due to ionisation of gas and that emitted by
cathode due to collision of positive ions.
(1) Properties of cathode rays
(i) Cathode rays travel in straight lines (cast shadows of objects placed in their path)
(ii) Cathode rays emit normally from the cathode surface. Their direction is independent of the position of the
anode.
(iii) Cathode rays exert mechanical force on the objects they strike.
(iv) Cathode rays produce heat when they strikes a material surface.
(v) Cathode rays produce fluorescence.
(vi) When cathode rays strike a solid object, specially a metal of high atomic weight and high melting
point X-rays are emitted from the objects.
(vii) Cathode rays are deflected by an electric field and also by a magnetic field.
(viii) Cathode rays ionise the gases through which they are passed.
(ix) Cathode rays can penetrate through thin foils of metal.
1 1
(x) Cathode rays are found to have velocity ranging th to th of velocity of light.
30 10
(2) J.J. Thomson's method to determine specific charge of electron
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 3
It's working is based on the fact that if a beam of electron is subjected to the crossed electric field E and
magnetic field B , it experiences a force due to each field. In case the forces on the electrons in the electron
beam due to these fields are equal and opposite, the beam remains undeflected.
S
P
A X +
C
P
Y –
P
L.T. Magnetic
(H.T.) field S
+
y
E e–
–
l
Positive Rays.
Positive rays are sometimes known as the canal rays. These were discovered by Goldstein. If the cathode
of a discharge tube has holes in it and the pressure of the gas is around 10 –3
mm of Hg then faint luminous glow comes out from each hole on the
backside of the cathode. It is said positive rays which are coming out from
the holes.
Positive rays
(i) These are positive ions having same mass if the experimental gas does not have isotopes. However if
the gas has isotopes then positive rays are group of positive ions having different masses.
(ii) They travels in straight lines and cast shadows of objects placed in their path. But the speed of the
positive rays is much smaller than that of cathode rays.
(iii) They are deflected by electric and magnetic fields but the deflections are small as compared to that for
cathode rays.
(iv) They show a spectrum of velocities. Different positive ions move with different velocities. Being heavy,
their velocity is much less than that of cathode rays.
(v) q /m ratio of these rays depends on the nature of the gas in the tube (while in case of the cathode rays
q/m is constant and doesn't depend on the gas in the tube). q/m for hydrogen is maximum.
(vi) They carry energy and momentum. The kinetic energy of positive rays is more than that of cathode
rays.
(vii) The value of charge on positive rays is an integral multiple of electronic charge.
(viii) They cause ionisation (which is much more than that produced by cathode rays).
Mass Spectrograph.
It is a device used to determine the mass or (q/m) of positive ions.
(1) Thomson mass spectrograph
It is used to measure atomic masses of various isotopes in gas. This is done by measuring q/m of singly
ionised positive ions of the gas.
Y
S Screen or
Cathode – Photo plate S
Low P
– y
pressure
+q v
gas
Z
Q z
+
The positive ions are produced N in the bulb at the left hand side. These ions are accelerated towards
N
cathode. Some of the positive ions pass through the fine hole in the cathode. This fine ray of positive
Pump
ions
is
D
subjected to electric field E and magnetic field B and then allowed to strike a fluorescent screen ( E || B but
E or B v ).
If the initial motion of the ions is in x direction and electric and magnetic fields are applied along y
axis then force due to electric field is in the direction of y-axis and due to magnetic field it is along z-direction.
qELD
The deflection due to electric field alone y .........(i)
mv 2
qBLD
The deflection due to magnetic field alone z .........(ii)
mv
From equation (i) and (ii)
q B 2 LD
z 2 k y , where k ; This is the equation of parabola. It means all the charged particles moving
m E
with different velocities but of same q/m value will strike the screen placed in yz plane on a parabolic track as
shown in the above figure.
Note : All the positive ions of same. q/m moving with different velocity lie on the same
parabola. Higher is the velocity lower is the value of y and z. The ions of different specific charge
will lie on different parabola.
Y q/m q/m q/m q/m
large small small large
V2 V3 V4 V1>V2>V3>V4 Light mass
V1
Heavy mass
Z
The number of parabola tells the number of isotopes present in the given ionic beam.
(2) Bainbridge mass spectrograph
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 5
In Bainbridge mass spectrograph, field particles of same velocity are selected by using a velocity selector
and then they are subjected to a uniform magnetic field perpendicular to the velocity of the particles. The
particles corresponding to different isotopes follow different circular paths as shown in the figure.
(i) Velocity selector : The positive ions having a certain velocity v gets isolated from all other velocity
particles. In this chamber the electric and magnetic fields are so balanced that the particle moves undeflected.
E
For this the necessary condition is v .
B
(ii) Analysing chamber : In this chamber magnetic field B is applied perpendicular to the direction of
motion of the particle. As a result the particles move along a circular
path of radius Velocity spectrum
mE q E r1 m1 v
r also m
B
qBB ' m BB ' r r2 m 2 +q
2r1
In this way the particles of different masses gets deflected on
2r2 m1
Photographic plate
Note : Separation between two traces
2v(m 2 m 1 )
d 2r2 2r1 d .
qB '
Matter waves (de-Broglie Waves).
According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.
or
A wave is associated with moving material particle which control the particle in every respect.
The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in
the form of wave packets with group velocity.
h
E kT ; where k = Boltzman's constant = 1.38 10 23 Joules/kelvin , T = Absolute
2mkT
temp.
6.62 10 34 30 .83
So Thermal Neutron Å
17 23
2 1.07 10 1.38 10 T T
Slope = h Small m
Large m
p 1/b v
Small Small
m m
Large Large
m m
Small m
Large m
1/V E 1 E
Note : A photon is not a material particle. It is a quanta of energy.
When a particle exhibits wave nature, it is associated with a wave packet, rather then a wave.
(3) Characteristics of matter waves
(i) Matter wave represents the probability of finding a particle in space.
(ii) Matter waves are not electromagnetic in nature.
(iii) de-Brogile or matter wave is independent of the charge on the material particle. It means, matter
wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).
(iv) Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order
of the size of the particles is nature.
(v) Electron microscope works on the basis of de-Broglie waves.
(vi) The electric charge has no effect on the matter waves or their wavelength.
(vii) The phase velocity of the matter waves can be greater than the speed of the light.
(viii) Matter waves can propagate in vacuum, hence they are not mechanical waves.
(ix) The number of de-Broglie waves associated with nth orbital electron is n.
(x) Only those circular orbits around the nucleus are stable whose circumference is integral multiple
of de-Broglie wavelength associated with the orbital electron.
(4) Davision and Germer experiment
It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of
electrons emitted by electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni
crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from
electron gun.
F Electron
gun
Detector
Incident beam
of electrons
Diffracted beam
of electrons
The diffracted beam of electrons is received by the detector which can be positioned at any angle by
Nickel crystal
rotating it about the point of incidence. The energy of the incident beam of electrons can also be varied by
changing the applied voltage to the electron gun.
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 7
According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be
same but Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but
different at different angles of scattering.
Incident beam
Incident beam
Incident beam
Incident beam
50o
44 V 48 V 54 V 64 V
Intensity is maximum at 54 V potential difference and 50o diffraction angle.
If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg's
formula 2d sin n , we can determine the wavelength of these waves.
(180 )
Where d = distance between diffracting planes, = glancing angle
2 =65° =50°
for incident beam = Bragg's angle. D
The distance between diffraction planes in Ni-crystal for this experiment is d = d
0.91Å and the Bragg's angle = 65o. This gives for n = 1, Atomic
planes
2 0.91 10 10 sin 65 o 1.65 Å
12 .27 12 .27
Now the de-Broglie wavelength can also be determined by using the formula 1 .67 Å .
V 54
Thus the de-Broglie hypothesis is verified.
Heisenberg Uncertainty Principle.
According to Heisenberg's uncertainty principle, it is impossible to measure simultaneously both the
position and the momentum of the particle.
Let x and p be the uncertainty in the simultaneous measurement of the position and momentum of the
h
particle, then x p ; where and h = 6.63 10–34 J-s is the Planck's constant.
2
If x = 0 then p =
and if p = 0 then x = i.e., if we are able to measure the exact position of the particle (say an
electron) then the uncertainty in the measurement of the linear momentum of the particle is infinite. Similarly,
if we are able to measure the exact linear momentum of the particle i.e., p = 0, then we can not measure the
exact position of the particle at that time.
Photon.
According to Eienstein's quantum theory light propagates in the bundles (packets or quanta) of energy,
each bundle being called a photon and possessing energy.
(1) Energy of photon
hc
Energy of each photon is given by E h ; where c = Speed of light, h = Plank's constant = 6.6 10–34 J-
sec, = Frequency in Hz, = Wavelength of light
hc 12375 12400
Energy of photon in electron volt E(eV )
e ( Å) ( Å)
(2) Mass of photon
Actually rest mass of the photon is zero. But it's effective mass is given as
E h h
E mc 2 h m 2 2 . This mass is also known as kinetic mass of the photon
c c c
(3) Momentum of the photon
E h h
Momentum p m c
c c
genius PHYSICS
8 Electron, Photon, Photoelectric Effect and X-rays
Example: 1 The ratio of specific charge of an -particle to that of a proton is [BCECE 2003]
(a) 2 : 1 (b) 1 : 1 (c) 1 : 2 (d) 1 : 3
q (q / m ) q m 1
p
Solution : (c) Specific charge ; Ratio .
m (q / m ) p q p m 2
Example: 2 The speed of an electron having a wavelength of 10 10 m is [AIIMS 2002]
(a) 7.25 10 6 m/s (b) 6.26 10 6 m / s (c) 5.25 10 6 m / s (d) 4.24 10 6 m / s
h h 6 . 6 10 34
Solution : (a) By using electron v 7 .25 10 6 m /s.
mev m e e 9 .1 10 31 10 10
Example: 3 In Thomson experiment of finding e/m for electrons, beam of electron is replaced by that of muons
(particle with same charge as of electrons but mass 208 times that of electrons). No deflection condition
in this case satisfied if [Orissa (Engg.) 2002]
(a) B is increased 208 times (b) E is increased 208 times
(c) B is increased 14.4 times (d) None of these
2
e E
Solution : (c) In the condition of no deflection . If m is increased to 208 times then B should be increased by
m 2VB 2
208 14 .4 times.
Example: 4 In a Thomson set-up for the determination of e/m, electrons accelerated by 2.5 kV enter the region of
crossed electric and magnetic fields of strengths 3.6 10 4 Vm 1 and 1.2 10 3 T respectively and go
through undeflected. The measured value of e/m of the electron is equal to [AMU 2002]
(a) 1.0 10 C-kg
11 -1 (b) 1.76 10 C-kg
11 -1 (c) 1.80 10 C-kg
11 -1 (d) 1.85 10 C-kg-1
11
e E2 e (3 .6 10 4 ) 2
Solution : (c) By using 1 .8 10 11 C / kg .
m 2VB 2 m 2 2 .5 10 3 (1 .2 10 3 ) 2
Example: 5 In Bainbridge mass spectrograph a potential difference of 1000 V is applied between two plates distant 1
cm apart and magnetic field in B = 1T. The velocity of undeflected positive ions in m/s from the velocity
selector is
(a) 10 7 m /s (b) 10 4 m /s (c) 10 5 m /s (d) 10 2 m /s
E V 1000 10 5
Solution : (c) By using v ; where E 10 5
V / m v 10 5 m /s .
B d 1 10 2 1
Example: 6 An electron and a photon have same wavelength. It p is the momentum of electron and E the energy
of photon. The magnitude of p/ E in S.I. unit is
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 9
(a) 3.0 108 (b) 3.33 10–9 (c) 9.1 10–31 (d) 6.64 10–34
h h hc
Solution : (b) (for electron) or p and E (for photon)
p
p 1 1
3.33 10 9 s / m
E c 3 10 8 m / s
Example: 7 The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let 1 be
the de-Broglie wavelength of the proton and 2 be the wavelength of the photon. The ratio 1/2 is
proportional to
[UPSEAT 2003; IIT-JEE (Screening) 2004]
(a) E 0 (b) E 1 / 2 (c) E 1 (d) E 2
hc h
Solution : (b) For photon 2 ……. (i) and For proton 1 …….(ii)
E 2mE
1 E1 / 2
Therefore 1 E1 / 2 .
2 2m c 2
Example: 8 The de-Broglie wavelength of an electron having 80eV of energy is nearly ( 1eV 1.6 10 19 J , Mass of
electron 9 10 31 kg and Plank's constant 6.6 10 34 J-sec)
(a) 140 Å (b) 0.14 Å (c) 14 Å (d) 1.4 Å
h 12 .27
Solution : (d) By using . If energy is 80 eV then accelerating potential difference will be 80 V. So
2mE V
12 .27
1 .37 1 .4 Å.
80
Example: 9 The kinetic energy of electron and proton is 10 32 J . Then the relation between their de-Broglie
wavelengths is
(a) p e (b) p e (c) p e (d) p 2e
h 1
Solution : (a) By using E = 10–32 J = Constant for both particles. Hence
2mE m
Since m p m e so p e .
Example: 10 The energy of a proton and an particle is the same. Then the ratio of the de-Broglie wavelengths of the
proton and the is [RPET 1991]
(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1
h 1 proton m 2
Solution : (b) By using (E – same) .
2mE m particle mp 1
Example: 11 The de-Broglie wavelength of a particle accelerated with 150 volt potential is 10 10 m. If it is accelerated
by 600 volts p.d., its wavelength will be [RPET 1988]
(a) 0.25 Å (b) 0.5 Å (c) 1.5 Å (d) 2 Å
1 1 V2 10 10 600
Solution : (b) By using 2 2 = 0.5 Å.
V 2 V1 2 150
Example: 12 The de-Broglie wavelength of an electron in an orbit of circumference 2r is [MP PET 1987]
(a) 2r (b) r (c) 1 / 2r (d) 1 / 4r
h h
Solution : (a) According to Bohr's theory mv r n 2 r n n
2 mv
For n = 1 = 2r
Example: 13 The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is
(taking h 6 10 34 J-sec) [Kerala (Engg.) 2002]
(a) 100 (b) 1000 (c) 3 10 20
(d) 3 10 18
9
P 100 540 10
Solution : (c) By using n 3 10 20
hc 6 .6 10 34 3 10 8
Example: 14 A steel ball of mass 1kg is moving with a velocity 1 m/s. Then its de-Broglie waves length is equal to
(a) h (b) h / 2 (c) Zero (d) 1 / h
h
Solution : (a) By using h.
mv 1 1
genius PHYSICS
10 Electron, Photon, Photoelectric Effect and X-rays
Example: 15 The de-Broglie wavelength associated with a hydrogen atom moving with a thermal velocity of 3 km/s will
be
(a) 1 Å (b) 0.66 Å (c) 6.6 Å (d) 66 Å
h 6 .6 10 34
Solution : (b) By using 0 .66 Å
mv rms 2 1 .67 10 27 3 10 3
Example: 16 When the momentum of a proton is changed by an amount P0, the corresponding change in the de-Broglie
wavelength is found to be 0.25%. Then, the original momentum of the proton was [CPMT 2002]
(a) p0 (b) 100 p0 (c) 400 p0 (d) 4 p0
1 p p p 0 .25 1
Solution : (c) 0 p = 400 p0 .
p p p p 100 400
Example: 17 If the electron has same momentum as that of a photon of wavelength 5200Å, then the velocity of electron
in m /sec is given by
(a) 103 (b) 1.4 103 (c) 7 10–5 (d) 7.2 106
34
h h 6 .6 10
Solution : (b) v v = 1.4 103 m/s.
mv m 9 .1 10 31 5200 10 10
Example: 18 The de-Broglie wavelength of a neutron at 27oC is . What will be its wavelength at 927oC
(a) / 2 (b) / 3 (c) / 4 (d) / 9
1 1 T2 (273 927 ) 1200
Solution : (a) neutron 2 2 .
T 2 T1 2 (273 27 ) 300 2
Example: 19 The de-Broglie wavelength of a vehicle is . Its load is changed such that its velocity and energy both are
doubled. Its new wavelength will be
(a) (b) (c) (d) 2
2 4
h 1 hv
Solution : (a) and E mv 2 when v and E both are doubled, remains unchanged i.e. ' = .
mv 2 2E
Example: 20 In Thomson mass spectrograph when only electric field of strength 20 kV/m is applied, then the
displacement of the beam on the screen is 2 cm. If length of plates = 5 cm, distance from centre of plate to
the screen = 20 cm and velocity of ions = 106 m/s, then q/m of the ions is
(a) 106 C/kg (b) 107 C/Kg (c) 108 C/kg (d) 1011 C/kg
qELD
Solution : (c) By using y ; where y = deflection on screen due to electric field only
mv 2
q yv 2 2 10 2 (10 6 ) 2
10 8 C / kg .
m ELD 20 10 3 5 10 2 0 .2
Example: 21 The minimum intensity of light to be detected by human eye is 10 10 W / m 2 . The number of photons of
wavelength 5.6 10 7 m entering the eye, with pupil area 10 6 m 2 , per second for vision will be nearly
(a) 100 (b) 200 (c) 300 (d) 400
P
Solution : (c) By using I ; where P = radiation power
A
nh c n IA
P I A IA
t t hc
10
n 10 10 6 5 .6 10 7
Hence number of photons entering per sec the eye = 300.
t 6 .6 10 34 3 10 8
Example 1. A particle of mass M at rest decays into two particles of masses m 1 and m 2 , having non-zero
velocities. The ratio of the de-Broglie wavelengths of the particles, 1 / 2 is [IIT-JEE 1999]
The curve drawn between velocity and frequency of photon in vacuum will be a
[MP PET 2000]
(a) Straight line parallel to frequency axis
(b) Straight line parallel to velocity axis
(c) Straight line passing through origin and making an angle of 45 o with frequency axis
(d) Hyperbola
Solution : (a) Velocity of photon (i.e. light) doesn’t depend upon frequency. Hence the graph between velocity of
photon and frequency will be as follows.
Velocity of
photon (c)
Frequency
()
Photo-electric Effect.
It is the phenomenon of emission of electrons from the surface of metals, when light radiations
(Electromagnetic radiations) of suitable frequency fall on them. The emitted electrons are called
photoelectrons and the current so produced is called photoelectric current.
This effect is based on the principle of conservation of energy.
(1) Terms related to photoelectric effect
(i) Work function (or threshold energy) (W0) : The minimum energy of incident radiation,
required to eject the electrons from metallic surface is defined as work function of that surface.
hc
W0 h 0 Joules ; 0 = Threshold frequency; 0 = Threshold wavelength
0
hc 12375
Work function in electron volt W0(eV)
e 0 0 ( Å)
Note : By coating the metal surface with a layer of barium oxide or strontium oxide it's work
function is lowered.
(ii) Threshold frequency (0) : The minimum frequency of incident radiations required to eject the
electron from metal surface is defined as threshold frequency.
If incident frequency < 0 No photoelectron emission
(iii) Threshold wavelength (0) : The maximum wavelength of incident radiations required to eject the
electrons from a metallic surface is defined as threshold wavelength.
If incident wavelength > 0 No photoelectron emission
(2) Einstein's photoelectric equation
According to Einstein, photoelectric effect is the result of one to one inelastic collision between photon
and electron in which photon is completely absorbed. So if an electron in a metal absorbs a photon of energy E
(= h), it uses the energy in three following ways.
(i) Some energy (say W) is used in shifting the electron from interior to Incident
photon
the surface of the metal. K
(ii) Some energy (say W0) is used in making the surface electron free Work
function
from the metal. e–
W0
(iii) Rest energy will appear as kinetic energy (K) of the emitted W
photoelectrons. e– Metal
Hence E = W + W0 + K
For the electrons emitting from surface W = 0 so kinetic energy of
emitted electron will be max.
Hence E = W0 + Kmax ; This is the Einstein's photoelectric equation
(3) Experimental arrangement to observe photoelectric
effect Radiation
e – e–se–
When light radiations of suitable frequency (or suitable wavelength e –
e– e– e– e–
and suitable energy) falls on plate P, photoelectrons are emitted from P. P e– e– e– e– Q
V
mA
Batter
genius PHYSICS
12 Electron, Photon, Photoelectric Effect and X-rays
(i) If plate Q is at zero potential w.r.t. P, very small current flows in the circuit because of some electrons
of high kinetic energy are reaching to plate Q, but this current has no practical utility.
(ii) If plate Q is kept at positive potential w.r.t. P current starts flowing through the circuit because more
electrons are able to reach upto plate Q.
(iii) As the positive potential of plate Q increases, current through the circuit increases but after some
time constant current flows through the circuit even positive potential of plate Q is still increasing, because at
this condition all the electrons emitted from plate P are already reached up to plate Q. This constant current is
called saturation current.
(iv) To increase the photoelectric current further we will have to increase the intensity of incident light.
Photoelectric current (i) depends upon
(a) Potential difference between electrodes (till saturation) I
(b) Intensity of incident light (I)
(c) Nature of surface of metal
(v) To decrease the photoelectric current plate Q is maintained at negative potential w.r.t. P, as the anode Q is
made more and more negative, fewer and fewer electrons will reach the cathode and the photoelectric current
decreases.
(vi) At a particular negative potential of plate Q no electron will reach the plate Q and the current will
become zero, this negative potential is called stopping potential denoted by V0.
(vii) If we increase further the energy of incident light, kinetic energy of photoelectrons increases and
more negative potential should be applied to stop the electrons to reach upto plate Q. Hence eV 0 K max .
Note : Stopping potential depends only upon frequency or wavelength or energy of incident
radiation. It doesn't depend upon intensity of light.
We must remember that intensity of incident light radiation is inversely proportional to the
1 1
square of distance between source of light and photosensitive plate P i.e., I 2 so I i 2 )
d d
Important formulae
h h 0 K max
1 2 h( 0 )
K max eV 0 h( 0 ) 2
mv max h( 0 ) v m ax
2 m
1 1 1 2hc 0
K m ax mv m2 ax eV 0 hc hc 0 v max
2 0 0 m 0
h hc 1 1 1 1
V0 ( 0 ) 12375
e e 0 0
(4) Different graphs
(i) Graph between potential difference between the plates P and Q and photoelectric current
i i
I3
I2
I1 3 > 2 > 1
3
2 1
– V0 – V0 – V 0 – V0 V
(ii) Graph between maximum V kinetic energy / stopping potential
1
of2 photoelectrons
3
and frequency of
For different intensities of incident For different Frequencies of
incident light light incident light
V0
Kmax
–W0
–
Photoelectric Cell. Slope = tan = h W0/e Slope = tan = h/e
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 13
A device which converts light energy into electrical energy is called photoelectric cell. It is also known as
photocell or electric eye.
Photoelectric cell are mainly of three types
Photo-emissive cell Photo-conductive cell Photo-voltaic cell
It consists of an evacuated glass or It is based on the principle that It consists of a Cu plate coated with
quartz bulb containing anode A conductivity of a semiconductor a thin layer of cuprous oxide
and cathode C. The cathode is increases with increase in the (Cu2O). On this plate is laid a semi
semi-cylindrical metal on which a intensity of incident light. transparent thin film of silver.
layer of photo-sensitive material is
coated.
Transparent
Surface film film of silver
AC
Light
A
Output
C
Galvanometer In this, Selenium
Selenium
a thin layer R ofOutput
some WhenSemiconducting
light fall,
layer of Cuthe
2 O electrons
R
or semiconductor
Metal layer(as selenium) is emitted from
Metal the
layer of layer
Cu of Cu2O and
When light incident Microon the
ammeter placed below a transparent foil of move towards the silver film. Then
cathode, it emits photo-electrons
some metal. This combination is the silver film becomes negatively
which are attracted by the
A anode.
fixed over an iron plate. When light charged and copper plate becomes
The photoelectrons
+ – constitute a
is incident on the transparent foil, positively charged. A potential
small current which flows through
the electrical resistance of the difference is set up between these
the external circuit.
semiconductor layer is reduced. two and current is set up in the
Hence a current starts flowing in external resistance.
the battery circuit connected.
Note : The photoelectric current can be increased by filling some inert gas like Argon into the
bulb. The photoelectrons emitted by cathode ionise the gas by collision and hence the current is
increased.
Compton effect
The scattering of a photon by an electron is called Compton effect. The energy and momentum is
conserved. Scattered photon will have less energy (more wavelength) as compare to incident photon (less
wavelength). The energy lost by the photon is taken by electron as kinetic energy.
The change in wavelength due to Compton effect is called Compton shift. Compton shift
h
f i (1 cos )
m0c Compton scattering
–
Target electron
at rest Recoil
h electron
–
i h
Incident
Note : Compton effect shows that photon
photon have momentum.
Scattered photon
f
X-rays.
X-rays was discovered by scientist Rontgen that's why they are also called Rontgen rays.
Rontgen discovered that when pressure inside a discharge tube kept 10 –3 mm of Hg and potential
difference is 25 kV then some unknown radiations (X-rays) are emitted by anode.
(1) Production of X-rays
There are three essential requirements for the production of X-rays
(i) A source of electron
(ii) An arrangement to accelerate the electrons
(iii) A target of suitable material of high atomic weight and high melting point on which these high speed
electrons strike.
(2) Coolidge X-ray tube
It consists of a highly evacuated glass tube containing cathode and target. The cathode consist of a
tungsten filament. The filament is coated with oxides of barium or strontium to have an emission of electrons
even at low temperature. The filament is surrounded by a molybdenum cylinder kept at negative potential
w.r.t. the target.
genius PHYSICS
14 Electron, Photon, Photoelectric Effect and X-rays
The target (it's material of high atomic weight, high melting point and high thermal conductivity) made of
tungsten or molybdenum is embedded in a copper block.
The face of the target is set at 45o to the incident electron stream.
V
Lead
chamber Anode
C
Water
T
The filament is heated by passing the currentF through it. A high potential difference ( 10 kV to 80 kV) is
W
Filament
applied between the target and cathode to accelerate the electrons
Target which are emitted by filament. The stream of
X-rays
highly energetic electrons are focussed on theWindow
target.
Most of the energy of the electrons is converted into heat (above 98%) and only a fraction of the energy of
the electrons (about 2%) is used to produce X-rays.
During the operation of the tube, a huge quantity of heat is produced in this target, this heat is conducted
through the copper anode to the cooling fins from where it is dissipated by radiation and convection.
(i) Control of intensity of X-rays : Intensity implies the number of X-ray photons produced from the
target. The intensity of X-rays emitted is directly proportional to the electrons emitted per second from the
filament and this can be increased by increasing the filament current. So intensity of X-rays Filament
current
(ii) Control of quality or penetration power of X-rays : Quality of X-rays implies the penetrating
power of X-rays, which can be controlled by varying the potential difference between the cathode and the
target.
For large potential difference, energy of bombarding electrons will be large and hence larger is the
penetration power of X-rays.
Depending upon the penetration power, X-rays are of two types
Hard X-rays Soft X-rays
More penetration power Less penetration power
More frequency of the order of 1019 Less frequency of the order of 1016 Hz
Hz
Lesser wavelength range (0.1Å – 4Å) More wavelength range (4Å – 100Å)
I
Incident X-
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 15
Note : Continuos X-rays are produced due to the phenomenon called "Bremsstrahlung". It
means slowing down or braking radiation.
Minimum wavelength
When the electron looses whole of it's energy in a single collision with the atom, an X-ray photon of
1 hc
maximum energy hmax is emitted i.e. mv 2 eV h max
2 min
where v = velocity of electron before collision with target atom, V = potential difference through which
electron is accelerated, c = speed of light = 3 108 m/s
eV
Maximum frequency of radiations (X-rays) max
h
hc 12375
Minimum wave length = cut off wavelength of X-ray min Å
eV V
Note : Wavelength of continuous X-ray photon ranges from certain minimum (min) to infinity.
max logemax logemax
min
X-ray photon
e–
e–
genius PHYSICS
16 Electron, Photon, Photoelectric Effect and X-rays
K, L, M, …… series
If the electron striking the target eject an electron from the K-shell n=5
O
of the atom, a vacancy is crated in the K-shell. Immediately an electron N n=4
from one of the outer shell, say L-shell jumps to the K-shell, emitting an M M
X-ray photon of energy equal to the energy difference between the two M M-series
n=3
shells. Similarly, if an electron from the M-shell jumps to the K-shell, X- L L L
ray photon of higher energy is emitted. The X-ray photons emitted due L L-series
n=2
K K K
to the jump of electron from the L, M, N shells to the K-shells gives K,
K n=1
K, K lines of the K-series of the spectrum. K-series
If the electron striking the target ejects an electron from the L-
shell of the target atom, an electron from the M, N ….. shells jumps to the L-shell so that X-rays photons of
lesser energy are emitted. These photons form the lesser energy emission. These photons form the L-series of
the spectrum. In a similar way the formation of M series, N series etc. may be explained.
Intensit K
Intensity-wavelength graph y
K
At certain sharply defined wavelengths, the intensity of X-rays is very L
L L
large K-
series
L-series
as marked K, K …. As shown in figure. These X-rays are known as
min Wavelengt
characteristic X-rays. At other wavelengths the intensity varies gradually and h
Concepts
Nearly all metals emits photoelectrons when exposed to UV light. But alkali metals like lithium, sodium, potassium,
rubidium and cesium emit photoelectrons even when exposed to visible light.
Oxide coated filament in vacuum tubes is used to emit electrons at relatively lower temperature.
Conduction of electricity in gases at low pressure takes because colliding electrons acquire higher kinetic energy due to
increase in mean free path.
Kinetic energy of cathode rays depends on both voltage and work function of cathode.
Photoelectric effect is due to the particle nature of light.
Hydrogen atom does not emit X-rays because it's energy levels are too close to each other.
The essential difference between X-rays and of -rays is that, -rays emits from nucleus while X-rays from outer part of
atom.
There is no time delay between emission of electron and incidence of photon i.e. the electrons are emitted out as soon as the
light falls on metal surface.
If light were wave (not photons) it will take about an year take about an year to eject a photoelectron out of the metal
surface.
Doze of X-ray are measured in terms of produced ions or free energy via ionisaiton.
Safe doze for human body per week is one Rontgen (One Rontgon is the amount of X-rays which emits 2.5 104 J free
energy through ionization of 1 gm air at NTP
Example
s
Example: 22 The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron
emission from this substance is approximately [AIEEE 2004]
(a) 540 nm (b) 400 nm (c) 310 nm (d) 220 nm
12375 12375
Solution : (c) By using 0 0 = 3093.7 Å ~– 310 nm
W0 (eV ) 4
Example: 23 Photo-energy 6 eV are incident on a surface of work function 2.1 eV. What are the stopping potential
[MP PMT 2004]
(a) – 5V (b) – 1.9 V (c) – 3.9 V (d) – 8.1 V
Solution : (c) By using Einstein's equation E = W0 + Kmax 6 2.1 K max K max 3.9 eV
K max
Also V0 3.9 V .
Example: 24 When radiation of wavelength is incident on a metallic surface the stopping potential is 4.8 volts. If the
same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes
1.6 volts. Then the threshold wavelength for the surface is
(a) 2 (b) 4 (c) 6 (d) 8
hc 1 1
Solution : (b) By using V0
e 0
1
hc 1 hc 1 1
4 .8 …… (i) and 1 .6 …… (ii)
e 0 e 2 0
From equation (i) and (ii) 0 4 .
Example: 25 When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 volts. If e/m
for the electron is 1.8 10 11 Ckg 1 the maximum velocity of the ejected electrons is
(a) 6 10 5 ms 1 (b) 8 10 5 ms 1 (c) 1.8 10 6 ms 1 (d) 1.8 10 5 ms 1
1 e
Solution : (c) m v m2 ax eV 0 v m ax 2 . V0 2 1.8 10 11 9 1.8 10 6 m / s .
2 m
Example: 26 The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal
(for which work function is 1.65 eV) will be [JIPMER 2002]
10
(a) 4 10 Hz
10
(b) 4 10 Hz
11
(c) 4 10 Hz
14
(d) 4 10 Hz
12375 12375
Solution : (c) Threshold wavelength 0 7500 Å.
W0 (eV ) 1 .65
c 3 10 8
so minimum frequency 0 4 10 14 Hz .
0 7500 10 10
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 19
Example: 27 Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminates a
metal of work function 0.5 eV. The ratio of maximum kinetic energy of the emitted electron will be
(a) 1 : 5 (b) 1 : 4 (c) 1 : 2 (d) 1 : 1
(K max )1 1 0 .5 0.5 1
Solution : (b) By using K max E W0 .
(K max ) 2 2.5 0.5 2 4
Example: 28 Photoelectric emission is observed from a metallic surface for frequencies 1 and 2 of the incident light rays
(1 2 ) . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1
: k, then the threshold frequency of the metallic surface is [EAMCET (Engg.) 2001]
1 2 k1 2 k 2 1 2 1
(a) (b) (c) (d)
k 1 k 1 k 1 k 1
Solution : (b) By using h h 0 k max h( 1 0 ) k 1 and h( 1 0 ) k 2
1 0 k 1 k 1 2
Hence 1 0
2 0 k 2 k k 1
Example: 29 Light of frequency 8 1015 Hz is incident on a substance of photoelectric work function 6.125 eV. The maximum
kinetic energy of the emitted photoelectrons is [AFMC 2001]
(a) 17 eV (b) 22 eV (c) 27 eV (d) 37 eV
34
Solution : (c) Energy of incident photon E h 6 .6 10 8 10 15 5 .28 10 18 J 33 eV .
From E W0 K max K max E W0 33 6.125 26.87 eV 27 eV .
Example: 30 A photo cell is receiving light from a source placed at a distance of 1 m. If the same source is to be placed at a distance
of 2 m, then the ejected electron [MNR 1986; UPSEAT 2000, 2001]
(a) Moves with one-fourth energy as that of the initial energy
(b) Moves with one fourth of momentum as that of the initial momentum
(c) Will be half in number
(d) Will be one-fourth in number
1
Solution : (d) Number of photons Intensity
(distance) 2
2
N1 d2
2
N1 2 N1
N2 .
N 2 d 1 N2 1 4
Example: 31 When yellow light incident on a surface no electrons are emitted while green light can emit. If red light is incident on
the surface then [MNR 1998; MH CET 2000; MP PET 2000]
(a) No electrons are emitted (b) Photons are emitted
(c) Electrons of higher energy are emitted (d) Electrons of lower energy are emitted
Solution : (a) Green Yellow Red
According to the question Green is the maximum wavelength for which photoelectric emission takes place. Hence no
emission takes place with red light.
Example: 32 When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm the maximum velocities of the
photoelectrons ejected are v and 2v respectively. The work function of the metal is (h = Planck's constant, c = velocity
of light in air) [EMCET (Engg.) 2000]
Example: 33 The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies f and 2f are incident on the surfaces
of A and B respectively, the ratio of the maximum kinetic energies of photoelectrons emitted is (f is greater than
threshold frequency of A, 2f is greater than threshold frequency of B) [EAMCET (Med.) 2000]
(a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 4
Solution : (b) By using E W0 K max E A hf W A K A and EB h (2 f ) WB KB
1 WA K A WA 1
So, ……(i) also it is given that ……..(ii)
2 WB K B WB 2
KA 1
From equation (i) and (ii) we get .
KB 2
Example: 34 When a point source of monochromatic light is at a distance of 0.2m from a photoelectric cell, the cut-off voltage and
the saturation current are 0.6 volt and 18 mA respectively. If the same source is placed 0.6 m away from the
photoelectric cell, then [IIT-JEE 1992; MP PMT 1999]
genius PHYSICS
20 Electron, Photon, Photoelectric Effect and X-
rays
(a) The stopping potential will be 0.2 V (b) The stopping potential will be 0.6 V
(c) The saturation current will be 6 mA (d) The saturation current will be 18 mA
1
Solution : (b) Photoelectric current (i) Intensity . If distance becomes 0.6 m (i.e. three times) so current becomes
(distance) 2
1
times i.e. 2mA.
9
Also stopping potential is independent of intensity i.e. it remains 0.6 V.
Example: 35 In a photoemissive cell with exciting wavelength , the fastest electron has speed v. If the exciting wavelength is
changed to 3 / 4 , the speed of the fastest emitted electron will be [CBSE 1998]
1/2
(a) v (3 / 4 )1 / 2 (b) v (4 / 3) (c) Less then v (4 / 3)1 / 2 (d) Greater then v (4 / 3)1 / 2
1 2 E 2W0 hc
Solution : (d) From E W0 mv m2 ax v max (where E )
2 m m
3
If wavelength of incident light charges from to (decreases)
4
Let energy of incident light charges from E to E ' and speed of fastest electron changes from v to v then
2 E 2W0 2 E' 2W0
v …..(i) and v' …….(ii)
m m m m
4
2 E 1/2
1 4 3 2W0 4 2E 2W0
As E E ' E hence v ' v'
3 m m 3 m 4
1/2
m
3
1/2 1/2
4 2E 2W0 4
v' X v so v ' v.
3 m 4
1/2
3
m
3
Example: 36 The minimum wavelength of X-rays produced in a coolidge tube operated at potential difference of 40 kV is
[BCECE 2003]
(a) 0.31Å (b) 3.1Å (c) 31Å (d) 311Å
12375
Solution : (a) m in 0 . 309 Å 0 .31 Å
40 10 3
Example: 37 The X-ray wavelength of L line of platinum (Z = 78) is 1.30 Å. The X –ray wavelength of La line of Molybdenum (Z = 42)
is [EAMCET (Engg.) 2000]
(a) 5.41Å (b) 4.20Å (c) 2.70Å (d) 1.35 Å
1 1 1 1
Solution : (a) The wave length of L line is given by R(z 7 .4 )2 2 2
2 3 (z 7 .4 )2
Tricky example: (z 7 .4 )2 1 .30 (42 7 .4 )2
1 3 2 2 5 . 41 Å .
2 (z1 7 .4 )2 2 (78 7 .4 )2
Example: 38 The cut off wavelength of continuous X-ray from two coolidge tubes operating at 30 kV but using different target
materials (molybdenum Z= 42 and tungsten Z = 74) are
(a) 1Å, 3Å (b) 0.3 Å, 0.2 Å (c) 0.414 Å, 0.8 Å (d) 0.414 Å, 0.414 Å
Solution : (d) Cut off wavelength of continuous X-rays depends solely on the voltage applied and does not depend on the material of
the target. Hence the two tubes will have the same cut off wavelength.
hc hc 6 .627 10 34 3 10 8
Ve h or m 414 10 10 m 0 .414 Å.
Ve 30 10 1 .6 10
3 19
genius PHYSICS
Electron, Photon, Photoelectric Effect and X-rays 21
Tricky example: 4
Two photons, each of energy 2.5eV are simultaneously incident on the metal surface. If
the work function of the metal is 4.5 eV, then from the surface of metal
(a) Two electrons will be emitted (b) Not even a single electron
will be emitted
(c) One electron will be emitted (d) More than two electrons
will be emitted
Solution : (b) Photoelectric effect is the phenomenon of one to one elastic collision between incident
photon and an electron. Here in this question one electron absorbs one photon and
gets energy 2.5 eV which is lesser than 4.5 eV. Hence no photoelectron emission takes
place.
Tricky example: 5
In X-ray tube when the accelerating voltage V is halved, the difference between the
wavelength of K line and minimum wavelength of continuous X-ray spectrum
(a) Remains constant (b) Becomes more than two
times
(c) Becomes half (d) Becomes less than two
times
Solution : (c) K min when V is halved min becomes two times but Ka remains the same.
' K 2 min 2( ) K a
' 2 ()
Tricky example: 6
Molybdenum emits K-photons of energy 18.5 keV and iron emits K photons of
energy 34.7 keV. The times taken by a molybdenum K photon and an iron K photon
to travel 300 m are
(a) (3 s, 15 s) (b) (15 s, 3s) (c) (1 s, 1 s) (d) (1
s, 5s)
Solution : (c) Photon have the same speed whatever be their energy, frequency, wavelength, and
origin.
300
time of travel of either photon 10 6 s 1 s
3 10 8
genius PHYSICS Pradeep Kshetrapal
Current Electricity 1
genius PHYSICS
vd
+
A
–
R R
R R
R R
Determination of resistance
R R R
R R R
O
2 Current Electricity
Current Electricity 3
4 Current Electricity
28 Energy consumed E = I.V.∆T ∆T is time duration
Current Electricity 5
Electric Current.
(1) Definition : The time rate of flow of charge through any cross-section is called current. So if
Q
through a cross-section, Q charge passes in time t then iav and instantaneous current
t
ΔQ dQ Q
i Lim . If flow is uniform then i . Current is a scalar quantity. It's S.I. unit is ampere (A) and
Δt 0 Δt dt t
C.G.S. unit is emu and is called biot (Bi), or ab ampere. 1A = (1/10) Bi (ab amp.)
(2) The direction of current : The conventional direction of current is taken to be the direction of
flow of positive charge, i.e. field and is opposite to the direction of flow of negative charge as shown below.
i i
E
E
Though conventionally a direction is associated with current (Opposite to the motion of electron), it is
not a vector. It is because the current can be added algebraically. Only scalar quantities can be added
algebraically not the vector quantities.
(3) Charge on a current carrying conductor : In conductor the current is caused by electron
(free electron). The no. of electron (negative charge) and proton (positive charge) in a conductor is same.
Hence the net charge in a current carrying conductor is zero.
(4) Current through a conductor of non-uniform cross-section : For a given conductor
current does not change with change in cross-sectional area. In the following figure i1 = i2 = i3
i1 i2 i3
– t t t
dc Inverter ac
ac Rectifier dc
(ii) Shows heating effect, chemical effect and magnetic
(ii) Shows heating effect only
effect of current
+ –
(iii) It’s symbol is ~ (iii) It’s symbol is
+ +
+ +
genius PHYSICS Pradeep Kshetrapal
6 Current Electricity
nq
In n particle each having a charge q, pass through a given area in time t then i
t
If n particles each having a charge q pass per second per unit area, the current associated with cross-
sectional area A is i nqA
If there are n particle per unit volume each having a charge q and moving with velocity v, the current
thorough, cross section A is i nqvA , for electrons i= neavd
(ii) Due to rotatory motion of charge
If a point charge q is moving in a circle of radius r with speed v (frequency , angular speed and time
q qv q q
period T) then corresponding currents i q r
T 2πr 2π
V
(iii) When a voltage V applied across a resistance R : Current flows through the conductor i
R
R
P
also by definition of power i i
V
(7) Current carriers : The charged particles whose flow in a definite direction constitutes the
electric current are called current carriers. In different situation current carriers are different.
(i) Solids : In solid conductors like metals current carriers are free electrons.
(ii) Liquids : In liquids current carriers are positive and negative ions.
(iii) Gases : In gases current carriers are positive ions and free electrons.
(iv) Semi conductor : In semi conductors current carriers are holes and free electrons.
Current density (J).
In case of flow of charge through a cross-section, current density is defined as a vector having
magnitude equal to current per unit area surrounding that point. Remember area is normal to the direction
di
of charge flow (or current passes) through that point. Current density at point P is given by J n
dA
ˆ
dA
dA
i J
i P J
n
dA cos
If the cross-sectional area is not normal to the current, the cross-sectional area normal to current in
accordance with following figure will be dA cos and so in this situation:
di
J
dA cos
i.e. di JdA cos or di J .dA i J dA
Note :
If current density J is uniform for a normal cross-section A then : i J ds J ds [as
J = constant]
i
or i J A JA cos 0 JA J
A
[as dA A and = 0 ]
o
genius PHYSICS Pradeep Kshetrapal
Current Electricity 7
(1) Unit and dimension : Current density J is a vector quantity having S.I. unit Amp/m2 and
dimension.[L–2A]
(2) Current density in terms of velocity of charge : In case of uniform flow of charge through a
i
cross-section normal to it as i nqvA so, J n (nqv) n or J nq v v ( ) [With
A
charge
nq ]
volume
i.e., current density at a point is equal to the product of volume charge density with velocity of charge
distribution at that point.
(3) Current density in terms of electric field : Current density relates with electric field as
E
J E ; where = conductivity and = resistivity or specific resistance of substance.
(i) Direction of current density J is same as that of electric field E .
(ii) If electric field is uniform (i.e. E constant ) current density will be constant [as = constant]
(iii) If electric field is zero (as in electrostatics inside a conductor), current density and hence current will
be zero.
Conduction of Current in Metals.
According to modern views, a metal consists of a ‘lattice’ of fixed positively charged ions in which
billions and billions of free electrons are moving randomly at speed which at room temperature (i.e. 300 K)
3 kT 3 (1 . 38 10 23 ) 300 ~
in accordance with kinetic theory of gases is given by v rms 31
– 10 5 m / s
m 9 . 1 10
The randomly moving free electrons inside the metal collide with the lattice and follow a zig-zag path as
shown in figure (A).
– +
(A) (B)
However, in absence of any electric field due to this random motion, the number of electrons crossing
from left to right is equal to the number of electrons crossing from right to left (otherwise metal will not
remain equipotential) so the net current through a cross-section is zero.
A motion of charge is possible by motion of electron or a current carrier.
8 Current Electricity
Relaxation Time :The time to travel mean free path is called Relaxation Period or Relaxation Time, denoted by Greek
letter Tau “τ”. If t1, t2, …tn are the time periods for n collisions then Relaxation Time τ = (t1+t2+ . . .tn)
Drift Velocity :When Electric Field is applied across a conductor, the free electrons experience a force in the
direction opposite to field. Dur to this force they start drifting in the direction of force. The Velocity of this drift is
called drift velocity “Vd”. During the drift they maintain their thermal velocity.
The drift velocity can be calculated as averaged velocity of all the electrons drifting.
i i
E
E
If n electrons are having initial speeds u 1, u2,…un and their time to travel free path is t1, t2, …tn then final velocities are
v1 =u1+ αt1,
V2 = u2+ αt2 ,
Vn = un+ αtn and so on.
Drift velocity is average of these velocities of charged particles. Therefore
Vd = (v1+ V2+…….. Vn)
= (u1+ αt1+ u2+ αt2+ . . . . . . un+ αtn )
= (u1+ u2+. . . un + αt1+αt2+ . . . αtn )
= (u1+ u2+. . . un ) + (αt1+αt2+ . . . αtn )
=0+α (t1+t2+ . . .tn)
= ατ
or Vd = (α = ) Vd =
-------------------------------------------------------------------
Relation of Current and Drift velocity : When an electric field is applied, inside the conductor
due to electric force the path of electron in general becomes curved (parabolic) instead of straight lines and
electrons drift opposite to the field figure (B). Due to this drift the random motion of electrons get modified
and there is a net transfer of electrons across a cross-section resulting in current.
Drift velocity is the average uniform velocity acquired by free electrons inside a metal by the
application of an electric field which is responsible for current through it. Drift velocity is very small it is of
the order of 10–4 m/s as compared to thermal speed (~– 10 5 m / s) of electrons at room temperature.
If suppose for a conductor l
Current Electricity 9
Note : The direction of drift velocity for electron in a metal is opposite to that of applied electric
field (i.e. current density J ).
v d E i.e., greater the electric field, larger will be the drift velocity.
When a steady current flows through a conductor of non-uniform cross-section drift
1
velocity varies inversely with area of cross-section v d
A
If diameter of a conductor is doubled, then drift velocity of electrons inside it will not
change.
(2) Relaxation time () : The time interval between two successive collisions of electrons with the
mean free path
positive ions in the metallic lattice is defined as relaxation time with
r.m.s. velocity of electrons v rms
rise in temperature vrms increases consequently decreases.
v
(3) Mobility : Drift velocity per unit electric field is called mobility of electron i.e. d . It’s unit is
E
2
m
.
volt sec
Concepts
Human body, though has a large resistance of the order of k (say 10 k), is very sensitive to minute currents even as low as
a few mA. Electrocution, excites and disorders the nervous system of the body and hence one fails to control the activity of the
body.
1 ampere of current means the flow of 6.25 1018 electrons per second through any cross-section of the conductors.
dc flows uniformly throughout the cross-section of conductor while ac mainly flows through the outer surface area of the
conductor. This is known as skin effect.
It is worth noting that electric field inside a charged conductor is zero, but it is non zero inside a current carrying
V
conductor and is given by E where V = potential difference across the conductor and l = length of the conductor.
l
Electric field out side the current carrying is zero.
+ + + + + + l
+ –
Ein = 0 Ein = V/l
+ + + + + +
1
For a given conductor JA = i = constant so that J i.e., J1 A1 = J2 A2 ; this is called equation of continuity
A
i
J1 J2
i
A1 A2
If cross-section is constant, I J i.e. for a given cross-sectional area, greater the current density, larger will be current.
The drift velocity of electrons is small because of the frequent collisions suffered by electrons.
The small value of drift velocity produces a large amount of electric current, due to the presence of extremely large
number of free electrons in a conductor. The propagation of current is almost at the speed of light and involves
electromagnetic process. It is due to this reason that the electric bulb glows immediately when switch is on.
In the absence of electric field, the paths of electrons between successive collisions are straight line while in presence of
electric field the paths are generally curved.
NAx d
Free electron density in a metal is given by n where NA = Avogrado number, x = number of free electrons per
A
atom, d = density of metal and A = Atomic weight of metal.
genius PHYSICS Pradeep Kshetrapal
10 Current Electricity
Example
s
Example: 1 The potential difference applied to an X-ray tube is 5 KV and the current through it is 3.2 mA. Then
the number of electrons striking the target per second is
(a) 2 1016 (b) 5 106 (c) 1 1017 (d) 4 1015
q ne it 3 .2 10 3 1
Solution : (a) i n 2 10 16
t t e 1 .6 10 19
Example: 2 A beam of electrons moving at a speed of 10 6 m/s along a line produces a current of 1.6 10–6 A. The
number of electrons in the 1 metre of the beam is [CPMT 2000]
q q qv nev ix 1 .6 10 6 1
Solution : (b) i n 10 7
t ( x / v) x x ev 1 .6 10 19 10 6
Example: 3 In the Bohr’s model of hydrogen atom, the electrons moves around the nucleus in a circular orbit of
a radius 5 10–11 metre. It’s time period is 1.5 10–16 sec. The current associated is
(a) Zero (b) 1.6 10–19 A (c) 0.17 A (d) 1.07 10–3 A
q 1 .6 10 19
Solution : (d) i 1 .07 10 3 A
T 1 .5 10 16
Example: 4 An electron is moving in a circular path of radius 5.1 10–11 m at a frequency of 6.8 1015
revolution/sec. The equivalent current is approximately
(a) 5.1 10–3 A (b) 6.8 10–3 A (c) 1.1 10–3 A (d) 2.2 10–3 A
1 Q
Solution : (c) 6.8 10 15 T sec i 1 . 6 10 19 6 . 8 10 15 = 1.1 10–3 A
6.8 10 15 T
Example: 5 A copper wire of length 1m and radius 1mm is joined in series with an iron wire of length 2m and
radius 3mm and a current is passed through the wire. The ratio of current densities in the copper and
iron wire is
[MP PMT 1994]
(a) 18 : 1 (b) 9 : 1 (c) 6 : 1 (d) 2 : 3
2
r
2
i 1 Jc A 3 9
Solution : (b) We know J when i = constant J i i
A A Ji A c rc
1 1
Example: 6 A conducting wire of cross-sectional area 1 cm2 has 3 1023 m–3 charge carriers. If wire carries a
current of 24 mA, the drift speed of the carrier is [UPSEAT 2001]
(a) 5 10–6 m/s (b) 5 10–3 m/s (c) 0.5 m/s (d) 5 10–2 m/s
i 24 10 3
Solution : (b) vd 5 10 3 m / s
neA 3 10 23 1 .6 10 19 10 4
Example: 7 A wire has a non-uniform cross-sectional area as shown in figure. A steady current i flows through it.
Which one of the following statement is correct
A B
(a) The drift speed of electron is constant (b) The drift speed increases on moving from
A to B
genius PHYSICS Pradeep Kshetrapal
Current Electricity 11
(c) The drift speed decreases on moving from A to B (d) The drift speed varies randomly
1
Solution : (c) For a conductor of non-uniform cross-section v d
Area of cross - section
Example: 8 In a wire of circular cross-section with radius r, free electrons travel with a drift velocity v, when a
current i flows through the wire. What is the current in another wire of half the radius and of the some
material when the drift velocity is 2v
(a) 2i (b) i (c) i/2 (d) i/4
2
r ne r 2 v i
Solution : (c) i neAv d = ner2v and i' ne .2v
2 2 2
Example: 9 A potential difference of V is applied at the ends of a copper wire of length l and diameter d. On
doubling only d, drift velocity
(a) Becomes two times (b) Becomes half (c) Does not change (d) Becomes one fourth
Solution : (c) Drift velocity doesn’t depends upon diameter.
Example: 10 A current flows in a wire of circular cross-section with the free electrons travelling with a mean drift
velocity v. If an equal current flows in a wire of twice the radius new mean drift velocity is
v v
(a) v (b) (c) (d) None of these
2 4
i 1 v
Solution : (c) By using v d vd v'
neA A 4
Example: 11 Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4
: 1. The ratio of drift speeds of electrons in A and B is
(a) 16 : 1 (b) 1 : 16 (c) 1 : 4 (d) 4 : 1
i1 A vd r 2 vd vd 16
Solution : (a) As i neA v d 1 1 12 . 1 1
i2 A 2 v d2 r2 v d 2 v d2 1
Tricky example: 1
In a neon discharge tube 2.9 1018 Ne+ ions move to the right each second while 1.2 1018
electrons move to the left per second. Electron charge is 1.6 10–19 C. The current in the discharge
tube [MP PET 1999]
(a) 1 A towards right (b) 0.66 A towards right (c) 0.66 A towards left (d) Zero
Solution: (b) Use following trick to solve such type of problem.
Trick : In a discharge tube positive ions carry q units of charge in t seconds from anode to cathode
and negative carriers (electrons) carry the same amount of charge from cathode to anode in t
q q'
second. The current in the tube is i .
t t'
2.9 10 18 e 1.2 10 18 e
Hence in this question current i 0.66 A towards right.
1 1
Tricky example: 2
If the current flowing through copper wire of 1 mm diameter is 1.1 amp. The drift velocity of
electron is (Given density of Cu is 9 gm/cm3, atomic weight of Cu is 63 grams and one free
electron is contributed by each atom)
(a) 0.1 mm/sec (b) 0.2 mm/sec (c) 0.3 mm/sec (d) 0.5 mm/sec
Solution: (a) 6.023 1023 atoms has mass = 63 10–3 kg
genius PHYSICS Pradeep Kshetrapal
12 Current Electricity
6 .023 10 23
So no. of atoms per m3 = n 9 10 3 8 .5 10 28
63 10 3
i 1.1
vd 0.1 10 3 m / sec 0.1 mm / sec
neA 8.5 10 28 1.6 10 19 (0.5 10 3 ) 2
Ohm’s Law.
If the physical circumstances of the conductor (length, temperature, mechanical strain etc.) remains
constant, then the current flowing through the conductor is directly proportional to the potential difference
across it’s two ends i.e. i V
V
V iR or R ; where R is a proportionality constant, known as electric resistance.
i
(1) Ohm’s law is not a universal law, the substance which obeys ohm’s law are known as ohmic
substance for such ohmic substances graph between V and i is a straight line as shown. At different
temperatures V-i curves are different.
V V
T1
1
T2
2
1 2
i i
V Here tan1 > tan2
Slope of the line = tan R
i So R1 > R2 i.e. T1 > T2
(2) The device or substances which doesn’t obey ohm’s law e.g. gases, crystal rectifiers, thermoionic
valve, transistors etc. are known as non-ohmic or non-linear conductors. For these V-i curve is not linear. In
these situation the ratio between voltage and current at a particular voltage is known as static resistance.
While the rate of change of voltage to change in current is known as dynamic resistance.
i Crystal
V 1 rectifier
Rst
i tan
V 1
while Rdyn V
I tan A V
A V V V V
(A) (B) (C) (D)
Resistance.
(1) Definition : The property of substance by virtue of which it opposes the flow of current through it,
is known as the resistance.
(2) Cause of resistance of a conductor : It is due to the collisions of free electrons with the ions or
atoms of the conductor while drifting towards the positive end of the conductor.
genius PHYSICS Pradeep Kshetrapal
Current Electricity 13
(4) Unit and dimension : It’s S.I. unit is Volt/Amp. or Ohm (). Also 1 ohm
1volt 10 8 emu of potenti al
= 109 emu of resistance. It’s dimension is [ML 2 T 3 A 2 ] .
1 Amp 10 1 emu of current
1 1
(5) Conductance (C) : Reciprocal of resistance is known as conductance. C It’s unit is or –1
R
i
or “Siemen”.
i 1
Slope = tan C
R
V
V
(6) Dependence of resistance : Resistance of a conductor depends on the following factors.
(i) Length of the conductor : Resistance of a conductor is directly proportional to it’s length i.e. R l
R
e.g. a conducting wire having resistance R is cut in n equal parts. So resistance of each part will be .
n
(ii) Area of cross-section of the conductor : Resistance of a conductor is inversely proportional to it’s
1
area of cross-section i.e. R
A
l
l
Note : If R1 and R2 are the resistances at t1oC and t2oC respectively then
R1 1 t1
.
R2 1 t2
genius PHYSICS Pradeep Kshetrapal
14 Current Electricity
b c b c b c
a a a
b a c
Resistance R Resistance R Resistance R
ac bc ab
(7) Variation of resistance of some electrical material with temperature :
(i) Metals : For metals their temperature coefficient of resistance > 0. So resistance increases with
temperature.
Physical explanation : Collision frequency of free electrons with the immobile positive ions increases
(vi) Alloys : For alloys has a small positive values. So with rise in temperature resistance of alloys is
almost constant. Further alloy resistances are slightly higher than the pure metals resistance.
Alloys are used to made standard resistances, wires of resistance box, potentiometer wire, meter bridge
wire etc.
Commonly used alloys are : Constantan, mangnin, Nichrome etc.
(vii) Super conductors : At low temperature, the resistance of certain substances becomes exactly zero.
(e.g. Hg below 4.2 K or Pb below 7.2 K).
These substances are called super conductors and phenomenon super conductivity. The temperature
at which resistance becomes zero is called critical temperature and depends upon the nature of substance.
Current Electricity 15
l
(1) Definition : From R ; If l = 1m, A = 1 m2 then R ρ i.e. resistivity is numerically equal to
A
the resistance of a substance having unit area of cross-section and unit length.
(2) Unit and dimension : It’s S.I. unit is ohm m and dimension is [ML3 T 3 A 2 ]
m
(3) It’s formula :
ne 2
(4) It’s dependence : Resistivity is the intrinsic property of the substance. It is independent of shape
and size of the body (i.e. l and A). It depends on the followings :
(i) Nature of the body : For different substances their resistivity also different e.g. silver =
minimum = 1.6 10–8 -m and fused quartz = maximum 1016 -m
(ii) Temperature : Resistivity depends on the temperature. For metals t = 0 (1 + t) i.e. resitivity
increases with temperature.
Metal Semiconductor Superconductor
TC
T T T
increases with temperature decreases with temperature decreases with temperature and
becomes zero at a certain temperature
(iii) Impurity and mechanical stress : Resistivity increases with impurity and mechanical stress.
(iv) Effect of magnetic field : Magnetic field increases the resistivity of all metals except iron, cobalt
and nickel.
(v) Effect of light : Resistivity of certain substances like selenium, cadmium, sulphides is inversely
proportional to intensity of light falling upon them.
(5) Resistivity of some electrical material : ρinsulator > ρalloy > ρsemi-conductor >ρconductor
1
Reciprocal of resistivity is called conductivity () i.e. with unit mho/m and dimensions
[M 1 L3 T 3 A 2 ] .
Stretching of Wire. If a conducting wire stretches, it’s length increases, area of cross-section decreases
so resistance increases but volume remain constant.
Suppose for a conducting wire before stretching it’s length = l1, area of cross-section = A1, radius = r1,
l
diameter = d1, and resistance R1 1
A1
Before stretching l1 l2
After stretching
l2
After stretching length = l2, area of cross-section = A2, radius = r2, diameter = d2 and resistance R 2
A2
2 2 4 4
R1 l1 A2 l1 A r d
Ratio of resistances 2 2 2
R2 l 2 A1 l 2 A1 r1 d1
2
R l
(1) If length is given then R l 1 1
2
R2 l2
genius PHYSICS Pradeep Kshetrapal
16 Current Electricity
4
1 R r
(2) If radius is given then R 4 1 2
r R2 r1
Note : After stretching if length increases by n times then resistance will increase by n2 times
1
i.e. R2 n 2 R1 . Similarly if radius be reduced to times then area of cross-section decreases
n
1
2
times so the resistance becomes n4 times i.e. R 2 n 4 R1 .
n
After stretching if length of a conductor increases by x% then resistance will increases by 2x %
(valid only if x < 10%)
Various Electrical Conducting Material For Specific Use.
(1) Filament of electric bulb : Is made up of tungsten which has high resistivity, high melting
point.
(2) Element of heating devices (such as heater, geyser or press) : Is made up of nichrome
which has high resistivity and high melting point.
(3) Resistances of resistance boxes (standard resistances) : Are made up of manganin, or
constantan as these materials have moderate resistivity which is practically independent of temperature so
that the specified value of resistance does not alter with minor changes in temperature.
(4) Fuse-wire : Is made up of tin-lead alloy (63% tin + 37% lead). It should have low melting point
and high resistivity. It is used in series as a safety device in an electric circuit and is designed so as to melt
and thereby open the circuit if the current exceeds a predetermined value due to some fault. The function of
a fuse is independent of its length.
Safe current of fuse wire relates with it’s radius as i r 3/2 .
(5) Thermistors : A thermistor is a heat sensitive resistor usually prepared from oxides of various
metals such as nickel, copper, cobalt, iron etc. These compounds are also semi-conductor. For thermistors
is very high which may be positive or negative. The resistance of thermistors changes very rapidly with
change of temperature.
i
Thermistors are used to detect small temperature change and to measure very low temperature.
Concepts
In the absence of radiation loss, the time in which a fuse will melt does not depends on it’s length but varies with radius as t r 4 .
l2
If length (l) and mass (m) of a conducting wire is given then R .
m
V
Macroscopic form of Ohm’s law is R , while it’s microscopic form is J = E.
i
Example
s
genius PHYSICS Pradeep Kshetrapal
Current Electricity 17
Example: 12 Two wires of resistance R1 and R2 have temperature co-efficient of resistance 1 and 2 respectively.
These are joined in series. The effective temperature co-efficient of resistance is
1 2 1 R1 2 R 2 R 1 R 2 1 2
(a) (b) 1 2 (c) (d)
2 R1 R 2 R 12 R 22
Solution : (c) Suppose at toC resistances of the two wires becomes R1t and R 2 t respectively and equivalent
resistance becomes Rt. In series grouping Rt = R1t + R2t, also R1t = R1(1 + 1t) and R2t = R2(1 + 2t)
R R 2 2
Rt = R1(1 + 1t) + R2(1 + 2t) = (R1 + R2) + (R11 + R22)t = (R 1 R 2 )1 1 1 t .
R1 R 2
R1 1 R 2 2
Hence effective temperature co-efficient is .
R1 R 2
Example: 13 From the graph between current i & voltage V shown, identity the portion corresponding to negative
resistance
[CBSE PMT 1997]
i
(a) DE C E
(b) CD B
D
(c) BC
A
(d) AB V
V
Solution : (b) R , in the graph CD has only negative slope. So in this portion R is negative.
I
Example: 14 A wire of length L and resistance R is streched to get the radius of cross-section halfed. What is new
resistance
[NCERT 1974; CPMT 1994; AIIMS 1997; KCET 1999; Haryana PMT 2000; UPSEAT 2001]
(a) 5 R (b) 8 R (c) 4 R (d) 16 R
4
R 1 r2
4
R r/2
Solution : (d) By using R' 16 R
R 2 r1 R' r
Example: 15 The V-i graph for a conductor at temperature T1 and T2 are as shown in the figure. (T2 – T1) is
proportional to
T2
(a) cos 2 V
T1
(b) sin
(c) cot 2
i
(d) tan
Solution : (c) As we know, for conductors resistance Temperature.
From figure R1 T1 tan T1 tan = kT1 ……. (i) (k = constant)
and R2 T2 tan (90o – ) T2 cot = kT2 ……..(ii)
From equation (i) and (ii) k(T2 T1 ) (cot tan )
Example: 16 The resistance of a wire at 20oC is 20 and at 500oC is 60. At which temperature resistance will be
25
[UPSEAT 1999]
18 Current Electricity
R1 (1 t1 ) 20 1 20 1
Solution : (d) By using
R 2 (1 t 2 ) 60 1 500 220
1
1 20
20 220 t = 80oC
Again by using the same formula for 20 and 25
25 1
1 t
220
Example: 17 The specific resistance of manganin is 50 10–8 m. The resistance of a manganin cube having length
50 cm is
l 50 10 8 50 10 2
Solution : (a) R 10 6
A (50 10 2 ) 2
Example: 18 A rod of certain metal is 1 m long and 0.6 cm in diameter. It’s resistance is 3 10–3. A disc of the
same metal is 1 mm thick and 2 cm in diameter, what is the resistance between it’s circular faces.
(a) 1.35 10–6 (b) 2.7 10–7 (c) 4.05 10–6 (d) 8.1 10–6
l R disc l A R disc 10 3 (0 .3 10 2 ) 2
Solution : (b) By using R . ; disc rod Rdisc = 2.7 10–7.
A R rod l rod A disc 3 10 3 1 (10 2 ) 2
Example: 19 An aluminium rod of length 3.14 m is of square cross-section 3.14 3.14 mm2. What should be the
radius of 1 m long another rod of same material to have equal resistance
(a) 2 mm (b) 4 mm (c) 1 mm (d) 6 mm
Current Electricity 19
Example: 23 Equal potentials are applied on an iron and copper wire of same length. In order to have same current
r
flow in the wire, the ratio iron of their radii must be [Given that specific resistance of iron = 1.0
rcopper
10–7 m and that of copper = 1.7 10–8 m] [MP PMT 2000]
(a) About 1.2 (b) About 2.4 (c) About 3.6 (d) About 4.8
Solution: (b) V = constant., i = constant. So R = constant
Pi li l i li Cu lCu
Cu Cu
Ai A Cu ri2 2
rCu
ri i 1 . 0 10 7 100
2 .4
rCu Cu 1 . 7 10 8 17
Example: 24 Masses of three wires are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratio of
their electrical resistance is
(a) 1 : 3 : 5 (b) 5 : 3 : 1 (c) 1 : 15 : 125 (d) 125 : 15 : 1
l l2 l2 m
Solution: (d) R
A V m V
l12 l2 l2 9 1
R1 : R 2 : R 3 : 2 : 3 25 : : 125 : 15 : 1
m1 m 2 m 3 3 5
Example: 25 Following figure shows cross-sections through three long conductors of the same length and material,
with square cross-section of edge lengths as shown. Conductor B will fit snugly within conductor A,
and conductor C will fit snugly within conductor B. Relationship between their end to end resistance is
3a
2a
(a) RA = RB = RC a
Solution : (a)
All the conductors have equal lengths. Area of cross-section of A is ( 3 a) 2 ( 2 a) 2 a 2
Similarly area of cross-section of B = Area of cross-section of C = a2
l
Hence according to formula R ; resistances of all the conductors are equal i.e. RA = RB = RC
A
Example: 26 Dimensions of a block are 1 cm 1 cm 100 cm. If specific resistance of its material is 3 10–7 ohm-m,
then the resistance between it’s opposite rectangular faces is
(a) 3 10–9 ohm (b) 3 10–7 ohm (c) 3 10–5 ohm (d) 3 10–3 ohm
10 2
1 cm
20 Current Electricity
Note : In the above question for calculating equivalent resistance between two opposite square
faces.
1
l = 100 cm = 1 m, A = 1 cm2 = 10–4 m2, so resistance R = 3 10–7 = 3 10–3
10 4
3. Two rods A and B of same material and length have their electric resistances are in ratio 1 : 2. When both the rods
are dipped in water, the correct statement will be [RPMT 1997]
(a) A has more loss of weight (b) B has more loss of weight
(c) Both have same loss of weight (d) Loss of weight will be in the ratio 1 : 2
L R A A R
Solution: (a) R 1 2 (, L constant) 1 2 2
A R2 A1 A2 R1
The V-i graph for a conductor makes an angle with V-axis. Here V denotes the voltage and i denotes
current. The resistance of conductor is given by
(a) sin (b) cos (c) tan (d)
cot
Solution: (d) At an instant approach the student will choose tan will be the right answer. But it is to be seen
here the curve makes the angle with the V-axis. So it makes an angle (90 – ) with the i-axis. So
resistance = slope = tan (90 – ) = cot.
V
i
Colour Coding of Resistance.
The resistance, having high values are used in different electrical and electronic circuits. They are
generally made up of carbon, like 1 k, 2 k, 5 k etc. To know the value of resistance colour code is used.
These code are printed in form of set of rings or strips. By reading the values of colour bands, we can
estimate the value of resistance.
The carbon resistance has normally four coloured rings or strips say A, B, C and D as shown in
following figure.
A B C D
Colour band A and B indicate the first two significant figures of resistance in ohm, while the C band
gives the decimal multiplier i.e. the number of zeros that follows the two significant figures A and B.
Last band (D band) indicates the tolerance in percent about the indicated value or in other ward it
represents the percentage accuracy of the indicated value.
The tolerance in the case of gold is 5% and in silver is 10%. If only three bands are marked on
carbon resistance, then it indicate a tolerance of 20%.
genius PHYSICS Pradeep Kshetrapal
Current Electricity 21
The following table gives the colour code for carbon resistance.
Letters as an aid to Colour Figure Multiplier Colour Tolerance
memory
(A, B) (C) (D)
O Orange 3 103
Y Yellow 4 104
G Green 5 105
B Blue 6 106
V Violet 7 107
G Grey 8 108
W White 9 109
Note : To remember the sequence of colour code following sentence should kept in memory.
Grouping of Resistance.
Series Parallel
(1) R1 R2 R3 (1) i1
R1
i2
V1 V2 V3
i3 R2
i
i
R3
+ –
V V
(2) Same current flows through each resistance but (2) Same potential difference appeared across each
potential difference distributes in the ratio of resistance but current distributes in the reverse ratio
resistance i.e. V R 1
of their resistance i.e. i
Power consumed are in the ratio of their resistance R
i.e. P R P1 : P2 : P3 R1 : R 2 : R 3 Power consumed are in the reverse ratio of
1 1 1 1
resistance i.e. P P1 : P2 : P3 : :
R R1 R 2 R 3
22 Current Electricity
i2 R2
+ –
V
R1 R2 R2 R1
V1 V and V2 V i1 i and i2 i
R1 R 2 R1 R 2 R1 R 2 R1 R 2
(6) If n identical resistance are connected in series (6) In n identical resistance are connected in parallel
V R i
R eq nR and p.d. across each resistance V ' R eq and current through each resistance i'
n n n
Note : In case of resistances in series, if one resistance gets open, the current in the whole circuit
become zero and the circuit stops working. Which don’t happen in case of parallel gouging.
Decoration of lightning in festivals is an example of series grouping whereas all household
appliances connected in parallel grouping.
Using n conductors of equal resistance, the number of possible combinations is 2n – 1.
If the resistance of n conductors are totally different, then the number of possible combinations will be 2n.
R R
R R R R
2R R
2R 2R R R R
3R/2
R 2R R 2R R 2R RA R B A B
A R B A R B A R B
(2) Method of equipotential points : This method is based on identifying the points of same
potential and joining them. The basic rule to identify the points of same potential is the symmetry of the
network.
(i) In a given network there may be two axes of symmetry.
(a) Parallel axis of symmetry, that is, along the direction of current flow.
genius PHYSICS Pradeep Kshetrapal
Current Electricity 23
(b) Perpendicular axis of symmetry, that is perpendicular to the direction of flow of current.
For example in the network shown below the axis AA is the parallel axis of symmetry, and the axis BB
is the perpendicular axis of symmetry.
B2
R R
6 7
R R R
A R A
1 O R 3
R R R
5 8
R R
B4
(ii) Points lying on the perpendicular axis of symmetry may have same potential. In the given network,
point 2, 0 and 4 are at the same potential.
(iii) Points lying on the parallel axis of symmetry can never have same potential.
(iv) The network can be folded about the parallel axis of symmetry, and the overlapping nodes have
same potential. Thus as shown in figure, the following points have same potential
(a) 5 and 6 (b) 2, 0 and 4 (c) 7 and 8
2,
R 4 R
R/2 R/2 3R/
R R
5, 7, 1 2 3
R 6 8 R R/2 R/2
R R/2 R/2
R R
R R R 1 3
1 O 3
Note : Above network may be split up into two equal parts about the parallel axis of symmetry as
R'
shown in figure each part has a resistance R, then the equivalent resistance of the network will be R .
2
2
R R
R = 3R
R R
1 3
1 R R 3
A A
(1) Equivalent resistance between points A and B in an unbalanced Wheatstone’s bridge as shown in the
diagram.
(i) P Q (ii) P Q
G G
A B A B
R S Q P
PQ(R S ) (P Q)RS G(P Q)(R S ) 2 PQ G(P Q)
R AB R AB
G(P Q R S ) (P R)(Q S ) 2G P Q
(2) A cube each side have resistance R then equivalent resistance in different situations
5 H G
(i) Between E and C i.e. across the diagonal of the cube R EC R
6
E
7 F
(ii) Between A and B i.e. across one side of the cube R AB R
12 D
C
A B
genius PHYSICS Pradeep Kshetrapal
24 Current Electricity
3
(iii) Between A and C i.e. across the diagonal of one face of the cube R AC R
4
R3 R3 R3 R3 R2 R2
R2 R2
B B
R2 R2 R2 R2
R AB
1 1
(R 1 R 2 ) (R 1 R 2 ) 2 4 R 3 (R 1 R 2 ) 1/2
R AB
1 R
R1 1 1 4 2
2 2 2
R1
Concepts
If n identical resistances are first connected in series and then in parallel, the ratio of the equivalent resistance is given by
Rp n2
.
Rs 1
1
If equivalent resistance of R1 and R2 in series and parallel be Rs and Rp respectively then R1 R s R s 4 R s R p
2
2
1
and R 2 R s R s 4 R s R p
2
.
2
If a wire of resistance R, cut in n equal parts and then these parts are collected to form a bundle then equivalent resistance
R
of combination will be 2 .
n
Example
s
Example: 27 In the figure a carbon resistor has band of different colours on its body. The resistance of the following
body is
(a) 2.2 k Red Silver
(b) 3.3 k
(c) 5.6 k
(d) 9.1 kl White Brow
n
,
Solution : (d) R = 91 102 10% 9.1 k
Example: 28 What is the resistance of a carbon resistance which has bands of colours brown, black and brown [DCE 1999]
(a) 100 (b) 1000 (c) 10 (d) 1
Solution : (a) R = 10 101 20% 100
Example: 29 In the following circuit reading of voltmeter V is [MP PET 2003]
(a) 12 V 4 16
(b) 8 V
V
(c) 20 V 2A
(d) 16 V 16 4
1A
and p.d. between X and Z is VXZ = VX – VZ = 1 16 = 16 V …. (ii) 2A
V
X
1A
16 Z 4
genius PHYSICS Pradeep Kshetrapal
Current Electricity 25
On solving equations (i) and (ii) we get potential difference between Y and Z i.e., reading of voltmeter
is VY VZ 12 V
Example: 30 An electric cable contains a single copper wire of radius 9 mm. It’s resistance is 5 . This cable is
replaced by six insulated copper wires, each of radius 3 mm. The resultant resistance of cable will be [CPMT 198
l l
R ….. (i)
(9 10 3 ) 2
Finally : Resistance of each insulated copper wire is 9 mm
l
R' l
(3 10 3 ) 2
Hence equivalent resistance of cable
R' 1 l
R eq ….. (ii)
6 6 (3 10 3 ) 2
R1 R 2 R (R R 2 ) 2
Solution : (d) Series resistance R S R1 R 2 and parallel resistance R P S 1 n
R1 R 2 RP R1 R 2
R1 R 2 R 12 R 22 R1 R2
n n n
R1 R 2 R1 R 2 R1 R 2 R2 R1
Example: 32 Five resistances are combined according to the figure. The equivalent resistance between the point X
and Y will be
10
(a) 10
(b) 22 20
X 10 10 Y
(c) 20
10
(d) 50 A
10
Solution : (a) The equivalent circuit of above can be drawn as Y
C D
10 10 10
genius PHYSICS Pradeep Kshetrapal
26 Current Electricity
(c) 30
(d) 40
Solution : (c) The equivalent circuit of above fig between A and D can be drawn as
10 10 Balanced wheatstone
10
A bridge A Series
10 10 10
D D
10 10
10
So R eq 10 10 10 30
Example: 34 In the network shown in the figure each of resistance is equal to 2. The resistance between A and B is
C [CBSE PMT 1995]
(a) 1
(b) 2
O A
B
(c) 3
(d) 4 D E
Solution : (b) Taking the portion COD is figure to outside the triangle (left), the above circuit will be now as
resistance of each is 2 the circuit will behaves as a balanced wheatstone bridge and no current flows
through CD. Hence RAB = 2
A C
O
D E
B
Example: 35 Seven resistances are connected as shown in figure. The equivalent resistance between A and B is [MP PET 2000]
(a) 3 10
3
(b) 4 A 10 B
(c) 4.5 5 8 6 6
(d) 5
Solution : (b)
10 Parallel (10||10) = 5
A 10 3 B A 5 3 B
5 8 6 6 5 8 3 P R
Q S
Parallel (6||6) = 3
(a) Infinite 1 1 1
A
(b) 2
1 1 1
B
genius PHYSICS Pradeep Kshetrapal
Current Electricity 27
1 5
(c)
2
(d) Zero
Solution : (c) Suppose the effective resistance between A and B is Req. Since the network consists of infinite cell. If
we exclude one cell from the chain, remaining network have infinite cells i.e. effective resistance
between C and D will also Req R R
A C
R R eq 1
So now R eq R o (R| | R eq ) R R eq [1 5 ] R Req
R R eq 2
B D
Example: 37 Four resistances 10 , 5 , 7 and 3 are connected so that they form the sides of a rectangle AB,
BC, CD and DA respectively. Another resistance of 10 is connected across the diagonal AC. The
equivalent resistance between A & B is
(a) 2 (b) 5 (c) 7 (d) 10
Solution : (b)
Series Parallel
7 (7 S 3) = 10 (10 ||10) = 5 Series
D C C C
10 (5 S 5) = 10
10 10 5
3 5 5 5
10 10 10
A B A B A B
10
10
10 10 A B
So R eq 5
10 10
Example: 38 The equivalent resistance between A and B in the circuit will be r
5 6 7 8 r
(a) r (b) r (c) r (d) r r
4 5 6 7 r r
A B
Solution : (d) In the circuit, by means of symmetry the point C is at zero potential. So the equivalent
r C circuit
r can be
drawn as
Series
r (r S r) = r Paralle
2r l
r r r r 2r r
r
A B A r r B
r r
Series
2 Series
8
rr r r
3 3
2
r
r 3 r
8r 8
R eq | | 2r r
3 7
A 2r B A 2r B
Example: 39 In the given figure, equivalent resistance between A and B will be [CBSE PMT 2000]
14 3 3 4
(a) (b)
3 14 A B
7
9 14
(c) (d) 6 8
14 9
genius PHYSICS Pradeep Kshetrapal
28 Current Electricity
P R
Solution : (a) Given Wheatstone bridge is balanced because . Hence the circuit can be redrawn as follows
Q S
14
Parallel R eq
3 4 Series 3+4=7 7 3
A B A B
Series 6 + 8 = 14
6 8 14
Example: 40 In the combination of resistances shown in the figure the potential difference between B and D is zero,
when unknown resistance (x) is B
4
x
(a) 4 12
(b) 2 A C
1
1
(c) 3 3 1
D
(d) The emf of the cell is required
Solution : (b) The potential difference across B, D will be zero, when the circuit will act as a balanced wheatstone
P R 12 4 1 3
bridge and x = 2
Q S x 1/2
Example: 41 A current of 2 A flows in a system of conductors as shown. The potential difference (VA – VB) will be
A [CPMT 1975, 76]
(a) + 2V 2 3
2A
(b) + 1V D C
(c) – 1 V 3 2
B
(d) – 2 V
Solution : (b) In the given circuit 2A current divides equally at junction D along the paths DAC and DBC (each path carry 1A
current).
Potential difference between D and A, VD – VA = 1 2 = 2 volt …. (i)
Potential difference between D and B, VD – VB = 1 3 = 3 volt ….. (ii)
On solving (i) and (ii) VA – VB = + 1 volt
Example: 42 Three resistances each of 4 are connected in the form of an equilateral triangle. The effective
resistance between two corners is
3 8
(a) 8 (b) 12 (c) (d)
8 3
Solution : (d) Series 4 + 4 = 8
4 4 8
On Solving further we get equivalent resistance is
3
4
Example: 43 If each resistance in the figure is of 9 then reading of ammeter is [RPMT 2000]
+
(a) 5 A (b) 8 A 9V
–
(c) 2 A (d) 9 A A
9
Solution : (a) Main current through the battery i 9 A . Current through each resistance will be 1A and only 5
1
resistances on the right side of ammeter contributes for passing current through the ammeter. So
reading of ammeter will be 5A.
Example: 44 A wire has resistance 12 . It is bent in the form of a circle. The effective resistance between the two
points on any diameter is equal to
(a) 12 (b) 6 (c) 3 (d) 24
Solution : (c) Equivalent resistance of the following circuit will be 6
6
genius PHYSICS Pradeep Kshetrapal
Current Electricity 29
6
R eq 3
2
Example: 45 A wire of resistance 0.5 m–1 is bent into a circle of radius 1 m. The same wire is connected across a
diameter AB as shown in fig. The equivalent resistance is
(a) ohm
A B
(b) ( + 2) ohm
(c) / ( + 4) ohm i i
(d) ( + 1) ohm
Solution : (c) Resistance of upper semicircle = Resistance of lower semicircle
0.5
= 0.5 (R) = 0.5
Resistance of wire AB = 0.5 2 = 1
A 1 B
Hence equivalent resistance between A and B
1 1 1 1 i i
R AB
R AB 0 .5 1 0 .5 ( 4 ) 0.5
Example: 46 A wire of resistor R is bent into a circular ring of radius r. Equivalent resistance between two points X
and Y on its circumference, when angle XOY is , can be given by
R R X
(a) (2 ) (b) (2 )
4 2 2
W O Z
4
(c) R (2 – ) (d) (2 )
R Y
R R l R R
Solution : (a) Here R XWY (r ) and R XZY r(2 ) (2 )
2r 2 r 2r 2
R R
(2 )
R
2 2
R XWY R XZY
R eq (2 )
R XWY R XZY R R (2 ) 4 2
2 2
Example: 47 If in the given figure i = 0.25 amp, then the value R will be [RPET 2000]
(a) 48 i R 60
(b) 12 20
12 V
(c) 120 10
(d) 42
V 12
Solution : (d) i = 0.25 amp V = 12 V R eq 48 Parallel
i 0 . 25 i R 60
Now from the circuit R eq R (60 | | 20 | | 10) 20
12 V
=R+6 10
R = Req – 6 = 48 – 6 = 42
Example: 48 Two uniform wires A and B are of the same metal and have equal masses. The radius of wire A is twice
that of wire B. The total resistance of A and B when connected in parallel is [MNR 1994]
(a) 4 when the resistance of wire A is 4.25 (b) 5 when the resistance of wire A is 4
(c) 4 when the resistance of wire B is 4.25 (d) 5 when the resistance of wire B is 4
Solution : (a) Density and masses of wire are same so their volumes are same i.e. A1l1 = A2l2
genius PHYSICS Pradeep Kshetrapal
30 Current Electricity
2 4
R l A A r
Ratio of resistances of wires A and B A 1 2 2 2
RB l 2 A1 A1 r1
RA 1
Since r1 = 2r2 so RB = 16 RA
RB 16
R A RB 16 R A
Resistance RA and RB are connected in parallel so equivalent resistance R , By
R A RB 17
checking correctness of equivalent resistance from options, only option (a) is correct.
Tricky Example: 5
The effective resistance between point P and Q of the electrical circuit shown in the figure is
2R 2R [IIT-JEE 1991]
r 2R r
P Q
2R
2R 2R
2 Rr 8 R ( R r) 5R
(a) (b) (c) 2r + 4R (d) 2r
Rr 3R r 2
Solution : (a) The points A, O, B are at same potential. So the figure can be redrawn as follows
A Series
2R 2R
r Series
r
P
O
Q P Q
Series
2R 2R
B
(II)
(I)
4R
2 Rr 2r
R eq 4 R | | 2r | | 4 R P Q
Rr
4R
Tricky Example: 6
In the following circuit if key K is pressed then the galvanometer reading becomes half. The
resistance of galvanometer is + –
R G
K
S = 40
(a) 20 (b) 30 (c) 40 (d) 50
Solution : (c) Galvanometer reading becomes half means current distributes equally between galvanometer and
resistance of 40 . Hence galvanometer resistance must be 40 .
Cell.
The device which converts chemical energy into electrical energy is known as electric cell.
+
Anode Cathod A
–
e
+ –
–
+
Electrolyte
genius PHYSICS Pradeep Kshetrapal
Current Electricity 31
(1) A cell neither creates nor destroys charge but maintains the flow of charge present at various parts
of the circuit by supplying energy needed for their organised motion.
(2) Cell is a source of constant emf but not constant current.
(3) Mainly cells are of two types :
(i) Primary cell : Cannot be recharged
(ii) Secondary cell : Can be recharged
(4) The direction of flow of current inside the cell is from negative to positive electrode while outside
the cell is form positive to negative electrode.
(5) A cell is said to be ideal, if it has zero internal resistance.
(6) Emf of cell (E) : The energy given by the cell in the flow of unit charge in the whole circuit
W
(including the cell) is called it’s electromotive force (emf) i.e. emf of cell E , It’s unit is volt or
q
The potential difference across the terminals of a cell when it is not given any current is called it’s emf.
(7) Potential difference (V) : The energy given by the cell in the flow of R
unit charge in a specific part of electrical circuit (external part) is called potential
difference. It’s unit is also volt or i
E Pmax =
(v) Internal resistance of the cell r 1 R E2/4r
V P
2
E
2
V
(vi) Power dissipated in external resistance (load) P Vi i 2 R .R R=r
R R r
R
E2
Power delivered will be maximum when R r so Pmax .
4r
This statement in generalised from is called “maximum power transfer theorem”.
genius PHYSICS Pradeep Kshetrapal
32 Current Electricity
(vii) Short trick to calculate E and r : In the closed circuit of a cell having emf E and internal
resistance r. If external resistance changes from R1 to R2 then current changes from i1 to i2 and potential
difference changes from V1 to V2. By using following relations we can find the value of E and r.
i1 i 2 i R i1 R1 V2 V1
E ( R 1 R 2 ) r 2 2
i 2 i1 i1 i 2 i1 i 2
Note : When the cell is charging i.e. current+isVgiven to the cell then E = V – ir and E < V.
–
i
E, r
E, r E, r
(i) Current through the circuit i = 0 (i) Maximum current (called short circuit current)
E
flows momentarily isc
r
(ii) Potential difference between A and B, VAB = E (ii) Potential difference V = 0
(iii) Potential difference between C and D, VCD = 0
imax =E/r ; V = 0 i
Concepts
It is a common misconception that “current in the circuit will be maximum when power consumed by the load is
maximum.”
Actually current i E /(R r) is maximum (= E/r) when R = min = 0 with PL (E / r)2 0 0 min . while power
consumed by the load E2R/(R + r)2 is maximum (= E2/4r) when R = r and i (E / 2r) max( E / r).
Emf is independent of the resistance of the circuit and depends upon the nature of electrolyte of the cell while
potential difference depends upon the resistance between the two points of the circuit and current flowing
through the circuit.
Whenever a cell or battery is present in a branch there must be some resistance (internal or external or both)
present in that branch. In practical situation it always happen because we can never have an ideal cell or battery
with zero resistance.
Example
s
genius PHYSICS Pradeep Kshetrapal
Current Electricity 33
Example: 49 A new flashlight cell of emf 1.5 volts gives a current of 15 amps, when connected directly to an
ammeter of resistance 0.04 . The internal resistance of cell is
(a) 0.04 (b) 0.06 (c) 0.10 (d) 10
E 1 .5
Solution : (b) By using i 15 r = 0.06
Rr 0 . 04 r
Example: 50 For a cell, the terminal potential difference is 2.2 V when the circuit is open and reduces to 1.8 V,
when the cell is connected across a resistance, R = 5. The internal resistance of the cell is
10 9 11 5
(a) (b) (c) (d)
9 10 9 9
Solution : (a) In open circuit, E = V = 2.2 V, In close circuit, V = 1.8 V, R = 5
E 2.2 10
So internal resistance, r 1 R 1 5 r
V 1.8 9
Example: 51 The internal resistance of a cell of emf 2V is 0.1 . It’s connected to a resistance of 3.9 . The voltage
across the cell will be [CBSE PMT 1999; AFMC 1999; MP PET 1993; CPMT 1990]
(a) 0.5 volt (b) 1.9 volt (c) 1.95 volt (d) 2 volt
E 2
Solution : (c) By using r 1 R 0 .1 1 3 .9 V 1.95 volt
V V
Example: 52 When the resistance of 2 is connected across the terminal of the cell, the current is 0.5 amp. When
the resistance is increased to 5 , the current is 0.25 amp. The emf of the cell is
(a) 1.0 volt (b) 1.5 volt (c) 2.0 volt (d) 2.5 volt
i1 i2 0.5 0.25
Solution : (b) By using E (R1 R 2 ) (2 5) 1.5 volt
(i2 i1 ) (0.25 0.5)
Example: 53 A primary cell has an emf of 1.5 volts, when short-circuited it gives a current of 3 amperes. The
internal resistance of the cell is
(a) 4.5 ohm (b) 2 ohm (c) 0.5 ohm (d) 1/4.5 ohm
E 1 .5
Solution : (c) isc 3 r = 0.5
r r
Example: 54 A battery of internal resistance 4 is connected to the network of resistances as shown. In order to
give the maximum power to the network, the value of R (in ) should be [IIT-JEE 1995]
(a) 4/9 R R
(b) 8/9 R 6R R
E
R
(c) 2
4R
(d) 18
Solution : (c) The equivalent circuit becomes a balanced wheatstone bridge
R 2R
R 2R 3R
6R
6R 2R 4R
2R 4R 6R
4R 4R 4
For maximum power transfer, external resistance should be equal to internal resistance of source
genius PHYSICS Pradeep Kshetrapal
34 Current Electricity
(R 2 R)(2 R 4 R) 3R 6R
4 i.e. 4 or R = 2
(R 2 R) (2 R 4 R) 3R 6R
Example: 55 A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The emf of the cell needed to
make the bulb glow at full intensity is
4.5 W, 1.5
(a) 4.5 V V
1
(b) 1.5 V
(c) 2.67 V
E (r =
(d) 13.5 V 2.67)
Solution : (d) When bulb glows with full intensity, potential difference across it is 1.5 V. So current through the bulb
and resistance of 1 are 3 A and 1.5 A respectively. So main current from the cell i = 3 + 1.5 = 4.5 A. By
using E V iR E = 1.5 + 4.5 2.67 = 13.5 V.
Tricky Example: 7
Potential difference across the terminals of the battery shown in figure is (r = internal resistance of
battery)
10 V r =1
(a) 8 V (b) 10 V
(c) 6 V (d) Zero
4
Solution : (d) Battery is short circuited so potential difference is zero.
Grouping of cell.
Group of cell is called a battery.
(1) Series grouping : In series grouping anode of one cell is connected to cathode of other cell and so
on.
(i) n identical cells are connected in series
(a) Equivalent emf of the combination E eq nE
E2
(g) Condition for maximum power R nr and Pmax n
4 r
(h) This type of combination is used when nr << R.
(ii) If non-identical cell are connected in series
Cells are connected in right order Cells are wrongly connected
E1, r1 E2, r2 E1, r1 E2, r2 (E1 > E2)
1 2
i R i
R
genius PHYSICS Pradeep Kshetrapal
Current Electricity 35
(2) Parallel grouping : In parallel grouping all anodes are connected at one point and all cathode
are connected together at other point. E, r
(i) If n identical cells are connected in parallel E, r
(a) Equivalent emf Eeq = E
E, r
(b) Equivalent internal resistance R eq r / n i
R
E
(c) Main current i
R r/n
(d) P.d. across external resistance = p.d. across each cell = V = iR
2
i E
(e) Current from each cell i' (f) Power dissipated in the circuit P .R
n R r/n
E2
(g) Condition for max power R r / n and Pmax n (h) This type of combination is used when nr
4r
>> R
(ii) If non-identical cells are connected in parallel : If cells are connected with right polarity as shown
below then
E1 r2 E 2 r1
(a) Equivalent emf E eq i1 E1, r1
r1 r2
E eq
(b) Main current i
r R eq i2
i E2, r2
E iR E iR R
(c) Current from each cell i1 1 and i2 2
r1 r2
Note : In this combination if cell’s are connected with reversed polarity as shown in figure
i1 E1,r1
then :
i i2 E2, r2
E r E 2 r1
Equivalent emf E eq 1 2
r1 r2 R
(3) Mixed Grouping : If n identical cell’s are connected in a row and such m row’s are connected in
parallel as shown.
(i) Equivalent emf of the combination E eq nE
E, E, E,
nr 1 r r r
(ii) Equivalent internal resistance of the combination req 1 2 n
m 2
i
m
genius PHYSICS Pradeep Kshetrapal
36 Current Electricity
nE mnE
(iii) Main current flowing through the load i
nr mR nr
R
m
(iv) Potential difference across load V iR
V
(v) Potential difference across each cell V '
n
i
(vi) Current from each cell i '
n
nr E2
(vii) Condition for maximum power R and Pmax (mn )
m 4r
(viii) Total number of cell = mn
Concepts
In series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive. If
dissimilar plates of cells are connected together their emf’s are added to each other while if their similar plates are
connected together their emf’s are subtractive.
E1 E2 E1 E2
Eeq E1 E2 & req r1 r2 Eeq E1 E 2 (E1 E 2 ) & req r1 r2
In series grouping of identical cells. If one cell is wrongly connected then it will cancel out the effect of two cells e.g. If in
the combination of n identical cells (each having emf E and internal resistance r) if x cell are wrongly connected then
equivalent emf Eeq (n 2 x ) E and equivalent internal resistance req nr .
E E
E eq E E eq 0
E E
When two cell’s of different emf and no internal resistance are connected in parallel then equivalent emf is
indeterminate, note that connecting a wire with a cell but with no resistance is equivalent to short circuiting. Therefore
the total current that will be flowing will be infinity. R
E1
E2
Example
s
Example: 56 A group of N cells whose emf varies directly with the internal resistance as per the equation EN = 1.5 rN
are connected as shown in the following figure. The current i in the circuit is [KCET 2003]
Current Electricity 37
Example: 57 Two batteries A and B each of emf 2 volt are connected in series to external resistance R = 1 .
Internal resistance of A is 1.9 and that of B is 0.9 , what is the potential difference between the
terminals of battery A [MP PET 2001]
A B
(a) 2 V
(b) 3.8 V
(c) 0
R
(d) None of these
E1 E 2 22 4 4
Solution : (c) i . Hence V A E A irA 2 1 .9 0
R r1 r2 1 1.9 0.9 3.8 3 .8
Example: 58 In a mixed grouping of identical cells 5 rows are connected in parallel by each row contains 10 cell.
This combination send a current i through an external resistance of 20 . If the emf and internal
resistance of each cell is 1.5 volt and 1 respectively then the value of i is
(a) 0.14 (b) 0.25 (c) 0.75 (d) 0.68
Solution : (d) No. of cells in a row n = 10; No. of such rows m = 5
nE 10 1 .5 15
i = 0.68 amp
nr 10 1 22
R 20
m 5
Example: 59 To get maximum current in a resistance of 3 one can use n rows of m cells connected in parallel. If
the total no. of cells is 24 and the internal resistance of a cell is 0.5 then
(a) m = 12, n = 2 (b) m = 8, n = 4 (c) m = 2, n = 12 (d) m = 6, n = 4
mr
Solution : (a) In this question R = 3, mn = 24, r = 0.5 and R . On putting the values we get n = 2 and m =
n
12.
Example: 60 100 cells each of emf 5V and internal resistance 1 are to be arranged so as to produce maximum
current in a 25 resistance. Each row contains equal number of cells. The number of rows should be [MP PMT 1
(a) 2 (b) 4 (c) 5 (d) 100
Solution : (a) Total no. of cells, = mn = 100 …….. (i)
nr n 1
Current will be maximum when R ; 25 n = 25 m …….. (ii)
m m
From equation (i) and (ii) n = 50 and m = 2
Example: 61 In the adjoining circuit, the battery E1 has as emf of 12 volt and zero internal resistance, while the
battery E has an emf of 2 volt. If the galvanometer reads zero, then the value of resistance X ohm is [NCERT 199
500 O
(a) 10 (b) 100 A G B
Solution : (b) For zero deflection in galvanometer the potential different across XDshould be E
P = 2V C
12 X
In this condition 2
500 X
X = 100
Example: 62 In the circuit shown here E1 = E2 = E3 = 2V and R1 = R2 = 4 . The current flowing between point A
and B through battery E2 is R1
E1
(a) Zero
E2
A B
(b) 2 A from A to B
E3
R2
genius PHYSICS Pradeep Kshetrapal
38 Current Electricity
(c) 2 A from B to A
(d) None of these
Solution : (b) The equivalent circuit can be drawn as since E1 & E3 are parallely connected
2V R = (R1 ||R2) = 2
22
So current i 2 Amp from A to B.
2 2V
A B
Example: 63 The magnitude and direction of the current in the circuit shown will be [CPMT 1986, 88]
7 7 a 1 e 2
(a) A from a to b through e (b) A from b and a through e b
3 3 10 V 4V
(c) 1.0 A from b to a through e (d) 1.0 A from a to b through e
10 4
Solution : (d) Current i 1 A from a to b via e d 3 c
3 2 1
Example: 64 Figure represents a part of the closed circuit. The potential difference between points A and B (VA –
VB) is
2A 2 1
(a) + 9 V (b) – 9 V
A 3V B
(c) + 3 V (d) + 6 V
Solution : (a) The given part of a closed circuit can be redrawn as follows. It should be remember that product of
current and resistance can be treated as an imaginary cell having emf = iR.
4V 3V 2V + 9V –
7 Hence VA – VB = +9 V
A B A B
Example: 65 In the circuit shown below the cells E1 and E2 have emf’s 4 V and 8 V and internal resistance 0.5 ohm
and 1 ohm respectively. Then the potential difference across cell E1 and E2 will be
(a) 3.75 V, 7.5 V E1 E2
(b) 4.25 V, 7.5 V
4.5 3
(c) 3.75 V, 3.5 V
6
(d) 4.25 V, 4.25 V
36
Solution : (b) In the given circuit diagram external resistance R 4 . 5 6 . 5 . Hence main current through
36
E 2 E1 84 1
the circuit i amp.
R req 6 .5 0 .5 0 .5 2
1
Cell 1 is charging so from it’s emf equation E1 = V1 – ir1 4 V1 0 . 5 V1 = 4.25 volt
2
1
Cell 2 is discharging so from it’s emf equation E2 = V2 + ir2 8 V2 1 V2 = 7.5 volt
2
Example: 66 A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to
this current, the temperature of the wire is raised by T in time t. A number N of similar cells is now
connected in series with a wire of the same material and cross-section but of length 2L. The
temperature of wire is raised by same amount T in the same time t. The value of N is
(a) 4 (b) 6 (c) 8 (d) 9
Tricky Example: 8
n identical cells, each of emf E and internal resistance r, are joined in series to form a closed
genius PHYSICS Pradeep Kshetrapal
Current Electricity 39
Kirchoff’s Laws.
(1) Kirchoff’s first law : This law is also known as junction rule or current law (KCL). According to it
the algebraic sum of currents meeting at a junction is zero i.e. i = 0.
In a circuit, at any junction the sum of the currents entering the junction
i1
must equal the sum of the currents leaving the junction. i1 i3 i2 i4
i4
Here it is worthy to note that :
(i) If a current comes out to be negative, actual direction of current at the i2 i3
Note : This law is also applicable to a capacitor through the concept of displacement current
treating the resistance of capacitor to be zero during charging or discharging and infinite in steady
state as shown in figure.
(2) Kirchoff’s second law : This law is also known as loop rule or voltage law (KVL) and according
to it “the algebraic sum of the changes in potential in complete traversal of a mesh (closed loop) is zero”, i.e.
V = 0
R2
B C
e.g. In the following closed loop. R1 i1 i2 + E1
–
A D
i4 i3
40 Current Electricity
(3) Sign convention for the application of Kirchoff’s law : For the application of Kirchoff’s
laws following sign convention are to be considered
(i) The change in potential in traversing a resistance in the direction of current is – iR while in the
opposite direction +iR
A R B A R B
i i
– iR + iR
(ii) The change in potential in traversing an emf source from negative to positive terminal is +E while
in the opposite direction – E irrespective of the direction of current in the circuit.
A E B A E B
i i
–E +E
(iii) The change in potential in traversing a capacitor from the negative terminal to the positive
q q
terminal is while in opposite direction .
C C
C C
A B A – + B
– +
q q
q q
C C
di
(iv) The change in voltage in traversing an inductor in the direction of current is L while in
dt
di
opposite direction it is L . A i
L B A i
L B
dt
di di
L L
dt dt
Note : The number of loops must be selected so that every element of the circuit must be included
in at least one of the loops.
While traversing through a capacitor or battery we do not consider the direction of current.
genius PHYSICS Pradeep Kshetrapal
Current Electricity 41
While considering the voltage drop or gain across as inductor we always assume current to be in
increasing function.
(5) Determination of equivalent resistance by Kirchoff’s method : This method is useful
when we are not able to identify any two resistances in series or in parallel. It is based on the two
Kirchhoff’s laws. The method may be described in the following guideline.
(i) Assume an imaginary battery of emf E connected between the two terminals across which we have
to calculate the equivalent resistance.
(ii) Assume some value of current, say i, coming out of the battery and distribute it among each branch
by applying Kirchhoff’s current law.
(iii) Apply Kirchhoff’s voltage law to formulate as many equations as there are unknowns. It should be
noted that at least one of the equations must include the assumed battery.
E
(iv) Solve the equations to determine ratio which is the equivalent resistance of the network.
i
e.g. Suppose in the following network of 12 identical resistances, equivalent resistance between point A
and C is to be calculated.
B
R R
E F
R R R
R
A C
R
R R R
H G
R R
D
B B
R R R R
i
E
R
F
R Ei F
R R i R R
A R C 2i R i 2i
A 2i C
R i R
R R R R i R 2i R
H G Hi G
R R R i R 4i
D 4i
D
E E
Concepts
Using Kirchoff’s law while dividing the current having a junction through different arms of a network, it will be same
through different arms of same resistance if the end points of these arms are equilocated w.r.t. exit point for current in
network and will be different through different arms if the end point of these arms are not equilocated w.r.t. exit point
for current of the network.
e.g. In the following figure the current going in arms AB, AD and AL will be same because the location of end points B, D
B i
C
2i
6i i i
A D
i i
i
genius PHYSICS Pradeep Kshetrapal
42 Current Electricity
and L of these arms are symmetrically located w.r.t. exit point N of the network.
Example
s
Example: 67 In the following circuit E1 = 4V, R1 = 2 [MP PET 2003]
E1 R1
E2 = 6V, R2 = 2 and R3 = 4. The current i1 is
i1
(a) 1.6 A R2
(b) 1.8 A
i2 R3
(c) 2.25 A
E2
(d) 1 A
4V 2
Solution : (b) For loop (1) 2i1 2(i1 i2 ) 4 0 2i1 i2 2 …… (i)
i1 1 i1
For loop (2) 4 i2 2(i1 i2 ) 6 0 3i2 i1 3 …… (ii) 2 (i1 – i2)
2 i2
After solving equation (i) and (ii) we get i1 1 .8 A and i2 1.6 A i2
6V 4
Example: 68 Determine the current in the following circuit
10 V 2
(a) 1 A
(b) 2.5 A
(c) 0.4 A
(d) 3 A 5V 3
Solution : (a) The current in the circuit are assumed as shown in the fig. 6 i1 B 3 i1 – i2
A C
Applying KVL along the loop ABDA, we get
– 6i1 – 3 i2 + 15 = 0 or 2i1 + i2 = 5 …… (i) 15 V 3 30 V
Applying KVL along the loop BCDB, we get i2
– 3(i1 – i2) – 30 + 3i2 = 0 or – i1 + 2i2 = 10 …… (ii)
i1 D
Solving equation (i) and (ii) for i2, we get i2 = 5 A
Example: 70 The figure shows a network of currents. The magnitude of current is shown here. The current i will be [MP
PMT 1995]
15 A
3A
(a) 3 A
(b) 13 A
(c) 23 A
i
(d) – 3 A 5A
genius PHYSICS Pradeep Kshetrapal
Current Electricity 43
Solution : (c) i 15 3 5 23 A
Example: 71 Consider the circuit shown in the figure. The current i3 is equal to [AMU 1995]
28 54
(a) 5 amp
(b) 3 amp 6V
i3
(c) – 3 amp
8V 12 V
(d) – 5/6 amp
Solution : (d) Suppose current through different paths of the circuit is as follows.
After applying KVL for loop (1) and loop (2) 28 54
1 6V
We get 28 i1 6 8 i1 A 1 2
2 i3
1
and 54 i2 6 12 i2 A 12 V
3 8V
5
Hence i3 i1 i2 A
6
Example: 72 A part of a circuit in steady state along with the current flowing in the branches, with value of each
resistance is shown in figure. What will be the energy stored in the capacitor C0
1A
3
(a) 6 10–4 J 3 A 5 D
2A i1
(b) 8 10–4 J 4F 1
i2 i3
B C
(c) 16 10–4 J 2A 1 2 4
3
1A
(d) Zero
Solution : (b) Applying Kirchhoff’s first law at junctions A and B respectively we have 2 + 1 – i1 = 0 i.e., i1 = 3A
and i2 + 1 – 2 – 0 = 0 i.e., i2 = 1A
Now applying Kirchhoff’s second law to the mesh ADCBA treating capacitor as a seat of emf V in open
circuit
1 1
So, energy stored in the capacitor U CV 2
(4 10 6 ) (20 ) 2 8 10 4 J
2 2
Tricky Example: 9
As the switch S is closed in the circuit shown in figure, current passed through it is
20 V 2 4 5V
A B
2
S
genius PHYSICS Pradeep Kshetrapal
44 Current Electricity
(2) Ammeter : It is a device used to measure current and is always connected in series with the
‘element’ through which current is to be measured.
(i) The reading of an ammeter is always lesser than actual current in the circuit.
(ii) Smaller the resistance of an ammeter more accurate will be its reading. An ammeter is said to be
ideal if its resistance r is zero.
(iii) Conversion of galvanometer into ammeter : A galvanometer may be converted into an
ammeter by connecting a low resistance (called shunt S) in parallel to the galvanometer G as shown in
figure.
S
GS
(a) Equivalent resistance of the combination i – ig
GS i
G
(b) G and S are parallel to each other hence both will have equal ig
potential difference i.e. ig G (i ig )S ; which gives Ammeter
ig
Required shunt S G
(i i g )
genius PHYSICS Pradeep Kshetrapal
Current Electricity 45
i
(c) To pass nth part of main current (i.e. ig ) through the galvanometer, required shunt
n
G
S .
(n 1)
(3) Voltmeter : It is a device used to measure potential difference and is always put in parallel with
the ‘circuit element’ across which potential difference is to be measured.
V
(i) The reading of a voltmeter is always lesser than true value.
(ii) Greater the resistance of voltmeter, more accurate will be its R
i
reading. A voltmeter is said to be ideal if its resistance is infinite, i.e., it draws
no current from the circuit element for its operation. +
V
–
+ –
(iii) Applications of wheatstone bridge : Meter bridge, post office V
box and Carey Foster bridge are instruments based on the principle of
wheatstone bridge and are used to measure unknown resistance.
(5) Meter bridge : In case of meter bridge, the resistance wire AC is 100 cm long. Varying the
position of tapping point B, bridge is balanced. If in balanced position of bridge AB = l, BC (100 – l) so that
Q (100 l) P R (100 l)
. Also S R S
P l Q S l R R.B.
P B Q
A C
l cm (100 – l) cm
genius PHYSICS Pradeep Kshetrapal
46 Current Electricity
Concepts
Wheatstone bridge is most sensitive if all the arms of bridge have equal resistances i.e. P = Q = R = S
If the temperature of the conductor placed in the right gap of metre bridge is increased, then the balancing length
decreases and the jockey moves towards left.
In Wheatstone bridge to avoid inductive effects the battery key should be pressed first and the galvanometer key
afterwards.
The measurement of resistance by Wheatstone bridge is not affected by the internal resistance of the cell.
Example
s
Example: 74 The scale of a galvanometer of resistance 100 contains 25 divisions. It gives a deflection of one
division on passing a current of 4 10–4 A. The resistance in ohms to be added to it, so that it may
become a voltmeter of range 2.5 volt is
(a) 100 (b) 150 (c) 250 (d) 300
Solution : (b) Current sensitivity of galvanometer = 4 10–4 Amp/div.
So full scale deflection current (ig) = Current sensitivity Total number of division = 4 10–4 25 =
10–2 A
To convert galvanometer in to voltmeter, resistance to be put in series is
V 2 .5
R G 100 150
ig 10 2
Example: 75 A galvanometer, having a resistance of 50 gives a full scale deflection for a current of 0.05 A. the
length in meter of a resistance wire of area of cross-section 2.97 10–2 cm2 that can be used to convert
the galvanometer into an ammeter which can read a maximum of 5A current is : (Specific resistance of
the wire = 5 10–7 m) [EAMCET 2003]
(a) 9 (b) 6 (c) 3 (d) 1.5
Solution : (c) Given G = 50 , ig = 0.05 Amp., i = 5A, A = 2.97 10–2 cm2 and = 5 10–7-m
i G G.ig l Gig Gig A
By using 1 S l on putting values l = 3 m.
ig S (i ig ) A (i ig ) (i ig )
Example: 76 100 mA current gives a full scale deflection in a galvanometer of resistance 2 . The resistance
connected with the galvanometer to convert it into a voltmeter of 5 V range is
[KCET 2002; UPSEAT 1998; MNR 1994 Similar to MP PMT 1999]
(a) 98 (b) 52 (c) 80 (d) 48
V 5
Solution : (d) R G 2 50 2 48 .
Ig 100 10 3
Example: 77 A milliammeter of range 10 mA has a coil of resistance 1 . To use it as voltmeter of range 10 volt, the
resistance that must be connected in series with it will be
(a) 999 (b) 99 (c) 1000 (d) None of these
V 10
Solution : (a) By using R G R 1 999
ig 10 10 3
Example: 78 In the following figure ammeter and voltmeter reads 2 amp and 120 volt respectively. Resistance of
voltmeter is
X 75 Y
(a) 100 A
V
genius PHYSICS Pradeep Kshetrapal
Current Electricity 47
(b) 200
(c) 300
(d) 400
75 R V
Solution : (c) Let resistance of voltmeter be RV. Equivalent resistance between X and Y is R XY
75 R V
75 R V
Reading of voltmeter = potential difference across X and Y = 120 = i RXY = 2 RV =
75 R V
300
Example: 79 In the circuit shown in figure, the voltmeter reading would be
1 2
(a) Zero
(b) 0.5 volt A V
(c) 1 volt
+ –
3V
(d) 2 volt
Solution : (a) Ammeter has no resistance so there will be no potential difference across it, hence reading of
voltmeter is zero.
Example: 80 Voltmeters V1 and V2 are connected in series across a d.c. line. V1reads 80 V and has a per volt
resistance of 200 , V2 has a total resistance of 32 k. The line voltage is
(a) 120 V (b) 160 V (c) 220 V (d) 240 V
Solution : (d) Resistance of voltmeter V1 is R1 = 200 80 = 16000 and resistance of voltmeter V2 is R2 = 32000
R'
By using relation V ' V ; where V = potential difference across any resistance R in a series
R eq
R1 R2
grouping.
V1 V2
So for voltmeter V1 potential difference across it is 80 V
R1 + –
80 .V V = 240 V
R1 R 2
V
Example: 81 The resistance of 1 A ammeter is 0.018 . To convert it into 10 A ammeter, the shunt resistance
required will be
(a) 0.18 (b) 0.0018 (c) 0.002 (d) 0.12
i 4 10 0 . 018
Solution : (c) By using 1 1 S = 0.002
ig S 1 S
Example: 82 In meter bridge the balancing length from left and when standard resistance of 1 is in right gas is
found to be 20 cm. The value of unknown resistance is [CBSE PMT 1999]
X 20 r
Solution: (a) The condition of wheatstone bridge gives , r- resistance of wire per cm, X- unknown
R 80 r X R = 1
resistance
20 1
X R 1 0 . 25
80 4
P = 20 r Q = 80 r
Example: 83 A galvanometer having a resistance of 8 is shunted by a wire of resistance 2 . If the total current is
20 cm 80 cm
1 amp, the part of it passing through the shunt will be [CBSE PMT 1998]
genius PHYSICS Pradeep Kshetrapal
48 Current Electricity
(a) 0.25 amp (b) 0.8 amp (c) 0.2 amp (d) 0.5 amp
Solution: (b) Fraction of current passing through the galvanometer
ig S ig 2
or 0.2
i S G i 28
So fraction of current passing through the shunt
is ig
1 1 0.2 0.8 amp
i i
Example: 84 A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a
shunt of resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of
resistance r connected across it. What is the maximum current which can be through this
galvanometer if no shunt is used
[MP PMT 1996]
(a) 0.01 A (b) 0.02 A (c) 0.03 A (d) 0.04 A
ig
Solution: (b) For ammeter, S G ig G (i ig )S
(i ig )
So ig G (0.03 ig )4 r …… (i) and ig G (0.06 ig )r …… (ii)
(0 .03 ig ) 4
Dividing equation (i) by (ii) 1 0 .06 ig 0 .12 4 ig
0 .06 ig
3ig = 0.06 ig = 0.02 A
Tricky Example: 10
The ammeter A reads 2 A and the voltmeter V reads 20 V. The value of resistance R is [JI
R
A
V
(a) Exactly 10 ohm (b) Less than 10 ohm
(c) More than 10 ohm (d) We cannot definitely say
20
Solution: (c) If current goes through the resistance R is i then iR = 20 volt R . Since i < 2A so R >
i
10.
Potentiometer.
Potentiometer is a device mainly used to measure emf of a given cell and to compare emf’s of cells. It is
also used to measure internal resistance of a given cell.
(1) Superiority of potentiometer over voltmeter : An ordinary voltmeter cannot measure the
emf accurately because it does draw some current to show the deflection. As per definition of emf, it is the
potential difference when a cell is in open circuit or no current through the cell. Therefore voltmeter can
only measure terminal voltage of a give n cell.
Potentiometer is based on no deflection method. When the potentiometer gives zero deflection, it does
not draw any current from the cell or the circuit i.e. potentiometer is effectively an ideal instrument of
infinite resistance for measuring the potential difference.
(2) Circuit diagram : Potentiometer consists of a long resistive wire AB of length L (about 6m to 10
m long) made up of mangnine or constantan. A battery of known voltage e and internal resistance r called
supplier battery or driver cell. Connection of these two forms primary circuit.
genius PHYSICS Pradeep Kshetrapal
Current Electricity 49
One terminal of another cell (whose emf E is to be measured) is connected at one end of the main
circuit and the other terminal at any point on the resistive wire through a galvanometer G. This forms the
secondary circuit. Other details are as follows
e, r K Rh
J = Jockey
Primary
K = Key circuit J
A B
R = Resistance of potentiometer wire, Secondary
circuit E
G
= Specific resistance of potentiometer wire.
Rh = Variable resistance which controls the current through the wire AB
(3) Points to be remember
(i) The specific resistance () of potentiometer wire must be high but its temperature coefficient of
resistance () must be low.
(ii) All higher potential points (terminals) of primary and secondary circuits must be connected
together at point A and all lower potential points must be connected to point B or jockey.
(iii) The value of known potential difference must be greater than the value of unknown potential
difference to be measured.
(iv) The potential gradient must remain constant. For this the current in the primary circuit must
remain constant and the jockey must not be slided in contact with the wire.
(v) The diameter of potentiometer wire must be uniform everywhere.
(4) Potential gradient (x) : Potential difference (or fall in potential) per unit length of wire is called
V volt e V iR iρ e R
potential gradient i.e. x where V iR . R . So x .
L m R Rh r L L A (R R h r) L
(i) Potential gradient directly depends upon
(a) The resistance per unit length (R/L) of potentiometer wire.
(b) The radius of potentiometer wire (i.e. Area of cross-section)
(c) The specific resistance of the material of potentiometer wire (i.e. )
(d) The current flowing through potentiometer wire (i)
(ii) x indirectly depends upon
(a) The emf of battery in the primary circuit (i.e. e)
(b) The resistance of rheostat in the primary circuit (i.e. Rh)
(5) Working : Suppose jocky is made to touch a point J on wire then potential difference between A
and J will be V xl e, r K Rh
G G
genius PHYSICS Pradeep Kshetrapal
50 Current Electricity
(i) V due to battery e and
(ii) E due to unknown cell
Application of Potentiometer.
(1) To determine the internal resistance of a primary cell
(i) Initially in secondary circuit key K' remains open and balancing length (l1) is obtained. Since cell E
is in open circuit so it’s emf balances on length l1 i.e. E = xl1 ……. (i)
genius PHYSICS Pradeep Kshetrapal
Current Electricity 51
(E1 E 2 ) xl 1 (E1 E 2 ) xl 2
E1 E 2 l E1 l1 l 2
1 or
E1 E 2 l2 E 2 l1 l 2
(3) Comparison of resistances : Let the balancing length for resistance R1 (when XY is connected)
is l1 and let balancing length for resistance R1 + R2 (when YZ is connected) is R
K
l2. Then iR1 = xl1 and i(R1 + R2) = xl2 h
J
A B
R 2 l 2 l1 G
X Y Z
R1 l1
i R1 R
2
K1 Rh
1
(i) The value of thermo-emf in a thermocouple for ordinary temperature difference is very low (10–6
volt). For this the potential gradient x must be also very low (10–4 V/m). K Rh
+ –
Hence a high resistance (R) is connected in series with the potentiometer
wire in order to reduce current. R A
A HRB
B
(ii) The potential difference across R must be equal to the emf of G
+ –
E0 G
standard cell i.e. iR = E0 i E0 1 2 3
R
Cold ice Hot
sand
(iii) The small thermo emf produced in the thermocouple e = xl
iR iRl
(iv) x i e where L = length of potentiometer wire, = resistance per unit length, l
L L
= balancing length for e
genius PHYSICS Pradeep Kshetrapal
52 Current Electricity
(5) To calibrate ammeter and voltmeter
Calibration of ammeter
(i) If p.d. across 1 resistance is measured by potentiometer, then current through this (indirectly measured) is
thus known or if R is known then i = V/R can be found.
(ii) Circuit and method
(a) Standardisation is required and per formed as already described earlier. (x = E0/l0)
Calibration of voltmeter
(i) Practical voltmeters are not ideal, because these do not have infinite resistance. The error of such practical
voltmeter can be found by comparing the voltmeter reading with calculated value of p.d. by potentiometer.
(ii) Circuit and procedure e K1 Rh
+ –
(a) Standardisation : If l0 is balancing length for E0 the emf of +
A B
standard cell by connecting 1 and 2 of bi-directional key, then x = E0/l0. E0 C
+ – 1
2
(b) The balancing length l1 for unknown potential difference V is given + – G
V 3
by (by closing 2 and 3) V ' xl 1 (E0 / l0 )l1 . RB
+ – K2
If the voltmeter reading is V then the error will be (V – V) which
Rh
may be +ve, – ve or zero.
Concepts
In case of zero deflection in the galvanometer current flows in the primary circuit of the potentiometer, not in the
galvanometer circuit.
Example
s
Example: 85 A battery with negligible internal resistance is connected with 10m long wire. A standard cell gets
balanced on 600 cm length of this wire. On increasing the length of potentiometer wire by 2m then
the null point will be displaced by
(a) 200 cm (b) 120 cm (c) 720 cm (d) 600 cm
L1 l 10 600
Solution : (b) By using 1 l 2 720 cm .
L2 l2 12 l2
Current Electricity 53
+ – R1
i
5m
A
B
G
Hot Cold
Junction Junction
E E 2 .4 10 3
Solution : (a) E x l i l i 4 10 4 A .
l l 1 .2 5
Example: 87 The resistivity of a potentiometer wire is 40 10–8 m and its area of cross section is 8 10–6 m2.
If 0.2 amp. Current is flowing through the wire, the potential gradient will be
(a) 10–2 volt/m (b) 10–1 volt/m (c) 3.2 10–2 volt/m (d) 1 volt/m
V iR iL i 0.2 40 10 8
Solution : (a) Potential gradient 6
10 2 V/ m
L L AL A 8 10
Example: 88 A deniel cell is balanced on 125 cm length of a potentiometer wire. When the cell is short circuited
with a 2 resistance the balancing length obtained is 100 cm. Internal resistance of the cell will
be [RPMT 1998]
l1 l2 125 100 1
Solution: (b) By using r R' r 2 0 .5
l2 100 2
Example: 89 A potentiometer wire of length 10 m and a resistance 30 is connected in series with a battery of
emf 2.5 V and internal resistance 5 and an external resistance R. If the fall of potential along
the potentiometer wire is 50 V/mm, the value of R is (in )
(a) 115 (b) 80 (c) 50 (d) 100
e R
Solution : (a) By using x .
(R R h r) L
50 10 6 2 .5 30
3
R = 115
10 (30 R 5) 10
Example: 90 A 2 volt battery, a 15 resistor and a potentiometer of 100 cm length, all are connected in series. If
the resistance of potentiometer wire is 5 , then the potential gradient of the potentiometer wire is [AIIMS 1982
(a) 0.005 V/cm (b) 0.05 V/cm (c) 0.02 V/cm (d) 0.2 V/cm
e R 2 5
Solution : (a) By using x . x 0.5 V /m 0.005 V /cm
(R R h r) L (5 15 0) 1
Example: 91 In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the
balance point is at a length of 2 m when the cell is shunted by a 5 resistance; and is at a length of 3
m when the cell is shunted by a 10 resistance. The internal resistance of the cell is, then
54 Current Electricity
l l2 l 2
Solution : (b) By using r 1 R ' r 1 5 …… (i)
l2 2
l 3
and r 1 10 ……. (ii)
3
10 V, 1
3m
5m
E 5
G
E
Example: 94 In the following circuit the potential difference between the points B and C is balanced against 40 cm
length of potentiometer wire. In order to balance the potential difference between the points C and D,
where should jockey be pressed
Rh
+ –
40 cm
X Y
10 G
C 4
B 10 D
+ – r = 1 K
A F
6V
1 1 1 2 1
Solution : (a) or R1 = 5
R 10 10 10 5
R2 40 4
R2 = 4, l1 = 40 cm, l2 = ? l 2 l1 or l 2 32 cm
R1 5
genius PHYSICS Pradeep Kshetrapal
Current Electricity 55
Example: 95 In the following circuit diagram fig. the lengths of the wires AB and BC are same but the radius of AB
is three times that of BC. The ratio of potential gradients at AB and BC will be
(a) 1 : 9
+ –
(b) 9 : 1
(c) 3 : 1
C
(d) 1 : 3 A B
2
1 x1 r2 r 1
Solution : (a) x Rp 22
r2 x2 r1
3 r 9
Example: 96 With a certain cell the balance point is obtained at 0.60 m from one end of the potentiometer. With
another cell whose emf differs from that of the first by 0.1 V, the balance point is obtained at 0.55 m.
Then, the two emf’s are
(a) 1.2 V, 1.1 V (b) 1.2 V, 1.3 V (c) – 1.1 V, – 1.0 V (d) None of the above
E1 0.6
Solution : (a) E1 = x (0.6) and E2 = E1 – 0.1 = x (0.55)
E1 0.1 0.55
6
or 55 E1 = 60 E1 – 6 E 1 1 . 2 V thus E2 = 1.1 V
5
Tricky Example: 11
A cell of internal resistance 1.5 and of emf 1.5 volt balances 500 cm on a potentiometer wire. If a
wire of 15 is connected between the balance point and the cell, then the balance point will shift
[MP PMT 1985]
56 Current Electricity
.
genius PHYSICS
Atomic structure 1
(i) Most of the -particles pass through the foil straight away undeflected.
(ii) Some of them are deflected through small angles.
(iii) A few -particles (1 in 1000) are deflected through the angle more than 90o.
(iv) A few -particles (very few) returned back i.e. deflected by 180o.
(v) Distance of closest approach (Nuclear dimension)
The minimum distance from the nucleus up to which the -particle approach, is called the distance of
1 Ze 2 1
closest approach (r0). From figure r0 . ; E mv 2 K.E. of -particle
4 0 E 2
(vi) Impact parameter (b) : The perpendicular distance of the velocity vector ( v ) of the -particle from
the centre of the nucleus when it is far away from the nucleus is known as impact parameter. It is given as
Ze 2 cot( / 2)
b b cot( / 2)
1
4 0 mv 2
2
genius PHYSICS
2 Atomic Structure
Note : If t is the thickness of the foil and N is the number of -particles scattered in a
N N t
particular direction ( = constant), it was observed that constant 1 1 .
t N 2 t2
After Rutherford's scattering of -particles experiment, following conclusions were made as regard as
atomic structure :
Atom
(a) Most of the mass and all of the charge of an atom concentrated in a
very small region is called atomic nucleus. Nucleus
+
(b) Nucleus is positively charged and it's size is of the order of 10–15 m1
10–15 m
Fermi.
(c) In an atom there is maximum empty space and the electrons revolve 10–10 m
around the nucleus in the same way as the planets revolve around the sun. Size of the nucleus = 1 Fermi = 10–15 m
Size of the atom 1 Å = 10–10 m
Draw backs
(i) Stability of atom : It could not explain stability of atom because according to classical electrodynamic
theory an accelerated charged particle should continuously radiate energy. Thus an electron moving in an
circular path around the nucleus should also radiate energy and thus move into
smaller and smaller orbits of gradually decreasing radius and it should
ultimately fall into nucleus. e–
Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a
single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom)
Bohr's model is based on the following postulates.
(i) The electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for
h
which total angular momentum of the revolving electrons is an integral multiple of ( )
2
h
i.e. L n mvr ; where n = 1, 2, 3, ……..= Principal quantum number
2
(ii) The radiation of energy occurs only when an electron jumps from one permitted orbit to another.
When electron jumps from higher energy orbit (E1) to lower energy orbit (E2) then difference of energies
of these orbits i.e. E1 – E2 emits in the form of photon. But if electron goes from E2 to E1 it absorbs the same
amount of energy. E
E1 1
E1 – E2 = h E1 – E2 = h
E2 E2
Emission Absorption
Note : According to Bohr theory the momentum of an e revolving in second orbit of H 2 atom
h
will be
For an electron in the nth orbit of hydrogen atom in Bohr model, circumference of orbit n ;
where = de-Broglie wavelength.
genius PHYSICS
Atomic structure 3
1 (Ze )e mv 2 nh
i.e. ……. (i) also mvr …….(ii)
4 0 r 2 r 2
n
n2
rn
Z
Note : The radius of the innermost orbit (n = 1) hydrogen atom (z = 1) is called Bohr's radius a0
i.e. a 0 0 .53 Å .
Note : The ratio of speed of an electron in ground state in Bohr's first orbit of hydrogen atom to
e2 1
velocity of light in air is equal to (where c = speed of light in air)
2 0 ch 137
(4) Energy
(i) Potential energy : An electron possesses some potential energy because it is found in the field of
(Ze )(e ) kZe 2
nucleus potential energy of electron in nth orbit of radius rn is given by U k .
rn rn
(ii) Kinetic energy : Electron posses kinetic energy because of it's motion. Closer orbits have greater
kinetic energy than outer ones.
mv 2 k . (Ze )(e ) kZe 2 | U |
As we know Kinetic energy K
rn rn2 2rn 2
(iii) Total energy : Total energy (E) is the sum of potential energy and kinetic energy i.e. E = K + U
kZe 2 n 2 h 2 0 me 4 z2 me 4 z2 Z2 Z2
E also rn . Hence E . ch R ch 13 . 6 eV
2rn mze 2 8 2 h 2 n2 8 2 ch 3 n2 n2 n2
0 0
me 4
where R = Rydberg's constant = 1.09 107 per metre
8 02 ch 3
large distances to form a hydrogen atom, 13.6 eV energy will be released. The binding energy of a hydrogen
atom is therefore 13.6 eV.
Note : In hydrogen atom excitation energy to excite electron from ground state to first
excited state will be 3.4 (13 .6) 10 .2 eV .
and from ground state to second excited state it is [ 1.51 (13 .6) 12 .09 eV ].
In an H 2 atom when e makes a transition from an excited state to the ground state it’s kinetic
energy increases while potential and total energy decreases.
(6) Transition of electron
When an electron makes transition from higher energy level having energy E2(n2) to a lower energy level
having energy E1 (n1) then a photon of frequency is emitted
(i) Energy of emitted radiation E2 n2
Rc h Z 2 Rch Z 2 1 1
E E 2 E1 13 .6 Z
2
n2 n2
E,,
n 22 n 2
1 1 2 E1 n1
(ii) Frequency of emitted radiation Emission
E E 2 E1 1 1
E h Rc Z 2 2 2
h h n1 n 2
(iii) Wave number/wavelength
1
Wave number is the number of waves in unit length
c
1 1 1 13 .6 Z 1
2
1
RZ 2 2 2 2
n1 n 2
2
hc n1 n 2
(iv) Number of spectral lines : If an electron jumps from higher energy orbit to lower energy orbit it
emits raidations with various spectral lines.
If electron falls from orbit n2 to n1 then the number of spectral lines emitted is given by
(n n1 1)(n 2 n1 )
NE 2
2
If electron falls from nth orbit to ground state (i.e. n2 = n and n1 = 1) then number of spectral lines emitted
n (n 1)
NE
2
genius PHYSICS
6 Atomic Structure
Note : Absorption spectrum is obtained only for the transition from lowest energy level to
higher energy levels. Hence the number of absorption spectral lines will be (n – 1).
(v) Recoiling of an atom : Due to the transition of electron, photon is emitted and the atom is recoiled
h 1 1
Recoil momentum of atom = momentum of photon hRZ 2 2 2
n1 n 2
2 2
p h
Also recoil energy of atom (where m = mass of recoil atom)
2 m 2m 2
(7) Drawbacks of Bohr's atomic model
(i) It is valid only for one electron atoms, e.g. : H, He+, Li+2, Na+1 etc.
(ii) Orbits were taken as circular but according to Sommerfield these are elliptical.
(iii) Intensity of spectral lines could not be explained.
(iv) Nucleus was taken as stationary but it also rotates on its own axis.
(v) It could not be explained the minute structure in spectrum line.
(vi) This does not explain the Zeeman effect (splitting up of spectral lines in magnetic field) and Stark
effect (splitting up in electric field)
(vii) This does not explain the doublets in the spectrum of some of the atoms like sodium (5890Å &
5896Å)
Emission spectra
Spectral series
The spectral lines arising from the transition of electron forms a spectra series.
(i) Mainly there are five series and each series is named after it's discover as Lymen series, Balmer series,
Paschen series, Bracket series and Pfund series.
(ii) According to the Bohr's theory the wavelength of the radiations emitted from hydrogen atom is given
by
1 1 1
R 2 2
n1 n 2
where n2 = outer orbit (electron jumps from this orbit), n1 = inner orbit (electron falls in this orbit)
(iii) First line of the series is called first member, for this line wavelength is maximum (max)
genius PHYSICS
Atomic structure 7
(iv) Last line of the series (n2 = ) is called series limit, for this line wavelength is minimum (min)
Spectral Transition n12 n 22 n12 max (n 1) 2 Region
series Wavelength ()
(n 22 n12 )R 2 min (2n 1)
1 n1 R
n 22
Maximum Minimum
wavelength wavelength
n1 n and n 2 n 1 n 2 , n1 n
n 2 (n 1)2 n2
max min
(2n 1)R R
1. Lymen n2 = 2, 3, 4 … (1)2 (1 1)2 4 n1 = n = 1 4 Ultraviolet
series n1 = 1 max 1 region
(2 1 1)R 3 R min
3
R
2.Balmer n2 = 3, 4, 5 … n1 = n = 2, n2 = 2 + 1 = 3 4 9 Visible
series min region
n1 = 2 36 R 5
max
5R
3. Paschen n2 = 4, 5, 6 … n1 = n = 3, n2 = 3 + 1 = 4 n1 = n = 3 16 Infrared
series n1 = 3 144 9 7 region
max min
7R R
4. Bracket n2 = 5, 6, 7 … n1 = n = 4, n2 = 4 + 1 = 5 n1 = n = 4 25 Infrared
series n1 = 4 400 16 9 region
max min
9R R
5. Pfund n2 = 6, 7, 8 … n1 = = 5, n2 = 5 + 1 = 6 25 36 Infrared
series min region
n1 = 5 900 R 11
max
11 R
Quantum Numbers.
An atom contains large number of shells and subshells. These are distinguished from one another on the
basis of their size, shape and orientation (direction) in space. The parameters are expressed in terms of
different numbers called quantum number.
Quantum numbers may be defined as a set of four number with the help of which we can get complete
information about all the electrons in an atom. It tells us the address of the electron i.e. location, energy, the
type of orbital occupied and orientation of that orbital.
(1) Principal Quantum number (n) : This quantum number determines the main energy level or shell
in which the electron is present. The average distance of the electron from the nucleus and the energy of the
1
electron depends on it. E n 2 and rn n 2 (in H-atom)
n
The principal quantum number takes whole number values, n = 1, 2, 3, 4,…..
(2) Orbital quantum number (l) or azimuthal quantum number (l)
This represents the number of subshells present in the main shell. These subsidiary orbits within a shell
will be denoted as 1, 2, 3, 4 … or s, p, d, f … This tells the shape of the subshells.
h
The orbital angular momentum of the electron is given as L l(l 1) (for a particular value of n).
2
For a given value of n the possible values of l are l = 0, 1, 2, ….. upto (n – 1)
(3) Magnetic quantum number (ml) : An electron due to it's angular motion around the nucleus
generates an electric field. This electric field is expected to produce a magnetic field. Under the influence of
external magnetic field, the electrons of a subshell can orient themselves in certain preferred regions of space
around the nucleus called orbitals.
genius PHYSICS
8 Atomic Structure
The magnetic quantum number determines the number of preferred orientations of the electron present
in a subshell.
The angular momentum quantum number m can assume all integral value between – l to +l including
zero. Thus ml can be – 1, 0, + 1 for l = 1. Total values of ml associated with a particular value of l is given by (2l +
1).
(4) Spin (magnetic) quantum number (ms) : An electron in atom not only revolves around the
nucleus but also spins about its own axis. Since an electron can spin either in clockwise direction or in
anticlockwise direction. Therefore for any particular value of magnetic quantum number, spin quantum
1 1
number can have two values, i.e. m s (Spin up) or ms (Spin down)
2 2
This quantum number helps to explain the magnetic properties of the substance.
Note : The maximum number of electrons in a subshell with orbital quantum number l is 2(2l +
1).
(2) Aufbau principle
Electrons enter the orbitals of lowest energy first.
As a general rule, a new electron enters an empty orbital for which (n + l ) is minimum. In case the value
(n l) is equal for two orbitals, the one with lower value of n is filled first.
Thus the electrons are filled in subshells in the following order (memorize)
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ……
(3) Hund's Rule
When electrons are added to a subshell where more than one orbital of the same energy is available, their
spins remain parallel. They occupy different orbitals until each one of them has at least one electron. Pairing
starts only when all orbitals are filled up.
Pairing takes place only after filling 3, 5 and 7 electrons in p, d and f orbitals, respectively.
genius PHYSICS
Atomic structure 9
Concepts
With the increase in principal quantum number the energy difference between the two successive energy level decreases,
while wavelength of spectral line increases.
E' E' ' E' ' ' n=4
E,
' ' ' ' ' ' n=3
E, E,
E E ' E ' ' E ' ' '
n=2
1 1 1 1
E,
' ' ' ' ' '
n=1
Rydberg constant is different for different elements
R( =1.09 107 m–1) is the value of Rydberg constant when the nucleus is considered to be infinitely massive as compared to
the revolving electron. In other words, the nucleus is considered to be stationary.
R
In case, the nucleus is not infinitely massive or stationary, then the value of Rydberg constant is given as R '
m
1
M
where m is the mass of electron and M is the mass of nucleus.
Atomic spectrum is a line spectrum
Each atom has it's own characteristic allowed orbits depending upon the electronic configuration. Therefore photons
emitted during transition of electrons from one allowed orbit to inner allowed orbit are of some definite energy only. They
do not have a continuous graduation of energy. Therefore the spectrum of the emitted light has only some definite lines
and therefore atomic spectrum is line spectrum.
Just as dots of light of only three colours combine to form almost every conceivable colour on T.V. screen, only about 100
distinct kinds of atoms combine to form all the materials in the universe.
B
a
92U 92U
235 236
K
r
genius PHYSICS
10 Atomic Structure
Rutherford's -scattering experiment established that the mass of atom is concentrated with small positively
charged region at the centre which is called 'nucleus'.
Nuclei are made up of proton and neutron. The number of protons in a nucleus e–
(called the atomic number or proton number) is represented by the symbol Z. The
number of neutrons (neutron number) is represented by N. The total number of
neutrons and protons in a nucleus is called it's mass number A so A = Z + N.
Neutrons and proton, when described collectively are called nucleons. e–
Note : Energy of thermal neutron is about 0.025 eV and speed is about 2.2 km/s.
Nucleus.
(1) Different types of nuclei
The nuclei have been classified on the basis of the number of protons (atomic number) or the total number of
nucleons (mass number) as follows
(i) Isotopes : The atoms of element having same atomic number but different mass number are called isotopes. All
isotopes have the same chemical properties. The isotopes of some elements are the following
1 H 1, 1H 2, 1H 3 8O
16
, 8 O 17 , 8 O 18 2 He 3 , 2 He
4
17 Cl
35
, 17 Cl
37
92 U
235
, 92 U
238
(ii) Isobars : The nuclei which have the same mass number (A) but different atomic number (Z) are called isobars.
Isobars occupy different positions in periodic table so all isobars have different chemical properties. Some of the examples
of isobars are
1 H 3 and 2 He
3
, 6C
14
and 7N
14
, 8 O 17 and 9F
17
(iii) Isotones : The nuclei having equal number of neutrons are called isotones. For them both the atomic number
(Z) and mass number (A) are different, but the value of (A – Z) is same. Some examples are
4 Be 9 and 5B
10
, 6 C 13 and 7N
14
, 8 O 18 and 9F
19
, 3 Li 7 and 4 Be
8
, 1 H 3 and 2 He
4
genius PHYSICS
Atomic structure 11
(iv) Mirror nuclei : Nuclei having the same mass number A but with the proton number (Z) and neutron number
(A – Z) interchanged (or whose atomic number differ by 1) are called mirror nuclei for example.
3
1H and 2 He 3 , 3 Li 7 and 4 Be
7
4 4
(ii) Nuclear volume : The volume of nucleus is given by V R 3 R 03 A V A
3 3
(iii) Nuclear density : Mass per unit volume of a nucleus is called nuclear density.
Massof nucleus mA
Nuclear density( )
Volume of nucleus 4
(R 0 A 1 / 3 )3
3
where m = Average of mass of a nucleon (= mass of proton + mass of neutron = 1.66 10–27 kg)
and mA = Mass of nucleus
3m
2 . 38 10 17 kg / m 3
4R 0
3
Density of a nucleus is maximum at it's centre and decreases as we move outwards from the nucleus.
- mesons are of three types – Positive meson (+), negative meson ( –), neutral meson (0)
The force between neutron and proton is due to exchange of charged meson between them i.e.
p n, n p
The forces between a pair of neutrons or a pair of protons are the result of the exchange of neutral meson ( o)
between them i.e. p p ' 0 and n n' 0
Thus exchange of meson between nucleons keeps the nucleons bound together. It is responsible for the nuclear
forces.
genius PHYSICS
12 Atomic Structure
Dog-Bone analogy
The above interactions can be explained with the dog bone analogy according to which we
consider the two interacting nucleons to be two dogs having a common bone clenched in between
their teeth very firmly. Each one of these dogs wants to take the bone and hence they cannot be
separated easily. They seem to be bound to each other with a strong attractive force (which is the
bone) though the dogs themselves are strong enemies. The meson plays the same role of the
common bone in between two nucleons.
(4) Atomic mass unit (amu)
The unit in which atomic and nuclear masses are measured is called atomic mass unit (amu)
1
1 amu (or 1u) = th of mass of 6 C 12 atom = 1.66 10–27 kg
12
Masses of electron, proton and neutrons
Mass of electron (me) = 9.1 10–31 kg = 0.0005486 amu, Mass of proton (mp) = 1.6726 10–27 kg = 1.007276 amu
Mass of neutron (mn) = 1.6750 10–27 kg = 1.00865 amu, Mass of hydrogen atom (me + mp) = 1.6729 10–27 kg = 1.0078 amu
Mass-energy equivalence
According to Einstein, mass and energy are inter convertible. The Einstein's mass energy relationship is given by E mc 2
If m = 1 amu, c = 3 108 m/sec then E = 931 MeV i.e. 1 amu is equivalent to 931 MeV or 1 amu (or 1 u) = 931 MeV
(5) Pair production and pair-annihilation
When an energetic -ray photon falls on a heavy substance. It is absorbed by some nucleus of the substance and an
electron and a positron are produced. This phenomenon is called pair production and may be represented by the following
equation h 1
0
1
0
( photon) (Positron) (Electron)
+1 0
Hence, for pair-production it is essential that the energy of -photon must be at least 2 0.51 = 1.02 MeV. If the
energy of -photon is less than this, it would cause photo-electric effect or Compton effect on striking the matter.
The converse phenomenon pair-annihilation is also possible. Whenever an electron and a positron come very close
to each other, they annihilate each other by combining together and two -photons (energy) are produced. This
phenomenon is called pair annihilation and is represented by the following equation.
1 1 h h
0 0
N
(i) Neutron-proton ratio Ratio
Z
The chemical properties of an atom are governed entirely by the number of protons (Z) in the nucleus, the stability of
an atom appears to depend on both the number of protons and the number of neutrons.
genius PHYSICS
Atomic structure 13
For lighter nuclei, the greatest stability is achieved when the number of protons and neutrons are approximately
N
equal (N Z) i.e. 1
Z
Heavy nuclei are stable only when they have more neutrons than protons. Thus heavy nuclei are neutron rich
compared to lighter nuclei (for heavy nuclei, more is the number of protons in the nucleus, greater is the electrical
repulsive force between them. Therefore more neutrons are added to provide the strong attractive forces necessary to keep
the nucleus stable.)
Stable nuclei
number (N)
(Line of
Neutron
stability)
N=Z
10 20 30405060708090
Proton number
(Z)
Figure shows a plot of N verses Z for the stable nuclei. For mass number upto about A = 40. For larger value of Z the
nuclear force is unable to hold the nucleus together against the electrical repulsion of the protons unless the number of
neutrons exceeds the number of protons. At Bi (Z = 83, A = 209), the neutron excess in N – Z = 43. There are no stable
nuclides with Z > 83.
A nuclide above the line of stability i.e. having excess neutrons, decay through emission (neutron
changes into proton). Thus increasing atomic number Z and decreasing neutron number N. In
N
emission, ratio decreases.
Z
Number (N)
Neutron
A nuclide below the line of stability have excess number of protons. It decays – N=Z
by emission, results in decreasing Z and increasing N. In emission, +
N
the ratio increases. Proton number (Z)
Z
(ii) Even or odd numbers of Z or N : The stability of a nuclide is also determined by the consideration whether it
contains an even or odd number of protons and neutrons.
It is found that an even-even nucleus (even Z and even N) is more stable (60% of stable nuclide have even Z and even
N).
An even-odd nucleus (even Z and odd N) or odd-even nuclide (odd Z and even N) is found to be lesser sable while
the odd-odd nucleus is found to be less stable.
2 6
Only five stable odd-odd nuclides are known : 1 H , 3 Li , 5 Be 10 , 7 N 14 and 75 Ta
180
(iii) Binding energy per nucleon : The stability of a nucleus is determined by value of it's binding energy per nucleon.
In general higher the value of binding energy per nucleon, more stable the nucleus is
Note : The mass of a typical nucleus is about 1% less than the sum of masses of nucleons.
m M A
Packing fraction (f ) where M = Mass of nucleus, A = Mass number
A A
Packing fraction measures the stability of a nucleus. Smaller the value
40
of packing fraction, larger is the stability of the nucleus.
30
(i) Packing fraction may be of positive, negative or zero value. 20
10
(iii) At A = 16, f Zero
0 Mass
(3) Binding energy (B.E.) A > 240 number (A)
– 10
The neutrons and protons in a stable nucleus are held together by – 20
nuclear forces and energy is needed to pull them infinitely apart (or the
same energy is released during the formation of the nucleus). This energy is called the binding energy of the nucleus.
or
The binding energy of a nucleus may be defined as the energy equivalent to the mass defect of the nucleus.
If m is mass defect then according to Einstein's mass energy relation
Binding energy = m c2 = [{mpZ + mn(A – Z)} – M] c2
(This binding energy is expressed in joule, because m is measured in kg)
If m is measured in amu then binding energy = m amu = [{mpZ + mn(A – Z)} – M] amu = m 931 MeV
(4) Binding energy per nucleon
The average energy required to release a nucleon from the nucleus is called binding energy per nucleon.
Total bind ing energy m 931 MeV
Binding energy per nucleon
Mass number (i.e . total number of nucleons) A Nucleon
Binding energy per nucleon Stability of nucleus
Binding Energy Curve.
It is the graph between binding energy per nucleon and total number of nucleons (i.e. mass number A)
Binding energy per
8.0 He 26Fe56
nucleon (MeV)
6.0
4.0 Li
2.0 H2
0
56 100 150 200
Mass number A
(1) Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour nuclei. For
example 2 He 4 , 4 Be 8 , 6 C 12 , 8 O 16 and 10 Ne 20 . These nuclei are more stable than their neighbours.
genius PHYSICS
Atomic structure 15
56
(2) The binding energy per nucleon is maximum for nuclei of mass number A = 56 ( 26 Fe ) . It's value is 8.8 MeV
per nucleon.
(3) For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the value of
binding energy per nucleon drops to 7.5 MeV.
Note : When a heavy nucleus splits up into lighter nuclei, then binding energy per nucleon of lighter
nuclei is more than that of the original heavy nucleus. Thus a large amount of energy is liberated in this
process (nuclear fission).
When two very light nuclei combines to form a relatively heavy nucleus, then binding energy per nucleon
increases. Thus, energy is released in this process (nuclear fusion).
B. E.
A +
Fusion Fission
+
Nuclear Reactions. A
The process by which the identity of a nucleus is changed when it is bombarded by an energetic particle is called
nuclear reaction. The general expression for the nuclear reaction is as follows.
X a C Y b Q
(Parent nucleus) (Incident particle) (Com pound nucleus) (Com pound nucleus) (Product particles) (Energy )
Here X and a are known as reactants and Y and b are known as products. This reaction is known as (a, b) reaction
and can be represented as X(a, b) Y
(1) Q value or energy of nuclear reaction
The energy absorbed or released during nuclear reaction is known as Q-value of nuclear reaction.
Q-value = (Mass of reactants – mass of products)c2 Joules
= (Mass of reactants – mass of products) amu
If Q < 0, The nuclear reaction is known as endothermic. (The energy is absorbed in the reaction)
If Q > 0, The nuclear reaction is known as exothermic (The energy is released in the reaction)
(2) Law of conservation in nuclear reactions
(i) Conservation of mass number and charge number : In the following nuclear reaction
2 He 4 7 N 14 8 O 17 1 H 1
Mass number (A) Before the reaction After the reaction
4 +14 = 18 17 + 1 = 18
Charge number (Z) 2+7=9 8+1=9
(ii) Conservation of momentum : Linear momentum/angular momentum of particles before the reaction is equal to
the linear/angular momentum of the particles after the reaction. That is p = 0
(iii) Conservation of energy : Total energy before the reaction is equal to total energy after the reaction. Term Q is
added to balance the total energy of the reaction.
(3) Common nuclear reactions
The nuclear reactions lead to artificial transmutation of nuclei. Rutherford was the first to carry out artificial
transmutation of nitrogen to oxygen in the year 1919.
2 He 4 7 N 14 9 F 18 8 O 17 1 H 1
genius PHYSICS
16 Atomic Structure
It is called (, p) reaction. Some other nuclear reactions are given as follows.
(p, n) reaction 1 H 1 5 B 11 6 C 12 6 C 11 0 n 1
(p, ) reaction 1 H 1 3 Li 11 4 Be 8 2 He 4 2 He 4
(p, ) reaction 1H
1
6 C 12 7 N 13 7 N 13
(n, p) reaction 0n
1
7 N 14 7 N 15 6 C 14 1 H 1
(, n) reaction 1 H 2 1 H 1 0 n1
92 U
235
0 n1 92 U
236
56 Ba
141
36 Kr
92
3 0 n1 Q
(unstable nucleus)
(ii) The energy released in U235 fission is about 200 MeV or 0.8 MeV per nucleon.
235
(iii) By fission of 92 U , on an average 2.5 neutrons are liberated. These neutrons are called fast neutrons and their
energy is about 2 MeV (for each). These fast neutrons can escape from the reaction so as to proceed the chain reaction they
are need to slow down.
(iv) Fission of U235 occurs by slow neutrons only (of energy about 1eV) or even by thermal neutrons (of energy about
0.025 eV).
(v) 50 kg of U235 on fission will release 4 × 1015 J of energy. This is equivalence to 20,000 tones of TNT explosion.
The nuclear bomb dropped at Hiroshima had this much explosion power.
(vi) The mass of the compound nucleus must be greater than the sum of masses of fission products.
Binding energy
(vii) The of compound nucleus must be less than that of the fission products.
A
(viii) It may be pointed out that it is not necessary that in each fission of uranium, the two fragments 56 Ba and
36 Kr are formed but they may be any stable isotopes of middle weight atoms.
Same other U 235 fission reactions are
92 U
235
0 n 1 54 Xe 140 38 Sr
94
2 0 n1
57 La 148 35 Br
85
3 0 n1
Many more
(ix) The neutrons released during the fission process are called prompt neutrons.
(x) Most of energy released appears in the form of kinetic energy of fission fragments.
Ba
Energy
Energy
Slow
Neutron
92U235 92U236
Energy
In nuclear fission, three neutrons are produced along with the release of large energy. Under favourable conditions,
these neutrons can cause further fission of other nuclei, producing large number of neutrons. Thus a chain of nuclear
fissions is established which continues until the whole of the uranium is consumed.
Kr
U
Kr Ba
n
U
Kr
Ba U
Ba
In the chain reaction, the number of nuclei undergoing fission increases very fast. So, the energy produced takes a
tremendous magnitude very soon.
Difficulties in chain reaction
(i) Absorption of neutrons by U 238 , the major part in natural uranium is the isotope U238 (99.3%), the isotope U 235
is very little (0.7%). It is found that U 238 is fissionable with fast neutrons, whereas U 235 is fissionable with slow neutrons.
238
Due to the large percentage of U , there is more possibility of collision of neutrons with U 238 . It is found that the
238
neutrons get slowed on coliding with U , as a result of it further fission of U238 is not possible (Because they are slow
and they are absorbed by U238). This stops the chain reaction.
235
Removal : (i) To sustain chain reaction 92 U is separated from the ordinary uranium. Uranium so obtained
92 U
235
is known as enriched uranium, which is fissionable with the fast and slow neutrons and hence chain reaction can
be sustained.
(ii) If neutrons are slowed down by any method to an energy of about 0.3 eV, then the probability of their absorption
238
by U becomes very low, while the probability of their fissioning U 235 becomes high. This job is done by moderators.
Which reduce the speed of neutron rapidly graphite and heavy water are the example of moderators.
(iii) Critical size : The neutrons emitted during fission are very fast and they travel a large distance before being
slowed down. If the size of the fissionable material is small, the neutrons emitted will escape the fissionable material
before they are slowed down. Hence chain reaction cannot be sustained.
Removal : The size of the fissionable material should be large than a critical size.
The chain reaction once started will remain steady, accelerate or retard depending upon, a factor called neutron
reproduction factor (k). It is defined as follows.
Rate of production of neutrons
k
Rate of loss of neutrons
If k = 1, the chain reaction will be steady. The size of the fissionable material used is said to be the critical size and
it's mass, the critical mass.
If k > 1, the chain reaction accelerates, resulting in an explosion. The size of the material in this case is super
critical. (Atom bomb)
If k < 1, the chain reaction gradually comes to a halt. The size of the material used us said to be sub-critical.
Types of chain reaction : Chain reactions are of following two types
Note : The energy released in the explosion of an atom bomb is equal to the energy released by 2000
tonn of TNT and the temperature at the place of explosion is of the order of 10 7 oC.
Nuclear Reactor.
A nuclear reactor is a device in which nuclear fission can be carried out through a sustained and a controlled chain reaction. It is also called an atomic
pile. It is thus a source of controlled energy which is utilised for many useful purposes.
Cadmium rods
Core
Coolant
Coolant out
Turbine
Concrete To electric
wall generator
Condenser
Moderator Water
Heat
Coolant in exchanger
Note : It may be noted that Plutonium is the best fuel as compared to other fissionable material. It is
because fission in Plutonium can be initiated by both slow and fast neutrons. Moreover it can be obtained
from U 238 .
Nuclear reactor is firstly devised by fermi. Apsara was the first Indian nuclear reactor.
(2) Uses of nuclear reactor
(i) In electric power generation.
(ii) To produce radioactive isotopes for their use in medical science, agriculture and industry.
(iii) In manufacturing of PU 239 which is used in atom bomb.
(iv) They are used to produce neutron beam of high intensity which is used in the treatment of cancer and nuclear
research.
Note : A type of reactor that can produce more fissile fuel than it consumes is the breeder reactor.
Nuclear fusion
genius PHYSICS
Atomic structure 19
In nuclear fusion two or more than two lighter nuclei combine to form a single heavy nucleus. The mass of single
nucleus so formed is less than the sum of the masses of parent nuclei. This difference in mass results in the release of
tremendous amount of energy
1 H 2 1 H 2 1 H 3 1 H 1 4 MeV
1 H 3 1 H 2 2 He 4 0 n 1 17 .6 MeV
or 1 H 2 1 H 2 2 He 4 24 MeV
For fusion high pressure ( 106 atm) and high temperature (of the order of 10 7 K to 108 K) is required and so the
reaction is called thermonuclear reaction.
Fusion energy is greater then fission energy fission of one uranium atom releases about 200 MeV of energy. But the
2 3
fusion of a deutron (1 H ) and triton (1 H ) releases about 17.6 MeV of energy. However the energy released per nucleon
in fission is about 0.85 MeV but that in fusion is 4.4 MeV. So for the same mass of the fuel, the energy released in fusion is
much larger than in fission.
Plasma : The temperature of the order of 108 K required for thermonuclear reactions leads to the complete
ionisation of the atom of light elements. The combination of base nuclei and electron cloud is called plasma. The
enormous gravitational field of the sun confines the plasma in the interior of the sun.
The main problem to carryout nuclear fusion in the laboratory is to contain the plasma at a temperature of 108K. No
solid container can tolerate this much temperature. If this problem of containing plasma is solved, then the large quantity
of deuterium present in sea water would be able to serve as in-exhaustible source of energy.
Note : To achieve fusion in laboratory a device is used to confine the plasma, called Tokamak.
Stellar Energy
Stellar energy is the energy obtained continuously from the sun and the stars. Sun radiates energy at the rate of
about 1026 joules per second.
Scientist Hans Bethe suggested that the fusion of hydrogen to form helium (thermo nuclear reaction) is continuously
taking place in the sun (or in the other stars) and it is the source of sun's (star's) energy.
The stellar energy is explained by two cycles
Proton-proton cycle Carbon-nitrogen cycle
1H
1
1 H 1 H 1 e Q1
1 2 0
1H
1
6 C 12 7 N 13 Q1
1H
2
1 H 1 2 He 3 Q 2 7 N 13 6 C 13 1 e
0
2 He 3 2 He 3 2 He 4 2 1 H 1 Q 3 1H
1
6 C 13 7 N 14
Q2
4 1 H 2 He 2
1 4
1 e
0
2 26 .7 MeV 1H
1
7 N 14
8O 15
Q3
8 O 15
7N 15
1e Q4
0
1H
1
7 N 15 6 C 12 2 He 4
4 1 H 1 2 He 4 2 1 e 0 24 .7 MeV
About 90% of the mass of the sun consists of hydrogen and helium.
Nuclear Bomb. Based on uncontrolled nuclear reactions.
Atom bomb Hydrogen bomb
Based on fission process it involves the fission of U235 Based on fusion process. Mixture of deutron and
tritium is used in it
In this critical size is important There is no limit to critical size
Explosion is possible at normal temperature and High temperature and pressure are required
pressure
Less energy is released compared to hydrogen bomb More energy is released as compared to atom bomb so
it is more dangerous than atom bomb
Concepts
A test tube full of base nuclei will weight heavier than the earth.
genius PHYSICS
20 Atomic Structure
The nucleus of hydrogen contains only one proton. Therefore we may say that the proton is the nucleus of hydrogen
atom.
If the relative abundance of isotopes in an element has a ratio n1 : n2 whose atomic masses are m1 and m2 then atomic
n m n2 m 2
mass of the element is M 1 1
n1 n 2
Examples
Example: 1 A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 8 : 1. The ratio
of radii of the fragments is
(a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 2 : 1
Solution : (a)
v1
By conservation of momentum m1v1 = m2v2
m1 v1 8 m 2
…… (i)
M v 2 1 m1
m2
1/3
r1 A1
1/3
1 1
v2 Also from r A1 / 3
r2 A2 8 2
Example: 3 If Avogadro’s number is 6 1023 then the number of protons, neutrons and electrons in 14 g of 6C14 are
respectively
(a) 36 1023, 48 1023, 36 1023 (b) 36 1023, 36 1023, 36 1021
(c) 48 1023, 36 1023, 48 1021 (d) 48 1023, 48 1023, 36 1021
14
Solution : (a) Since the number of protons, neutrons and electrons in an atom of 6C are 6, 8 and 6 respectively. As 14
gm of 6C
14
contains 6 1023 atoms, therefore the numbers of protons, neutrons and electrons in 14 gm
of 6 C 14
are 6 6 10 23 36 10 23 , 8 6 10 23 48 10 23 , 6 6 10 23 36 10 23 .
Example: 4 Two Cu64 nuclei touch each other. The electrostatics repulsive energy of the system will be
(a) 0.788 MeV (b) 7.88 MeV (c) 126.15 MeV (d) 788 MeV
Solution : (c) Radius of each nucleus R R0 ( A) 1/3
1 .2 (64 ) 1/3
4 .8 fm
k q2 9 10 9 (1 .6 10 19 29 ) 2
So potential energy U 126 .15 MeV .
r 2 4 .8 10 15 1 .6 10 19
235
Example: 5 When 92 U undergoes fission. 0.1% of its original mass is changed into energy. How much energy is
235
released if 1 kg of 92 U undergoes fission [MP PET 1994; MP PMT/PET 1998; BHU 2001; BVP 2003]
Example: 7 The binding energy per nucleon of deuteron (12 H ) and helium nucleus (42 He ) is 1.1 MeV and 7 MeV
respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
[MP PMT 1992; Roorkee 1994; IIT-JEE 1996; AIIMS 1997; Haryana PMT 2000; Pb PMT 2001; CPMT 2001; AIEEE 2004]
(a) 13.9 MeV (b) 26.9 MeV (c) 23.6 MeV (d) 19.2 MeV
Solution : (c) 1H
2
1 H 2 He Q
2 4
Energy required = Binding of O17 – binding energy of O16 = 17 7.75 – 16 7.97 = 4.23 MeV
Example: 11 A gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is 0.5 MeV
and the total kinetic energy of the electron-positron pair is 0.78 MeV, then the energy of the gamma ray
photon must be [MP PMT 1991]
(a) 0.78 MeV (b) 1.78 MeV (c) 1.28 MeV (d) 0.28 MeV
Solution : (b) Energy of -rays photon = 0.5 + 0.5 +0.78 = 1.78 MeV
Example: 12 What is the mass of one Curie of U234 [MNR 1985]
(a) 3.7 1010 gm (b) 2.348 1023 gm (c) 1.48 10–11 gm (d) 6.25 10–34 gm
Solution : (c) 1 curie = 3.71 1010 disintegration/sec and mass of 6.02 1023 atoms of U 234 234 gm
234 3 .71 10 10
Mass of 3.71 1010 atoms 1 .48 10 11 gm
6 .02 10 23
Example: 13 In the nuclear fusion reaction 2
1H 13 H 42 He n, given that the repulsive potential energy between the
two nuclei is 7.7 10 14 J , the temperature at which the gases must be heated to initiate the reaction is
nearly [Boltzmann’s constant k 1 .38 10 23 J /K ] [AIEEE 2003]
Solution : (c, d) Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus
20
is always less then the sum of masses of it's constituent particles 10 Ne is made up of 10 protons plus 10
neutrons. Therefore, mass of 20
10 Ne nucleus M 1 10 (m p m n )
Also heavier the nucleus, more is he mass defect thus 20 (mn m p ) M 2 10(m p mn ) M1
or 10 (m p m n ) M 2 M 1
M 2 M1 10 (m p mn ) M 2 M 1 M 1 M 2 2M 1
Tricky example: 1
Binding energy per nucleon vs mass number curve for nuclei is shown in the figure. W, X, Y and Z
are four nuclei indicated on the curve. The process that would release energy is [IIT-JEE 1999]
Y
(a) Y 2Z Binding energy
8.5 X
nucleon in
8.0 W
7.5
(b) W X + Z MeV
5.0 Z
(c) W 2Y
(d) X Y + Z 30 60 90 120
Mass number of nuclei
Solution : (c) Energy is released in a process when total binding energy of the nucleus (= binding energy per
nucleon number of nucleon) is increased or we can say, when total binding energy of products
is more than the reactants. By calculation we can see that only in case of option (c) this happens.
Given W 2Y
Binding energy of reactants = 120 7.5 = 900 MeV
and binding energy of products = 2 (60 8.5) = 1020 MeV > 900 MeV
Radioactivity.
The phenomenon of spontaneous emission of radiatons by heavy elements is called radioactivity. The elements which shows this phenomenon are called
radioactive elements.
(1) Radioactivity was discovered by Henery Becquerel in uranium salt in the year 1896.
(2) After the discovery of radioactivity in uranium, Piere Curie and Madame Curie discovered a new radioactive
element called radium (which is 106 times more radioactive than uranium)
(3) Some examples of radio active substances are : Uranium, Radium, Thorium, Polonium, Neptunium etc.
(4) Radioactivity of a sample cannot be controlled by any physical (pressure, temperature, electric or magnetic field)
or chemical changes.
(5) All the elements with atomic number (Z ) > 82 are naturally radioactive.
(6) The conversion of lighter elements into radioactive elements by the bombardment of fast moving particles is
called artificial or induced radioactivity.
(7) Radioactivity is a nuclear event and not atomic. Hence electronic configuration of atom don't have any
relationship with radioactivity.
Nuclear radiatons
genius PHYSICS
Atomic structure 23
According to Rutherford's experiment when a sample of radioactive substance is put in a lead box and allow the
emission of radiation through a small hole only. When the radiation enters into the external electric field, they splits into
three parts
– - -rays +
- -rays
Magnetic
– +
–
– rays -
+
+ rays - field
– + rays
rays
(i) Radiations which deflects towards negative plate are called -rays (stream of positively charged particles)
(ii) Radiations which deflects towards positive plate are called particles (stream of negatively charged particles)
(iii) Radiations which are undeflected called -rays. (E.M. waves or photons)
Note : Exactly same results were obtained when these radiations were subjected to magnetic field.
No radioactive substance emits both and particles simultaneously. Also -rays are emitted after the
emission of or -particles.
-particles are not orbital electrons they come from nucleus. The neutron in the nucleus decays into
proton and an electron. This electron is emitted out of the nucleus in the form of -rays.
atom (2He4)
2. Charge + 2e –e Zero
3. Mass 4 mp (mp = mass of 4 mp me Massless
proton = 1.87 10–27
4. Speed 107 m/s 1% to 99% of speed of light Speed of light
5. Range of kinetic energy 4 MeV to 9 MeV All possible values between Between a minimum
a minimum certain value to value to 2.23 MeV
1.2 MeV
6. Penetration power (, , 1 100 10,000
) (Stopped by a paper) (100 times of ) (100 times of upto 30
cm of iron (or Pb) sheet
7. Ionisation power ( > > ) 10,000 100 1
8. Effect of electric or Deflected Deflected Not deflected
magnetic field
9. Energy spectrum Line and discrete Continuous Line and discrete
10. Mutual interaction with Produces heat Produces heat Produces, photo-electric
matter effect, Compton effect,
pair production
11. Equation of decay decay
X A XA A
0
X A zXa
Z Z Z 1 Y 1 e z
A 4
2 He 4
n
Z 2 Y Z X A
Z' X
A
genius PHYSICS
24 Atomic Structure
X A
n A' n β (2 n α Z Z')
Z Z' Y
A' A
nα
4
Radioactive Disintegration.
(1) Law of radioactive disintegration
According to Rutherford and Soddy law for radioactive decay is as follows.
"At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that instant"
dN dN
i.e. N N . It can be proved that N = N0e–t
dt dt
This equation can also be written in terms of mass i.e. M = M0e–t
where N = Number of atoms remains undecayed after time t, N0 = Number of atoms present initially (i.e. at t = 0), M =
Mass of radioactive nuclei at time t, M0 = Mass of radioactive nuclei at time t = 0, N0 – N = Number of disintegrated nucleus in
time t
dN
= rate of decay, = Decay constant or disintegration constant or radioactivity constant or Rutherford Soddy's
dt
constant or the probability of decay per unit time of a nucleus.
Note : depends only on the nature of substance. It is independent of time and any physical or chemical
changes. N0
N N = N0e–t
(2) Activity 0 t
It is defined as the rate of disintegration (or count rate) of the substance (or the number of atoms of any material
dN
decaying per second) i.e. A N N 0 e t A0 e t
dt
where A0 = Activity of t = 0, A = Activity after time t
Note : Activity per gm of a substance is known as specific activity. The specific activity of 1 gm of radium –
226 is 1 Curie.
1 millicurie = 37 Rutherford
The activity of a radioactive substance decreases as the number of undecayed nuclei decreases with time.
1
Activity
Half life
Time (t) Number of undecayed atoms (N) Remaining fraction of Fraction of atoms
(N0 = Number of initial atoms) active atoms (N/N0) decayed (N0 – N) /N0
probability of survival probability of decay
t=0 N0 1 (100%) 0
t = T1/2 N0 1 1
(50%) (50%)
2 2 2
genius PHYSICS
Atomic structure 25
t = 2(T1/2) 1 N0 N 1 3
02 (25%) (75%)
2 2 (2) 4 4
t = 3(T1/2) 1 N0 N 1 7
0 (12.5%) (87.5%)
2 (2) (2)3 8 8
t = 10 (T1/2) N0 10 99.9%
1
10 0 .1 %
(2) 2
t = n (N1/2) N n
1 n
1
(2) 2 1
2 2
Useful relation
n t/T
1 1 1/2
After n half-lives, number of undecayed atoms N N 0 N 0
2 2
(4) Mean (or average) life ()
The time for which a radioactive material remains active is defined as mean (average) life of that material.
Other definitions
(i) It is defined as the sum of lives of all atoms divided by the total number of atoms
Sum of the lives of all the atoms 1
i.e.
Total number of atoms
N
ln N
t N0
(ii) From N N 0 e slope of the line shown in the graph ln
N0 Slope = –
t
N
i.e. the magnitude of inverse of slope of ln vs t curve is known as mean life ().
N0
t t
(iii) From N N 0 e
1 1
If t N N 0 e 1 N 0 0 .37 N 0 37 % of N0.
e
1
i.e. mean life is the time interval in which number of undecayed atoms (N) becomes times or 0.37 times or 37% of
e
original number of atoms. or
1
It is the time in which number of decayed atoms (N0 – N) becomes 1 times or 0.63 times or 63% of original
e
number of atoms.
0.693 1 1
(iv) From T1 / 2 . (t1 / 2 ) 1.44 (T1 / 2 )
0.693
i.e. mean life is about 44% more than that of half life. Which gives us > T(1/2)
Note : Half life and mean life of a substance doesn't change with time or with pressure, temperature etc.
Radioactive Series.
If the isotope that results from a radioactive decay is itself radioactive then it will also decay and so on.
The sequence of decays is known as radioactive decay series. Most of the radio-nuclides found in nature are
members of four radioactive series. These are as follows
Mass number Series (Nature) Parent Stable and Integer Number of lost
product n particles
4n Thorium (natural) Th 232 Pb 208 52 = 6, = 4
90 82
4n + 1 Neptunium Np 237
(Artificial)
93
83 Bi 209 52 = 8, = 5
genius PHYSICS
26 Atomic Structure
4n + 2 Uranium (Natural) 238
Pb 206 51 = 8, = 6
92 U 82
4n + 3 Actinium (Natural) Ac 227
Pb 207
51 = 7, = 4
89 82
Note : The 4n + 1 series starts from 94 PU 241 but commonly known as neptunium series because
neptunium is the longest lived member of the series.
The 4n + 3 series actually starts from 92 U 235 .
dN 1
Rate of disintegration of A 1 N 1 (which is also the rate of formation of B)
dt
dN 2
Rate of disintegration of B 2 N 2
dt
Net rate of formation of B = Rate of disintegration of A – Rate of disintegration of B
= 1N1 – 2N2
Equilibrium
In radioactive equilibrium, the rate of decay of any radioactive product is just equal to it's rate of production from the previous member.
1 N 2 2 (T )
i.e. 1N1 = 2N2 1/2
2 N 2 1 (T1 / 2 )1
Note : In successive disintegration if N0 is the initial number of nuclei of A at t = 0 then number of nuclei
1 N 0
of product B at time t is given by N 2 (e 1 t e 2 t ) where 12 – decay constant of A and B.
( 2 1 )
Uses of radioactive isotopes
(1) In medicine
(i) For testing blood-chromium - 51 (ii) For testing blood circulation - Na - 24
(iii) For detecting brain tumor- Radio mercury - 203
(iv) For detecting fault in thyroid gland - Radio iodine - 131
(v) For cancer - cobalt – 60 (vi) For blood - Gold - 189
(vii) For skin diseases - Phospohorous - 31
(2) In Archaeology
(i) For determining age of archaeological sample (carbon dating) C 14
(ii) For determining age of meteorites - K 40 (iii) For determining age of earth-Lead isotopes
(3) In agriculture
(i) For protecting potato crop from earthworm- CO 60 (ii) For artificial rains - AgI (iii) As fertilizers - P 32
(4) As tracers - (Tracer) : Very small quantity of radioisotopes present in a mixture is known as tracer
(i) Tracer technique is used for studying biochemical reaction in tracer and animals.
(5) In industries
(i) For detecting leakage in oil or water pipe lines (ii) For determining the age of planets.
Concept
If a nuclide can decay simultaneously by two different process which have decay constant 1 and 2, half life T1 and T2 and
mean lives 1 and 2 respectively then
1 1, T1, 1 = 1 + 2
TT
T 1 2
T
T 1 T2
2 2, T2, 2 1 2
1 2
Example: 16 When 90 Th
228
transforms to 83 Bi
212
, then the number of the emitted –and –particles is, respectively
[MP PET 2002]
(a) 8, 7 (b) 4, 7 (c) 4, 4 (d) 4, 1
genius PHYSICS
Atomic structure 27
Rest
genius PHYSICS
28 Atomic Structure
4v
According to conservation of momentum 4v ( A 4)v' v ' .
A4
Example: 24 The counting rate observed from a radioactive source at t = 0 second was 1600 counts per s econd and
at t = 8 seconds it was 100 counts per second. The counting rate observed as counts per second at t =
6 seconds will be [MP PET 1996; UPSEAT 2000]
(a) 400 (b) 300 (c) 200 (d) 150
n 8 / T1 / 2 8 / T1 / 2
1 1 1 1
Solution : (c) By using A A0 100 1600 T1 / 2 2 sec
2 2 16 2
6/2
1
Again by using the same relation the count rate at t = 6 sec will be A 1600 200 .
2
Example: 25 The kinetic energy of a neutron beam is 0.0837 eV. The half-life of neutrons is 693s and the mass of
neutrons is 1.675 10 27 kg. The fraction of decay in travelling a distance of 40m will be
(a) 10 3 (b) 10 4 (c) 10 5 (d) 10 6
2E 2 0 . 0837 1 . 6 10 19
Solution : (c) v = 4 103 m/sec
m 1 . 675 10 27
40
Time taken by neutrons to travel a distance of 40 m t' 10 2 sec
4 10 3
dN dN
N dt
dt N
N 0 . 693 0 . 693
Fraction of neutrons decayed in t sec in t t 10 2 10 5
N T 693
Example: 26 The fraction of atoms of radioactive element that decays in 6 days is 7/8. The fraction that decays in 10 days
will be
(a) 77/80 (b) 71/80 (c) 31/32 (d) 15/16
N0 N
t / T1 / 2 T1 / 2 log e log e 0
1 N N t N 1
Solution : (c) By using N N 0 t t log e 0 1
2 log e (2) N t2 N
log e 0
N 2
6 log e (8 / 1) N 10 N
Hence log e 0 log e (8 ) log e 32 0 32 .
10 log e ( N 0 / N ) N 6 N
1 31
So fraction that decays 1 .
32 32
Tricky example: 2
Half-life of a substance is 20 minutes. What is the time between 33% decay and 67% decay [AIIMS 2000]
(a) 40 minutes (b) 20 minutes (c) 30 minutes (d) 25 minutes
Solution : (b) Let N0 be the number of nuclei at beginning
Number of undecayed nuclei after 33% decay = 0.67 N0
and number of undecayed nuclei after 67% of decay = 0.33 N0
0 . 67 N 0
0.33 N0 ~ and in the half-life time the number of undecayed nuclei becomes half.
2
Example
s
Example: 1 The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen
atom is
(a) 16 : 1 (b) 18 : 1 (c) 4 : 1 (d) 2 : 1
Solution : (a) For a hydrogen atom
genius PHYSICS
Atomic structure 29
Example: 6 The ratio of the wavelengths for 2 1 transition in Li++, He+ and H is [UPSEAT 2003]
(a) 1 : 2 : 3 (b) 1 : 4 : 9 (c) 4 : 9 : 36 (d) 3 : 2 : 1
1 1 1 1 1 1 1
Solution : (c) Using RZ 2 2 2 Li : He : H : : 4 : 9 : 36
2 9 4 1
n1 n2 Z
13 .6
Example: 7 Energy E of a hydrogen atom with principal quantum number n is given by E eV . The energy of a
n2
photon ejected when the electron jumps n = 3 state to n = 2 state of hydrogen is approximately
genius PHYSICS
30 Atomic Structure
[CBSE PMT/PDT Screening 2004]
(a) 1.9 eV (b) 1.5 eV (c) 0.85 eV (d) 3.4 eV
1 1 5
Solution : (a) E 13 .6 2 2 13 .6 1 .9 eV
2 3 36
Example: 8 In the Bohr model of the hydrogen atom, let R, v and E represent the radius of the orbit, the speed of
electron and the total energy of the electron respectively. Which of the following quantity is proportional
to the quantum number n [KCET 2002]
(a) R/E (b) E/v (c) RE (d) vR
0n h
2 2
Solution : (d) Rydberg constant R
mZe 2
Ze 2 mZ 2 e 4
Velocity v and energy E 2 2 2
2 0 nh 8 0 n h
Now, it is clear from above expressions R.v n
Example: 9 The energy of hydrogen atom in nth orbit is En, then the energy in nth orbit of singly ionised helium atom
will be
(a) 4En (b) En/4 (c) 2En (d) En/2
2
Z
2
13 .6 Z 2 EH 1
Solution : (a) By using E H E He 4 E n .
n2 E He Z He 2
Example: 10 The wavelength of radiation emitted is 0 when an electron jumps from the third to the second orbit of
hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the
wavelength of radiation emitted will be [SCRA 1998; MP PET 2001]
16 20 27 25
(a) 0 (b) 0 (c) 0 (d) 0
25 27 20 16
Solution : (b) Wavelength of radiation in hydrogen atom is given by
1 1 1 1 1 1 1 1 5
R 2 2 R 2 2 R R …..(i)
n 1 n 2 0 2 3 4 9 36
1 1 1 1 1 3R
and R 2 2 R …..(ii)
2 4 4 16 16
5 R 16 20 20
From equation (i) and (ii) 0
36 3 R 27 27
Example: 11 If scattering particles are 56 for 90 o angle then this will be at 60 o angle [RPMT 2000]
(a) 4.16 10 34 J- s (b) 3.32 10 34 J-s (c) 1.05 10 34 J-s (d) 2.08 10 34 J-s
Solution : (c) Change in angular momentum
n 2 h n1 h h 6.6 10 34
L L 2 L1 L (n 2 n1 ) (5 4 ) 1.05 10 34 J-s
2 2 2 2 3.14
Example: 13 In hydrogen atom, if the difference in the energy of the electron in n = 2 and n = 3 orbits is E, the
ionization energy of hydrogen atom is
(a) 13.2 E (b) 7.2 E (c) 5.6 E (d) 3.2 E
genius PHYSICS
Atomic structure 31
1 1 1 1 5
Solution : (b) Energy difference between n = 2 and n = 3; E K 2 2 K K …..(i)
2 3 4 9 36
1 1
Ionization energy of hydrogen atom n1 1 and n 2 ; E K 2 2 K …..(ii)
1
36
From equation (i) and (ii) E E 7 .2 E
5
Example: 14 In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n = 2 and n = 1 orbits is
[EAMCET (Engg.) 2000]
(a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1
T2 n 23 2 3
8
Solution : (c) According to Bohr model time period of electron T n 3 T2 8T1 .
T1 n13 1 3 1
Example: 15 A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of
required radiation for emitting electron from first to third Bohr orbit in Li will be (Ionisation energy of
hydrogen atom is 13.6 eV)
(a) 182.51 Å (b) 177.17 Å (c) 142.25 Å (d) 113.74 Å
Solution : (d) Energy of a electron in nth orbit of a hydrogen like atom is given by
Z2
E n 13 .6 eV , and Z = 3 for Li
n2
Required energy for said transition
1 1 8
E E 3 E1 13 .6 Z 2 2 2 13 .6 3 2 108 .8 eV 108 .8 1 .6 10 19 J
1 3 9
hc hc 6 .6 10 34 3 10 8
Now using E 0 .11374 10 7 m 113 .74 Å
E 108 .8 1 .6 10 19
Example: 16 The absorption transition between two energy states of hydrogen atom are 3. The emission transitions
between these states will be
(a) 3 (b) 4 (c) 5 (d) 6
Solution : (d) Number of absorption lines = (n – 1) 3 = (n – 1) n = 4
n(n 1) 4 (4 1)
Hence number of emitted lines 6
2 2
Example: 17 The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of
wavelength is emitted for a transition 3 1. What will be the wavelength of emissions for transition 2
1
[CPMT 1996]
(a) /3 (b) 4/3 (c) 3/4 (d) 3
hc hc
Solution : (d) For transition 3 1 E 2 E E E …..(i)
4E hc 3hc
For transition 2 1 E E …..(ii)
3
From equation (i) and (ii) 3
Example: 18 Hydrogen atom emits blue light when it changes from n = 4 energy level to n = 2 level. Which colour of
light would the atom emit when it changes from n = 5 level to n = 2 level [KCET 1993]
1 1 1 1 5 2 47 . 2 36
E 13 . 6 Z 2 2 2 eV 47 .2 13 .6 Z 2 2 13 .6 Z Z2 24 . 98 ~ 25
n 2
36 13 . 6 5
1 n 2 2 3
Z=5
Example: 20 The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short
wavelength limit of Paschen series is [EAMCET (Med.) 2000]
(a) 1215 Å (b) 6560 Å (c) 8225 Å (d) 12850 Å
144
Solution : (c) First member of Paschen series mean it's max
7R
9
Short wavelength of Paschen series means m in
R
max 16 7 7
Hence m in m ax 18 ,800 8225 Å .
min 7 16 16
Example: 21 Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is
[EAMCET (Engg.) 1995; MP PMT 1997]
(a) 1 : 3 (b) 27 : 5 (c) 5 : 27 (d) 4 : 9
1 1 1 3R
Solution : (c) For Lyman series R 2 2 …..(i)
L1 1 2 4
1 1 1 5R
For Balmer series R 2 2 …..(ii)
B1 2 3 36
L1 5
From equation (i) and (ii) .
B1 27
Example: 22 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The
ground state energy of an electron of this ion will be [RPET 1997]
(a) 3.4 eV (b) 13.6 eV (c) 54.4 eV (d) 122.4 eV
1 1 1 1 1 1
Solution : (c) Using RZ 2 2 2 1 .1 10 7 Z 2 2 2
108 .5 10 9
2 5
n1 n2
1 21 100
9
1 .1 10 7 Z 2 Z2 4 Z=2
108 .5 10 100 108 .5 10 1 .1 10 7 21
9
Example: 23 Hydrogen (H), deuterium (D), singly ionized helium (He ) and doubly ionized lithium (Li ) all have one
electron around the nucleus. Consider n 2 to n 1 transition. The wavelengths of emitted radiations are
1 , 2 , 3 and 4 respectively. Then approximately [KCET 1994]
(a) 1 2 4 3 94 (b) 4 1 22 23 4 (c) 1 22 2 2 3 3 2 4 (d) 1 2 23 34
1 1 1 1 1 1
Solution : (c) Using R 2 2 1 .097 10 7 2 2 n = 4
n 1 n 2 975 10 10
1 n
n(n 1) 4 (4 1)
Now number of spectral lines N 6.
2 2
Example: 25 A photon of energy 12.4 eV is completely absorbed by a hydrogen atom initially in the ground state so that
it is excited. The quantum number of the excited state is
(a) n =1 (b) n= 3 (c) n = 4 (d) n =
Solution : (c) Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation
13 .6 13 .6 13 . 6
13 .6 12 .4 1 .2 n 2 12 n = 3.46 ≃ 4
2 2 1 .2
n n
Example: 26 The wave number of the energy emitted when electron comes from fourth orbit to second orbit in
hydrogen is 20,397 cm–1. The wave number of the energy for the same transition in He is
(a) 5,099 cm–1 (b) 20,497 cm–1 (c) 40,994 cm–1 (d) 81,998 cm–1
1 1
2
Z
2
1 Z
Solution : (d) Using RZ 2 2 2 Z 2 2 2 4 2 4 81588 cm 1 .
n n 2 1 Z1 1
1
Example: 27 In an atom, the two electrons move round the nucleus in circular orbits of radii R and 4R. the ratio of the
time taken by them to complete one revolution is
(a) 1/4 (b) 4/1 (c) 8/1 (d) 1/8
n3
Solution : (d) Time period T
Z2
For a given atom (Z = constant) So T n 3 …..(i) and radius R n 2 …..(ii)
3/2
T1 R 1
3/2
R 1
From equation (i) and (ii) T R 3 / 2 .
T 2 R 2 4R 8
Example: 28 Ionisation energy for hydrogen atom in the ground state is E. What is the ionisation energy of Li atom in the 2nd excited
state
(a) E (b) 3E (c) 6E (d) 9E
Z2
Solution : (a) Ionisation energy of atom in nth state E n
n2
For hydrogen atom in ground state (n = 1) and Z = 1 E E0 …..(i)
E0
For Li atom in 2nd excited state n = 3 and Z = 3, hence E 3 2 E0 …..(ii)
32
From equation (i) and (ii) E E .
Example: 29 An electron jumps from n = 4 to n = 1 state in H-atom. The recoil momentum of H-atom (in eV/C) is
(a) 12.75 (b) 6.75 (c) 14.45 (d) 0.85
Solution : (a) The H-atom before the transition was at rest. Therefore from conservation of momentum
Photon momentum = Recoil momentum of H-atom or
h E E1 0 .85 eV (13 .6 eV ) eV
Precoil 4 12 . 75
c c c c
Example: 30 If elements with principal quantum number n > 4 were not allowed in nature, the number of possible
elements would be
[IIT-JEE 1983; CBSE PMT 1991, 93; MP PET 1999; RPET 1993, 2001; RPMT 1999, 2003; J & K CET 2004]
(a) 60 (b) 32 (c) 4 (d) 64
Solution : (a) Maximum value of n = 4
genius PHYSICS
34 Atomic Structure
So possible (maximum) no. of elements
N 2 1 2 2 2 2 2 3 2 2 4 2 2 8 18 32 60 .
Tricky example: 1
m2 2
Solution : (d) (rm ) (0 . 53 Å) (n 0 . 53 Å) m n
Z
Z
(5) 2 1
n
100 4
Tricky example: 2
An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV)
required to remove both the electrons from a neutral helium atom is [IIT
22
So energy required to remove second electron from the atom E 13 .6 54 .4 eV
1
Total energy required = 24.6 + 54.4 = 79 eV