Physics by Two Teachers 2

1 PHYSICS BY TWO TEACHERS.
Written by Dr. Danso Kumi and Bright Forkuo.
PHYSICS BY TWO TEACHERS

A REVISION SUMMARY WITH ALTERNATIVE APPROACH
LIFE IS EVER REFRESHING AND REJUVINATING WHENEVER I GIVE A

PART OF ME TO LET OTHERS BE FILLED UP TO THEIR FAR REACHING POTENTIAL
1 CORINTHIANS 9; 19 -23
I MAKE MY SELF ALL THINGS FOR ALL MEN, THAT I MAY SAVE THEM FOR CHRIST, TO THE STUDIUS
I BECOME A TEACHER. TO THE SICK A DOCTOR
BY DR. DENNIS DANSO KUMI
AND BRIGHT FORKUO.
physics by two teachers © 2019

ELECTROSTATICS, MAGNETISM, ELECTROMAGNETIC INDUCTION
 Electromagnetic waves represent a transfer of energy due to a disturbance due to vibrating electric and magnetic fields at
right angles.
 Usually requires no medium
MAXWELL’S PREDICTIONS
 Electric field lines originate on positive charges and end on negative charges.
 Magnetic field lines always form close loops; they don’t begin or end anywhere.
 A varying magnetic field induces an emf.
 A varying charge creates a magnetic field.
Maxwell calculated the Speed of
Electromagnetic Wave to be;V= 1/√𝜺˳ ∪ ˳
Where𝜀˳ = Permittivity of free space

∪ ˳= Permeability of free space
In Vacuum, V= C= 3 x 108 m/s
Maxwell’s predictions were first confirmed by Heinrich Rudolf Hertz in 1887.
Resonant frequency of the electromagnetic wave:

f0= 1/2𝜋√𝐿𝐶
Hertz showed that circuits aside losing power through work done on the load such as resistors and back emfs, can also loose
energy through generation of electromagnetic waves.
NB: Bqv = Eq
v = E/ B
ELECTRIC FIELDS
This is the earliest branch of electricity known since 700 B.C. Greeks noted how Amber, rubbed against other nonconductors
could pick up pieces of items.
The concept of positive versus negative charges was introduced by Benjamin Franklin in 1909. Robert Millikan determined that
all charge was a fraction or multiple of fundamental unit charge = ℯ = 1.60219 x 10 -19 C
NB: Quarks have charges of -2/3 ℯ or +1/3 ℯ.
Concept of Conductors and Insulators Using Band Theory
__________________
Conduction Band
__________________
Valence Band
__________________
Region of nuclear attraction
a. For conductors, these are free electrons in conduction band.

b. For insulators, there are no free electrons in conduction band and the valence band is widely separated from the rest
of the electrons such that very high energy is needed to promote a ground state electron.
c. Semi- conductors do not have free electrons in conduction band but the valence band is very close to the conduction
band. Electrons can easily gain energy and move to the conduction band and create holes in the valence band. So
when semiconductor is conducting current, there is flow of electrons promoted into conduction band in one direction
and drifting of holes left in the valence band in the opposite direction.

Neutralization by
CHARGING MATERIALS
contact
1. By Conduction
Step 1, polarization of sphere final sphere has same charge as rod
PROCESS
i. You bring a charged rod close to an uncharged metal sphere.

ii. You induce polarization causing opposite charges to accumulate close to the charged rod.
iii. You then establish brief contact between the rod and the sphere.
iv. There is transfer of charges and neutralization of the charges at site of contact.
v. The sphere is left with a not charge similar to that of the rod.
2. By Induction
after inducing polarization, you earth the opposite side of sphere.
PROCESS
The initial steps are Similar to that achieved by conduction.
In this case no direct contact is established and no neutralization of opposite charges occurs.
However, the opposite side of the sphere is connected to earth and electrons are removed from the sphere leaving it with a
net charge opposite to that of the rod.
COULOMBIC FORCES
 Named after Charles Augustine de Coulomb, a French military officer and physicist whose landmark experiments in
1785 led to the law of electrostatic forces.
PROPERTIES OF COULOMB FORCES
1. The force is directed along a line joining the center of two particles.
2. It is attractive if charges are opposite and repulsive for similar charges.
3. It is proportional to the product of the charges.
4. It is inversely proportional to the square of the distance between them.
F𝛼Q1 Q2, Constant of proportionality is the coulombs constant.

R2
Ke= 8.9875 x 109𝑁. 𝑚2/c2
Ke= 1/4𝜋𝜀˳ , where𝜀˳ = Permittivity of free space
F = KeQ1 Q2
R2
In case Q1 = Q2 , then F = KeQ2
R2

Some Constants to Remember

 ℯ 2 = 2.56 x 10-38 C2
 [Me]2 = 8.3 x 10-16 kg2
 [Mp]2 = 2.79x 10-54 kg2
 [Mp. Me] = 1.52 x 10-57 kg
 Me = 9.11 x 10-31
 Proton Mass = 1.67 x 10-27
 Kℯ 2 = 2.301 x 10-28
SOME STRATEGIC APPROACHES TO ELECTROSTATIC PROBLEM SOLVING
(a) Where must a test charge be placed between 2 other charges?
When you are given two charges Q1 and Q2 separated by distance R and you are to find where a test charge q3 is to be placed
to have net electric force of 0 on it, assume distance of q3 from Q1 is x.
𝑄1 𝑥2
Then
𝑄2
= (𝑎−𝑥)2 ……………… Kumidees approach
Solve for the Quadratic equation to obtain the value of x.

. Example
3 charges lie along the x axis with q1 = 15uC at x=2m and q2 = 6uC at x=0. where will q3 be placed such that the net force on it is
zero
Ans. Let the distance between Q1 AND Q2 BE A, THEN Q3 IS X FROM Q1,
Q1/Q2= x2/(a-x)2 solve the quadratic to get x = 0.77m or 1.23m
(b) Finding the resultant force (non- zero) on a third charge between 2 other charge
𝑸𝟏 𝑸𝟐
𝑭 = 𝒌𝒒{ 𝟐 ± }
𝑿 (𝒂 − 𝒙)𝟐
Assign the right signs to Q1 and Q2. Use Q1– Q2 if test charge q is in between them and use Q1+ Q2 if q is located outside of
them.
𝟒𝒌𝒒
If it is exactly midway, the force 𝑭 = (𝑸𝟏 − 𝑸𝟐 )
𝑿𝟐
Example
A test charge of q= +2uC is placed halfway between 2 charges Q1=+6uC and Q2= +4uC which are 10cm apart. Find the resultant
force on the test charge
Ans. Q1 q Q2
½X ½X
resultant force on test charge q between Q1 an Q2 is F =Kq/( ½ X )2 x ( Q1 + Q2) , so F = 4Kq/x2 ( Q1 –Q2). Keeping the
signs of Q1 and Q2 in mind. F= 14.4N away from the +6uC charge. Note that the net direction of the force is determined by
the bigger of the two stationed charges.
CALCULATIONS
To be able to effectively solve electrostatic questions, you should understand the terms and properties of a charged body.
Every charged body creates around it, its own electric field (FORCE FIELD). Each point within this electric field has a property
called electric field strength E. The region beyond the reach of the electric field due to the charged body is called INFINITY. The
electric field strength at infinity is 0(zero).

What is the meaning of this electric field strength? It tells us what force another charged body of 1 coulomb will experience
when brought to that point within the electric field.
So when I say the electric field strength at a point is 0.3N/C, I mean that when a charge of 1 coulomb is brought to that field it
will experience 0.3N force.
Now if a 2C charged body is brought to that same point, what force will it experience?
You are right, 0.6N.
Hence the formula F=q×E.
Which means that, E= F/q. That means Electric field strength of a point in an electric field is defined as the force per charge
experienced by a charged body placed within that point.
2. Every CHARGE BODY creates around it another imaginary field, I will call, ENERGY FIELD. You can imagine that when we want
to move a positive charge to this CHARGED BODY (positively charged), I would have to do some work.
The work I do to bring a body from infinity to a point A in the electric field is what is called the electric potential energy.
The point A has a property we call the electric potential. Which is the work done per unit charge to move the test charge from
infinity to the point A.
Imagine two points A and B in the electric field. Both A and B have their electric potentials. Now whenever you are moving from
a higher potential to a lower potential, no external work is required. External work is required to move the test charge from a
lower potential to a higher potential. Now the difference between the potentials at A and B is called potential difference or
voltage.
Technically, potential difference is defined as the work done per unit charge to move a positive test charge from a point A of
lower potential to a point B of higher potential within an electric field.
 Work done (electrical potential energy) = (kq1*q2)/r
 Work done per charge (voltage or electric potential) = (kq)/r
Consider this equation

Work = force*distance
Think of voltage as work and electric field strength as force.
This brings another important equation
V = Ed.
This can be proved using the definitions.
If the potential at a is va, and that at b is vb . and the electric field strength is constant and is E. and the distance between A and
B is d.
Then VB – VA = Ed
ELECTRIC FIELDS AND ELECTRIC FIELD STRENGTH (ELECTRIC FLUX DENSITY)
The idea of field was introduced by Michael Faraday.

A charge Q produces a certain electric field around itself such that, when a test charge q0 is placed within this field, it will
experience a certain Force F. the measure of the Force (F) experience for a unit charge is called the electric field strength of Q.
E = F/ q
NB: The force between Q and qo→ 𝐹 = 𝐾𝑒𝑄𝑞
2
𝑅
E = F/q0 Hence 𝑬 =
𝑲𝒆 𝑸
𝑹𝟐
This shows that the electric field strength only depends on the magnitude and direction of Q and the distance as one move
away from Q.

PROPERTIES OF ELECTRIC FIELD

1. It only depends on magnitude of the charge producing it and totally independent of the test charge q0 placed within
it.
2. It decreases as one moves away from Q.
3. Its vector direction is always tangential to the average flux passing through a given area.
4. Its sign follows a Cartesian coordinate.
+ E
_ Q + -
NB: If the charge Q is a positive charge, E follows as above. If Q is negative, E is reverse in all directions as compared to what is
assigned to a positive charge.
NB: Force experiment by a test charge

F=Exq
MILLIKAN’S OIL DROP EXPERIMENT
Atomized oil drop
Field force = Weight

Eq = Vq/d = mg
Robert Millikan atomized oil to form tiny drops which he let down in a chamber of very low pressure almost like vacuum with
negligible air resistance or viscous drag. The process of atomizing the droplets made them charged. He applied an electric field
to the chamber and varied the field strength by changing the potential difference between the top and bottom plates. The
motion of the falling oil droplet was then monitored.
The charge on each droplet was estimated and when the charges were compared, he concluded that each charge was a ratio of
a single universal charge e = 1.6 x 10 -19C
1. If the oil drop is still Mg = Eq
2. If the drop accelerates upwards, Eq = M(g+a)
3. If falling down, Eq = M(g-a)
Example
The charge on an electron is 1.6 x 10 -19C. An oil drop has weight of 3.2 x 10 -13N and has an electric field of 5 x 10 5 V/m
between the plates of the Millikan’s oil drop apparatus. If the drop is observed to be essentially balanced, what is the charge on
the drop?
Ans = 6.4 x 10-19C
ELECTRIC FIELD LINES

Properties
1. Always start from positive charge and terminates negative charge.
2. Always seeks a field with less permittivity.
3. The density decreases as one moves further away from the charge.
4. They do not cross each other.
5. The number of lines is proportional to the magnitude of the charge.
ELECTRIC FLUX: The number of field lines permeating through space due to a certain charge Q is termed the Electric flux.
𝑸
∅𝑬 =
𝜺𝟎
The above is obtained from Gauss’ law which states that; for a given charge in a closed surface, the electric flux through the
surface is proportional to the amount of charge contained in the body.
A device used to measure electric fields is called the Field Mill.

A device used to measure amount of charge is the Electrometer.

Many experiments confirm that for a charged conductor, the charges reside on the surface. The electric field inside a charged
sphere is zero. The electric potential is however equal inside and on the surface of a charged sphere.
One experiment to confirm this was the Faraday’s ice-pail experiment
VAN DE GRAAF GENERATOR

The electrostatic generator was built in 1929 by Robert Van de Graaf.
+ + +
+ +
+B +
+
+
+
++
A
As charges build on the dome to a considerable point, the air around starts to get ionized and charges can leak out giving a
lightning bolt. To prevent this, the radius of the dome can be increased or the system put in pressurized gas.
Protons can be accelerated in nuclear reactors by introducing them into the generator. The high repulsion from the dome gives
this energy. Electrons could be accelerated using this generator for the production of x rays and for production of fast speed
protons for nuclear collisions.
LOOKING AT ELECTRIC FIELDS IN TERM OF FLUX

Electric fields are established in an area proportional to the amount of flux that is passing perpendicularly through that area.
ØE/A𝑐𝑜𝑠 𝜃 = E
ØE=EA cos 𝜃
Likewise ØE = Q/Ɛ0 , E= (Q/Ɛ0)/𝐴𝑟𝑒𝑎
For a spherical, find Area= 4𝜋𝑟 2
Therefore, E = (Q/Ɛ0)/4𝜋𝑟 2
𝑄 1
𝐸 = 4𝜋𝜀 2 ………….(1) but since the Electric field constant ,𝐾𝑒 = 4𝜋𝜀
0𝑅 0
𝐾𝑒 𝑄
𝐸= 𝑅2
…………….(2)
NB: Electric flux Density = Electric field strength

𝑸
Electric charge density = D=
𝑨𝑹𝑬𝑨
𝑫
E=
𝜺𝟎
Electric Potential
One electron volt is the work done or gain in kinetic energy to move an electron through a potential difference of 1 Volt.
1ev = 1.6 x 10 -19 C.V or 1.6 x 10 -19 J. the rest mass of an electron is often given as 0.511 MeV.
Electric potential (V) is the work done when a unit charge is moved from infinity into a point in an electric field. The value of V
is always given negative sign meaning potential increases with distance up to infinity where it becomes zero.
V =r∫∞ E. dr
V= E x d
V= 𝑹
−𝑲𝑸
E = -V/d, and the expression V/d is called the potential gradient.

This parameter is not easily measured but rather measured relative to other chosen field hence we need to discuss potential
difference.
EQUIPOTENTIAL SURFACES
The surface of a charged sphere is said to be equipotential because the potential difference of any 2 points on the surface is
zero. Remember that electric fields are conservative and work done between two points do not depend on the path taken. The
inside and the surface of a hollow sphere are all equipotential. Work done in an equipotential space is zero.
POTENTIAL DIFFERENCE
The difference in electric potential between 2 points. It is the same as the voltage
When the initial point is at infinity, the potential difference is referred to as the Absolute Potential V=
𝑲𝑸
𝑹
The electric potential is usually stated as the absolute potential since using infinity as a reference is easier. Remember that
absolute potential Is actually a form of potential difference
However, to find Potential difference between points a and b, where non is at infinity
𝟏 𝟏
𝑽𝒂𝒃 = 𝑲𝑸{ − }
𝒂 𝒃
ELECTRIC POTENTIAL ENERGY

This is the work done in moving a test charge Q from one point to another within an electric field.
Given 2 or more Absolute Potentials or Potential difference, the resultant is a vector sum of all of them.
Some Facts about Electric Potentials

1. The electric potential or potential energy decreases with decreasing distance from the source charge.
2. The work done is said to be negative if there is an attractive force on the test charge and positive if there is a
repulsive force on the charge.
3. For a hollow sphere, the surface and any point inside it is equipotential.
4. The potential difference for an equipotential surface is zero between any 2 points.
5. Potential difference is a conservative field parameter and it is independent of the path traced by the charge.
WORK ENERGY THEOREM IN ELECTROSTATIC FIELDS
½ mv2 = electric force (Eq) x distance travel (d)

½ mv2 = Eqd. Note that E.d = V
½ mv2 = Vq
Velocity of a charged particle in an electric field can thus be determined as;
𝟐𝒒𝑬𝒅
𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 = √
𝒎
𝟐𝒒𝑽
𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 = √ = √ [ 2 x voltage x charge to mass ratio]
𝒎
Remember the electric potential energy PE = Vq

Example.
A proton is released from rest at x = -2cm into a constant electric field of 1.5 x 10 3 N/c which points in positive x direction.
Determine the change in potential energy when the charges reach x=5cm.
Ans. Δ PE = -Eq x ΔX = -1.68 x 10 -17J

CAPACITORS AND DIELECTRICS
V
Q Q
A
d
𝑸
𝑪=
𝑽
Unit for capacitance is the Farad.

𝑨𝜺𝟎
𝑪=𝑲
𝒅
The circuit symbol of capacitor is
COMBINATION OF CAPACITORS
C1
C2
a. In Parallel, different charges build on each capacitor but same potential difference across all is the same.
Q= CV
Total Charge, QT = Q1 + Q2 + Q3
QT = C1V + C2V1 + C3V
QT = V (C1 + C2 + C3)
𝑸𝑻
= CT = C1 + C2 + C3]. So to find the total capacitance, you simply add.
𝑽
C1 C2
b. For series connection, it is same charge but different Voltages
CT = 1/C1 + 1/C2 + 1/C3, etc. so in series, you add the reciprocal capacitances.
c. Energy stored in a capacitor
W = ½ QV = ½ CV.V
W = ½ CV2
𝑸𝟐
𝑾=
𝟐𝑪
When a material that can be polarized is placed between the plates of a capacitor, the material is called Dielectric.
The capacitance is increased by a factor K called the Dielectric Constant.
The Voltage across is thus reduced by K. the charge is also increased by K.
𝑲𝑨𝜺𝟎
𝑪= , for Air K = 1 C = KC0
𝒅
The maximum electric field beyond which a dielectric begins to conduct charges is called the Dielectric Strength of the material.
COMBINATION OF DIELECTRICS
NB: If 2 dielectrics K1 and K2 are used, then KT= Combination of K1 and K2 as follows:
(a) k1
K2

Total dielectric constant kT = ½ (k1 + k2)

(b)
K1 k2
Total dielectric constant kT = 2k1k2 / (k1 + k2)

Examples
. A certain parallel plate capacitor consists of plates each of area of 200cm2 separated by a distance of 0.4m of air. Find the
capacitance.
Ans C=ε0A/d =ε0 is 8.85 x 10 -12., C = 44pF
3 capacitors (2uF, 3uF and 4uF) are connected in series to a 6V battery. When the current stops, what is the charge on the 3uF
capacitor?
Ans. 5.5Uc
DEBYE.
It is the unit for molecular electric dipole moment. 1 Debye = 3.339 x 10-30 C/m. the unit is named after peter Debye an
American-Dutch physicist and chemist (1884 to 1966). He became head of the max plank institute after Einstein. He is credited
with the development of the concept of Dielectric constant and molecular dipole moment
DETERMINATION OF ELECTRIC ENERGY DENSITY

Electric energy density(mu) = Energy due to electric charge per unit volume.
In a parallel plate capacitor, Energy due to the charge (W) = ½ CV2 and C = ε0A/d, also Voltage (V) = Ed, so substituting gives
W = ½ ε0A.d E2, Energy density (μ) thus is given as
μ = ½ ε0 Ad E2 / Volume, notice Ad is also volume hence cancelling out leaves us with the formula, μ = ½ε0E2.
ε0 = 8.85 x 10-12fad/m, hence μ = 4.425 x 10-12 E2
question. teaser…
A boy is 2 years older than his sister and 15 years ago, the sum of their ages was 20, how old are they… Answer in 5sec,
26years and 24years.

MAGNETIC FIELD( B )
The first magnets came to light in the city of Magnesia in Turkey. Soft magnetic materials are easily magnetized but also easily
lose magnetism.
Hard magnets retain their magnetism over a long term hence become permanent magnets.
Soft magnets include Nickel, Iron, Ferrites, and Cobalt.
The earth crust is abundant with basalt which is responsible for most of its magnetism.
AURORA BOREALIS: Plasma waves from the sun approach earth. Due to the dense charges moving at high speed, the waves
travel with a strong magnetic field around it. At the earth’s poles, there is deflection of the magnetic fields and this causes the
dissipation of energy in form of photons giving the bright lights at the poles.
THE EARTH’S FIELD

Magnetic south pole
Geo axis
equator
Magnetic axis geographic south pole.
a. Declination: The angle between the magnetic south and geographic north pole
b. Inclination: Angle the magnetic axis makes with the horizontal.
BIOPHYSICS: It is believed that anaerobic bacteria survive by following magnetic dipoles of the earth in direction of
anaerobic environments.
MAGNETIC FIELDS
PROPERTIES OF FIELD LINES
1. They do not cross

2. Within magnet, move from south to north and outside magnet, they move from north to south.
3. They are denser closer to the poles.
MAGNETIC FIELD FORCES ON MOVING CHARGES
For a charged particle
A charged particle of rest will not be experiencing any force due to a magnetic field. A force is only created when there is a
movement of a charge within a field.
F= B.q.v.
Where q is charge, B = field strength, v= drift velocity of the charge.
The force is in the direction perpendicular to the force. Hence F = Bqv 𝐬𝐢𝐧 𝜽
 The force is only like a centripetal force with a turning effect but no work done.

for a charge carrying or current carrying metal rod
If there is a charge carrying conducts or current streaming in a conductor, then the force on that conductor is an average
of the force on all the individual charges moving.
Each charge experience force = Bqv. For a given volume of conductor A.l, the total charge flow = ∩Al
Total force experienced is therefore, Bqv. ∩Al
∩ 𝑞𝑣𝐴 = 𝐼
F = 𝜷𝑰𝑳 𝐬𝐢𝐧 𝜽
Example : A proton moves with the speed of 1 x10 5m/s through a magnetic field of 55uT at a particular location. When
the proton moves eastward, it is deflected upwards by the force due to the field but when the proton moves directly
upwards, there is no deflection. Calculate the magnitude and determine the direction of the magnetic field.
Ans , F = Bqvsinθ, F = 8.8 x 10 -19N and points northward
For a current carrying metal loop
If the conductor is turned into a loop, there it experiences a turning effect with a torque.
T=FxL
T= 𝐵IL x L sin 𝜃
T = 𝑩IA 𝐬𝐢𝐧 𝜽
For structure will multiple loops

If the loop has N turns, then
T = 𝑩𝑰𝑨𝑵 𝐬𝐢𝐧 𝜽
IAN is called magnetic moment = 𝜇
T = 𝝁𝑩 𝐬𝐢𝐧 𝜽
NB: The part of a loop that is parallel to the field has no force on it.
Example: A wire rectangular coil of sides 2m and 3m have a current of 2A passing through it. The plane of the coil is 30 0
angle with the magnetic field of magnitude 0.5T. Find the torque on the coil
Ans. T = BIAN sin θ == 9.00 Nm
FINDING THE DIRECTION OF MOTION

The force on a moving charge in a magnetic field is always centripetal force, with acceleration directed to the center of a
circular path.
mv2/r = 𝐵qv
r= mv/q𝑩=m/q. V/𝐵 This is the Cyclotron eqution.
Mv (p) =𝑩qr(momentum)
Use the right hand rule to estimate the direction of the motion.
Thumb = direction of deflecting force due to external field
Index finger = direction of particle velocity
Rest of curled fingers = direction of field(B)

Hold out your right hand such that the thumb, index finger and the rest of three fingers are separated and almost at right
angles. Orient your hand in a posture such that the index finger faces the direction of the moving particle and the rest of the
fingers curled together facing the direction of the external magnetic field. The thumb points then to the direction of the
deflecting force on the charged particle due to the field.
MAGNETIC FIELD CREATED BY MOVING CHARGES THEMSELVES
Magnetic field is measured in Tesla or Wb/m2 or G(gauss). Moving charges experience a force in a magnetic field because they
themselves start to act as magnets themselves as they speed. When a charge is moving, it creates a magnetic field around itself
and it is this field that either repel or attract an external magnetic field with the characteristic forces initially described. These
observations were initially made by Hans Christian Oested in 1820 and he stated that current- carrying conductors produced
magnetic fields.
1 Tesla =1x 104G

Let’s assume XY is a very small portion of a current carrying wire with the distance of its
LAW OF BIOT- SAVART length XY = δL, a radius r and current flowing through it given by I. The change in
magnetic field at a point B due to current flow from the wire of length XY is given by
B the BIOT-SAVART LAW as
𝝁𝟎 𝝏𝑳𝒔𝒊𝒏𝜽 𝝁𝟎 𝑰 𝝏𝑳𝒔𝒊𝒏𝜽
X Y 𝜹𝑩 = ×𝑰× , which is same as 𝜹𝑩 =
𝟒𝝅 𝑹𝟐 𝟒𝝅𝑹𝟐
for an infinitely long wire, if we need to find the entire magnetic field B at any point a
fixed distance around the wire, then we must find the integral sum of δB which on
integration leaves us with the formula
𝝁𝟎 𝑰
𝑩=
𝟐𝝅𝒂
APPLYING THE CONCEPT TO LOOPS
where a is the distance from the wire. A point located along the direction of current
flow has no magnetic field
A B X C D
The wire ABCD is bent into a hemi loop with radius r at its midpoint between B and C and the center of the semicircle is marked
by the blue dot. with current I flowing from A to D. you will realize that within the loop portion and more so at its center, the
current in the straight portions AB and CD are acting at 0 degree and hence sin0 = 0, so these areas have zero magnetic field
within the loop.
we are thus left with only the loop portion. By extrapolation, any cross section of the loop will be at right angle to the point X.
using the Biot-savart law we will obtain the field due to any short length δ L to be,
𝝁𝟎 𝑰 𝝏𝑳𝒔𝒊𝒏𝜽 𝝁𝟎 𝑰 𝝏𝑳
𝜹𝑩 = but since sin 90 =1, we get 𝜹𝑩 =
𝟒𝝅𝑹𝟐 𝟒𝝅𝑹𝟐
. To get the total magnetic field of the entire hemi loop at point X, we will need to integrate the expression. it is seen that the
radius remains constant hence it is only δ L that is variable. integrating δ L in a semicircle is equal to the circumference of a
semi-circle which is πr.
the formula thus becomes,
𝝁𝟎 𝑰
𝑩=
𝟒𝒂
where a is usually the radius of the loop of the distance from the center of loop.
For a complete loop

The only difference is that the integral of δ L will yield the circumference of a full circle which is 2 πa. putting that into the
equation will yield
𝝁𝟎 𝑰
𝑩=
𝟐𝒂
is
A
AMPERES POINT OF VIEW
Andre Ampere, then demonstrated that the magnitude of this induced magnetic field was proportional to the current in the
conductor and inversely proportional to the distance away from the conductor.
a. Field of Straight Long Wire Carrying Currents
The field created is a circular pattern.
current The distance around the rod = 2𝜋𝑟
circular field a distance r from the straight rod.
According to Amperes law, B 𝜶 𝑰/𝟐𝝅𝒓.
The constant of proportionality is 𝜇0 = Permeability of free space
So for a straight rod carrying current I, the field B a distance of r from the rod is given by, B = 𝝁0I/2𝝅𝒓 . 𝜇0= 4𝜋 x 10-7T.M/A
𝑰. 𝟐 × 𝟏𝟎−𝟕
𝑩=
𝑹
b. The force between 2 Current Carrying Conductors
L1 L2
2 current carrying conductors close to
B1 B2 each other will attract if current is in
F same direction and repel if the
currents are in opposite directions
F=BIL
Where d is distance between the two,
𝝁𝟎 𝑰𝟏 𝑰𝟐
𝝁𝟎 𝑰𝟏 𝑰𝟐 𝑰𝟏 𝑰𝟐 .𝟐×𝟏𝟎−𝟕
𝑭= 𝟐𝝅𝒅
𝑳
=𝑭= 𝟐𝝅𝒅
𝑳 𝑭=
𝒅
𝑳
𝝁𝟎 𝑰𝟏 𝑰𝟐 𝑰𝟏 𝑰𝟐 .𝟐×𝟏𝟎−𝟕
f/l( force per unit length) = 𝑭 = 𝑭=
𝟐𝝅𝒅 𝒅
example

Estimate the magnitude of the magnetic field per unit length between 2 parallel wires 2m apart if they both carry a 3A
current.
Ans , F = U0 I1 I2 / 2π d but since it is same charge , F/l = U0 I2 / 2πd = 9 x 10 -7 N/m
c. Current in a loop:
𝝁 𝑰
𝑩= 𝟎
𝟐𝒂
d. Loop with N turns:

𝑵𝝁𝟎 𝑰
𝑩=
𝟐𝒂
e. For a solenoid:
𝑵𝝁𝟎 𝑰
𝑩= ,N/L = 𝑛
𝑳
B = U0𝑛𝐼 n = turns per unit length
example
A solenoid consists of 100 turns of wire and has a length of 10cm. If it carries a current of 0.5T, find the magnitude of the
magnetic field around this solenoid.
Ans B = U0 n I where n= N/L B = 6.28 x 10 -4 T
The resultant field between two conductors, A and B.
The field Ba and Bb are added to give the resultant. If the 2 carry current in the same direction, it is repulsive so Ba – Bb. if in
opposite direction, it is attractive so Ba + Bb.
𝝁𝟎 (𝑰𝟏 ±𝑰𝟐 )
𝑩𝒂𝒃 = Use + if attractive and – if repulsive.
𝟐𝝅𝒅
MAGNETIC DOMAINS
Magnetic properties of matter are due to electron movement around the nucleus or spin of electrons in orbitals. Usually, the
magnetic field created by one electron is cancelled by an opposite field of a nearby electron. Certain materials such as iron,
nickel, Cobalt, however do retain some residual magnetic properties and are termed ferromagnets. In an external magnetic
field, they align in the same direction as the field and the magnetization is very strong. The ordered arrangement of charges
such that a non-magnetic field is created is termed Magnetic Domain.
Ferromagnets have several magnetic domains that are aligned in the same direction. Susceptibility is positive and large.
Paramagnets have domains that are haphazardly arranged and only align in an external field. A weak magnetization is achieved
in the same direction as the field. The susceptibility is positive but small and non-zero.
Common paramagnets include:
 Free radicals
 Atoms with incomplete inner shells
 Atoms with unpaired valence electrons
 Metals with free to move electrons
The inducibility of a paramagnet for it to attain orderly arranged domain is called MAGNETIC SUSCEPTIBILITY = 𝜇
Ferromagnets can be heated above a certain temperature to introduce enough disorderliness into its domain hence it starts
behaving as a paramagnet. This temperature above which ferromagnets behave as paramagnets is called The Curie
Temperature Tc.
GROUPS OF MAGNETIC MATERIALS

Ferromagnet.
Ferrimagnet
paramagnet
CURIE- WEISS LAW
The curie law states that the magnetization of a material is directly proportional to an applied magnetic field above a threshold
temperature but with further heating, the magnetization is inversely proportional to the temperature.
Definition of Symbols and Terminologies
1 𝝁 = 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑠𝑢𝑠𝑐𝑒𝑝𝑡𝑖𝑏𝑖𝑙𝑡𝑦
2 M = magnetization (magnetic moment per unit volume) in A/m
3 B Magnetic field strength in Tesla or Gauss
4 H macroscopic magnetic field of the material due to its magnetic moment in A/m or oested
5 C curie constant
6 f Flux in Weber or Maxwell
7 λ Weiss field constant
8 u0 = permeability of vacuum
THE PROPERTIES OF CURIE TEMPERATURE
It is depended on:
1. Particle size
2. Pressure
3. Heating
4. Composite material
Curie Weiss law
B =u0(H + M) …………….1,
𝑪
𝝁= , where T0 is the initial temperature of the material
(𝑻𝟎 −𝑻𝒄 )
μ = M/H
𝑪λ
𝑻𝒄 =
𝝁
DIAMAGNETISM

It is a quantum phenomenon and a property of all matter
First analyzed in 1778 by Sebald and co whiles observing how Antimony and Bismuth were repelled by external fields. The term
diamagnetism was coined by Michael Faraday. In such materials, an applied magnetic field induces a field that is opposite and
hence repelled. Materials which have predominant diamagnetic properties are called Diamagnets. Most so called non magnets
are diamagnetic. E.g. Water, wood, Mercury, Gold. Most elements with full core electrons.
Permeability is <𝜇0
Susceptibility ≤ 0
The susceptibility is small and negative or 0.
ELECTROMAGNETIC INDUCTIONS
Based on Oested’s idea that electric current produced magnetic field, Michael Faraday and Joseph Henry independently had
investigated the possibility of the reverse. i.e. a magnetic field could produce current.
FARADAY’S EXPERIMENT
Primary coil secondary coil
+ - A Ammeter
Switch (S) Metal loop
When the switch (s) was closed, the Ammeter deflected in one direction and stopped. When the switch was opened the
ammeter again deflected but in the opposite direction and stopped. It was really not about current flow but about sudden
change in current either increasing from zero to maximum or suddenly reducing from maximum to zero that did the trick. The
changing current in primary loop creates a magnetic flux and thus this magnetic flux also changes. It is the changing flux that
induces emf in the secondary coils. And the direction of the flux change determines the direction in which the ammeter
deflects.
A constant magnetic field does not produce current. Either the field strength or direction must be changing from 0 to max or
max to 0 for induction of emf and thus current to be produced.
The magnetic flux(∅B) refers to the number of magnetic field lines.
Flux density (B) refers to the field strength. I.e. number of flux per unit area.
∅B / 𝐀 cos 𝜃 = B
∅B = BA cos 𝜃
∅B is in units of Weber
FARADAY’S LAW OF INDUCTION

Whenever there is a change in magnetic flux linking a circuit, an emf is induced. The instantaneous emf induced in a loop is
directly proportional but negative to the rate of change of flux through the loop.
∆∅𝑩
𝑬=−
∆𝐭
For a loop with N forms,
𝑵∆∅𝑩
𝑬=−
∆𝐭
Lenz law explains why the negative sign was observed by Faraday.

The current due to induced emf travels in a direction so as to produce a magnetic field with flux opposing the original flux
through the circuit.
MOTIONAL EMF CALCULATIONS
a. For any current carrying conductor or charge in translational motion in an external field
Eq = Bqv
[E (electric field) = Bv], where v is the velocity.
Voltage = El (remember Voltage=Ed in electrostatics??)

Emf = Voltage = Blv
Motional or induced Emf = Blv
Or
∅ = BA
Rate of change in flux ∅/ t= BA / t = Bl. l/t, notice L/t = v
∆∅
So we obtain = Blv = Emf
𝒕
b. For a loop moving in translational motion, we maintain E= BLV.

If the loop is however in rotation,
Velocity = Rω𝐬𝐢𝐧 𝛚𝒕
Emf = BLv = BL x Rω𝐬𝐢𝐧 𝛚𝒕, notice that R x L for a loop of wire = Area
so
Emf = B𝑨𝝎 𝒔𝒊𝒏 𝝎𝒕 or B𝑨𝝎 𝒔𝒊𝒏𝜽
If loop has N turns, Emf = N B𝑨𝝎 𝒔𝒊𝒏 𝝎𝒕
Example
An AC generator consist of 8 turns wire forming a loop with area 0.09m2. The coil rotates in a magnetic field of 0.5T at a
constant frequency of 30/πHz. Find the maximum induced emf
Ans. W = 2πf = 60rads/s, E = NBAw = 21.6V
Example.
Answer true or false, two coils A and B are wound separately around a solid metallic conductor. Coil A is connected to a cell
and a switch to form one circuit. Coil B is only closed with a voltmeter and the two coils are separated by a distance D.
(a) There is no magnetic field when a steady current is maintained in A

(b) There is a magnetic flux in coil B when a steady current flows in A
(c) There is zero voltage in B if a steady current flows
(d) Turning off the current will cause a zero magnetic flux in B
(e) Turning off the current will cause a voltage reading in the B circuit
DEALING WITH BACK EMF

IR = E – E back

SELF INDUCTANCE(INDUCTOR)
Inductors(reactors) are usually coils with a number of turns N that has significant self-inductance. Inductors store up part of the
circuits energy in magnetic fields which it creates as current passes through it. Any change in the current such that this
magnetic field is disturbed will be opposed by the inductor temporarily and thus they are said to be reactionary. A current will
always build a magnetic field but it is only if the field strength is changing that an emf will be built.
Direction of current I/A

+++ whenever current(I) flows across an inductor(L) with (N)
turns, it sets up a voltage across itself parallel to the plane of
V the coil V. This voltage becomes the back emf to the current
flow initially until a steady is achieved. Any change to this
---
steady state is again impeded. The ability for the coil to build
LET THIS BE COIL 1
up its own emf is due to a property called SELF INDUCTANCE
Voltage ∝ Rate of current change
The Voltage across the inductor is directly proportional to the
∆𝐈
V∝ ∆𝐭
rate of change in current through it. The constant of
proportionality is the inductance L
∆𝐈
𝑽=𝑳
∆𝐭
This means is current is constant, V=0

∆𝐈
𝑬 𝒃𝒂𝒄𝒌 = −𝑳 …..(1) where L is the self- inductance of the material measured in Henry after Joseph Henry. 1Henry =
∆𝐭
1Volt.s/A
Another concept to keep is that the self-inductance is not fixed but varies with the number of turns and inversely as the current
flow. The higher the self-inductance, the lesser current will flow through a coil. The self-inductance also directly relates with how
much the magnetic field will be changing with changing current. Mathematically;
∆∅𝑩
𝑳=𝐍 ……(2)
∆𝐈
we can generate a third formula by substituting (2) into (1)
𝑵∆∅𝑩
𝑬𝒃𝒂𝒄𝒌 = − …. (3)
∆𝐭
CONCEPT OF MUTUAL INDUCTANCE

It is the property of a circuit element or coil which allows the
buildup of a back emf when there is changing current in a
coupled coil. The 2 circuits must not have direct electric The changing flux in the nearby coil due to proximity in space
contact (to avoid short circuiting) but must be in close will pass through the nearby coil and will cause it to also have
proximity. The mutual inductance is the property of the self- its own field. This induced magnetic field thus will creat and
inductance of the two coils as well as the proximity in space. If induced emf assuming it keeps changing. The second induced
the two insulated coils are wound around a core of iron, due
to the ferromagnetism of iron, it helps to concentrate the flux
emf is given by; Emf(back emf)2 = -M∆I1 / ∆t but the mutual
inductance is dependent on
lines and thus increase the rate of change of the magnetic
field and by so doing will increase the mutual inductance. ∆∅𝑩 𝟐
𝑴 = 𝐍𝟐
∆𝐈𝟏
𝑵𝟐 ∆∅𝑩 𝟐
𝑬𝒃𝒂𝒄𝒌 𝟐 = −
∆𝐭
OR
𝑵𝟐 ∆𝐁𝟐 × 𝐀𝐫𝐞𝐚𝟐
𝑬𝒃𝒂𝒄𝒌 𝟐 = −
∆𝐭

1 2
Example
A solenoid with 900 turns has a total flux of 1.33 X 10-7 Wb through its air core when the coil current is 100 mA. If the flux takes
75 ms to grow from zero to its maximum level, calculate the inductance of the coil. Also, determine the counter-emf induced in
the coil during the flux growth.
Solution
(1) 1.2mH
(2) 1.6mV
HENRY = WEBER/AMPERE
In principle, the mutual inductance from coil 1 to 2 is same as the connection from 2 to 1.
If the initial current is in the coil 2, the induced emf in coil 1 is simply the product of changing flux in 1 and number of turns in 1
divided by the current in 2 times the common mutual inductance.
EMF induced in 1 due to current in 2

∆𝐈𝟏 𝑵𝟏 ∆∅𝑩 𝟏
𝑬 𝒃𝒂𝒄𝒌 𝟏 𝒅𝒖𝒆 𝒕𝒐 𝟐 = −𝑴 ∆𝐭
𝑬𝒃𝒂𝒄𝒌𝟏 𝒅𝒖𝒆 𝒕𝒐 𝟐 =− ∆𝐭
∆∅𝑩 𝟏
𝑴𝟏 𝒅𝒖𝒆 𝒕𝒐 𝟐 = 𝐍𝟏 note that M1 due to current in 2 = M2 due to current in 1 hence the relation below can be
∆𝐈𝟐
made.
N1/N2 = ∅B2 / ∅B1
Remember that due to the negative sign in the formula, we can conclude that an increasing current causes a decreasing EMF to
zero and a decreasing current causes an increasing EMF.
NON INDUCTIVE COILS

circuit components that require high precision must avoid changes to the energy due to induction. This is achieved by ensuring
the coil is doubled on itself but in opposite directions. every magnetic field is cancelled but the flux in the opposite direction.
ENERGY STORED IN MAGNETIC FIELD

PE = ½ LI2
A moving charge accelerated into a magnetic field from an electric field.
KE = ½ mv2 = Bqv = 𝓮 𝑽
(𝟐𝒒𝑽
The velocity when it is in the Electric field is V = √
𝒎
Velocity in the magnetic field, v= 2Bq/m

Combining both fields, v = E/B.
Example
A coil of 25 turns forms a 2quare loop with a side of 1.8cm. If an applied field perpendicular to the loop changes from 0.00
T to 0.50 T 0ver 0.8s, determine the induced emf
Ans. E = -N change in flux/ change in time. Change in flux = Area x change in magnetic field, so flux change = 1.62 x 10 -4wb,
emf = -5.06 x 10 -3 V
Example
A current in a 5H inductor decreases uniformly at a rate of 2A/s. What is the voltage drop across the circuit?
Ans, E = -10V
WAVES, S.H.M., SPRINGS, RESONANCE, MODERN PHYSICS
A wave is the transfer of energy due to a vibrating body which causes disturbance usually in medium without the displacement
of the medium from one point to the other.
Waves can generally be described as:
1. Mechanical Waves; always need a matter medium.
2. Electromagnetic Waves; can propagate in vacuum.
Waves that move from one point to the other in a unidirectional pattern are called TRAVELLING or PROGRESSIVE WAVE.
Waves that appear to be still are called STANDING OR STATIONARY WAVES. These are formed when two similar waves interact
whiles in 1800 out of plastic or simply when the same wave is reflected on itself after bouncing off a hard point.
Remember all waves can be reflected. If reflection is against a hard surface, the reflected wave has 1800 phase difference.
Depending on the spatial relationship between direction of wave propagation and amplitude of vibration, waves can be
classified as:
a. Longitudinal waves
b. Transverse waves
c. Surface waves
A soliton is a peculiar type of single circular wave that travels from one point to the other undiminished.
Examples of Disturbances and the Wave they generate
i. Seismic waves --- plate tectonic shift
ii. Radio waves --- vibrating electrons
iii. Sound Waves --- Compression and relaxation of matter
GENERAL PROPERTIES OF WAVES

i. Reflection ii. Refraction iii. Interference
iv. Dispersion v. Polarization (only transverse waves)
GENERAL FORMULAR OF
a. Travelling waves:
+ crest
amplitude
- λ Trough
The particle at t0 may be at origin or at a certain angle ∅ called the Initial Phase.
Y = a 𝐬𝐢𝐧(∅ ± 𝒘𝒕) --------- 1
If the wave is travelling towards the right Y = asin(∅ + 𝑤𝑡).
If travelling towards left, Y= a sin(∅ − 𝑤𝑡)

∆∅ is called the phase Difference and is given by ∆∅ = 𝑲 (𝒙1 – 𝒙2) where x1and x2 are displacements at 2 different times. The
variable (x1 – x2) is called the path difference. The constant K is called the wave constant which is given by K =2π/λ.
The relation 1/λ is called the wave number.
Example
The equation of the motion of a mass at the end of a spring is (y=0.3 cos 0.5t) meters. Find the velocity and acceleration at
t=0
Ans. V=dy/dt=-0.15sin 0.5t, at t=0, V=0

A=dV/dt=-0.075 cos 0.5t, at t=0, a= -0.075m/s2
SUPER POSITION PRINCIPLE
2 wave with similar waves properties such as amplitude and Frequency Passing through a common medium in close
proximity can interact to form a resultant wave whose amplitude is an algebraic sum of the 2.
During this process of interaction, interference of waves occurs and this could be constructive or deconstructive. The
composite wave formed is called a standing wave when the 2 interacting waves are1800out of phase or so to sat the same wave
reflecting on itself in the opposite direction.
point of constructive interference from large amplitude and are called antinodes, whiles point at destructive interference have
zero amplitudes and are called nodes.
EQUATION OF A STATIONARY WAVE
consider 2 waves, y1 = a sin (∅ + wt)……1 and y2 = a sin (∅ – wt)….2
If these two interact, the resultant wave called the stationary wave is given by the equation Y = 2acos kx sin wt, the expression
2acos kx =A( resultant amplitude)
concepts of nodes and antinodes
At nodes, the amplitude is zero, so cos kx =0, k = 2π/λ, at nodes, cos2πx/λ =0 , hence x= nλ/4 where n= 1,3,5,7 etc. At antinodes,
A = 2a, hence x=nλ/2 where n= 0,1,2,3 etc
WAVE FUNCTION CURVES
displacement ,Amplitude, displacement in y direction
displacement(x) = A cos ωt. = A cos2πft
velocity =± Aω sin ω t
acceleration a = -Aω cos ωt
other formulas to remember include
𝑽 = 𝝎√𝑨𝟐 − 𝑿𝟐

𝑲 𝒈 𝟐𝝅
𝝎=√ , 𝝎=√ , 𝝎=
𝒎 𝒍 𝑻
𝟐𝝅 𝒎 𝒍
𝑻= 𝑻 = 𝟐𝝅√ 𝑻 = 𝟐𝝅√
𝝎 𝑲 𝒈
Example
A particle is attached to a spring and undergoes SHM with a maximum acceleration of 18cm/s 2 and maximum velocity of
3m/s. find the frequency of vibration.
Ans. . V=ωR and a = ω2R, therefore a/v=ω , ω=6rad/s

F=ω/2 π = 3/πHz or 0.95Hz.
SOUND WAVES
Speed of sound depends on the properties of the medium in which it is travelling.
generally, V= √ (Elastic property/inertial property)
a. in liquids,
𝐵 ∆𝑷
𝑉=√
𝜌
,B is the bulk modulus and it is given by 𝑩 = − ∆𝑽
⁄𝑽
𝟎
b. in gases
𝐵
𝑉=√ , but for a gas B can be given by yP where y is called the adiabatic constant which is derived from ratio of the
𝜌
molar heat capacity at constant pressure Cp to the molar heat capacity ta constant volume Cv.
𝑪𝑷
.𝜸 = , y = 1.4 for a mono atomic gas
𝑪𝑽
c. in solids
𝑇
d. 𝑉=√ where μ is the linear density of the material given by m/L
𝜇
the speed is also affected by the ambient temperature.
𝑻𝑲
𝑽𝑻 = 𝟑𝟑𝟏√ or VT = 331 + 0.606(T0c)
𝟐𝟕𝟑
+++++++++++++++++++++++++++++++++++++-
Example
Helium is a monoatomic gas with a density of 0.179kg/m3 at a pressure of 76cm of mercury and temperature of
00 c. find the bulk modulus of the gas and determine the speed of sound in this gas. The adiabatic constant of
monoatomic gas is 1.67
Ans, pressure ρhg =1.031 x 10 5Pa, = B for a gas ===yP, B= 1.723x105
V= √ (B/ρ)√ (9.62 x 10 5) = 980m/s
STATIONARY SOUND
generally, there are a series of frequencies that characterize sound instruments. These are called the harmonics of the sound.
The characteristics frequencies that constitute the harmonics of a particular sound leads to the TIMBRE or QUALITY. the

perception of each frequency by the ear is called the PITCH. in a given instrument, the first harmonic is called the fundamental
note and any others after are called over tones. the 3rd harmonic is thus called the 2nd overtone.
when stationary waves are pictured on diagrams, a loop represent ½λ whiles half loop is ¼λ. given the fundamental
frequency f0, any extra frequency adds a complete loop to the shape hence the wavelength changes by n + ½.
The distance between a successive Node to Antinode is equal to ¼ λ or a ½ loop. therefore, a full wavelength is equal to 4 Node-
Antinode pairs.
(A) waves in stretched string
N A N
A
the picture above represents the fundamental note on a stretched string between 2 fixed points with a distance of L. The
fundamental has one complete loop hence the length L correspond to one loop vibration which is ½ wavelength. The
fundamental note wavelength is thus given by λ = 2L.Note that for f0, n=1
𝑽 𝑽
The frequency of the wave is thus given by 𝒇𝟎 = ., hence 𝒇𝟎 = .
𝛌 𝟐𝑳
𝒏𝑽
General equation for the frequency series is 𝒇𝒏 = where n is 1,2,3,4 ……
𝟐𝐋
For a string set up, the harmonics number n always matches the number of antinodes and the number of nodes is in excess of
one. Example the third harmonics has 3 antinodes and 4 nodes.
For the same string length L, the wavelength decreases with increasing harmonics but the frequency increases.
𝝀𝟎 𝒏 𝒏
fn = nf0 but 𝛌𝒏 = , for the nth harmonic 𝑳 = 𝟐𝛌 thus 𝛌𝒏 = 𝟐𝐋
𝒏 𝒏
(B) opened tube
A N A
The fundamental note as above has 2N-A pairs that are turned inside out. Meaning we have a fundamental tube length that is
½λ0
All the formulas are similar to the string situation as described earlier.
The difference is that with the opened tube, the harmonic number n, corresponds to n Nodes and n +1 Antinodes.

(C ) Closed tube
A N we have only one N-A pair so the
fundamental note corresponds to a

length L = ¼ λ0
𝑽
The general formula becomes 𝒇𝟎 = Fn =nf0 where n is only odd n =1,3,5,7
𝟒𝑳
𝛌𝟎 𝒏 .𝛌𝒏
𝛌𝒏 = where n= 1,3,5,7etc, and for a given length L, the nth harmonic, 𝑳=
𝒏 𝟒
The harmonic number n corresponds with equal number antinodes and nodes.
Examples
To what tension must a string of mass 0.01kg and length 2.5m be tightened to allow sound to travel on it at 125m/s
Ans V=√(m/k (T/μ), where μ=m/l v=62.5N
A metal ball is dropped from a height of 10m vertical onto a spring below. The mass of the ball is 1300kg and the
spring constant is 1x106N/m. determine the depth to which the spring is compressed.
Ans. X=√(2mgh/k), x=0.51m or =√(26)/10m
DOPPLER EFFECT
It is the apparent change in frequency heard by an observer due to relative motion between source and observer
𝑽𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒔𝒐𝒖𝒏𝒅 ± 𝑽𝒐𝒃𝒔𝒆𝒓𝒗𝒆𝒓 𝑽 ± 𝑽𝒐
𝒇𝟎 = 𝒇𝒔 𝒘𝒉𝒊𝒄𝒉 𝒐𝒇𝒕𝒆𝒏 𝒘𝒓𝒊𝒕𝒕𝒆𝒏 𝒂𝒔 𝒇𝟎 = 𝒇𝒔
𝑽𝒔𝒑𝒆𝒆𝒅 𝒐𝒇 𝒔𝒐𝒖𝒏𝒅 ± 𝑽𝒔𝒐𝒖𝒓𝒄𝒆 𝑽 ± 𝑽𝒔
Three situations are often encountered.
(a). static source and a moving observer.
Note that the observer component is above and the source component is below in our initial general formula. So for moving
𝑽 ± 𝑽𝒐
observer and still source, the Doppler frequency 𝒇𝟎 = 𝒇𝒔 𝑽
where v is the speed of sound. Use + if the wave-front is
getting closer and use – when the wave-front are getting further
(b) static observer and a moving source.

𝑽
In this situation the observed frequency 𝒇𝟎 = 𝒇𝒔 . in the situation of the moving observer use – when wave-front
𝑽 ± 𝑽𝒔
are getting closer and use + when they are getting further.
(c) both source and observer are moving

𝑽 ± 𝑽𝒐
𝒇𝟎 = 𝒇𝒔 𝑽 ± 𝑽𝒔
use the signs as described in the individual situations above
(d) Cosmologic shift.
Doppler shift can result in the change in frequency of electromagnetic waves from outer space as they approach the earth.
𝑽𝟐
𝒇𝟎 = 𝒇𝒔 (𝟏 ± 𝑪𝟐 ). When the wavelength gets smaller it is often termed as a red shift. Red shift is an evidence that the
universe is expanding.

The power of the emitter stays the same but

the distribution of the power per unit volume
LOUDNESS
which we call the loudness will decrease with
time because as the sound spreads from the
source, it enters into a bigger spherical area.
Intensity = POWER/ AREA.
This formula is for all waves including

electromagnetic waves like light.
𝑃
𝑰 = 4𝜋𝑅2 . The power is dependent on the frequency of the wave emitter and for the same emitter, the power doesn’t
chance but rather the intensity varies as the sound or wave spreads from one region to the other.
Because power stays the same. Power at point 1, P1 is equal to power at point 2 P2. Hence I1A1 = I2A2 or I1R1 = I2R2.
Intensity is measured in watts/m2. The intensity between 2 points can be compared using the decibel scale.
Loudness is the intensity level and it is measured on a logarithmic scale in decibels. For a given intensity I1, the loudness is
found by using a reference point which is usually the threshold of hearing which is 1 x 10 -12w/m2.
dB = 10 log10X, where x is the ratio of I1/I0. if 2 intensities I1 and I2 are given, the loudness level of one can be compared
with the other by making one the reference. In that cases X = I1/I2.
Given the power of a source P, the intensity reduces as the area of the sphere away from the source is increasing hence I ∝
1/r2. In a similar way, for 2 listeners at positions r1 and r2 away from the same source, the loudness level between them can be
found by
Using X= I1/I2 = r22/r12 and slotting the expression for x into our decibel formula
Examples:1 A sound source emits waves with a power output of 80W. find the intensity at 3m from the source.
Ans [ I=P/4 π r2] =0.707W/m2 , know 4 π =12.6
2 At what distance will the intensity of a sound from a 60W source become one fourth the intensity at 6m from the power
source.
Ans [ I1 R1 = I2 R2],R=12m
3 An ambulance travelling at 33.5m/s sounds 4x102Hz sound wave and is heard by a passenger in a car moving at24.6m/s
in the opposite direction. What is the frequency heard by the passenger as the 2 vehicles pass each other and move away?
Ans f = f(v-24.6/v+33.5) = 339Hz
4 The distance between crest and the next trough of water wave is 2m. If the wave frequency is 2Hz, Find the speed of
the wave.
INTERFERENCE PHENOMENON AND DETERMINATION OF MAXIMA AND MINIMA IN SOUNDS
When 2 coherent waves interfere, the interference may be constructive if they are in phase, meaning path difference is zero or
if they are out of phase by a path difference (r2 – r2) = mλ, where m is an integer.

If the path difference is equal to half the wavelength of (m + ½) λ where m is an integer, then we have destructive
interference. In all cases m= 0,1,2 etc
If 2 sources produce waves that travel
source screen different distances r1 and r2 towards an
r1 y observer. The observer may hear an amplified
sound if he stands exactly midway such that
r1 = r2 or if he moves a displacement y along
the vertical such that r1 – r2 will be = mλ
d x observor
The path difference r2-r1 can also be
r2 y
given by d sin θ, where sin θ = y/x
POSITION OF MAXIMA AND MINIMA
At position of center line where y = 0, the path difference r2 – r1 = 0. so this is our first maxima. To determine the others, use
𝟏
𝑑𝑌 𝒎𝒙𝝀 (𝒎+ )𝒙𝝀
𝟐
dsinθ = mλ which is same as = 𝑚𝛌, so 𝒚 = , where m is an integer. Similarly 𝒚 = is a position
𝑋 𝒅 𝒅
for destructive interference hence a minima.
Example
2 speakers are driven by a common oscillator of frequency 2.00x 102 Hz. They face each other and separated by a distance of
2m. An observer stands 10m away and at exactly midway between the speakers he hears an amplified sound from a
combination of the two speakers. Determine a position away from the midline such that he hears the first minimum sound.
V=340m/s
ans . y = Xλ(1 + ½ )/d

λ = v/f = 1.7
y= + or -12.75m
LIGHT AND OTHER ELECTROMAGNETIC WAVES
Geometric Optics.
For a given light wave travelling from medium 1 to 2,
n1V1 = n2V2, if medium1 is air/ then v1 = c and n1 =1, so the formula becomes
𝑪
𝒏 = 𝑽 where n is the refractive index of any medium aside air and v is the speed of light in any medium aside air.
similarly, n1λ1 = n2 λ2.
given the light is moving between 2 media where one is air, then the formula becomes
𝝀𝑨𝒊𝒓
𝝀𝒏 = where λn is the wavelength of light in a certain medium and n is the refractive index of that medium.
𝒏
Example Light with wavelength of 589nm in vacuum passes through a piece of quartz of refractive index of 1.458.
find the speed of this light through the quartz.
Ans V= C/n = 2.06x108 m/s
SPHERICAL AND A PLANE MIRROR REFLECTIONS
some properties of plane mirrors.

reflection could be specular or diffused

𝑉 𝐻𝑖
in plane mirror ho = hi and v=u. m= 1. the formula for magnification is M = − or 𝑀 = −
𝑈 𝐻𝑜
minimum length of mirror to capture full object height is at least ½ image height.
CURVED MIRRORS
sign conventions. for mirrors; the rays from the object are reflected back towards the side where the object is
found. so it is only such reflected rays that can actually intersect to forma real image.
any magnification above 1 is enlarged and below 1 is diminished. if the magnification is assigned +, it is upright and if
it is assigned -, it is inverted. this means magnification in plane mirrors and convex mirrors are always +. if the image
is formed on the same side as the object for a mirror is assigned a positive value whiles that formed on the other side
due to apparent intersection of rays is assigned a negative value. if the mirror is such that the image forms on the
same side as an object placed at infinity, its focal length is positive where as if the image of an object placed at infinity
forms on the opposite side, the focal length is negative.
convex mirrors
the image is always virtual, diminished and upright as well as behind the mirror. The magnification is thus less than
+1
f is negative and v is always negative. v has values between –f and center of mirror V regardless of where the object
is placed.
concave mirrors. there are 3 scenes with this mirrors that should be looked at carefully
(a) object placed from infinity approaching (+f)

object
the largest image though still smaller than the object is
v= +f +f>v>V
formed at the focal point when the object is at infinity.
V as the object is moved closer towards +f, the image gets
smaller and moves closer towards V. all images are
u=+∞ +∞>u>f +f
small, inverted and real. all magnifications are less than
-1

(b) object now at +f

when the object finally reaches +f, things flip
around and the mirror starts to behave like a
convex mirror. it forms its first largest virtual
image. the difference here is that the is
u=+f V v =-∞ magnified against a diminished virtual image as
found in convex mirrors but like all virtual
images, it is upright and on the opposite side.
this is the largest image of any concave mirror.
the images are larger, upright and virtual and

keep reducing as object approach center of
mirror V. at a point where the object is half
( c ) object between + and center of mirror V focal point, there will be a magnified virtual
upright image at focal point on the opposite
side which is designated -f
+f>u u= ½ f v=-f -f<v<-∞
SPHERICAL THIN LENSES
the main difference with thin lenses is that light passes through and refract through lenses to form real images on the other
side hence real inverted images are formed on the side opposite where the object is located and v is positive for such images
unlike in mirrors where due to reflection the opposite happens.
real image on other side
-∞ -f +f +∞
virtual on same side
concave lenses are converging lenses and they behave just as convex mirrors discussed earlier. the image is formed by apparent
intersection of rays, it is virtual on the same side of the object and upright. it is diminished. magnification is thus M<+1.
convex lens
they are converging lenses and rays from any object from infinity and beyond the focal point come to form real images on the
other side of the lens by actual intersection of converge rays. all the sign conventions and the image characteristics for convex
lenses are similar to that discussed for concave mirrors except that the real image is on the other side of the lens. the 3
scenarios will also apply here
(a) object from infinity and approaching -f (when speaking in terms of focal length and the image characteristics, every
parameter on the same side of object except object itself and U is deemed to be virtual and thus negative)

the image is always small, real (on the other side thus v is
V< v <+f v=+f positive) and inverted. the largest real image is at +f
when the object is at infinity on the other side. as the
u=-∞ -∞>u>-f
object gets closer but beyond –f, the image also
approaches center of lens V and keeps getting smaller
but still real and inverted.
(b) object exactly at –F

things start a sudden flip around where the
lens begins to behave like a concave lens or
convex mirror. it forms its first large virtual
image at this point and further images as the
V object moves closer to the mirror stay
magnified, upright and virtual but with
decreasing size. the image is on the same side
v=-∞ u=-f as the object and it is due to apparent
intersection of rays since lenses refract and
not reflect to form images. v is negative, M is
( c) object between –f and V >+1
-∞>v>-f v= -f the image is virtual erect and magnified but

gets smaller as the object approaches the
center of mirror V. at a point where object is
½ focal length, the virtual image forms at f,
this means a magnified virtual image which is
-f>u>V u= ½f V twice the size of object forms at focal point on
the same side of object when the object is at
half focal point.
SIMPLE MAGNIFIERS
Angular magnification is the ratio of the angle the rays from the image subtend
at the eye to the angle the rays subtend at the lens.
𝜽𝑬𝒚𝒆
𝑴=
𝜽𝑳𝒆𝒏𝒔
𝟐𝟓
By calculations and assuming normal near point vision, this formula comes to 𝑴 =
𝒇
This formula means that the image distance v of the lens is at the eyes near point thus the eye is in most relaxed state.
𝟐𝟓
The maximum angular magnification attainable for the lens is however given by M= 1 + and this is not obtained
𝒇
when the eye is most relaxed.
COMBINATION OF THIN LENSES.

𝑳
compound microscopes. the magnification is in 2 folds. the objective lens gives lateral magnification ML = − where L is
𝒇𝟎
the separation between the lenses and fo is the focal length of the objective lens.
𝟐𝟓
Angular magnification is found from the eye piece lens similar to the simple magnifier in relaxed state M = . The total
𝒇𝒆
𝟐𝟓𝑳
magnification the multiplication of the two magnifications. 𝑴 = −𝒇
𝒆 𝒇𝒐

TELESCOPES
𝒇𝒐
The magnification M =
𝒇𝒆
The lens maker’s equation helps you to determine the focal length of a thin curved surface with uneven curvatures on the 2
sides.
𝟏 𝟏 𝟏
𝒇
= (𝒏 − 𝟏) {𝒓 − 𝒓 } where r1 and r2 are the 2 uneven focal lengths of the 2 sides of the material and n is the refractive
𝟏 𝟐
index.
if the material with refractive index n1 as above is being observed whiles embedded in another material with refractive index
𝟏 𝒏 𝟏 𝟏
n2 say a piece of glass coated with vegetable oil. the equation becomes
𝒇
= (𝒏𝟏 − 𝟏){𝒓 − 𝒓 }
𝟐 𝟏 𝟐
THICK REFRACTING SURFACES.
The observer located in medium 1 with refractive index n2
refracted rays forming virtual image with distance v
thick refracting surface with refractive index n2
virtual image at a distance v from the surface of the refracting
medium as seen by the observer
the actual object which is distance u from the surface of
refracting surface
The magnification of such a surface is given by

𝑛1 𝑣
𝑀= − 𝑛2 𝑢
For a flat surface such as the surface of steady water or flat thick glass, the magnification = 1, hence
-n1V = n2U, where n1 = refractive index of the surface in which the object is situated and n2 is the refractive index
of the medium in which the observer is viewing.
lens equation for thick refracting surfaces is given as follows

𝒏𝟏 𝒏𝟐 𝒏𝟐 −𝒏𝟏
𝒖
+ 𝒗
= 𝑹
where n1 is medium with object and n2 is medium with observer.
IMAGE ABNORMALITIES OF OPTICAL DEVICES
SPHERICAL ABERATIONS.
For a perfect spherical object, it is expected that all rays from a common point source will focus at a common point to form a
definite and sharp image.
however, most spherical objects like lenses and curved mirrors are not perfect. the perfect focus of all rays is only achieved for
rays close to the principal axis. this is because they incident on the curved surface within the APLANATIC point (found by
dividing the radius of curvature by the refractive index of the material= R/n) of the curved surface and will all be brought to a
fixed focus.

these rays close to the principal axis are called the PARAXIAL rays. rays that are close to the edge of the curved surface often
incident outside the aplanatic point of the mirror and will focus in a slightly different location. these rays that are a bit distant
from the principal axis are called the MARGINAL rays.
PERFECT SPHERICAL REFRACTING SURFACE
marginal ray
paraxial ray
common focus
the production of an imperfect image focus due the fact that different rays are brought to different foci after passing through a
curved optical surface is called SPHERICAL aberration. this can happen for both spherical lenses and curved mirrors
EXAMPLE OF NEGATIVE CHROMATIC ABERATION
spherical aberrations are usually due to the problems with the marginal rays not being able to come to a common focus. if the
marginal rays are refracted more that the paraxial rays, it results in positive aberrations and if the marginal rays are refracted
less than paraxial rays it results in negative aberration.
CORRECTION
for concave lenses, using an Aspherical or Aplanatic lens. this is a lens where one of the surface has a curve inwards towards the
margins.
for concave mirrors, by using a parabolic mirror.
CHROMATIC ABERATIONS
this is the optical problem produced when different wavelengths are focused at different points after passing through a curved
lens. this is a problem due to diffraction. the resulting image has colored edges. the phenomenon of chromatic aberration is
also called color fringing.
There are 2 forms, namely axial/longitudinal aberration and lateral/transverse aberration.
CORRECTION
by use of achromatic lenses. This is a lens in which different wavelength and thus different dispersions are all assembled into
one-point focus.
using a combination of different lenses.
THE HUMAN EYE LENS PROBLEMS.
SHORT SIGHTEDNESS.

This results when the person can only bring object into focus when they are closer. parallel rays from infinity focus in front of
the retina. This is because in most of such individuals,1. the lens has a short focal length or 2.the eye ball is too large such that
the retina is located beyond the focal point.
Retina
parallel rays from infinity (far object)
Focal point
LENS
They can see object closer because those rays are not parallel and will often focus beyond the focal point and thus are likely to
incident on the retina. persons with short sightedness are said to have MYOPIA and they have a normal near vision of 25cm.
CORRECTION. The aim is to allow such a person to be able to focus on far objects and hence extend their vision to infinity. They
need a lens that will help to extend the focal point by diverging the rays towards the retina hence short sightedness is corrected
with a DIVERGING lens. The focal length of such lenses is usually longer than the individuals natural lens and have a negative
focal length hence the power is negative. They are therefore referred to as minus lenses.
DEALING WITH CALCULATIONS ON MYOPIC VISION
KUMIDEEs APPROACH
There are 2 questions to ask. 1 what is my problem? and 2 what is my expected results after corrections?
the answer to the problem at hand is taken as the virtual image distance -V and the answer to the expected solution is taken
as the object distance U. put these in the lens equation
𝟏 𝟏 𝟏
+ =
𝒖 −𝒗 𝒇
for short sightedness, our problem is a certain abnormal far point say X. and our aim is to achieve far point at infinity ∞.
so our equation is
= , we will realize that f = -X. so the required lens is a diverging lens with a focal length
𝟏 𝟏 𝟏
+
∞ −𝑿 𝒇
equal to negative of the defective far point of the person.

LONG SIGHTEDNESS
also called hypermetropia. this means that distance object can focus well on the retinal but objects closer will focus behind the
retina. such persons may have a long focal length or a short eyeball.
FOCAL POINT
these individuals have a perfect far point which is infinity but have a problem with near point. if an object is at or nearer than
the normal near point of 25cm, it focuses behind the retina and hence will not be properly seen. they can therefore only see
properly objects that are beyond 25cm. this means they have a longer near point.
CORRECTION
Using the same approach as per the near sighted case.
we need a lens that will converge the rays into focus when viewing at near things. in terms of calculation, we should seek to
answer the same questions above.
this time our problem is a certain defective near length say Y and our aim of correction is to achieve a normal near length of
25cm
𝟏 𝟏 𝟏
+ =
𝟐𝟓 −𝒀 𝒇
𝟐𝟓𝒀
𝒇= where Y is the defective near point of the person.
𝒀+𝟐𝟓
we realize the negatives cancel out so the focal length needed will be positive and that is a converging lens. The power of such
lenses is thus positive and these lenses are called Plus lenses
ASTIGMATISM
this is a problem with spherical aberration due to irregularity of the surface of the lens. we correct it with a cylindrical lens.
PRESBYOPIA
The human lens is suspended in place between the ciliary body by the suspensory ligaments of the eye. These connective
tissues help the ciliary muscles to change the curvature of the lens and therefore to adjust its focal length for accommodation.
For this to also happen, the lens must be elastic enough. An inability of the lens to adjust its shape and thus focal length in
other do achieve near vision due to weakening of elastic tissues of the lens as a results of aging is termed presbyopia. A
similar loss of connective tissues and sensory cells in the cochlear of the ear is called presbyacusis.
Fiber-air boundary OPTICAL FIBRES
cc C
α

Core-cladding boundary
Optical fibers have less energy losses through heating and are efficient due to
less interaction with circuit component, eddy currents and back EMF. There are however a few limiting properties of these
cables
ENERGY LOSSES/ATTENUATION
𝑷 𝒊𝒏
This is the ratio of optical power input to the optical power output expressed on the decibel scale, 𝑨 = 𝟏𝟎 𝐥𝐨𝐠 𝟏𝟎 𝑷 . (dB)
𝒐𝒖𝒕
𝑨
Attenuation rate or attenuation coefficient is defined as the attenuation per kilometer of fiber optic cable. 𝜶 = (dB/km)
𝑳
ANGLE OF INCIDENCE
To reduce energy losses by refraction and ensure that most of the light is propagated by total internal reflection, the light through
the core must incident at the core-cladding boundary at the critical angle. Given core refractive index to be n1 and cladding
refractive index to be n2, the critical angle is given by
𝒏
𝒔𝒊𝒏𝑪 = 𝒏𝒄 where nf is always greater than nc.
𝒇
To achieve the above, that is to make sure the light through the core achieves critical angle at the core-cladding boundary, the
light entering the core at the air-core boundary must be small enough. Beyond a certain angle, it is impossible to achieve C within
the core-cladding interface.
The maximum Angle of incidence at the Air to core interface can be determined as 𝜶 = √𝒏𝟐𝒇 − 𝒏𝟐𝒄 where nf and nc are the
refractive indices of fiber core and cladding respectively.
TYPES OF OPTICAL FIBRES.
1 MONOMODAL FIBRE. Small core diameter relative to cladding
2 MULTIMODAL FIBRES, much larger core relative to cladding, which could be step mode (every part of the core has same
refractive index) or graded mode (refractive index of core decreases as you move towards the periphery and gradually blends in
with cladding).
HOW ATTENUATION OCCURS IN OPTICAL TRANSMISSIONS
Scattering, bending, absorption

ILLUMINATION PHYSICS
THE CONCEPT OF SOLID ANGLE.
The dimensionless unit of plane angle is the radian, with 2π radians in a full circle.
A solid angle, ω, made up of all the lines from a closed curve meeting at a vertex, is defined by the surface area of a
sphere subtended by the lines and by the radius of that sphere, as shown below.


ω = AREA OF SPHERE/RADIUS2, 

the unit is the steridian


 
2
ω = A/R 



Luminous Flux
Luminous flux is the portion of total radiant power that is capable of affecting the sense of sight The unit for luminous
flux is the lumen which will be given a quantitative definition later
Example 1. What solid angle is subtended at the center of a sphere by an area of 1.6 m?The radius of the sphere is 5
m.
Ans. = 0.00640 sr
The Lumen as a Unit of Flux
One lumen (lm) is the luminous flux emitted from a 1/60 cm cm2 opening in a standard source and included in a solid
angle of one steradian (1 sr).
One lumen is also equal to 1/680 W of yellow-green light of wavelength 555 nm.
Luminous Intensity
The luminous intensity I for a light source is the luminous flux per unit solid angle.
I = F/ ω, its unit is the candela,Cd.
A source having an intensity of one candela emits a flux of one lumen per steradian
Total flux for Isotropic Source
An isotropic source emits equally in all directions; i.e., over a solid angle of 4πsteradians.

I = F/ω,Hence I= F/4π.
Total flux for isotropic source, F=4π *I

Flux in a confined area is generally given
by F= I*A
Example 2.
A 30 cd spotlight is located 3 m above a table. The beam is focused on a surface area of 0.4 m m2. Find the intensity of
the beam.
Ans. I = 8490 cd
Illumination of a Surface
The illumination E of a surface A is defined as the luminous flux per unit area (F/A) in lumens per square meter which
is renamed a lux (lx).
E=F/A
An illumination of one lux occurs when a flux of one lumen falls on an area of one square meter.
The illumination E of a surface is directly proportional to the intensity I
and inversely proportional to the square of the distance R
E = I/R2
Example 3.
A 400 cd light is located 2.4 m from a tabletop of area 1.2 m2. What is the illumination and what flux Ffalls on the table?
ANS E=I/R2 = 69.4lx,
F= E/A = 93.3lm
WAVE OPTICS
In geometric optics treated above, light is viewed as particulate and represented with coordinate known as a ray. In
the field of wave optics, light is considered as a pure wave and its dimensions are represented by semicircular curves known as
wave fronts. Wave optics will consider the wave properties of light such as polarization, diffraction and interference.
HYSTORY.
For so long Newtonian physics upheld the particulate postulate of light. Christian Huygens held a view that light was a
wave. At the time, diffraction could not be appreciated using light and thus its nature as a wave was not accepted.in 1801, It
was Thomas young who showed in his double slit experiment that light could undergo wave phenomena such as diffraction and
interference. The solid experimental and mathematic illustration of the nature of electromagnetic waves in general was later
revealed by James clerk Maxwell. It was Fredric Hertz who experimentally proved Maxwell’s electromagnetic concept
HUYGENS PRINCIPLE

Waves moves by forming expanding circular wave-fronts and each point traced on the wave-front represented a
wavelet on its own which formed secondary waves that could propagate forward
By Huygens idea, it becomes clearer that light

doesn’t travel as a straight line as geometric
optics depict in ray diagrams hence it makes
clear the understanding of how light can
spread around apertures and how this light
could lead to interference with neighboring
waves
YOUNGS DOUBLE SLIT EXPERIMENT
A coherent light source is placed behind the double slit and the
beam is allowed to pass through a double slit separated by d. if
r1
we consider this in the wave front form as above, we realize at
d y some points the 2 waves start to spread around the slits if the
slit is narrow enough in the proximity of the wavelength of
x r2 light. There are thus points where the 2 secondary wavelets
A geometric view of the double slit experiment emanating from the slits will overlap and interfere either
cancelling each other out (destructive interference) or
enhancing each other (constructive interference.
An image of the 2 waves will appear on the screen which is separated a distance of x from the slit. It was observed that the
image on the screen showed an interference pattern made up of alternating patterns of dark and bright bands. These bands
were called fringes. Maxima intensity occurred when there was constructive interference giving rise to a bright fringe and
minima intensity occurred when there was destructive interference giving rise to a dark fringe. At the center of the screen, a
bright fringe formed. It was observed that at this point, the 2 rays r1 and r2 travel the same length of path to reach the screen
hence path difference r1-r2 was zero. If the path difference was varied a bright fringe always form if the path difference was an
integer multiple of the wavelength. The position y on the screen from the center of screen can be determined.
Constructive interference. r1 – r2 =mλ , but r1 – r2 can be deduced to be equal to dsinθ, hence d sinθ= mλ.
𝒎𝒙𝝀
Sin θ = y/x hence position on screen from center where maxima occur is 𝒚 = where m is called the order number m= 0,
𝒅
1, 2, 3etc.
For instance, the first maxima or first bright fringe has order number m=1. A bright fringe always forms when we move a
𝒎𝒙𝝀
distance +y or -y from the central line corresponding to . The width of the central bright fringe is thus bounded by +y and –
𝒅
𝟐𝒎𝒙𝝀
y which gives +y- -y = 2y =
𝒅
Destructive interference.
Similarly, the formation of dark fringe has some conditions. r1 – r2= (m + ½) λ, d sin θ = ( m + ½ )λ. So at
𝟏
(𝒎+ )𝒙𝝀
𝟐
𝒚= is dark fringe will form.
𝒅
Some forms of diffraction patterns on a screen
Diffraction through slits like what has been discussed above will give a fraunhofer diffraction. With bands of bright
and dark pattern usually with a thick central band and the rest gradually narrowing as we move away from the center.

Fraunnhofer diffraction
Another type of diffraction pattern is obtained by placing a small opaque obstacle in the path of a beam. This
gives the Fresnel diffraction. The central bright fringe is tiny and the bands come off as circular bright and dark
fringes. The central spot is called the Fresnel bright spot. The obstacle must be close to the screen. If the distance
from screen is long it looks like the fraunhofer pattern. You can get the fraunhofer pattern by placing a converging
lens between the obstacle and the screen.
DIFFRACTION GRATING.
Generally, diffraction increases with increasing wavelength. For diffraction grating, the same formula applies as the double slit
diffraction discussed above.
THIN FILM INTERFERENCE AND DIFFRACTION.
Interference pattern could be studied using thin films that can refract light. Usually the light passes through the film and gets
refracted. Part of the light is also refracted back into air. The refracted and reflected ray interact to create diffraction patterns.
This is the phenomenon that is observed when light sets on a film of oil spill in water. Experimentally, thin film diffraction can
be observed by using a Plano- convex lens placed over a mirror. The pattern observed looks like the circular Fresnel’s pattern
described when one looks down the lens and it is called the newton’s rings.
Student should recognize that at the surface of any refracting

INCIDENT RAY R1 body such as lenses there is both refraction and some degree of
reflection.
lens R2
mirror
If our diagram was drawn to proper scale, it will be realized that at the convex surface of the lens which is the part over the
mirror, part of the ray gets reflected back into air R1 and part is refracted into the air space between the lens and mirror. That
refracted ray is then reflected over the surface of the mirror back into the lens and then back into air as R2. R2 travels extra
distance through the lens which equals twice the lens thickness 2t as compared withR1. The path difference R2 –R1 = 2t (twice
lens thickness). The interference pattern seen by the observer is due to the interplay of the 2 types of rays R1 and R2. note
however that because these 2 rays are all reflected rays, there is a phase reversal so what holds true for the direct incident ray
in terms of conditions for interference will hold true but in the opposite form.
Constructive interference
R2 – R1 =(m+ ½) λn, but r2 – r1 = 2t so 2t = (m+ ½) λn, where λn is the wavelength of light through the thin film material.
In most cases it is the wavelength in air or vacuum will be provided λ0. But we know λn =λ0/n where n is the
refractive index of the material.
The formula can thus be written as 2nt = (m+ ½) λ0

Destructive interference
2t = mλn or 2nt =mλ0

SINGLE SLIT DIFFRACTION
This is quite similar to the way we handle the interference /diffraction in double slit. We assume similar phase reversal hence
the formula is similar to double slit but the conditions are reversed. Also in single slit situations, the slit width a replaced the slit
separation d.
Constructive interference
Given a single slit with a width of a and separated from a screen by distance x, a sinθ = ( m + ½ )λ .
Destructive interference
𝒎𝑿𝝀
is given by a sin θ =mλ and at every position +y or –y on the screen such that 𝒀 = , a dark fringe forms. The
𝒂
central bright fringe is bounded on both sides by dark fringes at +y and –y so thickness of bright central maxima =
𝟐𝒎𝑿𝝀
2𝒀 =
𝒂
DIFFRACTION GRATING
COMMENT ON DISPERSION VRS DIFFRACTION
REYLEIGH SCATTERING
POLARIZATION
A progressive transverse wave such as light has different planes in which it vibrates as it moves. when the wave is obtained
such that it vibrates only in one plane, it is said to be plane polarized. this can be achieved by
(a) selective absorption. A polaroid is a material that can selectively absorb all light in other planes and allow only light in a
desired plane to be transmitted through it. The path along which the light exits is called the transmission axis and the
orientation of the polarizer forms the axis of polarizer. as the light is transmitted out, its intensity will drop from I0 to Ip.
the new intensity can be found using MALUS LAW. If the polarizer has its axis at right angle to the transmission axis, no
light passes through meaning the intensity is reduced to zero.
Ip = Io cos 2θ. where theta is the (polarizing angle)angle between transmission axis and the polarizer axis.
the average value of cos 2θ = ½ with maximum of 1 and minimum of 0.
so average attenuation in intensity after passing through polaroid is usually given by
I p = ½ Io
(b) by reflection
(c) by scattering
(d) optical activity
BIREFRINGENCE
The ability of a material to polarize light in two different planes due to the possession of two different refractive indices.
commonest birefringent material is CALCITE also called Iceland spar. They are employed in polarizing prism and quarter wave
plates
OPTICALLY ACTIVE SUBSTANCES
substances such as glucose can alter the plane of polarized light passing through them and are said to be optically active. you
can study the optical activity of any substance by using 2 polarizers and placing a test tube containing the material in between
the two polaroid with the light source at one extreme end of the setup. the two polaroids are arranged to have their axis
aligned and the light incident on a screen is observed without the test material. Then the test material is now introduced
between the 2 polaroid. if the light on the screen does not change, this material has no optical activity. If the light is no longer
seen on the screen, then the second polaroid on the opposite side of the light source will be rotated till the light is seen again.
The angle through which it was rotated corresponds to the angle through which the polarized light was deviated by the
optically active substance. if this angle is leftward or anticlockwise, it is termed Levorotatory and if it is rightward or clockwise it
is dextrorotatory. Optically active substances usually exist as optical isomers due to the possession of a chiral element. The
two isomers differ in their affinity towards receptors and as such have different enzyme kinetics. A perfect mixture that has
equal amount of both optical isomers is called a racemic mixture(RACEMATE) and this tends to nullify their optical activity.

QUARTER WAVE PLATE
It has birefringent materials arranged such that the plane of the larger refractive index is quarter wave (90o) in front of the
smaller index. when light incidents on the plate at 45o it will be polarized from linear to circular after the light has split into 2
and vice versa
VIBRATIONS. CIRCULAR MOTION AND S.H.M
MOTION IN CIRCULAR PATH
This type of motion can be termed uniform circular motion under a constant centripetal force or a non-uniform circular motion
in which case a non-constant force is applied
UNIFORM CIRCULAR MOTION.
The force is called a centripetal force which is constant and always directed at the center of the circular path. The body only
experience a torque but work done is zero since the tangential velocity is perpendicular to the applied force. The constant
acceleration always acts toward the center in the direction of the centripetal force and there is zero tangential velocity. there is
a constant angular frequency whose value does not depend on the distance from the center. the tangential velocity is however
non constant and depends on the distance from center of circle and also keeps changing with the changing direction caused by
the torque. angular acceleration is zero. There is a constant non changing centripetal acceleration but no tangential component
to it.
NON UNIFORM CIRCULAR MOTION
The main difference here is the that the applied force is non uniform and the acceleration is resolved into both a centripetal
component and a tangential component which acts along the direction of the tangential velocity. the total acceleration is thus
found using Pythagoras theorem.
aT =√ (ac2 + at2)
DEALING WITH PROBLEMS AND FORMULAS IN CIRCULAR MOTION
most equations here are similar to the Newtonian kinematic equations just apply an angular point to it.
Δθ = ωt + ½α t2………………1
where θ is the total angle subtended which is same as the angular or radial version of the displacement measured in
radians. ω is the angular velocity of angular frequency in rad/s.ω= 2πf = 2π/T.
conversion of rev/s to rad/s.
one revolution is a full circular turn =3600 = 2π rad.
so 1 rev/s = 2π rad/s
ωf = ω0 + αt…………………….2,
ωf2 = ω02 + 2αΔθ………………3,

𝝎𝒇 −𝝎𝒊
from equation 2, 𝜶 = or
𝒕
𝝎𝒇 𝟐 −𝝎𝒊 𝟐
from equation 3, 𝜶= 𝟐∆𝜽
comparing linear quantities and angular quantities

Δs = radius x Δθ, v = radius x ω and at ,(tangential) = radius x α
Remember at is not the same as ac which is the centripetal acceleration.

𝐯𝟐 𝒎𝐯𝟐
𝐚𝐜 = = ω2r . The centripetal force 𝐅= 𝐫
=mω2r
𝐫
The torque is F x perpendicular distance. for uniform motion this is = radius
torque T =mv2or mω2r2
different forces causing centripetal acceleration
(a) Friction.
the centripetal acceleration depends directly on the magnitude of the coefficient of static friction μs.
𝐯𝟐
mv2/r = μsmg, μs = and μs = a /g. A car will stop rolling its wheels and skid like a skate over the road
𝐫𝐠
when the velocity exceeds what the static friction can support. wetting a road for instance reduces the coefficient of static
friction and thus cars can easily skid if the velocity is increased on a rainy day. Banking the road also helps to increase the
𝐯𝟐
frictional force since coefficient os static friction varies as the banking angle. tan θ = = μs.
𝐫𝐠
(b) Tension in a cord

let us consider this in scenarios.
1 ,a cord swung to form a horizontal circular loop
F = mv2/r
U
R
Tension(T) weight(mg)
The tension in the cord T is resolved into a vertical component T cos θ which balances the weight of the spinning body Mg, thus
Tcos θ = mg whiles the horizontal component T sin θ contributes the centripetal force F to keep the body in the circular
path hence Tsin θ = mv2/r. To find the angle θ, we often employ tan θ = v2/rg. As the speed increases, more of the
tension is converted to centripetal force and vice versa.

2 a cord swung to form a vertical loop.
top of loop., we get minimum tension

𝟐
Tmin = F –W = 𝐦(𝐯𝐫 − 𝐠)
at the equator, we get median

tension which is the tension itself T.
𝒗𝟐
T = m𝒓
at the bottom of loop, we obtain the

we observe that the 𝟐 difference between minimum and
maximum tension is maximum tension T max = F + W = 𝐦(𝐯𝐫 + 𝐠) 2mg and between the tension and
either minimum or maximum tensions is ± mg.
CIRCULAR MOTION AND CONSERVATION OF MECHANICAL ENERGY IN A LOOP SYSTEM
let’s consider a toy train that is on a frictionless circular loop tract. if the train is pushed from a position x with a magnitude of
force, we need to consider with what velocity will the train need at the top of the loop in order to negotiate a full circle without
stopping.
At the top, the normal is zero so the centripetal

force must just be equal to the weight. mv2/r = mg.
The maximum speed with which it will negotiate the
curve at the top of the loop is thus v = √rg. similar to
the formula for orbital speed in circular orbit
without a normal reaction.
The kinetic energy in this state is ½ mgr and

potential energy = mg h where h = 2r hence PE =
2mgr. Total mechanical energy of the system in a
circular loop without a normal force =2.5mgr.
At the bottom of track, the centripetal force is the difference between the normal
force n and the weight W
mv2/r = n – mg. By estimation, we notice that due to conservation of energy the total
mechanical energy at the bottom of loop when the train returns is still 2.5mgr. But at SIMPLE
HARMONIC
this point, there is zero vertical height so PE is all converted to kinetic energy. the
MOTION
maximum speed with which it negotiates a curve at the bottom of loop will thus be
V= √5gr A simple
harmonic motion
describes the to
and fro motion of a body under the influence of a constant restoring force with a fixed frequency and amplitude. the restoring

force varies as the displacement from the equilibrium position and is in the same direction as the acceleration but opposite to
the displacement since it always back toward equilibrium. if damping is negligible and a body oscillates with small amplitudes,
the SHM is just a sinusoidal curve of a body in circular motion and most of the assumptions of non-uniform circular motion
hold. in this case the restoring force keeps changing with the displacement. the angular frequency and frequency are constant
everywhere on the path of motion.
positive displacement
maximum displacement =A and correspond to V= 0, a=max, F = max.
R Δx
0 , equilibrium correspond to Δx=0, v = max ,F=0 and a = 0
negative displacement
Displacement is Δ X, maximum displacement is called amplitude A.
Acceleration is given by a = -ω2x ……….1. a max occurs when x = A, hence a max= -ω2A. The value of a is
positive if displacement is negative and vice versa.
velocity 𝐕 = 𝛚√𝐀𝟐 − 𝐗 𝟐 ……….2. Maximum velocity occurs where displacement x=0 at the equilibrium position.
SPRING MASS SYSTEM
By Hooke’s law, F = -kx, ma = -kx, acceleration, 𝐚 = − .x. comparing to equation 1 above shows ω2 = k/m
𝐊
𝐦
𝐕= √ .x
𝐤 𝐤
therefore 𝛚 = √ similarly, we realize that
𝐦 𝐦
ROTATIONAL DYNAMICS
Torque T = F x R. since the perpendicular force causing the turning effect is similar to centripetal force, T =mv2 = mω2r2. If
the force is not exactly applied perpendicularly, then T =mω2r2 sinθ.
The quantities in Linear motion have some similarities with some of the quantities in rotational dynamics. Because you are
conversant with the linear motion and its equations, know such relations will help you remember the equations in rotational
motion.
Linear motion Rotational motion

force torque
mass Moment of inertia
momentum Angular momentum
velocity Angular velocity/ angular frequency
displacement Angular displacement
acceleration Angular acceleration
Example.
1. Force =mass * acceleration,
then Torque = moment of inertia * angular acceleration
2. Momentum= mass*velocity,
then angular momentum= moment of inertia * angular velocity

MOMENT OF INERTIA.
In circular motion the tendency for a body to resist change in its uniform circular motion is called the moment of inertia. In
dealing with calculations, it is the parameter, which is employed whenever you would have used mass.
for most point-like objects, the general formula for moment of inertia is I =mr2 and this also holds true for the
moment of inertial from any point located on a discrete body.
For a non-point like body, the total moment of inertia of the entire body can be found by adding the moment of
inertia of all discrete points I =Σmr21,mr22…………….mr2x
some regular bodies that are non-point like have their moments of inertia predetermined by experimentation and
calculations and it is worth noting some of these.
1. uniform rod pivoted in the center, I = 1/12 ML2

2. uniform rod pivoted at one end I = 1/3 ML2
3. hollow cylinder (appreciable wall thickness, I = ½ M (R12 + R22)
4. a hoop or thin cylindrical shell I = MR2
5. solid sphere I = 2/5 MR2
6. hollow sphere I = 2/3MR2
7. solid cylinder or a disc. I = ½ MR2
The torque causes an angular acceleration just as the applied parallel force would cause linear acceleration. since F = ma,
𝐈∆𝝎 𝛚𝟐 −𝛚𝟏
likewise T = Iα= 𝐓= or 𝐓=𝐈
𝒕 𝐭
The principle of conservation of momentum also applies here. the momentum in linear motion is given by P =mv. in a like
manner, angular momentum is given by p x radius = Pr. that can also be derived to be Iω. To prove this, let’s compare linear
and angular quantities. Iω = mvr, remember I = mr2 and ω = v/r
mr2 x v/r = mvr.
law of conservation of angular momentum
I1ω1 = I2ω2
m1v1r1 = m2v2r2
v1/v2 = r2/r1
rotational kinetic energy= ½ I ω2 = ½ mr2ω2
ΔKE = ½ ΔI x (ωf2- ωi2)
work done on a body due to a constant torque
W = Torque x angular rotation (θ)
W =T θ
total kinetic energy of a body rolling from one point to the other
such a body possess both angular and translational kinetic energies
total kinetic energy will consider summation of the translational kinetic energy and the rotational kinetic energy
= ½ I ω2+ ½mv2
= ½ m (v2 + r2ω2)
= ½ v2(m + I/r2)

HOOKES LAW AND RELATED FORMULAS
Named after sir Robert Hooke. F = -kx
K is called the spring constant and it measures the stiffness of the spring. Hooke’s law is only obeyed within the elastic limit of
a material. Glass and metals have large young’s modulus hence low elastic strain. Plastics have a low young’s modulus but do
not obey Hooke’s law.
DEFINITION OF SOME TERMS.
most solid materials could be classified as crystalline, glassy, amorphous or polymeric.
extension
point C
point B
point A
point O Force
the diagram above shows a plot of extension of a wire with increasing stretch force.
point A is called the proportional limit of the material. It is the amount of maximum extension within which a material obeys
Hooke’s law by undergoing elastic deformation. During this process the extension is only due to shear displacement of lattice
structure and the energy is stored as molecular potential energy. when the stretching force is relieved, all the energy is
returned to restore the lattice structure.
The force which extends the material to the proportional limit such that the material undergoes maximum elastic deformation
is called the elastic limit.
Beyond the elastic limit, extra extension at point B starts to cause rapid extension and at this stage, removing the force does
not return the material to original length. We describe this type of extension as plastic deformation and the level of extension
at which a material starts to show plastic deformation is called the yield point. During plastic deformation, lattices are not only
strained but start to slip pass each other. the energy is lost to heat through friction hence cannot fully be recovered.
If plastic deformation continuous to a certain limit, the material reaches a breaking point extension and any extra extension
will cause it to break. The force leading a breaking point extension is called the breaking stress. The breaking stress of a wire
corresponds to the maximum force per unit area of most extended (narrowest cross section) wire. All materials demonstrate
some level of elastic deformation. It is more obvious for materials that are easily stretch hence materials with a small young’s
modulus or springs with a small spring constants. However, when it comes to plastic deformation it is the preserve of some
materials and not all.
Materials demonstrate ductility if they show a lot of plastic deformation before breaking or are said to be brittle if they show
very little or no plastic deformation before breaking. Materials like Brass, Bronze appear to be super elastic as if they have no
yield point. They continue to obey hooks law over a large extension. A material with ductility can also show another feature
called malleability. An extensometer is a device to measure the extension in materials.
WORK HARDENING AND ANNEALING
As described earlier, materials undergo plastic deformation due to slipping of lattices at the boundary layer. If the boundary
layers are long enough and free to glide, the material will be ductile. Ductility allows a material to be molded into different
shapes without breaking and thus are said to be malleable. If a metal is stretched, it takes a large force to break it because it
undergoes considerable plastic deformation due to the dislocation of boundaries. If the metal is bent up and down repeatedly,
boundaries are moved haphazardly causing some to interlock at joints. The joints now act as barriers to the free slipping of
lattice boundaries and material behaves as though the material was brittle and hence will break easily. This is described as work

hardening. In a similar way if a hardened metal is heated, it expands and allow interlocked boundaries to slip back and regain
mobility hence the metal becomes more ductile on heating. this is called heat annealing.
Hardening of a material can also occur by chemical means through the introduction of impurity into the lattices. example iron is
usually very ductile but introducing Carbone impurities in the lattices turn them into hard steel. similarly, oxygen can react with
polymers like rubber band when left outside for too long and they lose their stretchiness.
TOUGHNESS AND HARDNESS
Toughness is the resistance to cracks whereas hardness on the other hand is the ability to resist any plastic deformation.
usually hard materials also have a large elastic modulus such as metals and glass.
POLYMERICS
Natural rubber is cross linked with sulfur to make it more useful. some rubber called vulcanite has 1% sulfur whiles hard rubber
called ebonite has 4% sulfur.
Thermosetting polymers are ones that decompose on heating and cannot be remolded back to the original structure on
cooling. These are polymers that form numerous cross links to ensure a stable structure and on heating, the crosslinks are
permanently broken, e.g. epoxide, Bakelite
Thermoplastic polymers have very few cross links and hence there is some inherent stability in the structure of the fibrils even
in the absence of the cross links. Thus on heating the few cross links broken only soften the material and on cooling the original
mold could be attained again. e.g., polyethene, polystyrene and PVC
HYSTERISIS
Most materials get stretchy with heating hence the young’s modulus decreases with heat. Y = stress/strain.
strain = extension /initial length, extension with heat can be determine as e = L αΔT
Rubber shows a unique behavior in that its young’s modulus rather increases with heating so rubber coils up on heating
STRESS TO STRAIN CURVE WITHING ELASTIC LIMIT
stress
For most metals the curve is similar when loading and during unloading
there is no elastic hysteresis
strain
stress
loading curve
unloading curve
energy loss = hysteresis loop
strain

SIGNIFICANCE OF ELASTIC HYSTERESIS
Due to hysteresis, mechanical compression could be dissipated as heat even within the elastic limit hence making rubber a
good material for padding metals ends to absorb mechanical shocks. In a similar light, rolling car tires are likely to suffer large
mechanical energy losses through hysteresis which is undesirable hence pure rubber is not used in making car tyres. Tyres are
made from synthetic rubber that suffer less losses through hysteresis and such materials are said to be resilient.
SPRING POTENTIAL ENERGY
total mechanical energy is = ½ KA2. The potential energy increases with displacement from equilibrium
position and spring potential energy isPE= average force x extension = ½ F.x, but F = Kx
PE =½ k x2.
½ k A 2 = ½ kx2 + ½ mv2
𝟏
spring kinetic energy can thus be given by the relation 𝐊𝐄 = 𝐊√𝐀𝟐 − 𝐗 𝟐
𝟐
𝟏
Or ∆𝐊𝐄 = 𝐊√𝐗 𝟐 𝟐 − 𝐗 𝟏 𝟐
𝟐
energy stored in the spring can also be given as
W = ½ YAx2/L where Y = young’s modulus
FINDING THE CHANGE IN MECHANICAL ENERGY WHEN OTHER FORCES ARE AT PLAY
𝐊 𝐊
V = √ (𝐗 𝟐 𝟐 − 𝐗 𝟏 𝟐 ) or V = √ (𝐗 𝟐 𝟐 − 𝐠𝐡) Remembering that x =√(gh). or V = √𝐠(𝟐𝐡 − 𝐦)
𝐦 𝐦
𝟐𝐦𝐠𝐡
if all the energy is given as potential energy, then ½ kx2 = mgh, maximum depth of compression 𝐗 𝐦𝐚𝐱 = √
𝐤
𝐊𝐗 𝟐
and maximum height 𝐡𝐦𝐚𝐱 =
𝟐𝐦𝐠
MOMENTUM
P = Mv
KE = P2/2m
for explosive burst where momentum is conserved, such as detonation of a bomb or a piece of a rocket splitting off
as it is cruising or pieces of shrapnel breaking off in a bomb blast.
Mt m1
m2

Mt =M1 + M2. the smaller piece m1 moves with very high speed. because kinetic energy is proportional to
the square of the velocity, it gains a bigger portion of the total kinetic energy.
let kinetic energy of the combined mass KEt. kinetic energy of the various masses is related to the total kinetic energy as
follows.
𝒎𝟐 𝒎𝟏 𝑲𝑬𝟏 𝒎𝟐
KE 1(smaller mass) = KEt x , KE2(bigger mass) = KEt x , Ratio of KE1 to KE2 , =
𝒎𝟐 +𝒎𝟏 𝒎𝟐 +𝒎𝟏 𝑲𝑬𝟐 𝒎𝟏
This stems from the fact that in explosions, where there is conservation of momentum, the kinetic energy is directly
proportional to the momentum and inversely proportional to the mass.
The same analysis is applied in dealing with energies in nuclear fission in which most of the energy released due to mass excess
of parent nucleus is taken away by speeding neutrons regardless of their small size.
similarly, if a gun was fired, the total momentum of the gun and bullet will be zero that means the momentum of the gun is
equal but opposite the momentum of the bullet. if the speed of bullet is Vb and mass of bullet is Mb. the recoil speed of gun is
𝐌𝐁𝐮𝐥𝐥𝐞𝐭
given by
𝐌𝐆𝐮𝐧
× 𝐕𝐁𝐮𝐥𝐥𝐞𝐭
ELASTIC AND INELASTIC COLLISIONS
(A) INELASTIC COLLISIONS. In this type of impact, momentum is conserved but energies are not conserved. some energy
is dissipated in deformation of materials, heat, sound etc.
the common examples of inelastic collisions are car crash scenarios. inelastic collisions can be said to perfectly
inelastic if the 2 bodies stick together form a common mass with a common velocity immediately after impact.
in calculations involving inelastic collisions, (m1v1) I + (m2v2) I = (m1v1) f + (m2v2) f……………..1
if the collision is perfectly inelastic, (m1v1) I + (m2v2) I = (m1 + m2) vf…………………………2
(B) ELASTIC COLLISIONS. in this type of collision, there is conservation of both mechanical energy and momentum. such
collisions are what we get when dealing with Ballard ball questions or collisions of gases, electrons and nucleons. if
the question is about such collisions, assume it is perfectly elastic. for other types of scenarios, the question usually
will specify.
in these cases, we use 2 equations to solve. the first is similar to equation one above.
(m1v1) I + (m2v2) I = (m1v1) f + (m2v2) f……………..1
the second is the kinetic energy conservation equation.

(½ m1v12) i+( ½ m2v22) I = (½ m1v12) f +( ½ m2v22)f ……………3
. we will notice that, for calculations involving elastic collisions, we usually deal with bodies of similar symmetry,
orientation and masses but at different speeds.
equation 3 above is thus often simplified as (V1 – V2) initial = - (V1 – V2)final.
the difference in velocities before collision is equal to the negative of the difference in velocities after impact.
(C) FINDING THE TOTAL FORCE EXERTED ON MULTIPLE IMPACT.

for instance, if a tennis ball strikes a wall N times and in each case recoils elastically, the momentum changes per
strike ΔP = mv - -mv = 2mv. The impulse is thus . the total force is given as 𝑵
𝟐𝐦𝐕 𝟐𝐦𝐕
𝐭 𝐭
.
(D) FINDING THE IMPULSE WHEN GIVEN MASS RATE OR VOLUME RATE
in some settings, problems may describe a fluid pouring over a surface at a volume of V per second or solid impacting a surface
at a rate of M(kg) per second.
Force = Mv/t, hence if you have mass flow rate m/t, simply multiply that by the velocity v. also if you have volume flow rate,
obtain mass flow by multiplying this by the density.
F = 𝑡 x velocity = x ρ x velocity
𝑚 𝑉
𝑡

FLUID DYNAMICS
Specific gravity of a substance is the ratio of the density of that substance to the density of water at 4 0c.
ELASTIC DEFORMATION.
the measure of the resistance of a substance to elastic deformation is called the elastic modulus of the substance. it
measures the toughness of a material.
ELASTICITY IN LENGTH. the elastic modulus in length deformation is called the Young’s modulus. which has units of
Pascal.
𝑬𝒍𝒂𝒔𝒕𝒊𝒄 𝒔𝒕𝒓𝒆𝒔𝒔 𝑭𝒐𝒓𝒄𝒆 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏(𝒆)=∆𝑳
Young modulus, 𝒀 = . 𝑺𝒕𝒓𝒆𝒔𝒔 = and 𝑺𝒕𝒓𝒂𝒊𝒏 =
𝑬𝒍𝒂𝒔𝒕𝒊𝒄 𝒔𝒕𝒓𝒂𝒊𝒏 𝑨𝒓𝒆𝒂 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉=𝑳𝒐
so
𝑭 × 𝑳𝒐
𝒀=
𝑨 × ∆𝑳
Extension(e) ∆𝑳 = which is 𝒆 =
𝑭×𝑳𝒐 𝑺𝒕𝒓𝒆𝒔𝒔 ×𝑳𝒐
𝑨×𝒀 𝒀
which is 𝑭𝒐𝒓𝒄𝒆 = 𝒔𝒕𝒓𝒂𝒊𝒏 × 𝑨𝒓𝒆𝒂 × 𝒀
𝑨×𝒀×𝑳 𝒐
𝑭𝒐𝒓𝒄𝒆 =
∆𝑳
Note that under compression force, the young modulus of a material is different from when the same material is
under extension force. for instance, cement blocks have a high young’s modulus of compression and a small young’s
modulus of extension. Young’s modulus is only applied to solids.
SHAPE ELASTICITY.
the elastic modulus regarding deformation of shape is the shear modulus.
S = Stress / strain. in this type of deformation, there is no change in volume. the applied force is parallel to the
cross sectional area as opposed to a perpendicular force applied in a length deformation (young’s modulus). only
apply to solids.
VOLUME ELASTICITY
this is measured with the Bulk modulus.
Δ pressure = -B ΔV/V0. the inverse of bulk modulus is the compressibility of the substance. In solids and fluids.
Breaking stress is the maximum force per unit area of a material beyond which it breaks. this occurs
beyond the yield point of any material.
SB = Fmax/Amin
PRESSURE MEASUREMENTS
absolute pressure is the gas pressure.
P0
mercury pressure = ρ hg
if the prevailing atmospheric pressure
is known, P0, then the manometer
connected could read the mercury
pressure = ρ hg. the Gas pressure also
called the absolute pressure
BOUYANT FORCES P= ρ hg + P0
upthrust experience in a fluid = weight in air – weight in fluid
FORCES ON A SUBMERGED BODY
, if the body is floating, then weight in Air < or = Upthrust in the fluid
W = B, B = weight of fluid displaced and B = ρVg = ρhg.A

if the body is fully submerged and released, how to determine if the body will float upwards or sink further down.
W – B = gVobject( ρobject – ρfluid )
if the answer is 0, then the body will just float inside the fluid with its full height submerged, if positive, then the body
will sink, if negative then the body will float above the fluid surface.
FLUID IN MOTION
an ideal fluid is non viscous, incompressible, moves with steady flow such that each molecule has only linear
acceleration and zero angular frequency about the center of fluid flow
equation of continuity
rate of flow at one end of a close tube is = rate of flow at the opposite end of the same tube.
𝐦𝟏 𝐦𝟐
in terms of mass flow rate, =
𝐭 𝐭
ρ1A1 x velocity1 = ρ2 A2 x velocity 2
𝐕𝐨𝐥𝐮𝐦𝐞𝟏 𝐕𝐨𝐥𝐮𝐦𝐞𝟐
in terms of volume flow rate, =
𝐭 𝐭
A1 x velocity 1 = A2 x velocity 2
Work done by a fluid in a confined tube. W = P (V2 – V1) or V (p2 – P1)

BERNOULI EQUATION
Named after the Swiss physicist, Daniel Bernoulli.
he observed that for an ideal fluid in a closed space, at any point in the fluid, the work done (PV) + the kinetic energy
+ the potential energy remained constant.
PV + ½ρ V. v2 + ρVgh = constant.
If we divide each parameter per unit volume of fluid, we getP + ½ρ v2 + ρgh = constant.
This is the mathematical relation of the statement of Bernoulli’s equation, that the pressure, the kinetic energy per unit
volume and potential energy per unit volume in an ideal fluid in a confined space remains constant. for fluid in a
horizontal space where h = zero or non-appreciable, we assume the pressure is directly related to the negative of the
kinetic energy such that a gain in kinetic energy equal a loss in pressure leading to the assertion that the pressure in
a fluid is inversely proportional to the velocity of flow.
TORICELLIS LAW
The speed of a jet of fluid emerging freely from a

hole is equal to the speed acquired by a freely
falling object from a similar vertical height
h jet of fluid v =√ 2gh
Some applications of Bernoulli’s equation
Atomizer
Carburetors

Use of lighter fuels for spacecraft
𝟐∆𝐏
½ρv2= ΔP, velocity of a propelled rocket engine is thus given by 𝐕 = √
𝛒
SURFACE TENSION
surface tension force F
A fluid possesses cohesive forces which attract each molecule to

one another. The forces of cohesion allow the fluid to assume the
most stable structure and that is the structure with least surface
weight W
area. When the surface of the fluid is compressed, the attempt to
spread out the stable configuration is resisted with an opposite
force. This allows the fluid to restore its stable surface
From the diagram, the weight W configuration. A force that acts as though the fluid surface was a
is balanced by the upward surface tension force F cosθ continuous elastic membrane is the surface tension force. The
surface tension force exerted per unit area is called the surface
Surface tension y = force / length or perimeter of contact tension.
𝑭 𝒄𝒐𝒔𝜽 𝑭 𝒄𝒐𝒔𝜽
𝒀= = The surface tension has unit N/m
𝑳 𝟐𝝅𝑹
F = y 2π r cos θ
WETTING OR NON WETTING SURFACES
obtuse contact angle acute contact angle
When the fluid assumes an obtuse contact angle with a surface, it means the strength of cohesion inside the fluid is
much greater that the adhesion for the surface hence it will not wet the surface. This is because the surface tension is high. To
make such fluid to wet a surface, one adds detergents to break the cohesion down a bit.
CAPILLARY ACTION
The word capillary is hair like. in capillary tubes it is the surface tension force that drives fluids up the tube. Surface
tension force = weight of fluid column in the tube.
Height of fluid. The weight of the height of fluid =ρVg = ρgπr2h
ρgπr2 h = y 2π r cos θ. Therefore, the height of fluid column

𝟐𝒀
𝑯𝒎𝒂𝒙 = 𝝆𝒈𝒓
VISCOUSITY

Surface A
Viscous gel
Surface B
To slide surface A over surface B, the resistance created by the internal friction of the gel between the surfaces must be
overcome. The frictional force produced due to viscosity of a fluid is given by
𝐀𝐯
𝐅=ŋ 𝐝
.
where A is the surface area occupied by the fluid, d = is the distance through which the surface is moved and ŋ = coefficient of
viscousity.
It is measured in N.s/m2 or dyne. s/cm2 = 1 poise.
1 poise = 10-1 N.s/m2. For practical purposes, coefficient of viscosity is often written in centipoises = 10-2 poise.
VISCOUS DRAG ON A BODY FALLING THROUGH FLUID
The viscous drag according to stokes law
Fr = 3 x circumference of the spherical body x coefficient of viscosity x velocity
Fr = 6π r ŋ v. where r is the radius of the body falling through the fluid
POISEULES LAW
When dealing with viscous fluids, we can determine flow rate using the poiseulles law, which is given mathematically
as;
𝑽 𝟏 𝛑 𝐫 𝟒 ∆𝐏
𝒇𝒍𝒐𝒘 𝒓𝒂𝒕𝒆 ( ) =
𝒕 𝟖 ŋ𝐋
REYNOLDS NUMBER
The velocity beyond which a streamline flowing fluid start to become turbulent is determined by a dimensionless quantity
called the Reynolds number. (RN).
RN below 2000 means the fluid stays streamline and above 3000 means the fluid turns turbulent. Between 2000 and 3000
however, the fluid is metastable and any disturbance will create turbulence.
𝛒𝐕𝐝
𝐑𝐍 = ŋ
FICKS LAW
When it comes to rate of flow of solute particles in a solvent medium, it is termed diffusion. The rate of diffusion can be
determined by the relation known as ficks law
𝑀𝑎𝑠𝑠 [ 𝐂𝟐 −𝐂𝟏 ]
= D. A [concentration gradient]. Concentration gradient = .
𝑡𝑖𝑚𝑒 𝐋
𝒎 ∆𝑪
= 𝑫 × 𝑨𝒓𝒆𝒂 × , where D = coefficient of diffusion.
𝒕 𝑳
THERMODYNAMICS
The best state to study thermal physics is one in which molecules act as separate and freely mobile units i.e. in gases. For this
purpose, most of the discussions in thermal physics are restricted to the behavior of gases

CONCEPT OF ENERGY
The total energy possessed by a system by the effect of its total kinetic energy of the constituent molecules is called INTERNAL
ENERGY(U). The internal energy of a system can be changed by
1. Adding or removing energy through heat (Q) or
2. Doing work on or by system (W).
Heat is the transfer of energy between a system and its environment due to temperature difference.The temperature difference
between 2 points as measure over a uniform distance of separation is called Temperature gradient.
[ 𝐓𝟐 −𝐓𝟏 ]
Temperature gradient = 𝐋
CONCEPT OF TEMPERATURE: It is subjectively the perception of hotness or coldness. It is objectively the property of a
system that determines if 2 bodies are in thermal equilibrium.
THERMAL EQUILIBRIUM: When there is no heat exchange between 2 bodies which are placed in thermal contact, the
bodies are set to be in thermal equilibrium.
THERMAL CONTACT: 2 objects are in thermal contact if they are placed such that heat energy can be exchanged between
them. The rate of this heat exchanged depends on the temperature between the 2 bodies. if the temperature of the 2 bodies is
equal, temperature gradient becomes zero. At this point the 2 bodies are in thermal equilibrium. 2 bodies in thermal equilibrium
have the same temperature.
ZEROTH LAW OF THERMODYNAMICS: When objects A and B are separately in thermal equilibrium with a third body
C, then A and B are also in thermal equilibrium with each other.
The concept of a thermometer is built on the Zeroth law. When a thermometer is placed in thermal contact with a liquid, it
changes its properties i.e. volume, pressure etc., till it is in thermal equilibrium. We assume that when the parameter becomes
steady, the thermometer is in thermal equilibrium with the liquid hence its temperature can be determined on a scale. This value
represents the temperature of the liquid too.
THERMOMETERS
The thermometer must be small compared with the system so that the internal energy of the system is not altered too much due to
heat transfer.
CALIBRATION OF THERMOMETERS:
 Obtain the ice or freezing point, using water in equilibrium with ice. This is 0º c or 273.15 k at 760 mmHg. The
boiling point or steam point is obtained when water is in equilibrium with steam. It is 100ºc or 373.15k at 760 mmHg.
 These values, especially the ice point is very sensitive to pressure changes and therefore not easily reproducible. There
is thus departure towards the use of triple point of water as the lower limit and this is 0.01ºc above the ice point =
273.16k at 4.6 mmHg
NB. Pure water devoid of any dissolved gases is used to obtain the triple point temperature.
ACURACY OF COMPARING DIFFERENT THERMOMETERS

 Different materials used in different thermal expansions hence this introduces inaccuracies.
CONSTANT VOLUME GAS THERMOMETER

This is a thermometer in which a gas is used and volume is kept constant. The variable measured is changing pressure
corresponding to equilibrium temperature to be measured.
atmospheric pressure P0
P
h, corresponds to manometer pressure hρ g

The gas in a glass bulb is immersed in a liquid to be measured. As the gas expands, the mercury in the manometer is adjusted till
the gas returns back to the zero line (i.e. the volume is made normal again). The amount of mercury pressure needed to keep the
volume constant is given as hρg which is the gauge pressure. The pressure of the gas called the absolute pressure. P =Pº +
hρg.
The ‘h’ is measured accurately using a cathetometer. The gas thermometer is the most accurate type of thermometer but because
it is cumbersome, it is used to calibrate other thermometers.
FORMULAS IN THERMOMETRY
1. On the Kelvin’s scale;

The set point is only triple point – 273.16k
𝐩𝐡𝐲𝐬𝐢𝐜𝐚𝐥 𝐩𝐚𝐫𝐚𝐦𝐞𝐭𝐞𝐫 𝐚𝐭 𝐓
𝐓𝐊 = 𝐩𝐡𝐲𝐬𝐢𝐜𝐚𝐥 𝐩𝐚𝐫𝐚𝐦𝐞𝐭𝐞𝐫 𝐚𝐭 𝐓𝐫𝐢𝐩𝐥𝐞𝐊𝐩𝐨𝐢𝐧𝐭 × 𝟐𝟕𝟑. 𝟏𝟔𝐊
E.g. if the constant volume gas is being used, the physical parameter that is changing is pressure so
𝐩𝐫𝐞𝐬𝐬𝐮𝐫𝐞 𝐚𝐭 𝐓𝐊
𝐓𝐊 = × 𝟐𝟕𝟑. 𝟏𝟔𝐊
𝐩𝐫𝐞𝐬𝐬𝐮𝐫𝐞 𝐚𝐭 𝐓𝐭𝐫𝐩
2. On celcius scale: We use 2 end points. P0º and P100º
𝑷𝑻 − 𝑷𝟎
𝑻𝟎𝒄 = × 𝟏𝟎𝟎𝟎𝒄
𝑷𝟏𝟎𝟎 − 𝑷𝟎
MORE ON THERMODYNAMIC SCALE (ALSO CALLED KEVIN SCALE)
If various gases are used, it is seen that provided the pressures are low and temperatures are not extreme, the values are all similar
and if extrapolated backwards, they all give a pressure of 0 at -273.15K. This value is therefore taken as absolute zero. Ideally
matter cannot exist at this point because it means there is no pressure and this temperature has only been approached but not
achieved.
In terms of quantum physics however, objects will still have energy at absolute zero due to the fact that quacks have some energy
due to rest mass. this energy is called zero – point energy
DIFFERENT SCALES FOR TEMPERATURE
 Celsius, Fahrenheit, Rankine, Kelvin etc.

TF = 9/5 Tc + 32
TC= 5/9(TF─32)
∆TF=9/5∆Tc
ELECTRIC THERMOMETERS
They are only less accurate than gas thermometers and are less cumbersome. The important aspect of the thermometer
is the thermoelectric junction which consists of 2 fine wires welded together.
A galvanometer instead of a potentiometer is used to measure the EMF change with temperature.
Tºc = ET─ E0× 100
E100 ─ E0
RESISTANCE THERMOMETER
Usually is more accurate at extreme temperature. Usually made of a wire of platinum wound around 2 strips of mica
arranged crosswise. The temperature scale ranges from ─259.34ºc and 630.74ºc. The platinum used must be stain free platinum.
PROBLEMS INVOLVING OTHER THERMOMETERS CALIBRATION WITH GAS THERMOMETERS

Usually an equation for the parameters will be given. for instance, the formula for the calibration of the resistance thermometer is
given as
RT = R0(1 + aθ + bθ2).
you will then be given the values of the constants a and b and the resistance at say 100 0c. to find the temperature θ that
corresponds to a certain resistance R when the gas thermometer record T
𝐚 + 𝐓𝐛
𝛉= 𝐓
𝐚 + 𝟏𝟎𝟎𝐛
the 100 is because we are using the Celsius scale and the upper limit is 100
THERMOCOUPLES
Between 630.74ºc and 1064.43ºc (called the freezing point of Gold), temperatures are measured in EMF of thermocouples
preferably.
Thermocouple consists of a single joint of platinum and platinum/Rhodium alloy. This is where the thermal EMF is set up. To
prevent thermal EMFs from setting up in the free ends of the metals, both ends are jointed to copper and the joints placed in Ice.
The copper is then connected to a potentiometer.
Electrical equivalence of this is like measuring potential difference between Ɵ and 0ºc. usually, the calibration of thermocouple is
given by
E = a + b θ + c θ2
Where a,b,c are constants determined by EMF at gold point(1064.43), silver point(960.8ºc) and antimony point (630.74)
respectively.
THERMAL EXPANSION
Heat content or thermal energy is given by Q = mcƟ and ∆Q = mc∆Ɵ
LINEAR THERMAL EXPANSION: (∆L)
= Lo α∆T
∆𝐋
 Fractional length change = 𝛂 ∆𝐓
𝐋𝟎
Where α is called Coefficient of linear expansion
 The new length L = Lo (1+α∆T)
AREA EXPANSION: (∆A)
∆A = Ao.Y∆T
Fractional change in area

∆𝐀
=Y∆T =2α∆T
𝐀𝐨
where Y= coefficient area expansion.
new area A = A0(1 + Y∆T)
VOLUME EXPANSION:
similar ideas as above.
β is the coefficient of volume expansion and it is 3α

β = 3/2Y
DENSITY CHANGE WITH THERMAL EXPANSION (∆ ρ)

𝟎 𝝆
𝝆 = 𝟏+ 𝜷∆𝑻
SOLVING COMPOSITE PROBLEMS
How to find the temperature change needed to cause a body A to expand to the size of body B. when coefficient of Area
expansion of A=Ya, B= Yb.
𝐀𝐫𝐞𝐚 𝐀−𝐀𝐫𝐞𝐚 𝐁 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒊𝒏 𝑨𝒓𝒆𝒂
∆𝐓 = 𝐘𝐁×𝐀𝐫𝐞𝐚𝐁−𝐘𝐀×𝐀𝐫𝐞𝐚 𝐀 = 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒊𝒏 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒐𝒇 𝒂𝒓𝒆𝒂 𝒂𝒏𝒅 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒂𝒓𝒆𝒂 𝒆𝒙𝒑𝒂𝒏𝒔𝒊𝒐𝒏
IDEAL GAS
A system made of a collection at atoms or molecules that move randomly and exert no long-range forces on each other. Each
particle acts like a point and hence occupies negligible volume.
MACROSCOPIC PROPERTIES OF IDEAL GASES
Most gases behave like ideal gases at moderate temperatures and low pressures. Below a certain temperature, a gas will deviate
from ideal gas behavior no matter the pressure. Gas starts to liquefy below this temperature called the critical temperature of that
gas.
MACROSCOPIC LAWS OF IDEAL GASES
 Boyles law: p ∝ 1/V at constant T

 Charles law: T ∝ V at constant P
 Gay-Lusacs law: P ∝ T at constant V

IDEAL GASES LAW
PV= nRT
Where n is the moles of a gas = N/NA
R is called the ideal gas
Constant R= 8.31J/mol.k
R = 0.0821 L. atm/mol.k
A mole of gas occupies 22.4L at 1 atm and 0ºc.

𝑁𝑅𝑇
𝑃𝑉 = where NA is Avogadro’s constant
𝑁𝐴
𝑁𝑅𝑇
𝑃𝑉 = but R/NA is called Boltzmann’s constant
𝑁𝐴
1.38 x 10 -23 J/K

Boltzmann’s constant KB=
∆𝑃. 𝑉 𝜌𝑅𝑇
∆𝑛 = 𝑃 =
𝑅𝑇 𝑀

OTHER WAYS OF WRITING THE IDEAL GAS EQUATIONS
PV = n RT
𝜌𝑅𝑇
PV = NKBT and 𝑃 = where M = molar mass and ρ = density
𝑀
All the above equations can be deduced from what happens at the microscopic level. The theory developed by Maxwell is called
kinetic theory of ideal gases.
The Kinetic theory of ideal gases is built on same postulates.
1. Attraction between the particles is negligible

2. Volume of individual particles is negligible compared with the volume of container.
3. The particles are perfect spheres with random motion
4. The molecules make elastic collisions with the walls
5. The time during collision is negligible compared with time between successive collisions.
CONNECTED VOLUMES AND PRESSURE CHANGE
P1V1=P2(V1+V)
If this system has volume v and is used to pump out or suction a given volume Vo then the number of suctions done is n. the new
𝑉
pressure of the system Pn 𝑃𝑛 = 𝑃𝑜 ( 𝑜 )𝑛
𝑉𝑜 +𝑉
THE IDEAL GAS AND KINETIC MODEL
P =mv,in elastic collision ∆P = 2mv
The force impacted on the wall is 2mv/t per collision.
The translational kinetic energy per molecule = ½ mv2. However, each molecule does not possess exactly the same speed v. so
we determine an average speed, which is often used
Kinetic translational energy per molecule is thus ½ mv2 where V2is called the mean square speed of the gas.
But for a monoatomic gas, it has not just translational kinetic energy but rotations and vibrational as well. The degree of freedom
is 3 and this also expresses that, there are 3 ways in which a monoatomic gas can store kinetic energy. So the total kinetic energy
per molecule is given ny
KE = 3 ( ½ mv2) = 3/2mv2…………..(1)
In a like manner, remember that the kinetic energy = work done as we learnt in work –energy theorem. For a gas , work done is
W = PV
Since ½ mv2 = PV, we conclude that the total work done or the total internal energy 3/2 PV
Internal energy of n moles of a gas is given by U = 3/2nRT for a monoatomic gas such as helium, U = 5/2nRT for a
diatomic gas, this is because the degree of freedom is five and U = 6/3nRT for a polyatomic gas such as CO2.
U = x/2nRT where x = 3, 5, 6 etc………………(2)

For one mole of a gas, the internal energy is U = 3/2RT or 5/2RT or 6/2RT. Within this one mole, each atom will have a total
kinetic energy given by U/Na

For instance, one mole of a monoatomic gas will have internal energy of U = 3/2RT and each individual gas molecule will have a
total kinetic energy of U/Na = 3/2RT/Na, but we realize here that R/Na = KB. Therefore, for any mole of a gas, the average
kinetic energy of an atom is given by 3/2KBT for monoatomic, 5/2KBT for diatomic and 6/2KBT for polyatomic.
The total kinetic energy per molecule in any number of moles of an ideal gas is given by
KE = X/2 KBT where x = 3, 5, 6 etc………..(3)
ROAT MEAN SQUARED SPEED
We notice from above that for a monatomic gas, the average kinetic energy is KE = 3/2K BT
So 2KE = 3KBT.
2( ½ mv2) = 3KBT = mv2 = 3KBT
= v2 = 3KBT/m =(3nRT/m = 3RT/M)
Root mean square speed Is thus √ (V2) =√ (3KBT/m)= √ (3RT/M) =√ (3P/ρ)where P is pressure of the gas
and M is the molar mass.
MAXWEL DISTRIBUTION
If you take a system of gases, all do not possess the rms velocity. There is an average velocity which most of the molecule
possess called the most probable velocity. Others possess what is called the average velocity and some possess the rms
velocity. The relative properties of molecules with those speeds are described by the Maxwell distribution.
Gas temperature
A<B<C
A = most probable velocity
B = average velocity
C = RMS velocity
As the gas gets heated, the dotted lines shift

more to the left.
A B C
velocity
The maxwell-boltzmann distribution curve
Average velocity; given by the relation Vav = √(8RT/πM)
Most probable velocity; given by Vmp = √(2RT/M)
Root mean square speed; given by Vrms=√(3RT/M)
It is clear from above that ; Vrms/√3 = Vmp/√2 or Vrms /√3 = Vav/(√8/π)
For calculation purposes, note that the common constant in all the formulas is V =√(QRT/M) or v=√(QKT/m)
the changing variable Q is given as; Q =8/π for average speed, Q = 2 for most probable speed and Q = 3 for root mean
squared speed.

USING THE KINETIC THEORY TO EXPLAIN MACROSCOPIC LAWS
 BOYLES LAW: The pressure of a gas represents the impulse acting per unit surface area.
Volume =Area × Length
If the length is increased, the time between successive impacts is increased but since the temperature is constant, the
translational kinetic energy is same so the impulse change is same mv- - mu=2mv. With increase time for impulse
change, the impulse or force will reduce F =2mv
The pressure even reduces further because the increased volume means a larger surface area. P = F/A
CHARIES LAW
PαT
The internal energy u=3/2KBT or 3/2nRT. The internal energy determines the speed of translation of molecules to the wall. This
speed v2 = 3KBT/m
So if temperature is increased, there is more collision with the wall and the time of impulse is shortened. Keeping the volume
constant means that the impulse force /unit area, which is the pressure, exerted on the walls increases.
THERMAL PROCESSES, HEAT TRANSFER AND LAWS OF THERMODYNAMICS.
Most of the work on energy transforms as James Prescott Joule did heat. He was trained by John Dalton.
The internal energy of a system consists of the kinetic energy, which is translational, vibrational, rotational as well as molecular
potential energy due to bond lattice energies.
UNITS OF HEAT: It is measured in joules. The calorie is the heat energy needed to raise the temperature of one gram of water
by 1ºc (from 14.5ºc to 15.5ºc). 1 calorie =4.186J
The British thermal unit is another unit. (btu)
CALORIMETRY
-Heat change in warm = + Heat change in cold

- -Q hot = +Q cold.
TO FIND THE EQUILIBRIUM TEMPERATURE
Tfinal= m1C1 + m2C2
m1 + m2
LATENT HEAT PROBLEMS
Q= ± mL
Where L is the latent heat given
ENERGY TRANSFER
1. CONDUCTION: Rate of heat change Q/t = Surface Area ×Temperature gradient.
Q/t α A. ∆T/L, introducing the constant of proportionality K gives
Q/t = KA. ∆T/L = KA ( T2 - T1)/L where the constant K is called the thermal conductivity of the material. It is
similar to the electric conductivity of the material.
Newton’s law of cooling states that the rate of loss of heat is directly proportional to the temperature difference between the
body and the environment.

2. RADIATION: Rate of radiation is proportional to the surface area, the emissivity of the object and the fourth power
of its temperature. This is STEFAN’S law.
Radiant energy/time (R/t) α A e T4. e is called the emissivity of the body radiating the energy.
Introducing the constant of proportionality δ gives
R/t = δ AeT4 where δ is called the Stefan Boltzmann constant with a value of
5.6696 x 10-8W/m2.K4
Emissivity of a material ranges from 1 to 0. A pure reflector does not absorb radiation hence emits non and has emissivity of 0. A
black body is perfect absorber and emitter of radiant heat with emissivity if 1.
Energy given off per unit time per given area = P/A= =
δeT4
For a black body power = δAT4 and energy rate per surface area (intensity) = δT4
WIEN’S DISPLACEMENT LAW
λm T is = constant. This tells us that for a radiation emitter. The wavelength emitted is inversely proportional to the temperature
of the body this was developed by Wilhelm Wien many years before Max Planck.
λm T = 2.9 x 10 -3 m.K
RADIATION EXCHANGE BETWEEN BODIES
When one body receives radiation from another body. Power radiated from A= power received by B.
T4aAa = T4b Ab …………………….. 1
If we are dealing with spherical bodies such as exchange of radiation between the sun and the earth,
We can represent the equation as T4a R2a = T4b R2b
Hence Ta/Tb = √ (Rb/Ra)……………….2
PREVOST’S THEORY OF EXCHANGES
A body absorbs or radiate heat based on its temperature and the nature of its surface and
when the body is at constant temperature, it loses and gains heat by radiation at the same
rate.
If a body is, cooler than its environment, it doesn’t mean it stops radiating. It means it is absorbing more than it is radiating to the
environment.
EMMISSION AND ABSORPTION SPECTRA
Kirchhoff law states that the spectra emissivity of a body for a given wave length is equal to its spectral absorption factor for the
same wavelength.
ABSORPTION SPECTRUM
When matter absorbs light from one direction, the energy is used to;
1. In solids warm up the body. 2. In gases excite atoms.
This absorbed energy is later radiated off as the solid gets hot or as the gaseous atoms de excite. The energy is however
randomly given off equally in all directions.
If one is observing from the part of the original radiation, it is as though the energy has been reduced (darker band).
For solids, the overall wavelength given off is longer because it is cooler whereas for gases, it is the same wavelength but the
intensity is reduced along the part of initial radiation.
TYPES OF SPECTRA

Gases such as H2, Ne give line spectrum. This is type of spectrum is one of the experimental evidence of quantized energy levels
existing in atoms. Usually line spectrum is obtained from atoms or molecules of gases at low pressure.
Molecules, diatomic or polyatomic such as C02 produce band spectrum. It consists of thick bands due to vibrational energy and
light fine lines due to rotational energy.
Heated solids, liquids, or plasma produces continuous spectrum including the sun and other star.
When the temperature of a gas is reduced and pressure is applied, the line spectrum starts to spread into a continuous
spectrum. It starts to behave like hot liquid under these conditions.
ABSORPTION SPECTRUM
When bright yellow light is passed through hot sodium gas and observed in the path of the incident light, the light is observed
to have dark bands within it. This is explained by the concept discussed earlier and is a sign of absorption and re- radiation of
yellow light by sodium.
EMMISSION SPECTRUM OF THE SUN
The dark bands in the sun’s emission spectrum are the result of hydrogen absorption spectrum and these are called Fraunhofor
lines.
Normally, hydrogen is transparent to visible light because only light in ultraviolet rang can cause excitation to a higher level due
to the strong nuclear attraction. Hence, the absorption spectrum of Hydrogen is normally not obvious in the eyes.
Near the sun however, the hydrogen gas is already energized by the enormous heat hence visible light spectrum can easily be
absorbed as a top up to cause excitations hence leading to these dark bands.
GLOBAL WARMING AND GREEN HOUSE GASES
Greenhouse gases include CO2, SO2, CH4, N2O, water vapour.
LAWS OF THERMODYNAMICS
The first law states that:

∆U = ∆Q + W
The change in internal energy of a system is equal to the heat change and the work done on the system.
If work is done on the system, it is positive and if work is done by the system, it is negative.
I.e. if work is done on the system, the system increases its internal energy whilst the environment loses energy. If the system
however does work in the environment such as expansion of a gas, then the environment gains energy.
Work done = - P∆𝑽
MOLAR HEAT CAPACITY
CV – Molar heat capacity at constant volume
Cp – Molar heat capacity at constant pressure
Cp/ Cv = y(Adiabatic index)
Cp – Cv = R
THERMAL PROCESS (Q, U, W)
1. Isothermal process (occurs at constant temperature) the only energy component that is zero at constant temperature is
Heat. So the thermodynamic equation becomes:
∆𝑈 = 𝑊
2. Isobaric process (occurs at constant pressure) if pressure is constant, then a change in volume will lead to work done. W =
P∆V
∆𝑈 = 𝑄 + 𝑊

3. Adiabatic process: The system is isolated such that no heat comes in or goes out.
Q=O
∆𝑈 = 𝑊
4. Isovolumetric/ Isochoric process:∆𝑉 𝑖𝑠 𝑧𝑒𝑟𝑜0
So no work done.
∆U = Q
5. General process: involves a combination of all.
HEAT ENGINE AND HEAT PUMPS
HEAT ENGINES.
This is a machine operating between two heat reservoirs using the second law of thermodynamics to do work by generating
mechanical energy.
Heat engines were improved to their best during the dawn of the industrial revolution by the works of James Watt. Heat
engines go through 4 thermal processes, Isothermal expansion to isothermal compression to adiabatic expansion to adiabatic
compression. These form a HEAT CYCLE and based on the engineering design, there are three types of heat cycles namely,
CARNOT CYCLE
OTTO CYCLE
DIESEL CYCLE
Heat engine takes in Heat into a system to perform work with little heat is given off as exhaust.
Work done = Qin - Qout
W = Qhot - Qcold
Efficiency = (W / Qin)x 100% = (Qhot - Qcold/ Q hot) x 100%
= 1 – (Qc / Qh) x 100% = 1 – (Tinitial / Tfinal) x 100%
HEAT PUMPS
This system uses some other form of work done to reverse the work of a heat engine. That is, small heat is put into the system
but much heat is taken out of the system. Example, refrigerators and air conditions.
Heat pumps have coefficient of performance, COP.
Any heat pump can operate in forward or reserve.
In one order, it may remove small heat and generate more heat. In the reverse, it may take more heat and remove any small
heat resulting in heating or cooling of the environment respectively.
COP (cooling mode) = Qc / W
COP (heating mode) = QH / W
SI unit of COP – dimensionless
THE SECOND LAW OF THERMO DYNAMICS
This can be stated in 2 to 3 statements.
1. No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely to perform work.
This is KELVIN – PLANCK STATEMENT.

2. If two systems are in thermal contact, net thermal energy is transferred spontaneously by heat from hotter to the cooler
system and not reversed.
This is the CLAUSIUS STATEMENT.
HEAT PIPES: An engineered pipe device that transfer heat energy between 2 solid interphases by combining the effect of
thermal conductivity of a conductor and the effect of phase transition of a liquid/gaseous substance. the common conductor
used is often a material like copper and the conducting fluid is often a Freon as in refrigerators. these devices are useful in
refrigerators, air conditioners and so on
THE CARNOT ENGINE : Sadi Carnotis considered the father of thermodynamics. He was a French engineer. He developed the
model of an ideal engine, which all other heat engines are compared with. The Carnot engine does not exist but it is the ideal
engine which no other engine can work better than.
CARNOT’S THEOREM
No real engine operating between two heat reservoirs can be more efficient than a Carnot engine operating between the same
reservoirs.
ENTROPY
Energy change during a reversible process = Qr
Entropy S = Qr / T
Rudolf Clausius, a German physicist, pioneered the idea.
THE THIRD LAW OF THERMODYNAMIC
1. The entropy of the universe tends to increase in all natural process.
OR
2. The entropy of a perfect crystal is zero.
SPECTRAL SERIES OF HYDROGEN ATOM AND THE FRANK HERTZ EXPERIMENT.
Wavelength given off in atomic transition is inversely proportional to the energy change.
𝜆α 1 / ΔE
 Borh’s idea of wavelength

𝝀 = 𝟐𝝅𝒓/𝒏 Where n is the principal quantum number
 Borh angular Momentum = nh / 2𝜋
FRANK HERTZ EXPERIMENT
T T e- e- e- e- e- e- G A I /(A)
V
V
B
X Y Z
S V / (volt)
The idea by Bohr that atoms were confined to specific discrete energy levels was proved by the Frank hertz experiment. Sodium
vapor at very low pressure was sealed in a tube with a tungsten filament (T). When T is heated, electrons are ejected through

thermionic emissions the distance T – G is such that it is far larger than the mean free path so that the electrons can have
enough collisions with the sodium atoms.
MEAN FREE PATH
Mean free part is the average distance a gaseous particle can travel without collision with nearby particles.
𝝀 = 𝟏/√𝟐𝒏𝝅𝒅2
When d is the diameter of a molecule or the distance between the centers of 2 close molecules and n is the number of moles.
Viscosity of a gas:
ŋ = 1/3 mnv𝝀, where m is the mass, v is the mean velocity and n is the number of moles.
GRAHAM’S LAW OF DIFFUSION
Rate of Diffusion of a gas is inversely proportional to the square root of its density.
1
Rate of Diffusion 𝛼
√𝜌
√𝑽12 / √𝑽22 = √ 𝝆2 / √ 𝝆1 OR
R1 /R2 = √𝑴2 / √M1 = V1 / V2
VAN DER WAALS CONNECTIONS
He noticed the following assumptions:
 The volumes of gases were not negligible.

 Attractive forces between gases were not negligible.
If the gases attracted each other, then there should be a deficit in pressure. He proposed that the ideal gas equation might have
understated the pressures.
The pressure connection was to add the factor a / V2.
He also noticed the gases occupied a certain volume, which must be subtracted. The volume connection was to subtract b.
Van der Waals equation is this:
(P + a/V2) (V-b) = nRT
THERMOELECTRICITY
The thermoelectric effect (Sometimes termed seebeck-peltier effect) describes the phenomenon of voltage generation across a
conductor as a result of temperature difference or the reverse.
in one form, the phenomenon can generate EMF through the effect of temperature difference. This is used in the creation of
the thermoelectric generator similar to but less cumbersome than a regular heat engine. in the reverse process, passing voltage
across a thermoelectric device can bring about temperature change as it happens in certain thermoelectric refrigerators.
There are 5 physical processes upon which the concept of thermoelectricity is based. the seebeck effect, Peltier effect and
Thomson effect are reversible whereas the Fourier effect and joule heating are irreversible.
common thermoelectric materials include.
 Graphene (this is the tiniest molecule visible to man and forms the basic hexagonal carbon lattice unit on which other
allotropes of carbon such as graphite are built), Bismuth tellurite, lead tellurite, silicon-germanium, sodium cobaltite

A device that operates on thermoelectric phenomenon to determine temperature is the thermocouple. it is not a sensitive
instrument and thus only accurate at high temperatures. several thermocouples reverted together in series form a more
sensitive thermometer called thermopile.
SEEBECK EFFECT.
The generation of E.M.F in a composite material (usually alloy) due to the temperature difference at the ends. Discovered by
Johannes seebeck
seebeck coefficient(s)/v/K = - change in Voltage/ change in temperature
S = - ΔV / ΔT
example.
 calculate the seebeck coefficient of an alloy that generates 300mV when the 2 ends are held at 25 oc and
120oc.
PELTIER EFFECT
The generation of temperature difference across a composite material due to the presence of electric potential difference
across it.
THOMSON EFFECT, states that when a current flow through an unequally heated metal, heat energy is either absorbed or
evolved at different parts throughout the body of the metal.
the concept can be divided into 2 namely positive and negative Thomson effects.
UNEVEN HEATING OF A METAL
A B C D E
Current
CELL(S) HEATER
Consider the set up above with a copper bar AB. without any current flow, the heating of region C will gradually spread in both
directions towards BA and DE. Initially C is hottest followed by regions B and D at same temperatures and finally regions A and E
also at same temperatures.
Positive Thomson effect
when the circuit is closed and current flows due to the emf of (S), it is observed that after a while the region AB is coldest,
followed by BC, then it gets warmer from CD and finally to DE. It is as though the current flow is causing heat to be absorbed
from regions before the heating sources and then given out or evolved in regions after the heat source. This is termed positive
Thomson effect and materials that demonstrate this include; antimony, copper, zinc, cadmium and silver.
Negative Thomson effect
In other situations, the incoming current rather causes the afferent part of the metal to heat up by giving it heat (heat is
evolved) whiles the heat is absorbed from the efferent regions making it colder. This is called negative Thomson’s effect and
materials that demonstrate it include; iron, platinum, bismuth and cobalt, nickel and mercury
Zero Thomson effect

Certain metals show no Thomson effect and regions equidistant from the midpoint of heating remain at equal temperatures.
classic example is Lead(Pb). This is the reason why Lead is used for the demonstration of thermoelectricity and used in
instruments such as thermocouples.
Thomson Coefficient
the amount of heat absorbed or evolved when one coulomb of charges flows over one second (1 ampere of current) between
two points on a metal with a temperature difference of 1degrere is called the Thomson’s coefficient (σ) with unit of volt/oC
JOULE -THOMSON EFFECT
In thermodynamics, the Joule–Thomson effect also known as the Joule–Kelvin effect, Kelvin–Joule effect describes the
temperature change of a real gas or liquid when it is forced through a valve or porous plug while keeping it insulated so that no
heat is exchanged with the environment. This procedure is called a throttling process or Joule–Thomson process
At room temperature, all gases except hydrogen, helium, and neon cool upon expansion by the Joule–Thomson process when
being throttled through an orifice; these three gases experience the same effect but will only do so at much lower
temperatures. The gas-cooling throttling process is commonly exploited in refrigeration processes such as air conditioners, heat
pumps, and liquefiers
Most liquids such as hydraulic oils will be warmed by the Joule–Thomson throttling process. In hydraulics, the warming effect
from Joule–Thomson throttling can be used to find internally leaking valves as these will produce heat which can be detected
by thermocouple or thermal-imaging camera.
Throttling is a fundamentally irreversible process. The throttling due to the flow resistance in supply lines, heat exchangers,
regenerators, and other components of (thermal) machines is a source of losses that limits the performance.
JOULE EXPANSION
The Joule expansion (also called free expansion) is an irreversible process in thermodynamics in which a volume of gas is kept
in one side of a thermally isolated container (via a small partition), with the other side of the container being evacuated. The
partition between the two parts of the container is then opened, and the gas fills the whole container.
Vf= the final combined volumes
for an ideal gas, the temperature is constant so Ti=Tf
P1V1 =nRT1
P2V2=nRT2 but since T1=T2, then nRT1 =nRT2
so P1V1=P2V2=nRT
The Joule expansion, treated as a thought experiment involving ideal gases, is a useful exercise in classical thermodynamics. It
provides a convenient example for calculating changes in thermodynamic quantities, including the resulting increase
The Joule expansion should not be confused with the Joule-Thompson effect.
The system in this experiment consists of both compartments; that is, the entire region occupied by the gas at the end of the
experiment. Because this system is thermally isolated, it cannot exchange heat with its surroundings. Also, since the system's
total volume is kept constant, the system cannot perform work on its surroundings. As a result, the change in internal
energy, is zero. Internal energy consists of internal kinetic energy (due to the motion of the molecules) and internal potential
energy (due to intermolecular forces). When the molecular motion is random, Temperature is the measure of the internal
kinetic energy. In this case, the internal kinetic energy is called heat. In the theoretical application of this concept to ideal gases,
since there is constant volume and constant temperature, no change in kinetic energy and no work is done thus temperature
remains same and thus change in internal energy is zero.

For real gases however, there is cooling of the gases as they expand regardless of zero work on the environment. This is
because as gases expand there is conversion of kinetic energy into molecular potential energy
Empirically, it is found that almost all gases cool during a Joule expansion at all temperatures investigated; the exceptions are
helium, at temperatures above about 40 K, and hydrogen, at temperatures above about 200 K. This temperature is known as
the inversion temperature of the gas. Above this temperature gas heats up during Joule expansion. Since internal energy is
constant, cooling must be due to the conversion of internal kinetic energy to internal potential energy, with the opposite being
the case for warming.
Intermolecular forces are repulsive at short range and attractive at long range. Since distances between gas molecules are large
compared to molecular diameters, the energy of a gas is usually influenced mainly by the attractive part of the potential. As a
result, expanding a gas usually increases the potential energy associated with intermolecular forces.
In liquids, where molecules are close together, repulsive interactions are much more important and it is possible to get an
increase in temperature during a Joule expansion.
HALL EFFECT
The production of potential difference (the hall voltage) across a conductor when a magnetic field is applied perpendicular to
the direction of current flow. named after Edwin Hall
PHONON
it is a quasiparticle which carries the quantum of energy transmitted by compressional waves or vibrations. (e.g., sound, bomb
blast, fire explosions etc.)
Most abundant rock mineral in the earth crust is feldspar followed by quartz. the element with the most allotropes is carbon
followed by Sulphur.
PIEZOELECTRIC EFFECT.
the accumulation of electric charge in a material in response to the buildup of mechanical stress within it. common
piezoelectric materials include,
Quartz, Topaz, sucrose, lead titanate, Berlinite(AlPO4). piezoelectricity is used in wrist watches, transducers of ultrasound
machines.
the most abundant mineral salt of the element is Feldspar followed by Quartz which are all different rock forms of SiO 2
ALLOTROPES.
oxygen…dioxygen, ozone ( bluish), Tetraoxygen ( metastable), octaoxygen( red)
Diphosphorous is gaseous, black phosphorous is a semiconductor

MECHANICS, FRICTION, WORK, ENERGY
Displacement, x = ut + ½ at2
PROJECTILES
1. If the path traced by the missile is symmetric either half circle or quarter circle, then the usual shorthand formulas are easy to
deal with.
Eg, U u
1 2
2. If the path is asymmetric,Eg

U
S2 s1
3 4
S1 s2
Then aim at finding which path either horizontal or vertical is the limiting path (i.e. the one whose distance is already
predefined in the question.) Use that path to find the time of flight and that will then be used to calculate other
parameters.
To determine the flight in such asymmetric scenes, the usual shorthand formulas are problematic. Use the standard
x = ut + ½ gt2.
If the initial velocity was projected downward like in 4 above, which is in the direction of g, then
X= ut + ½ gt2
If the initial velocity was projected upward against g as in 3 above, then
X= ut - ½ gt2
Remember X is the displacement between initial point of projection and final point of projection.
i.e. S2 – S1 = X=ut± ½ gt2
OTHER COMMON IDEAS ABOUT PROJECTILES
1. For symmetric analysis:
a. Hmax = Ux2/ 2g =U2 sin2 θ /2g = ½ gt2

b. Range = U2sin 2 θ /g
c. Time to maximum height = U sin 𝜽 /g

d. Time for range = 2U sin 𝜃 /g
Finding Different Projectile Angles and their Ranges
R = U2sin 2 𝜃 /g
Hence R𝛼 sin 2 𝜃
Given same U, Range is the same for any 2 complementary angles of projection.
Finding the Impact Speed
𝑈𝑥 𝑠𝑡𝑎𝑦𝑠 𝑠𝑎𝑚𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡
Uy increases Vx = Ux
Vy
Impact speed V2 = Vy2 + Vx2
Uimpact = √𝑈𝑥2 + Uy(max)2
Angle of impact 𝜃
tan 𝜃 = Uy / Ux
SOME CHARACTERISTICS OF PROJECTILES
1. The horizontal component of velocity remains same through.
2. The vertical component changes from Uymin to Uymax
3. At the peak of the height, the velocity in the vertical path becomes zero but acceleration becomes + g from previous –g.
4. The acceleration vector and velocity vector Vx are perpendicular at the peak height only.
5. The acceleration and velocity vector are only parallel if the projectile is straight up.
FRICTION, INERTIA AND DISSIPATIVE FORCES
Friction always does negative work. The force vector is always directed opposite the displacement vector = - W
STATIC FRICTION
μs mg Cos 𝜃
μs increases with increasing force till it attains its maximum static friction = mg where maximum μs = 1. Beyond this point, the
object starts to move and experience a dynamic friction whose magnitude is largely constant and a little less than the Weight of
the object(mg).
Coefficient of static friction μs= tan 𝜃
μs = v2 / rg
Coefficient of dynamic friction
μd = -a /g = v2 /rg

ATWOODS MACHINE
1. acceleration
m1
m2
1. The common acceleration with one mass on a horizontal surface where m2>m1
a = (m2 /m1 + m2) g, If the floor is frictionless.
or a = (m2 – Um1 /m1 + m2) g, if m1 experiences friction whose coefficient is U.
2. The common acceleration with both masses freely hanging where m2> m1
a = (m2 – m1 /m1 + m2) g
m1 m2
Common Tension
T = (2m2m1 /m1 + m2) g
WORK ENERGY THEOREM
1. States that the total work done is equal to the changes in kinetic energy.
2. Sum of work done by conservative force – work done by dissipative forces = Change in kinetic energy.
CHANGE IN ENERGY DUE TO FRICTION
During relative motion of a body, the dynamic friction is = μsmgCos𝜃 but if the resultant force is applied at an angle of 𝜃,
then the friction is reduced by ma 𝐬𝐢𝐧 𝜽
Resultant friction maximum = μ s (mg - ma 𝐬𝐢𝐧 𝜽)
W = (v2 – u2 / 2) m
S = v2 – u2 / 2a
a = v2 – u2 / 2s
FOR SPRINGS

∆v2 = 2W / m, where W = spring potential energy
∆v2 = 2(½ kx2) / m
FOR FALLING OBJECTS
Treat inclined planes as free fall if it is frictionless
gh1 + ½ v12 = gh2 + ½ v22
v= √2𝑔(∆ℎ)
INCLINES WITH FRICTION
½ ∆𝑣2 = -μg x ∆d
GRAVITY
Universal Gravitation Law
F = GM1M2 / R2 NB: GM1/R2 =g
Gravitational Potential Energy
-GM1M2 / R =
Gravitational kinetic Energy
√𝑮𝑴
½ mgR = GM1M2 / 2R, orbital speed = v= , V = √𝒈𝑹
𝑹
Total Orbital Energy
GM1M2 / R = PE total = m2gR
Escape Speed
√𝟐𝐆𝐌𝐌
V= = √𝟐𝒈𝑹
𝑹
Escape speed = √𝟐 x Orbital Speed
GRAVITATIONAL POTENTIAL
-Gm1/ R = -g
Gravitational potential increases with R up to infinity. It decreases up to g at the surface of the body. Then it remains constant
throughout the surface and inside of the body.
Equipotential = g
OUTSIDE INSIDE OUTSIDE
INSIDE
g ∝ -1/R g ∝ -1/R

the same applies to the gravitational potential energy. note that as the distance decreases from infinity towards the surface of
the body, the magnitude of potential gets bigger but because infinity is regarded as zero potential, every other potential is in
the negative. increasing magnitude therefore means a decreasing potential taking maximum potential at infinity to be zero.
GRAVITATIONAL FIELD STRENGTH
Gm / R2 = g/R
This decreases with distance in an inverse fashion till it gets to the surface. Between the surface and centre, it assumes a direct
linear correlation.
Gravitational Force and its Variation with Volume
For the same mass, an increased volume changes the Force 𝑏𝑦 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 R. where R is the increase in radius.
ρ 1R1 = ρ 2R2
OUTSIDE INSIDE OUTSIDE
E ∝ +1/R2 E∝ R E ∝ +1/R2
At the center, the field is zero.
the gravitational force of attraction follows a similar variation. a body falling through a crater in the earth will experience
decreasing attraction towards the center of earth till the force gets zero as it just passes the earth center.
Variations of Gravitational Field Between Two Bodies
Body A + Ea Eb Body B
- Va E = 0 at infinity
F = 0 at infinity

DEFINITION OF ELLIPSE An ellipse describes a geometric figure such that the sum of the distance from any point on the
perimeter to 2 separate points known as foci is always constant.
a1 X b1
A B major axis
a2 b2
Y
minor axis
consider the ellipse above with 2 foci A and B. if any point say X is chosen of the circumference and
straight lines are drawn from X to meet the 2 foci. call the length of AX =a1 and the length BX = b1. if another point
independent of X, say Y is also selected and lines from that point are drawn to meet the 2 foci, we obtain AY = a2 and BY = b2. it
will be realized that a1 + b1 = a2+ b2. and this will hold true for any 2 points chosen.
minor axis in y plane
major axis in x plane
a =semi major axis b = semi minor axis
if any point P on the perimeter has coordinates P(x,y), then the general equation of an ellipse is given by
x2/a2 + y2/b2 = 1 where a is the radius in the x axis (for our diagram that is the semi major axis) and b is the radius in the y axis (
for our diagram that is the semi minor axis. for a circle, the semi major axis = semi minor axis = r, so the equation above
becomes x2 + y2 = r2.
THE GRAVITATIONAL CONSTANT
This was experimentally determined by Henry Cavendish in 1708. The Cavendish apparatus consisted of 2 spheres on a spring.
THE DIFFERENT MODELS
THE Geocentric model of the universe was developed by the Greek astronomer Claudius Ptolemy and had been accepted for
1400 years.
In 1543, a polish astronomer Nicolaus Copernicus proposed the Heliocentric Universe Model. He got support from Galileo.
Danish astronomer Tycho Brahe had studied and made precise geometric calculations based on Copernicus ideas. After Brahe’s
death, his assistant named Johannes Keppler (a German astronomer) continued the work and developed the laws.
KEPPLERS LAWS
1. All planets move in elliptical orbits with the sun at one of the focal points.
2. All lines drawn from the sun to any planet sweeps out equal areas in equal time intervals.

3. The square of the orbital period of any planet is proportional to the cube of the average distance between the planet and the
sun.
T2𝜶 R3
The path of larger planetary body describes an almost circular orbit T2𝜶 𝒓3 but for smaller bodies like comets, it is a true ellipse.
T2𝜶 𝒂3 where a is the semimajor axis. The constant of proportionality is the kepplers constant K. if the body at the focus of the
orbit is the sun then we use Ks and if earth we use Ke.
T2 = KxR3 Kx = 4𝝅2/GMX
For the sun Ks = 2. 97 x 10-19s2m3
For the earth Ke = 9.89 x 10-14s2m3
Whenever time is measured in light years, distances are usually stated in Astromical units; 1 AU.
For the sun and its planets, asteroids and comets,
T2 = a3
V = 2𝝅R/ T where V is the orbital speed and T is the period.
R = GM/ V2 where R is radius of orbit, GMe = 3. 989 x 1014m3/s2
GEOSYNCHRONOUS SATELLITE
A satellite whose period of revolution is equal to the period of rotation of the earth = 24 hours. a satellite placed in a
geosynchronous orbit gives a 24hr feed to earth from the same location as if it is staying static at the same point. these orbits
are usually very far from earth surface and the main problem is with the poor quality of live feed such as satellite TV on live
football matches. nearer orbits give better pictures but are not geosynchronous. this can be by passed by placing 2 or more
satellites in lower orbits in such a way that their periods will overlap
DISTANCES ABOUT THE CENTRE OF REVOLUTION
Farthest point from focus is called Apocenter
Nearest point from focus is called pericenter.
Earth as a focus – GEO
So we get Perigee and Apogee
Sun as the center -Helios, we get Aphelion and Perihelion.
Satellites gain speed as they move into closer distance with the focus. They however loose total mechanical energy even though
they gain kinetic energy. This is because a gain in KE = twice lost potential energy.
Remember that the gravitational force is a centripetal force keeping these planets in motion hence it does zero work on the
planets.
MODERN PHYSICS
BLACK BODY RADIATION

Thermal equilibrium distributions may be described by any of the following.
 Bose-Einstein distribution
 Maxwell-Boltzmann distribution
 Femi-Dirac distribution.
Earlier analysis of black bodies followed the Wien displacement law but there arose a challenge when the graph approach very
high energy region close to the UV and this paradox was called the catastrophe. This eluded scientist until max Planck was able

to deduce the mathematical formula that showed why hot black bodies behaved that way. Out of his equation, we came to
accept that energy was not a continuous wave but given off in discrete forms and these were quantized.
RELATIVITY:
This concept is addressed at 3 levels.
1. CLASSICAL OR GALILEAN RELATIVITY: is the notion that all things are in constant relative motion. It was believed that the
earth moved relative to a wind of other.
The Galilean relativity states that;
All the laws of mechanics hold true under all inertial frames of reference. The idea of ether wind was disproved by the
Michelson and Morley experiment using the Michaelson’s interferometer.
Also the constancy of the speed of light in vacuum regardless of the speed of the observer meant that classical relativity was
somehow flawed.
2. SPECIAL RELATIVITY: Einstein postulated that: i. the laws of physics (not just mechanics) are true in all inertial frames AND ii.
The speed of light remains constant in vacuum.
3. Einstein further postulated the laws of gravitational attraction which is known as the law of general relativity.
Any 2 masses attract each other with a gravitational force F = GMM/R2.
RELATIVISTIC TRANSFORMATIONS ( The Lorentz transformations)
Using the Lorentz transformations, most physical parameters can be transformed into relativistic derivatives when objects
move close to the speed of light. If the parameter increases with approach to c, you multiply by 𝜸. If the parameter decrease
with approach to C, then divide by 𝛾. 𝛾called the relativistic factor is given by; 𝜸 = 1/ √1- v2/c2
Effect of Relativistic Transformation on Various Measures
 Length, Area, volume ---------------Reduces

 Mass, time Velocity ------------------Increase
 Momentum, kinetic energy --------Increase
In dealing with volume and area, it is only the Length in x x1 direction that is transformed. The final X is then
multiplied by b x h to get the new volume.
ELECTROMMAGNETIC WAVES
Waves can also have a change in frequency as a result of relative motion. Relativistic Doppler is also sometimes called
Cosmological Shift. If the frequency approaches the infrared spectrum, the wave is said to undergo a red shift.
F0 – fs (√𝟏 ± 𝒗2/ c2), it is plus (+) if wave fronts are increasing towards observer and negative when vice versa.
RELATIVISTIC RELATIVE VELOCITY
2 bodies A and B with velocities Va and Vb. if their speeds are approximating C, then Relative Velocity Vab = (Va±Vb) / (1
± [VaVb/C2])
It is plus if they approaching and negative if they are receding.
ENERGY AND MASS TRANSFORMATION
A body at rest possesses rest mass M0 and rest energy = M0C2, at rest KE = 0. When the body accelerates to proximity at C, it
attains relativistic mass given by:
MR =M0 x 𝜸 = M0/ √1 – V2/C2

Its new total energy MRC2 = 𝛾M0C2
Total energy MRC2 = KE + M0C2
KE = ½ (MR – M0) C2 = ½ (𝛾M0 – M0) C2
Momentum = P
PC = KE, |PC|2= E2 – E02
For an electron M0C2 = 0.511 MEV
SOME OTHER FORMULARS
P = E/C = h/ 𝜆; h= 4.136 x 10-15eV.S; hc= 1. 24 X 1014 eV
E= MC2 = hf = hc/ 𝜆 ; 1ev = 1. 602 x 10-19
Maxwell’s formula for speed of Electromagnetic Wave, V = 1/ √𝑼0 x E0
Also remember that for an electromagnetic wave, the force due to electric field = force due to magnetic field.
Eq = Bqv
V = E/ B
PHOTOELECTRIC EFFECT
Emitter Photon Receiver
e- e- e- e-
V A
+ -
The metal(emitter) has free electrons which do not ordinarily move away from the metal surface due to positive attraction.
The minimum energy to liberate these outer electrons is called the Work function of the metal or ∅ or W0 = hf0 where f0 is
the minimum frequency a photon needs called Threshold frequency.
The total energy or the photon that incident on the receiver is hf = W0 + eV where eV is the kinetic energy beyond the work
function that determines the energy or speed of the flow of electrons.
The voltage can be increased till Vs called the Stopping Voltage. Beyond Vs, no more electrons can flow regardless of how much
photon intensify.
The maximum kinetic energy attainable by photoelectrons is eVs.
eVs = hf -∅ = h(f – f0)

PROTERTIES OF PHOTOELECTRONS
1. Each photoelectron is formed by a single photon with minimum energy = W0.

2. The kinetic energy of photoelectrons depends on the frequency of light and of the work function and not intensity.
3. If light has f ≥ f0, electrons are emitted instantaneously regardless of the intensity.
COMPTON EFFECT
That idea was developed by Arthur Holly Campton.
When light incidents on a surface and gets reflected or refracted, the photons interact with the atomic structure of the
surface hence there is some attenuation to the frequency in terms of energy lost. The wavelength of the reflected or
refracted ray is thus longer by a certain value. This change in wavelength ∆ 𝝀 is called the COMPTON wavelength.
∆ 𝝀 = h/ M0C(1- 𝐜𝐨𝐬 𝜽)
In some situations, the incident wave can be fully absorbed with excitation of an electron and its antiparticle the positron,
this is described as PAIR PRODUCTION.
In other case, the incident wave can disturb an electron – position pair to de-excite and give off another photon thus 2
photons formed and this is called PAIR ANHILATION. Annihilation occurs whenever a particle meets its antiparticle.
MATTER WAVES
 Particles possess particulate properties such as position, Momentum, KE, Mass, Electric charge.
 Waves have properties such as frequency, wavelength, intensity, energy, momentum, etc.
In 1942, Louis de Broglie postulated that, if electromagnetic waves like light have wave-particle duality, then all other matter
whose speed approximates that of C and with mass small enough such as electrons could have similar behavior.
DE BROGLIES HYPOTHESIS
He proposed that if such matter passed through apertures with width comparable to their wavelength, they will undergo
diffraction just as photons and other waves do.
He proposed that the wavelength 𝜆 = h/mv; with mv called the momentum.
EXPERIMENTAL PROOF OF DE BROGLIE’S HYPOTHESIS
Proof of X-ray Scattering: In 1912, Max Von Laue suggested that because of their regular pattern of arrangements, atoms in
crystals could cause diffraction of x-rays. X- rays have wavelength = 1A0 which is approximate to lattice size of atypical crystal.
The detailed theory of X- ray diffraction was later developed by Sir William H. Bragg. He showed that a piece of crystal could
behave like a mirror reflecting x-rays.
The reflection occurred at atomic planes called Bragg’s planes.
incident X ray r1
r2 Braggs plane 1
d r3 Braggs plane 2
Braggs plane 3
By measuring the scattering angles at the different planes, the wavelength of the scattered particle could be found using;

2d 𝐬𝐢𝐧 𝜽 = n 𝝀; where d is the path difference and n is the diffraction order (n = 1,2,3,4, etc)
A crystal can be identified by the array of dotted arrangements that an incident x- ray creates on it. These spots are termed
LAUE SPOTS.
Evidence of Electron Scattering
In 1927, at bell telephone laboratory, C. J Davisson and L. H. Germer did the first experiment using electrons incident on a
crystal.
Another experimental proof of de Broglie’s hypothesis was the thermal Neutron Diffraction Experiment.
De Broglie’s hypothesis also raised another scientific question.
Beam of Electromagnetic Waves
Beam of electromagnetic radiation screen to locate position of rays
a metal shield with adjustable aperture
It was noticed that, narrowing the split made it easier or more precise to detect the exact location of the rays position on the
screen with more accuracy. However, narrowing the slit so much brought the aperture close to the wavelength of the ray and
caused diffraction. Thus, the momentum was altered due to the interference.
Werner Heisenberg based on above observations stated that, the product of the uncertainty in position and uncertainty in
momentum was never less than h/4𝜋, ∆P. ∆x ≥ h/4𝝅
This is the well-known HEISENBERG UNCERTAINTY PRINCIPLE.
ENERGY OF ELECTRONS
E = nhf, where n is the quantum number and also reflect the number of photons assuming each photon has E 0 = hf.
Angular Momentum
1. Bohr’s Model, Mvr = nh/ 2𝝅 where n = principal quantum number

2. Schrodinger’s Model, P =√𝒍(𝒍 + 𝟏) . hbar, where l is the Azimuthal quantum
Sometimes, excitation – dexcitation of electrons in atoms could liberate x- rays. These internally generated X- rays could eject
nearby electrons leading to photoelectric emissions. This is called the AUGUR EFFECT.
X’ RAYS
They were discovered in 1895 by Wilhelm Roentgen. Their nature as true electromagnetic waves were developed by max Von
Laue in 1912.
X’ rays are high energy photons emitted when high speed electrons are slowed down. If the electron is suddenly brought to
complete rest, then all of its kinetic energy eV is transferred into photon energy for the x’ ray = hfmax. hence the x’rays formed
has maximal frequency.
eV = hfmax = hc/ 𝜆min

X’RAY PATTERNS
Intermittent spikes = characteristic X rays
X ray frequency
smooth continuum of x rays ( Bremsstrahlung )
Increasing Voltage
a. The x’ rays have a series of continuously varying energy or frequency represented by the steady curve. These x’ rays
are formed by a process of BREMSSTRAHLUNG. As the electrons approach a metal, the positive charged centers
accelerate the electrons and they lose energy in the process. The electrons lose kinetic energy and are progressively
slowed till they stop. The energy lost comes as X rays.
b. Characteristic X’ rays: These show as discrete spikes of specific energies. These are also termed transition series x-
rays. They are formed when a specific amount of energy is absorbed from the electrons in inelastic collisions. The
energy excites a ground state electron into higher levels. When the excited electron de- excites, it gives off energy as
the characteristic x’ rays.
K series means the de- excited electron falls back to the Kshell. If it de- excites from L to K, it is K 𝜶, if from M → K, it is KB, etc. L
series means de-excited electrons return to L- shell. If it de- excites from M → L, it is L 𝛼 , it from N → L, it is LB, etc.
BOHR’S CORRESPONDING PRINCIPLE
Akin to the fact that when objects attain speed in proximity to c, Newtonian mechanics change to quantum mechanics, Bohr
postulated that:
‘Quantum mechanics is in agreement with classic physics when the energy differences between quantized levels are very small.’
→ This is Bohr’s corresponding Principle.
ATOMIC TRANSITIONS, EMMISSION SPECTRA AND LASERS
Absorption Spectra are only formed from stimulation of matter with incident waves.
Emission spectra are however spontaneous or stimulated. In spontaneous emissions, excited electrons stay in the transition
state briefly < 10-8 s before falling to ground state. The de- excited electron can fall to any ground level hence if there are many
electrons transitioning to ground state, the photons given off are of variable energies hence are said to be incoherent.
In 1917, Einstein developed the concept of stimulated emissions.
STIMULATED EMISSIONS
1. We need to keep the transition state longer so that we can adequately influence which particular steady state level
the electrons can de- excite to. The excited electrons are thus kept in METASTABLE state.
2. We need to provide another photon which will provide the disturbance to the metastable electrons so that they will
fall to an energy level such that the photon given off is similar to the passing photon that caused the disturbance.
Excited state
approaching photon an extra photon from DE- excited electron

Ground state
LASERS
When the switch is pressed, voltage generated is used to cause a lot of the atoms in the laser tube to go into metastable state.
The excited electrons are now more than the ground state electrons – This is called POPULATION INVERSION. A
monochromatic light is released from one end and this creates the disturbance to de- excite electrons which in turn give off
photons similar to the passing photon. Gases used in lasers include Argon, Krypton.
There is therefore amplification of the Single photon and all of them bear the same characteristics (in phase, same frequency,
same wavelength, same energy). The beam given off is thus amplified, energetic and congruent.
The other end of the tube where the beam emerges from is partially silvered so that not the entire beam goes out of the tube.
Part of it is reflected back to continue the process.
Switch Reflected ray Transmitted Beam
Initial photons when switch is closed Amplified photon from stimulated emission
ELEMENTARY PARTICLES
Modified Standard Model;
Elementary particles mediating forces and energy are called BOSONS. They obey (Bose- Einstein statistics). They are grouped in:
1. GUAGE BOSSONS 2. HIGGS BOSSON
GAUGE BOSSONS
1. GLUONS- mediate the strong force, which is the interaction between quacks.
2. VECTOR BOSONS (W+ W- Z0), mediate the weak force which is the interaction between protons and neutrons.
3. PHOTONS: they mediate the electromagnetic force. Remember that the normal force is due to the electrostatic
repulsion on the material when its atoms are squeezed together due the existence of the Pauli’s exclusion principle.
4. GRAVITONS: They mediate the gravitational force. These are yet to be experimentally found. for now, it is only
theoretical evidence.
HIGGS BOSSONS.
These are the particles that give a large rest mass to heavy elements
THE 4 FUNDAMENTAL FORCES OF NATURE
In terms of decreasing strength:
Strong force → Electromagnetic force → Weak force → Gravitational force
In terms of range of interaction (increasing range),
(10-3 fm) weak → → strong (1 fm) → Electromagnetic → Gravitation.
The higher the mass of a boson, the shorter its range of interaction. It believed that actually the 4 fundamental forces are
simply different expressions of the same single force of the universe in different dimensions and that as matter accelerates into
relativistic realms, the forces will unify. So at high speed, the electromagnetic and weak force becomes the electroweak force.
At even higher speeds, all will combine to give a single force.
THE GRAND UNIFIED THEORY

This theories states that at high speeds all the gauge bosons will combine to produce one single force.
Elementary Particles that Mediate Matter
These are called FERMIONS. They obey Femi- Dirac statistics. They include 2 groups:
1. LEPTONS: These are basically elementary in nature with spin of ½, charge of –e or +e, 0. They are six with their
corresponding antiparticles. Electron, Electron- Neutrino, Muon, Muon- Neutrino, Tau, Tau- Neutrino. The smallest is
the electron neutrino and largest is Tau.
2. HADRONS: These are grouped into lighter MESONS and heavier BARYONS. All Hadrons are made of composites of
quacks and are therefore not truly fundamental. The net color charge of any hadron is zero.
MESONS: Formed by a quark and another antiquark. NB: If a quark combines with its own antiquark, there is pair
annihilation.i.e .any fundamental particle annihilates its antiparticle. Mesons include Pion, Eta, Kaon, they have a mass less
than the mass of a proton. The lightest is the Pion with mass of 140MeV/ c2.
BARYONS: They have a mass equal to or bigger than the proton mass. Usually made of 3 quarks. Eg.Proton, Neutron, Lambda,
Sigma, Xi, Omega. Occasional , particles are formed by three antiquarks and these are called antibaryons.
Proton is the lightest and Sigma, the heaviest. They have a spin of ½ or 3/2 .
PARTICLE INTERACTIONS
When 2 particles interact, field quanta largely made of bosons are exchanged. This idea was championed by Richard Feynman,
an American physicist. His Feynman diagrams are still used to depict these exchanges of field quanta. E.g., when 2 electrons
interact through electrostatic forces during bonding, a proton is lost which is formed virtual photon.
INTEREST IN MUONS
It has a mass of 105.7 MeV/ c2 compared to electrons mass of 0.511 MeV/ c2 which means it is 207 times bigger. It has a half-
life of 2.2 μS. However, despite the short half-life, most Muons reach Earth because they ravel at relativistic Velocity which
causes time dilation or length constriction.
COSMIC RAYS
Around the stars like the sun, there is large plasma of Helium and Hydrogen ions, Muons and others. As these approach the
atmosphere, they bombard with atmospheric gases to lead to the ejection of other particles such as gamma rays and other
waves. Gamma Rays are high energy Photons that are ejected whenever an excited nucleon falls back to ground state.
THE CONCEPT OF NEUTRINO, BETA DECAY AND ELECTRON CAPTURE
Beta Decay: There is decay of neutron into proton.
An electron is thus ejected from the nucleus. This fast moving electron is called a Beta ray. It was observed that, when the
Kinetic energy of the beta ray and the new nucleus were added, there was still a mass deficit compared with the original
nucleus. Enrico Fermi postulated the idea that, another particle must be produced which carried part of this energy away hence
accounting for the mass deficit. This particle was small and neutral thus hard to detect. This particle was called Neutrino (small
Neutron) and its antiparticle called Antineutrino. There are 2 types of beta decay:
i. Beta + decay; n + +e(positron) + Neutrino →P
OR
ii. Beta – decay; n →P + -e (electron) + Antineutrino.
CARBON 14 FORMATION (ELECTRON CAPTURE)
Cosmic radiation can cause electrons to be trapped within the nucleus of Nitrogen atoms. The negative charge thus reduces the
positive charges by + 1. Hence a proton is converted into a neutron.
147 N (P + e) → n + (146 C)
This forms the basis of Carbone dating. The body always keeps a fixed ratio of Carbone- 14 to Carbone- 12 when in health. On
dying, the relative amounts start to decline and the amount remaining compared to Carbone- 12 can be determined to predict

age since one died. The accuracy of Carbone- 14 dating is affected by events that suddenly alter the stable carbon-14 to
Carbone- 12 ratio such as nuclear wars and explosions, depletion of ozone layer and hence more cosmic bombardments etc.
MATTER AND ANTIMATTER
This idea that every particle had an antiparticle with equal mass but opposite charge was developed by Paul Adriene Maurice
Dirac.
Exceptions are the Neutral Pion 𝝅0 and photons which exist as their own antiparticles. In 1932, Carl Anderson discovered the
Positron using a Cloud Chamber experiment.
DECAY OF PARTICLES
Both elementary and composite particles can decay (transform) into others that are more stable. The most stable particles are
Electron, Electron Neutrino, Muon Neutrino and Tau Neutrino all of lepton group as well as Protons in the Hadron group.
EXAMPLES OF SOME DECAYS
 All mesons degenerate into Electrons, Positrons, Neutrinos and Photons.

 A muon decays into an electron, Neutrino or Antineutrino.
 Baryons degenerate into other products but always include a proton. protons do not decay except in electron
capture.
THE QUARK MODEL
There are 6 quarks; up, down, bottom, top, strange and charm. baryons always have 3 quarks whiles mesons have one quark
and antiquark. each quack has a spin of ½ and each antiquark too. Each quack has a charge of -1/3 e and +2/3 e.
Most matter hadrons contain only u and d. E.g., Protons (UUd) and Neutrons (Udd)
OTHER QUANTUM PHENOMENA OF QUARKS
 Strangeness, Charmness, Bottomness, Topness
Charmonium is a particle made of Charm particle and anticharm particle. It is a type of meson.
Quacks also possess color charges which mediate what is termed the Color force.
We can conclude that the most fundamental particles of matter are leptons and Quacks.
LAWS OF CONSERVATION
Conservation of baryon number. The baryon number of a particle is always conserved. The baryon number is a quantum
number given as;
B = 1/3 {N quarks ---- N antiquarks} .It is 1/3rd of the difference in quarks and antiquarks in the particle. So baryons have a baryon
number of +1 whiles mesons have zero and antibaryons have -1. Effectively, each quark will have a baryon number of 1/3.
Conservation of lepton number. The total lepton number of a particle is conserved before and after an interaction. The three
lepton numbers are electron lepton number, muon lepton number and tau lepton number.
Others are conservation of strangeness and Chamness.

CELESTIAL BODY PHYSISCS
APSIS
Refers to the two extreme points in an elliptical orbit. With the orbited body as a reference center, the apse closest to the
orbited body is called the periapsis and the apse farthest is called the apoapsis. The two celestial bodies orbit around their
common centre of mass called the BARYCENTRE which becomes the focus of the orbit.
SOME COMMON APSIDAL NAMES
Earth orbit( Apogee and perigee)
Sun’s orbit( Aphelion and perihelion)
Stars ( Apastron and periastron)
Lunar orbit ( Pericynthion and apcynthion)
APSIDAL DISTANCES
The distances of periapsis and apoapsis together are called the apsidal distance of that orbit. The sum of the two apsidal
distances forms a straight line usually drawn through the center of mass of the orbiting celestial body and the body being
orbited and is referred to as the major axis. Exactly half the major axis is the semimajor axis usually designated as a. Another
straight line through the mid or center of the orbit which is perpendicular and bisects the major axis is called the minor axis and
half of it is also called the semi minor axis usually designated as b.
Mathematical relations
. a (semi major axis) = ½ Major axis
Major axis = Apoapsis + Periapsis, so a = ½ (Sum of apsidal distances)
ECCENTRICITY.
Describes the closeness of an elliptical body to being a perfect circle. The smaller the eccentricity the closer it is to a circle.
With a perfect circle the apsidal distances Rp and Ra are equal.
Periapsis = a (1 - E) and Apoapsis = a (1 + E)
E = Difference of apsidal distance / Sum of apsidal distances
For a circle, E= 0, for ellipse E < 1, for a parabola E =1 and for hyperbola E > 1

Another approach is to use the given semi major axis and semi minor axis in the following relation
E2 = 1 – a2/b2
ASTROPHYSICS
A star is a mass of gases that emits its own cosmic radiations. Stars that undergo gravitational collapse can form anything
including Red giant, white dwarf, neutron star or a black hole.
in most galaxies, stars orbit in twos and form a binary system. Binary stars are thus two stars that orbit around a common
center of mass. Many of the stars in the milky way form a binary system. in some binary system, we have a normal star or
sometimes a red giant (companion star) and the other is a collapsed star such as neutron star or white dwarf and this form X-
Ray binary system. The X-rays are form due to the heating up of gases at the accretion disk of the collapsed star and this leads
to formation of plasma and this causes fast moving free electrons bombard other atoms. when the collapsed star is a neutron
star we call this binary system a CATACLYSMIC VARIABLE. in this the companion is often a red Giant.
Galaxies are collection of gases within which stars are found in the universe. all galaxies emit electromagnetic waves and the
energy from these emissions is the sum from all the stars. Some galaxies called ACTIVE GALAXIES emits extra amount of energy
larger than what the sum of radiations from all it stars can give. These are galaxies that have a supermassive black hole at the
center. Gases speed up as they approach the accretion disc and this creates heat that gives of radiation. Examples are
QUASARS, BLAZARS, RADIO GALAXIES.
STELLAR EQUILIBRIUM
stars are made of gases and stays in hydrostatic equilibrium so far as the kinetic energy of gases balances the potential energy
due to the gravitational force. VIRIAL THEOREM says to maintain stellar equilibrium, the potential energy due to gravity must
be equal to twice the internal energy of the gases.
GRAVITATIONAL COLLAPSE refers to the contraction in size of an astronomical object under the influence of its own gravity.
The entire universe is formed by progressive gravitational collapse, from nebular to galaxy to star to planet. The size of a star is
due to the balance between the outward forces of expanding gases with their electrostatic repulsion due to plasma charges
and the gravitational force of attraction due to matter mass tending to collapse the system.
COLLAPSE OF STARS
At a point in the life of a star, most of the fuel is used up by fusion reaction so the expanding electromagnetic waves decrease
and gravity starts to condense the volume and the star
If a star collapses, a huge electromagnetic outburst is given out and this results in a Supernova. E.g., The Crab Nebula. The
dense matter formed within this Supernova is called a Neutron Star or sometimes a black hole.
Within the Neutron Star, there are occasional discharges of light called QUAZAR. Some quasars occur in pulses and are called
PULSARS.
A star is born when interstellar gases collapse to form a denser star and the compression leads to thermonuclear fusion and
generation of large amounts of radiation. The gravitational force is dependent on the mass of the star. If the mass is beyond a
certain minimum, the gravitational pull will be too great for the outward forces to compensate and the star will collapse. The
minimum mass that will bring about gravitational collapse is the JEAN MASS.
when a star burns all it fuels, it is said to be dead and it will contract sharply until a new equilibrium can be reached. A star may
go through several stages called stellar remnants whiles dying.
THE SUN
It is the closest star to earth. The inner most boundary that is visible to any astronomical device is called the PHOTOSPHERE.
There are scatter regions of high temperature gases called FLARES and low temperature gases called SOLAR SPOTS (due to
intense magnetic activity). Most of the visible spectrum comes from here. The next layer of the suns atmosphere above the
photosphere is called the CHROMOSPHERE. This layer is transparent to the light from the photosphere and hence usually not
seen. It is only obvious as a bright rim of light around the obscured sun during total solar eclipse. The outermost atmosphere is
called the CORONA. The temperature here is much higher than the inner atmospheric layers due to electromagnetic processes.

The plasma is so hot that it is not easily contained by the suns gravity hence flows off as what is termed SOLAR WIND. from the
inside out, the internal part of the sun includes the CORE, RADIATION ZONE and CONVECTION ZONE. The most abundant gases
are Hydrogen 98% and Helium 1%.
STELLAR REMNANTS
Red Giant. When a star burns all its energy, the core starts to contract. This allows the little hydrogen close to the core to fuse
and produce heat that will be given off to the atmosphere. The atmosphere thus starts to expand and therefore becomes
cooler hence radiating in the red spectrum leading to low temperature ball of burning gas called the RED GIANT. with time the
hydrogen is totally finished and the Red giant will die to become a White Dwarf.
White dwarf. This is the stable state of collapse for smaller stars and stability is due to the fact that gravity is opposed by
electron degeneracy pressure. The Red giant after exhausting all the hydrogen collapse to become like the size of the star at its
birth but doesn’t collapse further due to electron repulsion. The energy is provided also by the fusion of Helium to produce
other elements such as carbon. When there is nothing else to burn in the white dwarf, it becomes a black dwarf.
Neutron star. This is the stable state of most massive stars and stability is due to the fact that gravity is opposed by neutron
degeneracy pressure. They are formed when a massive star collapses after they have undergone a supernova. The neutron star
is able to produce minimal radiation due to the massive magnetic fields. Most exist as bodies spinning in space and causing
particles to accelerate under their massive magnetic field usually 1000 times that of the earth. The spinning of elements causes
them to intermittently give off radiations at timed intervals and such neutron stars are called PULSARS. Some neutron stars
have extreme magnetic fields are they are referred to as MAGNETARS. Other neutron stores exist quietly and only emit X-Rays
without pulsed lights.
Black hole. A body that is a perfect absorber and radiator of all electromagnetic waves. There is no outward force enough to
oppose gravity. it is the densest matter that exist.
SUPERNOVA
Star explosion leading to a final stable state of gravitational collapse. Two forms of supernova exist. Either 1. a massive star
explodes into a neutron star or black hole or 2. A white dwarf in binary with a Red giant explodes. supernova phenomenon are
means by which heavy elements are formed and distributed across the galaxy as well as the streaming of cosmic rays around
galaxies. When the star just exhausts its fuel, initially the outer gases swell up but with low temperature radiation to form a RED
GIANT. The core however contracts sharply under gravity. The core is made of iron so fusion cannot yield any energy. As the
core is compressed further repulsion of the charged nuclei leads to a massive explosion which extrudes electrons and free
protons with release of high energy photons such as gamma rays and X rays. The energy released is this explosion leads to a
supernova. If the original star is about 8 solar equivalents, its mass is not so huge hence the gravitational pull leads to the
formation of NEUTRON STAR. if the original star was massive about 15 solar equivalent, then a BLACK HOLE will form.
PROPERTIES OF BLACK HOLES
In terms of density of space matter, neutron stars are only superseded by black holes. They are so dense that even light cannot
escape from it. Any celestial body with density so great as not to allow any electromagnetic wave escape from its gravitational
field is a black hole. The radius of a body at which an object in its gravitational field has an escape velocity equal to the speed of
light is the SCHWARZSCHILD RADIUS. Any object that contracts in size below its Schwarzschild’s radius will become a black
hole. The center of the black hole is called the SINGULARITY and that is where the dense celestial mass is located and the
boundary from the center to the point where escape velocity is equal to speed of light is called the EVENT HORIZON. the
distance from the singularity to the event horizon is the Schwarzschild radius. Any object within the event horizon cannot
escape outward. Light can just escape from the zone of event horizon.
√𝟐𝑮𝑴 √2𝐺𝑀 √2𝐺𝑀

Escape speed = ; C= ; r=
𝒓𝟐 𝑟2 𝐶2
R = Schwarzschild radius
AN ACCRETION DISC is a region around a collapsed star or black hole where gases accelerate and heat up as they encircle the
body under its gravitational field.

ERGOSPHERE
The region within the gravitational influence of a black hole but outside the event horizon. Within its boundaries, the space
time of any object is distorted by the gravity of the black hole and an object cannot be stationary.
SPACE-TIME
This is a concept of modern physics which seeks to create a 4 D parameter out of interaction between time (1D) and space (3D).
It is believed that the effects of gravity on objects is due to the gravitational distortion of space time around the bodies in the
vicinity
since light cannot escape from a black hole, they can only be observed indirectly by studying evidence of its existence such as
presence of accretion disc, other stars orbiting an unseen center or observing materials being pulled into an unseen central
area. There are 2 types of black hole. Black holes are always formed from very large stars about 15 times and more the sun.
Black holes that are about 5 times the size of the sun are called STELLAR-MASS BLACKHOLES and are found scattered in most
galaxies. Those formed from extremely large stars are about a million the sun =s size and are called SUPERMASSIVE
BLACKHOLES and are usually sitting at the center of galaxies.
THE ELECTROMAGNETIC SPECTRUM
IN ORDER OF INCREASING WAVELENGTH BUT DECREASING ENERGY/FREQUENCY

THE EARTHS ATMOSPHERE
The Earth’s atmosphere has a series of layers and moving upward from ground level, these layers are named the troposphere,
stratosphere, mesosphere, thermosphere and exosphere. The exosphere gradually fades away into the realm of interplanetary
space thus beyond the thermosphere the boundaries are not well defined.
Troposphere
It is the lowest layer of our atmosphere. Starting at ground level, it extends upward to about 10 km above sea level. Humans
live in the troposphere, and nearly all weather occurs in this lowest layer. Most clouds appear here, mainly because 99% of the
water vapor in the atmosphere is found in the troposphere. Air pressure drops, Thus the air expands more and temperatures
get colder, as you climb higher in the troposphere. This is because the air is heated from radiations from the earth below.
occasionally traps greenhouse gases in the upper part of the stratosphere can absorb so much heat causes the upper part to be
warmer than the lower regions and this phenomenon is called thermal inversion.
Stratosphere
The stratosphere extends from the top of the troposphere to about 50 km above the ground. The ozone layer is found within
the stratosphere. Ozone molecules in this layer absorb high-energy ultraviolet (UV) light from the Sun, converting the UV
energy into heat. Unlike the troposphere, the stratosphere actually gets warmer the higher you go! That trend of rising
temperatures with altitude means that air in the stratosphere lacks the turbulence and updrafts of the troposphere beneath.
This area is heated from radiations above from space. Commercial passenger jets fly in the lower stratosphere, partly because
this less-turbulent layer provides a smoother ride.
Mesosphere
It extends upward to a height of about 85 km above our planet. Most meteors burn up in the mesosphere. Unlike the
stratosphere, temperatures once again grow colder as you rise up through the mesosphere. The coldest temperatures in Earth's
atmosphere, about -90° C, are found near the top of this layer. The air in the mesosphere is far too thin to breathe; air pressure
at the bottom of the layer is well below 1% of the pressure at sea level, and continues dropping as you go higher.
Thermosphere
The layer of very rare air above the mesosphere is called the thermosphere. High-energy X-rays and UV radiation from the Sun
are absorbed in the thermosphere, raising its temperature to hundreds or at times thousands of degrees. However, the air in
this layer is so thin that it would feel freezing cold to us! In many ways, the thermosphere is more like outer space than a part
of the atmosphere. Many satellites actually orbit Earth within the thermosphere! Temperatures in the upper thermosphere can
range from about 500° C to 2,000° C or higher. The aurora, the Northern Lights and Southern Lights, occur in the thermosphere.
Exosphere
Although some experts consider the thermosphere to be the uppermost layer of our atmosphere, others consider
the exosphere to be the actual "final frontier" of Earth's gaseous envelope. As you might imagine, the "air" in the exosphere is

very, very, very thin, making this layer even more space-like than the thermosphere. There is no clear-cut upper boundary
where the exosphere finally fades away into space.
Ionosphere
The ionosphere is not a distinct layer like the others mentioned above. Instead, the ionosphere is a series of regions in parts of
the mesosphere and thermosphere where high-energy radiation from the Sun has knocked electrons loose from their parent
atoms and molecules. The electrically charged atoms and molecules that are formed in this way are called ions, giving the
ionosphere its name and endowing this region with some special properties.
SOME HEAT MEASUREMENT DEVICES

A bolometer is a device for measuring the power of incident electromagnetic radiation via the heating of a material with a
temperature-dependent electrical resistance. It was invented in 1878 by the American astronomer Samuel Pierpont Langley.
Principle of operation
A bolometer consists of an absorptive element, such as a thin layer of metal, connected to a thermal reservoir (a body of
constant temperature). Any radiation incident on the absorptive element raises its temperature above that of the reservoir (the
greater the absorbed power, the higher the temperature). The difference in heat between the two materials leads to a change
in electrical resistance of the absorptive material and therefore in its thermal conductance and this leads to current flow that
records the rate of radiation detected. Most modern bolometers use semiconductor or superconductor absorptive elements
rather than metals. These devices can be operated at cooler temperatures, enabling significantly greater sensitivity.
Bolometers are directly sensitive to the radiant energy left inside the absorber. For this reason they can be used not only for
ionizing particles and photons, but also for non-ionizing particles, any sort of radiation, and even to search for unknown forms
of mass or energy (like dark matter); this lack of discrimination can also be a disadvantage.
Langley's bolometer
The first bolometer used by Langley consisted of two platinum strips covered with lampblack. One strip was shielded from
radiation and one exposed to it. The strips formed two branches of a Wheatstone bridge which was fitted with a
sensitive galvanometer and connected to a battery. Electromagnetic radiation falling on the exposed strip would heat it and
change its resistance. By 1880, Langley's bolometer was refined enough to detect thermal radiation from a cow a quarter of a
mile away. This instrument enabled him to thermally detect across a broad spectrum, noting all the chief Fraunhofer lines. He
also discovered new atomic and molecular absorption lines in the invisible infrared portion of the electromagnetic
spectrum. Nikola Tesla personally asked Dr. Langley if he could use his bolometer for his power transmission experiments in
1892.
CURRENT APPLICATIONS OF BOLOMETER
1. whiles there are better methods to detect other wavelength, bolometers remain the best for detecting sub-millimeter
wavelengths ( 200 µm to 1 mm, also known as the far-infrared or terahertz). To achieve the best sensitivity, they must be
cooled to a fraction of a degree above absolute zero. Notable example of its use is in the James Clerk Maxwell Telescope
2.They are used as particle detectors. In elementary particle physics, they are referred to as calorimeters since they measure
energy change when particles incident on the plate but unlike the usual calorimeter in thermodynamics, they measure minute
heat changes.
A pyrometer is a type of remote-sensing thermometer used to measure the temperature of a surface. In the modern usage, it is
a device that from a distance determines the temperature of a surface from the amount of the thermal radiation it emits, a
process known as pyrometry and sometimes radiometry.
The word pyrometer comes from the Greek word for fire(pyro) and meter, meaning to measure. The word pyrometer was
originally coined to denote a device capable of measuring the temperature of an object by its incandescence (visible light
emitted by a body which is at least red-hot) Modern pyrometers or infrared thermometers also measure the temperature of
cooler objects, down to room temperature, by detecting their infrared radiation flux.
The potter Josiah Wedgwood invented the first pyrometer to measure the temperature in his kilns. He used crude methods
observing changes in color of clay.

The first disappearing-filament pyrometer was built in 1901.This device had a thin electrical filament between an observer's
eye and an incandescent object. The current through the filament was adjusted until it was of the same color (and hence
temperature) as the object, and no longer visible; it was calibrated to allow temperature to be inferred from the current.
Brightness pyrometer is dependent on the emissivity of the object. With greater use of brightness pyrometers, it became
obvious that problems existed with relying on knowledge of the value of emissivity. Emissivity was found to change with surface
roughness, bulk and surface composition, and even the temperature itself.
To get around these difficulties, the ratio or two-color pyrometer was developed. They rely on the fact that Planck's law, which
relates temperature to the intensity of radiation emitted at individual wavelengths, can be solved for temperature if Planck's
statement of the intensities at two different wavelengths is divided. This solution assumes that the emissivity is the same at
both wavelengths and cancels out in the division. This is known as the gray body assumption. Ratio pyrometers are essentially
two brightness pyrometers in a single instrument.
After several uses of ratio pyrometers, it became clear that for most metals the emissivity didn’t fully cancel each other out. To
more accurately measure the temperature of real objects with unknown or changing emissivity, multiwavelength pyrometers
were developed.
APPLICATION OF PYROMETERS
Pyrometers are suited especially to the measurement of moving objects or any surfaces that cannot be reached or cannot be
touched. Temperature is a fundamental parameter in metallurgical furnace operations. Reliable and continuous
measurement of the metal temperature is essential for effective control of the operation. Smelting rates can be
maximized, slag can be produced at the optimum temperature, fuel consumption is minimized and refractory life may also be
lengthened. Thermocouples were the traditional devices used for this purpose, but they are unsuitable for continuous
measurement because they melt and degrade.
Salt bath furnaces operate at temperatures up to 1300 °C and are used for heat treatment. At very high working temperatures
with intense heat transfer between the molten salt and the steel being treated, precision is maintained by measuring the
temperature of the molten salt. Most errors are caused by slag on the surface which is cooler than the salt bath.[10]
A steam boiler may be fitted with a pyrometer to measure the steam temperature in the super heater.
A hot air balloon is equipped with a pyrometer for measuring the temperature at the top of the envelope in order to prevent
overheating of the fabric.
Pyrometers may be fitted to experimental gas turbine engines to measure the surface temperature of turbine blades. Such
pyrometers can be paired with a tachometer to tie the pyrometer output with the position of an individual turbine blade.
Timing combined with a radial position encoder allows engineers to determine the temperature at exact points on blades
moving past the probe.
The tasimeter, or microtasimeter, or measurer of infinitesimal pressure, is a device designed by Thomas Edison to
measure infrared radiation. In 1878, Samuel Langley, Henry Draper, and other American scientists needed a highly sensitive
instrument that could be used to measure minute temperature changes in heat emitted from the Sun's corona during the solar
eclipse.
An aethrioscope is a meteorological device invented by Sir John Leslie in 1818 for measuring the chilling effect of a clear sky.
The name is from the Greek word for clear]
It consists of a metallic cup standing upon a tall hollow pedestal, with a differential thermometer placed so that one of its bulbs
is in the focus of the paraboloid formed by the cavity of the cup. The interior of the cup is highly polished and is kept covered by
a plate of metal, being opened when an observation is made. The second bulb is always screened from the sky and so is not
affected by the radiative effect of the clear sky, the action of which is concentrated upon the first bulb. The contraction of the
air in the second bulb by its sudden exposure to a clear sky causes the liquid in the stem to rise ]
The device will respond in a contrary fashion when exposed to heat radiation and so may be used as a pyrometer too
ANGULAR MOMENTUM
Let us first revise our quantum numbers.

Principal quantum number n 1≤n
Azimuthal quantum number(angular momentum) ℓ 0≤ℓ≤n−1
Magnetic quantum number(projection of angular momentum) mℓ −ℓ ≤ mℓ ≤ ℓ
Spin quantum number ms −s ≤ ms ≤ s
Orbital angular momentum/Azimuthal quantum number is also simply referred to as the orbital quantum number.
Classical angular momentum of a particle is given by Mvr = Mr2 ω and its vector direction is given by the right hand rule.
Angular momentum of an isolated system is always conserved.
Quantum Theory of angular momentum. There are two models, based on the orbital path of either Bohr or Schrodinger.
Quantum theory stipulates that all angular momentum is quantized in magnitude and direction and magnitude ranges within
the limit given by
L2= ℓ(ℓ+1) ℏ2 for orbital angular momentum and S2 = s(s+1) ℏ2.
The projected angular momentum along any axis is also quantized.
Bohr orbit, Angular momentum
De Broglie wavelength is given by λ = h/mv, Bohr postulated that for the circular orbit, the wavelength must describe a path of
the circumference of a circle such that nth number of wavelength should be equal to the full circumference.
nλ=2πr
L (mvr) = nh/2 π
L (mvr) = hr/ λ
Schrodinger orbit, Angular momentum

L(mvr) = √ {ℓ (ℓ + 1) ℏ} Where L is the orbital angular momentum
S = √ {s (s + 1) ℏ} where S is the spin angular momentum and s is the spin quantum number.
where ℏ(hbar) is called the DIRAC constant or reduced Planck constant. It is obtained as h/2 π = 1.0545718 x 10 -34kgm2/s. ℓ is
the azimuthal or orbital angular momentum quantum number.
Resolution of Schrodinger’s orbit and angular momentum
Let consider the projection on the z- axis of the orbiting electron.
L z = m ℏ where m is the magnetic quantum number and an integer in the range -ℓ to + ℓ
The total number of values of angular momentum values projected in the Z-axis is thus 2L + 1, similar to the magnetic quantum
number
Example 1
Given the azimuthal quantum number of 3, determine (i) the angular momentum (ii) the least and highest values of the
projected angular momentum in the z-axis
solution:
(i) if you are given azimuthal quantum number, then use the Schrodinger orbit equation.
L= √ {ℓ (ℓ + 1) ℏ}
L =√ {3 (3 + 1) ℏ} = √ {3(4) ℏ} = 2√3ℏ

(ii) For a projected axis, the range of possible angular momentum values are given by L z = mℏ. Hence, Lz =
mℏ, with m ranging from -3 to +3.
The range of angular momentum values for y-axis has least value of -3 ℏ and greatest value of 3 ℏ
Example 2
Given the squared value of angular momentum to be 30ℏ2, determine the angular momentum quantum number.
Solution:
using L2= ℓ(ℓ+1) ℏ2., we obtain 30ℏ2 = ℓ(ℓ+1) ℏ2.
30 = ℓ(ℓ+1), ℓ = 5
Remember angular momentum can be added. Due to magnetic interaction, if given the orbital angular momentum quantum
number (azimuthal) and the spin quantum number, the total angular momentum quantum number J is given ass a vector sum
of the two.
J =S+L
Example 3
You have a system of two electrons whose orbital quantum numbers are l1 = 2 and l2 = 4 respectively.
(a) Find the possible values of l (total orbital angular momentum quantum number) for the system.
(b) Find the possible values of s (total spin angular momentum quantum number) for the system.
(c) Find the possible values of j (total angular momentum quantum number) for the system
Solution:
 Concepts: Addition of angular momentum

 Details of the calculation:
(a) The quantum numbers associated with the total orbital angular momentum will range from a maximum value
found by adding the individual quantum numbers together to a minimum value found by taking the absolute value of
the difference of the two numbers in integer steps.
lmax = l1 + l2 = 2 + 4 = 6
lmin = |l1 - l2| = |2 - 4| = 2
The possible values for L are 2, 3, 4, 5, and 6 .
(b) The quantum numbers associated with the total spin angular momentum will range from a maximum value found
by adding the individual quantum numbers together to a minimum value found by taking the absolute value of the
difference of the two numbers in integer steps.
smax = s1 + s2 = ½ + ½ = 1
smin = |s1 - s2| = |½ - ½| = 0
The possible values for S are 0 and 1.
(c) The largest possible value for j will be found by adding together the maximum values for both l and s. The
minimum value for j will be found by subtracting the largest possible value of s from the smallest possible value for l.
jsmax = lmax + smax = 6 + 1 = 7
jmin = |lmin - smax| = |2 - 1| = 1
j = 1, 2, 3, 4, 5, 6, 7

QUESTIONS FOR TRIAL
PHYSICS, KUMIDEE SERIES(MIXED)
1. A 13000N car starts from rest from the top an incline with a vertical height of 10m. it slides down and collides with a
spring downhill. Neglecting any losses due friction, calculate the impact speed and the maximum depth of
compression of the spring after impact (spring constant =1*10^6N/m)
2. A 0.5Kg object connected to a spring of constant 20N/m oscillates on a frictionless horizontal table. Calculate the total
energy of the spring mass system and determine the maximum speed at equilibrium position given the initial
amplitude of 3cm.
3. State if True or False and briefly explain your choice mathematically. Doubling the initial displacement doubles the
maximum speed of a simple harmonic oscillator
4. A 45kg boy jumps on a spring of mass 5kg which has a constant of 45000N/m. Find the angular speed, frequency and
period of the boy’s motion whiles bouncing up and down on the spring
5. If the Displacement of an object –spring system is described by the equation, x=0.25Cm (@/6) t. Find the maximum
velocity and acceleration of the system.
6. A swinging pendulum has its displacement defines by x=0.33m cos 1.5t. Find its Amplitude, Angular frequency and
Period.
7. A block-spring system vibrating on a frictionless horizontal floor has an amplitude of 6cm and Total energy of 12J. The
Block is then replaced with another block of twice the mass of the initial block. The amplitude is still maintained at
6cm and the system set into SHM once again. Determine the total energy of the new block-spring system
8. The distance between crest and the next trough of water wave is 2m. If the wave frequency is 2Hz, Find the speed of
the wave.
9. If a simple pendulum oscillates with small amplitudes and its length is doubled, what happens to the frequency.
10. A mass of 0.4kg hanging from a spring with constant of 80N/m attached to a ceiling is set into up and down vibration.
If at the start of vibration, the mass was pulled down by 0.1m from equilibrium position and released from rest, with
what speed does the mass pass through equilibrium position when in motion.
11. An object of mass 0.4kg hanging vertically down from a spring of constant 8N/m is set into S.H.M with a maximum
displacement of 0.1m. Find the maximum acceleration of the system.
12. A particle on top of a spring system vibrates in tandem with spring and the particle swings between positions X=85cm
and X=100cm as measure by a meter rule placed horizontally by the side of the spring. Determine the location of the
maximum speed of the particle.
13. The fundamental frequency of a pipe is 120Hz and two adjacent higher resonant frequencies are 360Hz and 600Hz.
Identify the resonant positions corresponding to the two higher frequencies and identify the type of tube.
14. A flute has a length of 60cm. if the speed of the sound is 340m/s, what is the frequency of the second overtone if the
flute is described as an open tube.
15. A 1kHz source of sound is moving towards a man at 50m/s whiles the listener is moving away at 30m/s. given the
speed of sound in air to be 340m/s, find the frequency of the sound heard by the man.

16. Quadrupling the power of a single frequency of sound changes the loudness by how much?
17. Ethyl Alcohol has a density of 0.806*10^3kg/m3 and a bulk modulus of 1*10^9Pa. Find the speed of sound waves in
Ethyl Alcohol.
18. The loudness of a certain jet engine is 40dB. What is the intensity of the sound from this engine?
19. A stone is dropped from rest into a well with the ambient temperature of the air above the well given to be 10 oc. If
the splash is heard exactly 2s after the stone was dropped, determine the depth of the well.
20. A sound wave has a frequency of 4KHz. Calculate the change in wavelength as the sound passes from the lab with
ambient temperature of 20 oc to the outside with temperature of 10 oc. Take speed of sound at 0. oc to be 331m/s.
21. A sound wave from a siren has an intensity of 100W/m2 at a certain location. A second sound from the horn of a car
has intensity of 10Db GREATER THAT THE SIREN. What is the intensity of the car horn?
22. At rest a car horn sound has a frequency of 440Hz. The car sounds a note whiles moving with a speed of Vs. a cyclist
moving in the same direction as the car with a speed 1/3rd that of the car hears the frequency of the cars sound as
415Hz. Determine if the cyclist is behind or in front of the car and determine the velocity of the car.
23. 2 speakers are driven by a common oscillator of frequency 2.00* 10^2 Hz. They face each other and separated by a
distance of 2m. An observer stands 10m away and at exactly midway between the speakers he hears an amplified
sound from a combination of the two speakers. Determine a position away from the midline such that he hears the
first minimum sound. V=340m/s
24. A 20cm long guitar string has a tension of 40N applied from a finger. If the linear density is 0.1kg/m, find the highest
resonant frequency that can be perceived by a person with presbyacusis who’s ears can only hear up to 6699Hz.
25. A steel wire has length of 0.7m and mass of 2.1*10^-3kg. What Tension must the wire be stretched with to give it a
fundamental frequency of 100Hz.
26. A stretched string has L and is observed to vibrate in five identical segments with a frequency of 630Hz. What forced
frequency will cause this string to vibrate with 3 loops instead
27. A stretched string fixed at both ends has a mass of 40g and a length of 9m. if the tension in the string is 40N,
determine the positions of all nodes and antinodes of the 3rd harmonic
28. A vertical steel beam in a building supports a load of 6*10^4N. if the length of the beam is 4m and its Area is
8*10^3m2, find the depth to which the beam is compressed by the load. Find the maximum Load this beam can
withstand if the maximum compressive stress is 5*10^8N/m
29. Given a bulk modulus of 4.2*10^10Pa, determine the compressibility of the material if its volume is 100m3 and a
pressure of 1 atm is applied
30. Calculate the pressure on the top lid of a treasure chest buried under 4meter thick mud under the floor of a lake of
depth 10m. density of mud =1.75*10^3kg/m3, density of lake water= 1*10^3kg/m3, 1atm=1.01*10^5Pa.
31. A certain fluid has density of 1080kg/m3 and is observed to rise up a height of 2.1cm within a capillary tube of radius
1mm. calculate the surface tension in the fluid.
32. Water jets out of a bullet hole in a can. The can is initially filled to a height of 20m from the floor. If the hole is 10m
below the top of the fluid level, determine the horizontal distance from the base of the can.

33. A needle is 3cm long and 0.3mm in diameter. What difference in pressure between the ends will cause water to flow
out of the needle at 1g/s
34. Find the length to which water will rise in a capillary tube with radius of 5*10^-5m assuming the contact angle is
negligible. Take density of water as 1000kg/m3. Surface tension of 0.073N/m
35. To lift a light wire loop of radius 1.75cm from the surface of a film of blood, a vertical force of 1.61*10^-2N is
required. Calculate the surface tension of blood.
36. The human eye can respond to light with total energy of as little as 10^-18J. If red light has a wavelength of 600nm,
what is the minimum number of photons the eye can perceive
37. What wave phenomenon occurs only for transverse waves?
38. A screen is separated from a double slit by 1.2m. the slits are 0.03m apart. The second bright fringe is measure 4.5cm
from the centerline between the 2 slits. Determine the wavelength of the light. Determine the position of the next
dark fringe from the centerline.
39. A glass sphere(n=1.5) with radius of 15cm has a tiny air bubble above its center. The sphere is viewed along its
principal axis. what is the apparent depth of the bubble below the surface of the sphere?
40. A lens and mirror are arranged in line with the lens to the left of the mirror and a common principal axis. The 2 are
separated by 1m and have focal lengths =80cm and -50cm respectively. If an object is placed 1m to the left of the
lens, where will the final image form and describe its physical features.
41. A space ship is approaching a station at a speed of 1.8* 10^5m/s. the station has a beacon that emits green light at
frequency of 6.0*10^14Hz. What is the observed frequency by the astronauts in the ship?
42. What is the maximum angular magnification of a lens with focal length of 10cm as viewed by the human eye?
43. A certain microscope has two lenses with the focal length of objective and eyepiece being 2cm and 5cm respectively.
If the separation of the lenses is 18cm, determine the lateral and angular magnification
VIBRATIONS, WAVES AND SIMPLE HARMONIC MOTION
1. A block of mass m at the end of a horizontal spring is drawn from equilibrium at x=0 to x=A and released into SHM. If
the spring constant is Find the energy of the system at equilibrium position and determine the total distance travelled
in 3/4T, where T is the period.
Ans. KE=1/2A2 and distance travelled is 3A
2. A 0.35kg object attached to a spring of constant 1.3x10 x 102N/m is free to move on a horizontal frictionless floor. If
the object is released from x=0.1m, find the force on it and the acceleration when it is at x=0.1 and x=0
Ans. F=-Kx ---=-13N, a0.1=-37.1m/s2 and a0=0m/s2
3. Define elastic potential energy and give the expression for the potential energy of a spring
4. A metal ball is dropped from a height of 10m vertical onto a spring below. The mass of the ball is 1300kg and the
spring constant is 1x106N/m. determine the depth to which the spring is compressed.
Ans. X=√(2mgh/k), x=0.51m
5. A 0.5kg object connected to a light spring with a spring constant of 20N/m oscillates on a horizontal frictionless floor.
If the amplitude is 3cm, find the velocity at a displacement of 2cm.

Ans. V = ± ω√(A2-X2), remember ω=√(K/m), V=±0.141m/s
6. An object of mass m attached to a horizontal spring is displaced through A from equilibrium position and made to
undergo SHM with a period T. if the mass is replaced with one whose mass is 4m, what happens to the period of the
motion?
T=2 π√(m/k) T final=2Tinitial
7. The position of a vibrating body with time is given by x=0.25m cos π t/8. Find the period and maximum acceleration of
the body.
Ans. T=1/f , f=0.0625Hz, T=16s, accelerations a=A ω2 = 0.0386m/s2
8. State the types of damping of oscillations

9. Using a pendulum of length 0.171m, a geologist counts 72complete swings in a time of 60s. what is the value of g in
this area.
Ans. T=2 π√(l/g), g=4 π 2l / T2 g=9.73m/s2
10. To what tension must a string of mass 0.01kg and length 2.5m be tightened to allow sound to travel on it at 125m/s
Ans V=√(m/k (T/μ), where μ=m/l v=62.5N
11. A runaway car of mass 3x105kg speeds down a track at 2m/s and collides elastically with a spring of constant
2x106N/m. what is the maximum depth of compression of the spring.
Ans. 1/2kx2=1/2mv2, x=√(m/k).V x=√15/5=0.775m
12. A block at the end of a spring moves to and fro along the x axis between x=100cm and x=140cm. What is the velocity
of the block when the system is at x=130cm. given the mass of the system to be 0.1kg and the spring constant to be
40N/m.
Ans. . V = + ω√(A2-X2), ω=√(K/m), v=200√3 = 346.4m/s
13. A load of 50N attached to a spring hanging vertically downwards stretches the spring by 5cm. the spring is now
placed horizontally and stretched by 11cm. determine what magnitude of force caused this stretch.
Ans. K=F1/x1 = 5x103N/m. F2= -kx2 =5.5x102N
14. A 50g object is attached to a spring horizontally placed with constant of 10N/m and released with an amplitude of
25cm. what is the velocity when the system is halfway back to equilibrium
Ans v= ω√(A2-X2), ω=√(K/m), v=√(K/m) x A√( ¾) = √15/4m/s =0.968m/s
15. Find the velocity of sound in air when the temperature is 266K
Ans V=331√(Tk/273) or 331+0.606(T0c) =327m/s
16. A noisy machine in a factory has sound intensity of 1x10-5W/m 2. Find the loudness in the decibel scale
Ans [dB =10log10X ] where X=intensity known/threshold of hearing. Threshold of hearing is given by 1x10 -12W/m2 ,
loudness=70dB
17. Find the new loudness if an extra machine is added to an already existing one with a decibel level of 70. Assuming
both machines operate together at the same intensity
Ans 73dB
18. A sound source emits waves with a power output of 80W. find the intensity at 3m from the source.
Ans [ I=P/4 π r2 ] =0.707W/m2
19. At what distance will the intensity of a sound from a 60W source become one fourth the intensity at 6m from the
power source.
Ans [ I1 R1 = I2 R2], R=12m

20. A train moving at 40m/s sounds its whistle with frequency of 5x102Hx. Determine the frequency heard by an
passenger waiting for the train on a bench at the train station. The ambient temperature of the air is 240 c
Ans Vat 24c=345m/s, fo = fs(v/v-40) =500( 69/61) = 566Hz
21. An ambulance travelling at 33.5m/s sounds 4x102Hz sound wave and is heard by a passenger in a car moving
at24.6m/s in the opposite direction. What is the frequency heard by the passenger as the 2 vehicles pass each other
and move away?
Ans f = f(v-24.6/v+33.5) = 339Hz
22. A spring is stretched 4cm when a mass of 50g is hung on it. If a total of 150g is hung on it and the mass is started in a
vertical SHM, what will be its period.
Ans. K=mg/x = 12.5N/m, T=2 π√(m/k) = π/25√(10) = 0.397s
23. A 0.5kg mass is undergoing SHM with a frequency of 2Hz and amplitude of 8mm.find the maximum velocity and the
restoring force
Ans. V= ω R , ω=2 π f, V= 0.032 π m/s = 0.101m/s, a= ω2R or V2/R = 1.26m/s2, F=ma = 0.632N
24. A body of mass 100g hangs from a long spring. When pulled down 10cm below equilibrium position and released it
vibrates with a period of 2s. determine the velocity t equilibrium position and the acceleration when 5cm above
equilibrium position.
Ans. . V= ω R , ω=2 π f, ω= π rad/s, v= 0.1 πm/s=0.324m/s, a = ω2R or V2/R = 0.05π2m/s2 =0.4935

m/s2
25. The equation of the motion of a mass at the end of a spring is( y=0.3 cos 0.5t) meters. Find the velocity and
acceleration at t=0
Ans. V=dy/dt=-0.15sin 0.5t, at t=0, V=0
A=dV/dt=-0.075 cos 0.5t, at t=0, a= -0.075m/s2
26. A particle is attached to a spring and undergoes SHM with a maximum acceleration of 18cm/s2 and amaximum
velocity of 3m/s.find the frequency of vibration.
Ans. . V= ω R and a = ω2R, therefore a/v= ω , ω=6rad/s

F=ω/2 π = 3/ π Hz or 0.95Hz.
27. If an applied force F=3i-4j moves a body in the direction of the force through distance R=2j+6k, find the work done W.
Ans. W=F.R , W= 6k-18j-24i, Newtons
28. Find the torque T if the separation distance is r=2j-6k meters and the perpendicular force is F=3i+4k Newtons.
Ans T=8i-18j-6k Nm
29. A tiny laser beam is directed from moon towards earth.it is expected that the beam should have a diameter of 2.5m
on earth. If the distance between the earth and the moon is 3.8x108m, how small must the divergence angle from the
laser source be?
Ans S= θ R., s= 2.5m, R=3.8x108m, θ = 6.6x10-9rad.
Preamblefor Q 30 to 31; A flywheel at rest is to reach an angular velocity of 36rad/s in 6s
30. what constant angular acceleration must it have?

Ans. α = ω/t , 6rad/s2
31. What total angle does it turn through within 6s?

Ans . Δ θ = ( ω1 +ω2/2) x time = 108rad
32. A wheel turning with angular speed of 30rev/s is brought to rest with a constant acceleration. It turns through an
angle of 60 revs before coming to a stop. Determine the angular acceleration and the time expended.
Ans α = ( ω2 final --- ω2 initial)/ 2 Δ θ === -47rad/s2

Time = ( ω final --- ω initial)/ α === 45 s
33. A spinning wheel has an angular velocity of 50rad/s eastwards. 20s later, its speed is 50ra/s westward. If the angular
acceleration is constant, find the magnitude of the constant angular acceleration and the total angle subtended in
20s.
Ans α = ( ω final --- ω initial)/ Δ time ==5rad/s

Δ θ = ( ω2 final --- ω2 initial)/ 2α === 0
34. 2 gears are mashed together and have radii of 0.5cm and 0.15cm. through how many revolutions will the smaller gear
move if the bigger gear moves 3revs.
R 1 θ1 = R2 θ2 R === 10revs
35. A car accelerates uniformly from rest to a speed of 15m/s in a time of 20s. find the angular acceleration of one of the
wheels and the number of revolutions turned by the wheel if its radius is 1/3m
Ans ω=V/R ==45rad/s and α = ω/t == 2.24rad/s2

θ = S/r = 450rad, number of revs = θ/2 π== 72revs
36. A flywheel having a moment of inertia of 900kgm2 rotates at a speed of 120rev/min.it is slowed by a brake to a
speed of 90rev/min. how much energy did the flywheel loose in slowing down.
Ans. ΔKE = ½ IΔ ω2 == 31J
37. Compute the rotational kinetic energy of a 25kg wheel rotating at 6rev/s if the radius of gyration is 0.22m
Ans ΔKE = ½ IΔ ω2 but I = Mass x radius of gyration2 (ml2) , I= 1.21kg.m2 and ω= 12 π rad/s
KE = 860 J
38. When 100j of work is done on a flywheel, its angular speed increase from 60rev/min to 180rad/s. what is its moment
of inertia.
Ans == 0.63kg.m2
39. An ice skater spins with arms opened at 1.9rev/s. her moment of inertia during this spin is 1.33kg/m2 . she pulls her
arms in on her next spin bringing her moment of inertia to 0.48kg/m2 . determine the angular speed of the second
spin.
Ans I1 ω1 = I2 ω2 , ω == 5.26rev/s or 10.52 π rad/s
40. The radius of the aphelion of the earth is Ra and that of the perihelion is Rp. Find the ratio of the earths aphelion
tangential speed to perihelion tangential speed in terms of the radii.
Ans. Ra/Rp
41. Helium is a monoatomic gas with a density of 0.179kg/m3 at a pressure of 76cm of mercury and temperature of 00 c.
find the bulk modulus of the gas and determine the speed of sound in this gas. The adiabatic constant of monoatomic
gas is 1.67
Ans B for a gas ===yP, B= 1.723x105

V= √(B/ ρ)
42. Hydrogen a diatomic molecular gas has an adiabatic index y=1.4 and a molar mass M of 2.1g/mol. Find the speed of
sound in hydrogen at 270 c.
Ans V=√(yP/ ρ) remember P= ρ RT/M, so V=√(yRT/M) = 1321m/s
43. 2 closed pipes are sounded simultaneously and give a 5beat/s between them when they both vibrate at their
fundamental frequencies. If the shorter pipe is 1.1m and the speed of sound is 340m/s, find the length of the longer
tube.
F beat = f shorter- f longer, for closed tubes, each frequency f0 = v/4L

F beat = ¼ V (| L1 - L2 | / L1L2), put L1 = 1.1m and vat 27oc = 340m/s
L2 = 1.18m
44. A wave on a string travels at an angular frequency of 40rad/s and has an amplitude of 16mm. how much energy exist
in an 80g of this vibrating wire.
KE= ½ MV2, but V= R ω so V2 == R2 ω2, KE is thus =m ω2 r2 = 32.8 m J
45. Standing waves are produced in a rubber tube 12m long. If the wave vibrates in 5 segments and the velocity of the
wave is 24m/s, what is the wavelength and the frequency.
For fundamental note, the wave vibrates in one loop, ½ λ = L. so on the fifth harmonic, there are 5 loops hence 5/2 λ = L ,
λ=4.8m
Frequency =5Hz.
OPTICS AND OTHER ELCTROMAGNETIC WAVES
State Huygens law.
Light propagate as waves in circular wave fronts with each point on the wave front acting as a secondary wavelet and
propagating a new wave forward.
mirrors make an angle of 1200 between them. If a ray incident on the mirror on the left at an angle of incidence of 65,
determine the angle of reflection in the second mirror due to double reflection of the same ray. Ans = 550
At point A, the initial angle of incidence is α=650.,

the reflected ray and the 2 mirrors form a triangle
with angles (250, 1200, 350).
C From the triangle ABC, we can get the angle of
incidence on the second mirror at C to be β=550
A
B
Light with wavelength of 589nm in vacuum passes through a piece of quartz of refractive index of 1.458. find the speed of this
light through the quartz.
Ans V= C/n = 2.06x108 m/s
For the same question above, find the wavelength of this light through the quartz.
Ans . λn = λc /n 404nm

The index of refraction for water is 4/3. what happens to the speed of light as it travels from air into water?
Ans. It becomes ¾ C
Find the energy of a photon having a wavelength of 3x102nm, express your answer in electron volts.
Ans. E= hc/ λ =6.663x10-19 J
A room contains air in which the speed of light is 343m/s. the walls are made of concrete through which sound travels at
1850m/s. in which medium must incident sound be travelling in other to undergo total internal reflection at the air-concrete
boundary and determine the critical angle of incidence.
Ans. in the air. This is because the refractive index for sound will be higher due to the lower speed of propagation.
N sin θ = n sin θ
c/v1 sin θ1 = c/v2 sin θ2 , taking θ1= C and θ2 = 900, sin C = v1/v2
C= sin -1 0.185 =10.7 0c
A man 1.8m tall stands in front of a plane mirror and sees his full height. If his eyes are 0.14m from the top of his head, what is
the minimum height of mirror.
Ans. mirror height = ½ height of the man = 0.9m
A concave mirror has focal length of 10cm. if an object is placed 25cm in front of it, find the image location, magnification and
its descriptions.
Ans V= uf/(u-f) = 16.7cm, M = -v/u = -0.668, so we get a diminished inverted image on the same side of the object hence
it is real.
When a woman stands with her face 40cm from a cosmetic mirror, the upright image is twice as tall as her face. What is the
focal length?
Ans. NB it is only concave mirrors that can form virtual upright and magnified images.
M= -v/u, v = -80cm. f = 80cm. NB, when u=1/2 f, v will be equal to –f.
A coin 2cm in diameter is embedded in a solid glass sphere with radius 30cm and refractive index 1.5. the coin is at a depth of
20cm from the spherical surface. Find the apparent depth observed when it is viewed from above.
Ans. n1/u + n2/v =( n2-n1)/R, where n2 is the refractive index of the medium with the object embedded. And R is the radius of
curvature of the sphere. V= -17.1m
for the same question above, find the magnification of the image
ans. M = –n1v/n2u , M = 1.28
A small fish in a swimming pool at a depth of 5m below the steady water surface. If the water has refractive index of 1.33. find
the apparent depth of the fish to an observer looking down directly over the fish.
Ans. for flat refracting surfaces, n1v=-n2u where N2 is the index of the medium containing the object. V= -3.76m.
For the same question above, if the fish has a length of 12cm, how will this be magnified.
Ans. M=0.

A converging lens of focal length 10cm forms an image of an object situated at 30cm from the lens. Locate the position of the
image and give its magnification and physical features.
Ans v= +15cm, M= -0.5, inverted, diminished, real.
Suppose the image of an object is upright and magnified 1.75 times when an object is placed 15cm from a lens, find the focal
length of this lens and give the image location
Ans. V= -26.3cm, f= 34.9cm.
A diverging lens of focal length 10cm has an object placed 5cm in front of it. Find the location, magnification and physical
features of the image formed.
Ans. V=-3.33cm
M=+0.666, it is upright, virtual, diminished and on the same side as the object
2 converging lenses are 20cm apart with an objected 30cm to the left of lens 1 with focal length of 10cm. if the 2nd lens has a
focal length of 20cm. Locate the position of the image due to lens 1 only
Ans. V1= =15cm
Locate the position of the final image from the combined lens as viewed by an observer on the right side of the setup.
Ans. the image of lens 1 is 5cm from lens 2 and serves as the new object for lens 2. V = -6.67cm
An object is placed 20cm to the right of a concave mirror with focal length of 12cm.the object is 30cm from the left of a convex
mirror of focal length 10cm. the 2 optical instruments have a common principal axis.
Find the location of the image formed due to lens and mirror alone respectively
Ans. VL=15cm, Vm=30cm
Find the location of the image formed by the combination as one looks through the lens from the right side of the set up
Ans Vcombined = 20cm
A diverging lens with n=1.5 has radius of one surface to be 15cm and the other to 10cm. find the focal length of this lens
Ans . 1/f= (n-1) ( 1/R12--- 1/R22)……..the lens makers equation
A screen is 1.2m from a double slit screen. The 2 slits are separated by 0.03mm. the second order bright fringe is measured
4.5c. from the center line on the screen.
Determine the distance between successive bright fringes
Ans , Location between each bright fringe is given by y0= λX/d where x is the horizontal distance from screen, λ is the
wavelength and d is the slit separation. Location of 2nd order bright fringe is Y2 = 2y0 , so y0 = 2.25cm.
For the same problem above, determine the wavelength of the light.
Ans . λ= 563nm
Calculate the minimum thickness of a soap bubble (n=1.33) that will result in constructive interference of a reflected light with
wavelength of 602nm in free space.
2nt= ½ λ , t = 113nm

Sate 2 types of visible light diffraction patterns as seen on a screen.
Ans,, Fraunhofer diffraction and Fresnel diffraction
Light of wavelength 5.8x102nm incidents on a single slit 0.3mm wide which is 2m from a screen. Find the position of the first
dark fringe.
Ans. a sin θ = ±λ , sin θ = y/X where y is the vertical position of fringes on the screen and X is the horizontal distance
between screen and the slit.
Therefore, y =±λX/a = ±3.86x10-3 m .
For this problem above, find the width of the bright central fringe.
Ans. the area is bordered by =y and –y, so its thickness is +y- -y = 2y, =7.72x10-3m.
State means of bringing about plane polarization of light.
Ans, Selective absorption with polaroid, reflection, optical active materials
Light of wavelength 5.4x102nm passes through a slit of width 0.2mm. find the width of the bright central maximum on a screen
1.5m from the slit
Ans width = 8.1x10-3m.
Write the relation for the behavior of stationary waves and state 2 values of X each for which we have nodes and antinodes
respectively.
Ans. Y = 2a cos kx sin wt , or Y = 2a cos 2 πx/ λ sin 2 πft , at nodes 2a cos 2 πx/ λ = 0 , so x= λ/4,
3λ/4, 5λ/4, antinodes will occur at 0, λ/2, λ
The second order dark fringe in a single slit diffraction pattern is 1.4mm from the midpoint of central maximum. If the slit is
0.8mm wide and located 85cm from the screen, determine the wavelength of the light.
Ans = 4x105nm
The near point vision of a person is 50cm. what is the problem suffered by this person and state the power of the corrective e
lens needed.
Ans. the person suffers long sightedness thus things at near point which is our expected goal of correction will be blurred.
Hint. Always use the persons impairment as –v and the preferred corrected distance as +u. ( his problem is 50cm, our aim of
correction is to achieve normal near sight at 25cn)
1/u+ 1/v= 1/f, use u=25 and v=-50, f= 50cm so power is +2 Diopters. NB in calculating the power, focal length
must be in meters)
A near sighted patient cant see objects clearly when they are beyond 25cm. what focal length should the prescribed lens have
Ans ( hint. Take u to be infinity where we want to extend his vision and v= -25cm which is the current problem)
Ans= f= -25cm NB for near sightedness, the corrective lens is (– the impairment limit) .
What power of lens will be needed by a person whose far point is 35cm?
Ans. -2.86D

What power of lens will be needed by a person with far point of 35cm assuming an eye-corrective lens distance of 2cm?
Ans. similar approach to the previous ones but P= 1/(f--- eye to. Lens distance)
Ans = -3.03D
What is the maximum angular magnification of a lens with focal length of 10cm?
Ans. M maximum = 1+ D/f for a normal eye unless otherwise stated, use D=25cm
M max = 3.5
What is the angular magnification when the eye is most relaxed using a magnifier with focal length of 10cm?
Ans. M relax =D/f = 2.5
A certain compound microscope has an objective focal length of 0.2cm and eye piece focal length of 0.5cm. if the lens
separation is 18cm. find the total magnification
Ans = -450
A telescope of length 13.2cm has objective of focal length 57.8cm. find the eye piece focal length is magnification is 7.23x10-3
Ans 7.23x10-3m
The near point of a person’s eye is 55cm. what is the focal length of the corrective lens needed to give him normal vision.
Ans ,45.33cm
Compare and contrast the modern camera and the human eye with features and function corresponding to iris, retinal, cornea
and pupil.
The overall magnification of a compound microscope is 140 times. The objective lens alone has a magnification of 12 times.
Determine the focal length of the eyepiece lens.
Ans. Total M = lateral M x angular M, lateral M is given by –L/fo , where L is the lens separation distance. angular M is the
magnification due to objective alone given by D/fe, where D is 25cm unless otherwise stated. The question give lateral M =
12, so angular M = 11.67times. fe= 2.143cm
State 2 forms of fiber optic transmissions
Ans , monomodal and multimodal transmissions
The refractive index of a core fiber is 1.52 and the refractive index of the cladding is 1.48. Find the maximum angle of incidence.
Ans. sin@= √( n core 2 – n clad 2), = sin -1 √3/5 = sin -1( 0.35) = 200
A plastic hemisphere has radius of curvature of 8cm and refractive index of 1.6. a dot is located on the central axis ½ way
between the plane surface and the curves surface. If the dot is view directly above the sphere, what will be its perceived depth?
Ans, R=8, N1=1.6, N2 = 1, U=4, V=?
Use the equation, n1/u + n2/v = ( n2-n1) R . V= -3.25cm

What is the luminous intensity of a 200W lamp whose efficiency is 18 lm/W.
Ans , luminous intensity = luminous flux / 4π steredians. Luminous flux = power x efficiency = 3600lm
Luminous intensity = 3600/4π = 900/π or 286 candela
Light from a lantern provides illumination of 12000 lm/m2 on a wall perpendicular to the beam and 5m from the source. What
intensity must an isotropic source have to give such illumination from this distance away.
Ans. Illumination (E) = intensity (I) / Radius2, I = 3 x 10 3 candela
Find the illumination of a small surface at a distance of 120cm from an isotropic source of luminous intensity 72Cd.
Ans. 50lm/m2
2 plane mirrors are placed parallel to each other and separated by a distance of 20cm. a luminous point is placed in between
them such that it is 5cm from mirror 1 and 15cm from mirror 2. Determine the distance of the 3 nearest images in each mirror
Ans.
Add the separation

Mirrorand
1, keep
5cmswinging between
35cm them after45
obtaining the first
75 positions.
Mirror 2, 15cm 25 55 65 etc
Where must an object be placed in the case of a converging lens of focal length 10cm if the image is virtual and 3 times
magnified
Ans. 6.66cm
A diverging lens forms an image 1/3rd the size of the object placed 24cm from the lens. Determine the focal length of this lens
Ans f = -12cm
Given the focal length f and magnification M, state a relation for the image distance v in terms of M and f.
Ans v= f(1+M)
ELECTROSTATICS AND CAPACITORS
A tiny droplet of oil acquires a small negative charge whiles dropping through vacuum. An electric field of magnitude 5.92 x 104
N/C points straight down. If it’s mass is 2.93 x10-15kg, find the charge if the drop remains static in the field.
Ans. Eq = mg, q=-4.85 x 10 -19 C
An oil drop of mass 2.93x 10 -15 C falls downward through an electric field of 5.92 x10 4 N/c. If the charge is negative and the
drop falls 10.3cm from rest in 0.25s, find the quantity of charge.
Ans. Y=1/2 at2 + ut, a= -3.3m/s2. Weight – electric force = net accelerating force
Mg – Eq = ma, q= m(g=a) / E = -3.22 x 10 -19 c

State 2 commercial uses of the van de Graaf generator.
Ans .To create high speed electron s for X ray generation and to produce high speed nucleons for nuclear reactions.
State 2 safety designs to avoid electric sparks in the van de Graaf generator
Increasing the diameter of the dome and rounding its surface ensures that there is reduced electric field .Using pressurized
gas around the dome reduces risk of corona discharges
State in words and state mathematical relation for Gauss law of electrostatics
For a closed surface, the total flux passing through the surfaces is proportional to the charge contained within the surface.
ФE = Q/ε0
An electron with speed of 3 x 10 6 m/s moves into an electric field of 1 x 10 3 N/c. The field is parallel to the moving charge. How
far does the charge move before coming to stop?
Ans. 1/2 mv2 = Vq, or ½ mv2 = Eqd, d = mv2/ 2Eq or d= m/q x v2/2E d = 0.026m
A uniform electric field of 1.0 N/c is set up by a charge in the x-y plane. A metal sphere is placed 0.5cm from the charge.
Determine the field strength inside the sphere due to this charge.
Ans ==0
The electric force between 2 protons is 2.3 x 10 -26 N. how far apart are these protons.
Ans. Student should memorize Ke2 = 2.3 x 10 -28
F = K e2 /R2. R2= 0.01 R= 0.1m
A proton is released from rest at x = -2cm into a constant electric field of 1.5 x 10 3 N/c which points in positive x direction.
Determine the change in potential energy when the charges reach x=5cm.
Ans. Δ PE = -Eq x ΔX = -1.68 x 10 -17J
Determine the electric force between 2 electrons that are 1 A0 apart
Ans = 23nN
A copper sphere has a mass of 2g and contains 2 x 10 22 atoms of copper. The charge on each nucleus of an atom is +29e. What
fraction of electrons must be removed to leave the sphere with a net charge of +2uC?
Ans. 2.16 x 10 -11
2 point charges q1 and q2 3m apart and with a combined charge of 20uC between them. If they repel with a force of 0.075N,
what are the individual charges?
Ans Kq1q2 /9 = 0.075, hence q1q2= 75 c2…..1 and q1+q2 = 20c…..2, solving 1 an 2 gives q1 and q2 to be either 5uC or 15uC
A test charge of q= +2uC is placed halfway between 2 charges Q1=+6uC and Q2= +4uC which are 10cm apart. Find the resultant
force on the test charge
Ans. Q1 q Q2
½X ½X resultant force on test charge q between Q1 an Q2 is F = Kq/( ½

X )2 x ( Q1 + Q2) , F = 4Kq/x2 ( Q1 –Q2). Keeping the signs of Q1 and Q2 in mind. F= 14.4N away from the +6uC charge.

3 charges are placed in succession along a straight line with 2m separation between them. If each charge is +2uC, calculate the
force on the rightmost charge.
Ans. Q1 x Q2 a q
in this scene the test charge is outside the 2 other charges. So the resultant is found by adding Q1 and Q2 but still keeping
note of their charges. F = Kq( Q1/(x+a)2 + Q2 / a2 ) =1.125N
3 charges are placed at the following locations on the x axis. + 2uC at x=0, -3uC at x=40cm and -5uC at x=120cm. find the
resultant force on the -3uC charge due to the other 2 charges.
Ans 0.55N towards the left ie negative x direction.
2 identical charged balls have charges of +3nC and -12nC. They are placed 3cm apart. Find the fall and the nature of this force
between them.
Ans F= 3.6 x 10 -4N, it is attractive towards the -12nC charge.
If the same balls as above are made to touch briefly and separated again by 3cm as before, determine the Force and its nature
due to their charges
Ans. After touching, excess charges neutralize till the 2 bodies have a common combined charge of -9nC. When they are
separated, each ball has a charge of -4.5nC, the force is thus 2x10-4N and is repulsive .
An electron (q=-e, m= 9.1 x 10 -31kg) is projected out along the +x axis with initial speed of 3 x 10 6 m/s. It goes 45cm and stops
due to constant field acting on it. Find the magnitude and direction of this field.
Ans. , find the acceleration, a= -1 x 10 13m/s2, using Ma = Eq, E = 57N/C and it acts in the positive x direction
Define an electron volt.
Ans , It is the work done or kinetic energy gained when one electron is accelerated through a constant potential difference of
1V
Define the electric potential.
Ans, It is the work done to bring a unit charge from infinity into an electric field
A point charge q1 = +2uC is at the origin on the x axis. A second charge q2 =-3uC is placed at x=100cm. at what point on the x
axis will the absolute potential be zero?
Ans. , Resultant potential V12 = V1 + V2 bearing in mind their charges, but V = Kq/r, let the distance between q1 and q2 be
a and the location of zero potential from q1 be x.
V12 = K( q1/x + q2/(a-x) ). Or if resultant is zero, then dividing both sides by k, q1/q2 = x/(a-x) For this case, 2uC/x – 3uC/(
1-x) = 0 , x = 40cm and -200cm
Determine the absolute potential in air at a distance of 3cm from a point charge of 50nC
Ans , V = 15kv
What is the potential at the midpoint between charges +2uC and =5uC which are 6m apart on the x axis.
Ans. V = K/3 ( q1 + q2) = 21kv

3 equal charges of +6nC are located at the corners of an equilateral triangle whose sides are 12cm long. Find the potential at
the centre of the base of the triangle.
Ans +6nc
hh h = 10.4cm
+6nC +6nC
P
L= 6cm v= K ( Q/0.06 + Q/ 0.06 + Q/ 0.104), KQ(
1/0.06+1/0.06+1/0.104), V= 2320v
A metal sphere 30cm in radius is positively charged with 2uC. Find the electric potential at the center of the sphere.
Ans. Everywhere on a sphere from surface to center is equipotential with V= kq/radius
V = 60kv
Supposed 2 metal plates are placed 0.5cm apart and are connected to a 90v battery, determine the electric field strength
between them.
Ans. E = V/d = 18kV
The charge on an electron is 1.6 x 10 -19C. An oil drop has weight of 3.2 x 10 -13N and has an electric field of 5 x 10 5 V/m
between the plates of the Millikan’s oil drop apparatus. If the drop is observed to be essentially balanced, what is the charge on
the drop?
Ans = 6.4 x 10 -19C
Determine the charge as above in electronic charge units
Ans. ECU = q/e = 4 electrons
In the Millikan’s experiment, an oil drop carries 4 electronic charges and has a mass of 1.8 x 10 -12g. it is held almost at rets
between 2 plates 1.8cm apart. What voltage is applied between the plates
ans , mg = Vq/d, V= mgd/q = 496V
It requires 50uJ of work to carry a 2uC charge from point R to S. what is the potential difference between the 2 points.
Ans. W = Vq, V = 25V
A proton ( q=+e, m= 1.67 x 10 -27kg) accelerates from rest through a potential difference of 1mV. What is the final speed?
Ans. Vq = ½ mv2, v = 14 x 10 6m/s
A proton is accelerated from rest in a van de Graaf accelerator by a potential difference of 0.9 MV. What is the kinetic energy of
the proton?
Ans = 1.44 x 10 -13J
A 1.2 uF capacitor is charged to 3KV. Find the energy stored
Energy stored in a capacitor = ½ CV2 = 5.4J
A capacitor with air between its plates has a capacitance of 8uF. Determine the capacitance when a dielectric with constant of
6.0 is introduced between the plates.
Ans. Introducing dielectric changes the capacitance and charge by a factor of k, ie c= kc, q= kq where k is the dielectric
constant. Air has a constant of 1. C = 48F
A certain parallel plate capacitor consists of plates each of area of 200cm2 separated by a distance of 0.4m of air. Find the
capacitance.

Ans C=ε0 A/d = ε0 is 8.85 x 10 -12., C = 44pF
If the same capacitor above is connected to a 500v source, what is the charge stored and determine the electric field created
between the plates
Ans. Q = CV, Q= 22nC, E =V/d E= 125KV/m
For the same capacitor as above, if the air is replaced with a dielectric of constant 2.6, how much additional charge will be
added?
Ans, new charge = Q x 2.6 = 57nC, so additional charges added is 35nC
3 capacitors ( 2uF, 3uF and 4uF) are connected in series to a 6V battery. When the current stops, what is the charge on the 3uF
capacitor?
Ans. 5.5uC
A parallel plate capacitor has an area of 2 x 10 -4 m2 and a plate separation distance of 1 x 10 -3 m. how much charge is on the
plates when connected to a 3V battery?
Ans = 5.31 x 10 -12 C
For the capacitor above, determine the charge density and the electric field strength.
Ans. D = Q/A = 2.66 x 10 -8 C/m2, E = D/ ε0 or V/d = 3 x 10 3 V/m
State 4 properties of electric field lines
3 charges lie along the x axis with q1 = 15uC at x=2m and q2 = 6uC at x=0. where will q3 be placed such that the net force on it is
zero
Ans. Let the distance between Q1 AND Q2 BE A, THEN Q3 IS X FROM Q1,
Q1/Q2 = x2/(a-x)2 solve the quadratic to get x = 0.77m or 1.23m
Describe a good dielectric material and give 3 examples.
a material that supports an electrostatic field but does not allow conduction through itself and hence dissipated no energy.
the material has high polarizability. Eg. air, mica, distilled water
What will happen if the charges on the plates of a capacitor keeps increasing nonstop.
the maximum dielectric strength might be overcome and the material may start to conduct charge. also the air within the
vicinity may get electrified and charges may pass through air leading to sparks.
MODERN PHYSICS AND RADIOACTIVITY

State the special relativity theory and state 2 consequences of these postulates.
All laws of mechanics are true in all inertial reference frames and the speed of light in vacuum is always constant. it leads to
time dilation and length contraction
The period of a simple pendulum is measured to be 3.00s in the inertial frame of the pendulum at earth’s surface. What is the
period as measured by an observer moving at 0.95c with respect to the pendulum?
Ans. T = Tproper / √ (1-v2/c2) = 9.61s, there is dilation of time
An observer on earth sees a spaceship at an altitude of 4350km which is moving back to earth surface with a speed of 0.97c
straight downwards. What is the distance from the. what is the distance of the spaceship to earth as measured by the captain
on the ship?
ans . L relativity = L proper x √ ( 1-v2/c2) = 1.06 x 10 3 km.
After firing his engine, a spaceship captain measures his altitude from earth to be 267km whereas an observer on earth
measured the altitude at same instant to be 625km. what is the speed of the spaceship
Ans , from L relativity = L proper x √ ( 1-v2/c2) , we obtain v2/c2 = 1 – ( Lr/ Lp) so v = √ ( 1- (Lr/Lp) ) . c
== 0.904c
Suppose an observer on a ship travelling downward towards earth from space measures an altitude of 50Km whereas an earth
stationed observer measures the altitude at same instance to be 125Km, how fast is this ship moving
= 0.917c
An electron with mass of 9.11 x 10 -31 kg moves with speed of 0.75c. find the classical momentum.
Ans. P =mv = 2.05 x 10 -22 Kgm/s
Determine the relativistic momentum.
Ans = mv/ √ ( 1- v2/c2) = 3.1 x 10 -23 kgm/s
According to measurements from an earth based observer, Alice space shift is moving at 0.6c in positive x direction whiles bobs
spaceship is moving at o.8c in the negative x direction. What is Alice velocity according to Bob’s measurement?
Ans , Relative velocity of Alice and Bob to earth frame will be Vab proper = Va – Vb, but due to relativistic speed,
Vab relativity = Va – Vb / ( 1- VaVb/c2 ) = 0.946c
An electron moves with speed of 0.85c. Find the total energy of the electron given rest mass of electron to be 9.11 x 10 -31 Kg.
Ans , total energy increases and is given by Rest energy E0 = m0c2 x relativistic factor Y = 1/ √ ( 1- v2/c2)
Et = Y m0c2 = m0c2/ √ ( 1- v2/c2) = 1.56 x 10 -3 J
If the total energy of an electron is given as 0.975MeV. Determine the kinetic energy of this electron.
Ans, Et = KE + m0c2, KE = Et - m0c2 ie KE = Y m0c2 - m0c2 = m0c2( Y -1), the student must memorize the rest energy of
an electron as 0.511MeV, hence KE = 0.464MeV.
A deep space probe moves away from earth with a speed of 0.8c. an antenna on the probe spends 3s in probe time to rotate
through 1 rev. how much time will this take according to an earth observer.
Ans …
A sodium surface is illuminated with light of wavelength 0.3um. the work function of sodium is 2.46eV. calculate the energy of
the incident photon in electron volts.

Ans. Ephoton = hf or hc/ λ = 4.4eV.
For the same question above, determine the kinetic energy of the most energetic photoelectron in joules.
Ans. KE = hf – W0 = 1.68eV
For the same electron above, determine the cutoff wavelength of light that could cause photoelectric emission
Ans, W0 = hf0, where f0 is the threshold or minimum frequency, hence W0 = hc/ λ max , where λ max is the maximum or
cutoff wavelength, λ max = 505nm
Determine the De Broglie wavelength of an electron ( Me = 9.11 x 10 -31kg) moving at a speed of 1 x 10 7 m/s.
Ans, λ = h/mv = 7.28 x 10 -11 m
Find the De Broglie wavelength of an electron moving at 0.99c
Ans , λ = λ 0 / √ ( 1- v2/c2) = 1. 09 x 10 -13 m
The speed of an electron is measured to be 5 x 10 3 m/s to an accuracy of 0.00300%. find the uncertainty in determining the
position of this electron.
Δ P . Δx = h/4 π, student should memorize the constant h/4 π as 5.273 x 10 -35
From the question, P = mv = 4.56 x 10 -27 kgm/s. so Δ P = 0.003% x P = 1.37 x 10 -31kgm/s, Δ x is then = 3.85 x 10 -4m
State Wien’s displacement law and write a relation for your statement stating the value of the constant.
The wavelength of a black body radiation is inversely proportional to the temperature, λ

-3
peak T = 2.898 x 10 mK
What is the surface temperature of a distant star having a peak wavelength of radiation of 475nm as measure close to the star
surface?
Ans, λ peak T = 2.898 x 10 -3, T = 6.1 x 103 K
If the work function of zinc metal is 4.31eV. What is the lowest frequency of photon that can cause zinc to discharge electrons?
Ans , W0 = hf0 , f0 = 1.04 x 10 15, when energy is in eV, use the[ h/e = 4.14 x 10 -15Js/c]
Calculate the minimum wavelength of x rays that can be produced by an electron that is accelerated through a potential
difference of 1 x 10 2 KV when it strikes a tungsten target.
Ans eV = hc/ λ , [memorize hc = 1.99 x 10 -25], λ = 1.24 x 10 -11 or 1.24 x 10 -2nm
In an experiment, light of wavelength 4 x 102nm is incident on lithium and beryllium. The work functions of both substances are
given as 2.3eV and 3.9eV respectively. Determine if any of them will or will not emit photoelectrons and explain your answer
briefly.
ans, for electron volts, [ hc/e =1.244 x 10 -6eVm], energy of light is = 3.11ev, so only lithium will discharge
An electron is located on a pinpoint having a diameter of 2.5um. What is the minimum uncertainty in the speed of the electron?
Find the speed of an electron having a De Broglie wavelength equal to its Compton wavelength
Ans h/mc = h/mv, v =c
How fast must an electron be moving if all its kinetic energy is lost to a single x ray photon of wavelength 1 x10 -13m?
Ans ½ mv2 = hc/y memorise hc/me = 2.18 x 10 5, so energy per unit mass of electron is =2.18 x 10 18 J/kg.
v = √4.4 x 10 18 m/s = 2.09 x 10 9m/s

An electron initially at rest recoils after a head on collision with a 6.2KeV photon. Determine the kinetic energy acquired by the
electron.
A light source of wavelength y illuminates a metal and ejects photoelectrons with maximum kinetic energy of 1eV. A second
light with wavelength ½ y ejects photoelectrons with maximum kinetic energy of 4eV. What is the work function of this metal?
h/ λ - KE = W0………1, h/ ½ λ - 4KE = W0 ….2, dividing(2) by 2 gives h/ λ - 2KE = ½ W0…….3
equation 1 – 3 = === -1eV = ½ W0 , W0 = -2eV
Give the names of the transition series returning to the following ground state, n3 , n4 , n5, n6
paschen series, Bracket series, Pfund series, Humphrey series
State Bohr’s corresponding model.
if the transition energy changes are small, then quantum changes are comparable to changes in classical physics.
Explain how fraunhofer lines are formed in the suns emission spectrum
they are due to visible absorption spectrum of hydrogen atoms in space close to the sun.
State with equations the different types of beta decay
beta + decay, n p+ + e+ + ve and beta – decay, , n p+ + e- + ve-
State 2 safety engineering designs of the fusion reactor and how they work
If the average energy released in fission reactor is 208MeV. Find the total number of fission events required to operate a 100W
bulb over 1Hr.
Ans Q = 208MeV, E = N x Q where N is number of fission event, find E = P x T =3.6 x 10 5J,
Q = 3.33x10 -11J N = 1.08 x 10 16 events
A typical nuclear reactor produces power of 1GW. Assume it is 40% efficient and each fission event in this reactor produces
200MeV of thermal energy. Calculate the mass of 235U consumed each day.
Ans. Energy in 24hr = 40% x 1 x 10 9W x 8.64 x 10 4s = 3.46 x 10 13 J =2.16 x 10 32 eV N = E/Q = 1.08 x 10 24
If the objective lens of a telescope has a diameter of 200mm and its mean wavelength of light from a distant star under
observation s 6 x 10 -7 m, find the resolving power of the lens.
α = 1.22y/ diameter = 4 x 10 -6 rad
Differentiate between a photovoltaic cell and a photoconductive cell.
Ans , photovoltaic cells produce change in emf due to light incident on it and leads to increasing current. Photoconductive cells
cause a varying resistance at constant voltage.
State 2 examples each of the photoelectric cells mentioned above and state one daily use.
Ans ,Example of photovoltaic cell consist of cu20/cu with gold film over the oxide surface. Used in solar batteries.
Example of photoconductive cell is gold on glass comb with a cadmium sulphide plating. Used in LDR (light dependent resistors)
used in street lights.
Determine the excitation energy of an electron in a hydrogen atom when it transit from n1 to n3
Ans. E3 – E1 En = -13.6eV/n2, excitation energy = -13.6eV/9 - -13.6eV/1 = 13.6eV ( -1/9 + 1)
13.6eV (8/9) =12.089eV

Calculate the ionization energy of a valence electron of hydrogen atom.
Ans. This is the energy to move the = 13.6eV ( -1/inf- -1) electron from n=1 to n= infinity, E ionization
= 13.6eV (-1/infi +1) = +13.6eV, NB ionization energy is the negative of ground state energy.
An electron at ground state energy E1 = -21.8 x 10-19 J is excited to a higher state with energy E2= -5.4 x 10 -19J. If it de-
excites shortly after and returns to ground state, Calculate the wavelength of the emitted radiation given off.
Ans hf = E1 – E2, y = hc/( E1-E2) = 1.2 x 10 -7 m
Explain the reason why the current rises in a discontinuous fashion even with a steady increase in voltage in the frank Hertz
experiment.
ans .As the voltage is increased the kinetic energy of electrons increase and this reflects in increasing current since more
electrons reach the collector plate, at some point when eV = nhf, the collisions with the gaseous atoms become inelastic and
the electron loose kinetic energy to excited electrons in the atoms. the current thus dips.
State Moseley law
Ans/ for a particular type of characteristic x-ray, the frequency of the photon varies with the atomic number
Medical radiation usually uses cobalt -60 which has a half-life of 5.25 years. How long after a sample is delivered will its activity
reduce to 1/8th its original value.
Ans. 3half lives = 15.75years
Polonium has a half-life of 24000 years and decays by the reaction, 23994Pu 235 U + X, If polonium is stored for 73200
92
years, determine the particle X and find what fraction of the initial amount will be lost over the period of time.
Ans , x = 4 He and fraction left after 3 half-lives is 1/8th so the amount lost is 7/8th
2
Tritium and Helium 3 have the same number of nucleons. Tritium has one proton and 2 neutron s whiles helium-3 has 2 protons
and one neutron. Which of them has a higher nuclear binding energy and why.
Tritium has higher binding energy because it faces less internal repulsion unlike Helium -3 in which the 2 protons create
internal repulsion thus reducing the binding energy.
The half life of the radioactive nucleus, 22688Ra is 1.6 x 10 3 years. If the sample initially contains 3 x 10 16nuclei, determine the
initial activity.
Ans R 0 = K N0, to find K, K = 0.693/T1/2 , T1/2 = 5 x 10 10s so K = 1.4 x 10 -11, R 0 = 4.2 x 10 5 decays/s
Assuming the initial activity of a radioactive substance is given as 4.2 x 10 5 decay/s and there are 3 x 10 16 nuclear at the
3
beginning. Given the half-life of the nuclei to be 1.6 x 10 3 yrs, determine the amount left after 4.8 x 10 years
Ans. N = N0 x (1/2 )T / T1/2 = 3.8 x 10 15 nuclei
For the same question above, determine the activity level after 4.8 x 10 3 years if the decay constant is 1.4 x 10 -11
Ans R = K N = 5.3 x 10 4 decay/s
Define 1 roentgen of radiation.
Ans. The amount of ionizing radiation that will produce 2.08 x 10 9 ion pairs in 1cm3 of air under standard conditions
Define 1 rad of radiation
Ans , the amount of radiation that deposits 10 -2 J of energy into 1 Kg of absorbing material
After 2 days the activity of an unknown type of radioactive material has reduced to 84.2% of the initial activity. What is the half
life if the material

ans. 8.061 days
FLUID DYNAMICS AND THERMAL PHYSICS
Define an ideal fluid
Ans. It is an incompressible non viscous fluid in which each unit moves with laminar flow and maintains zero angular
frequency about the center of the fluid.
State Bernoulli principle in a mathematical relation and state one useful practical demonstration of it.
Ans. P + ½ RhoV2 + Rhog = constant, it is seen in carburetors, Atomizers etc.
State Torricelli’s law.
Ans . Fluid flowing freely from a point on a containing vessel above ground behaves like a freely falling object under
gravitational acceleration.
A vertical steel beam in a building supports a load of 6 x 10 4N. if the length of the steel beam is 4m and it has an area of 8 x 10 -
3 m2, find the depth of compression given that it has a young’s modulus of 2 x 10 11 Pa.
Ans x = 1.5 x 10 -4 m
A heavy cable is used to carry a steel bar. If the cable supports a load of 2 x 10 6 N. what cross sectional area must the cable
have to avoid breaking if the maximum compressive stress of the metal used for the cable is 5 x 10 8 Pa.
Ans. Stress max = maximum load / Area , Area = 4 x 10 -3m2
A manometer is connected to a sealed gas vessel immersed in a fluid. The gas in the vessel expands when the fluid warms up. If
the mercury height is adjusted to 10cm to keep the volume of gas constant and the absolute pressure of the gas at this
temperature is 1.3695 x 10 5 Pa, Find the prevailing atmospheric pressure given the density of mercury to be 13.595 x 10 3
Kg/m3
Ans , Absolute gas pressure P = Atmospheric pressure P0 + mercury pressure Rho x H x g,
P0 = P – Rho h g = 1.235 x 10 5Pa
A hydraulic press has a smaller cylinder with radius of 5cm and a larger cylinder of radius of 15cm. what force is needed to lift a
load of 13300N placed over the larger cylinder
Ans. Pressure remains uniform, P1 = P2 means F1/A1 = F2/A2 hence F1/F2 = R12/R22 Force = 1480N
The weight of a piece of gold in air is 7.84N and the weight when immersed in water is 6.86N. if the gold is fully immersed
inside the water when weighed, determine its density.
Ans. Upthrust B = W air – W water = 0.98N, B Rho water x V gold x g, so V gold = 1 x 10 -4 m3, you know mass of gold
from its weight in air mg, density of golf is thus = 8 x 10 3 kg/m3
A floating wooden board with density of 6 x 10 2kg/m2 and Area of 5.7m2 and volume of 0.6m3 floats on water. Determine the
depth submerged.
Ans, weight of wood = weight of displaced water whose volume is the volume submerged. Since surface areas are same,
we eliminate it. Rho wood x Full height wood = Rho water x h, where h is the submerged height of wood. h = 0.0632m
Calculate how much in percentage of an iceberg is beneath the surface of an ocean given the density of ice = 917kg/m 3 and
density of salt water is 1025kg/m3.
Ans , weigth of ice is = weight of water displaced , Rho ice x height of full ice = Rho salt water x height submerged since
SA remains same. Hfull/hsub = Rho brine/ Rho ice = 89.5%

Water flows at 5525m3/s over a 670m wide cliff of the boti water falls in Ghana. The water is measured to be 2m deep as it
reaches the waterfall. Estimate its speed.
Ans. Volume flow rate = Area x Velocity, Area = 1340m2 , v = 4m/s2
A police man fires a pistol and mistakenly hits a 1m high metal tank filled with water to the brim. If the tank is uncovered and
the hole was exactly ob the half mark, determine the speed of the jet of water from the hole.
V = 3.13m/s
The wings of an airplane each have an area of 4m2. Air flows over the top of the wings at 245m/s and beneath them at 222m/s
when the plane is speeding to take off from the ground. Determine the mass of the air plane that can be lifted off the ground
by upthrust due to air of density 1.29kg/m3.
Force of air thrust on each wing F = Pressure difference x Area
From Bernoulli’s relation, before liftoff, no change in vertical height has occurred yet so ΔP = ρ (v12-V22) , Δ p
= 6.93 x 10 3 Pa, F total = 2 x Δ P /A , mass = 5.66 x 10 3 kg
Use the concept of surface tension to briefly explain why different fluids rise or fall in capillary tube relative to the height in the
container.
Ans. Fluid with high cohesion that adhesion for the material of the capillary tube will have a lower level in the capillary tube
and they have a convex meniscus due to the high contact angle and vice versa.
Find the height to which water will rise in a capillary tube with a radius of 5 x 10 -5m. take surface tension of water to be
0.073N/m and density of water to be 1 x 10 3 kg/m3
Ans h = 2Y/( ρ g r) = 0.3m
State 3 units for the coefficient of viscosity
Ans. Poise, dyne.s/cm2 , N .s/m2
A patient receives blood transfusion through a needle of radius 0.2mm and length of 2cm. the density of blood is 1050kg/m3.
The bottle containing the blood is hung 0.5m vertical above the patients arm. What will be the flow rate taking coefficient of
viscosity as 2.575.
Use poiselles formula, R =
Determine the velocity at which the flow in an artery of diameter 0.2m will become turbulent. Take coefficient of viscosity of
blood to be 2.7 x 10 -3 Ns/m2. And density of blood is 1050kg/m3
RN = Rho v d l/ @ , for turbulence to happen. RN = 300, v = 3.9m/s
An object weighs 300N in air , 265N in water and 275N in oil. If density of water is 1000kg/m 3. Determine the density of the
object. In each case the object gets fully immersed in the fluid.
Determine the density of oil
Ans. Since it is immersed, first find the volume of the object which will be = volume water displace = volume of water displaced.
Rho w g V object = upthrust in water ( 35N), V = 3.5 x 10 -3m 3.
Rho oil = upthrust in oil ( 25N) /(V object g) = 714.2kg/m3
A hypodermic needle is 3cm long and 0.3mm in diameter. What pressure difference between the ends will force water out at a
rate of 1g/s. take viscosity of water to be 1 x 10 -3 Pa.
Use poiseulles law.

To lift a wire ring of radius 1.75cm from the surface of a sample of blood, a vertical upward force of 9.94 x 10 -4N must be
applied. Find the surface tension of blood.
Define an ideal gas and state conditions under which most gases remain ideal.
Ans. Ideal gas is one in which the particles are free point like entities exerting no long range forces on each other and in
constant random motion with their collisions with the containing vessel being perfectly elastic. Most gases can be ideal in
moderate temperatures above the critical temperature the gas and at low pressures.
Why are some gases such as helium called perfect gases?
Ans They have a very low critical temperature such that it has ideal behavior over a wider temperature range. It is difficult thus
to liquefy them even under very high pressures.
State 4 assumptions behind the kinetic theory of ideal gases
Using the kinetic theory model, explain Boyle’s law
The kinetic energy per molecule of helium gas is estimated to be 6.07 x 10 -21J. Determine the temperature of the gas.
Ans . KE = 3/2 KBT, T = 293K
2mol of a monoatomic gas has a temperature of 200c. if the gas maintains ideal behavior, find its internal energy
Ans. U = 3/2 n RT = 7.3 x 103J.
The temperature of 5mol of argon gas is lowered from 3 x 10 2K to 2.4 x 10 2 K. find the change in internal energy.
Ans U = 3/2 nR( T2-T1) = -3.74 x 10 3J
A ags under pressure of 5 x 10 6 Pa at 25 0c expands to 3 times its initial volume and has a new pressure of 1.07 x 10 6 Pa.
determine the new temperature.
Ans P1/T1 = P2/T2 = 212K
If the temperature of an ideal gas increases from 200K to 600K. what will happen to its root mean square speed.
Ans. Increases sqrt of 3
The average coefficient of linear expansion of copper is 17 x 10 -6 /0c. the statue of Kwame Nkrumah is 93m tall on a morning
with temperature of 25 0c. what will be its height on a very cold night with temperature of 5 0c
Ans. DL = @Lo x D T DL = - 0.0316, new length = 92.268m
A container holds 0.5m3 of oxygen at pressure of 4atm. A valve is opened to let the gas expand until its pressure is 1atm. If the
temperature to remain constant, what is the new volume.
Ans V= 2m3
What is the internal energy of 26g of neon gas at a temperature of 152 0c?
Ans. U = 3/2nRT. N= m/M , U = 6830J
A rubber balloon with 1L of air at 1 atm and 300K is cooled to 100K. What happens to the volume of the balloon?
V = 0.33L
Most ordinary glasses break when warm substances are poured unto them. Pyrex glass however can withstand the heat
released from highly exothermic reactions. Give the structural formula for Borax and explain why Pyrex is efficient in containing
hot substances and why others are not.

The pressure in a constant volume gas thermometer is 0.7 atm at 100 0c and 0.212 atm at 0 0c. what is the temperature of a
substance that gives a gas pressure of 0.04 atm
Ans . Pt - P0 /( P100- P 0 ) x 100 0c = T , T = -241.1 0c
A pair of eye glasses frames are made of epoxy plastic which has coefficient of linear expansion of 1.3 x 10 -4/0c . at 20 0c, the
frames have a radius of 2.2cm. to what temperature must the frames be heated to allow a lens of radius 2.21cm to fit into the
frames.
Ans . the desired linear expansion is 0.01cm and this is = αRo x ΔT, T=
A spherical steel ball has a diameter of 2.54cm at 25 0c. what temperature will change its volume by 1%. Take the coefficient of
linear expansion of steel to be 7.2 x 10 -6 /0c
Ans. ΔV/V = B x ΔT, you have been given Δv/v = 0.01. B = 3α
Lead has a density of 11.3x10 3 kg/m3 at 0 0c. what will be its density at 90 0c
State the zeroth law of thermodynamics
A brass ring of diameter 10cm at 20 0c is heated and slipped onto an aluminum rod of diameter 10.01cm at 20 0c. Assuming the
coefficients of linear expansions remain constant, at what temperature can the two be cooled to get them separated
What is the average kinetic energy of oxygen gas at 300k.
Ans. KE = 5/2KBT…because oxygen is diatomic gas.
2 gases in a mixture pass through a filter at a rate proportional to their root mean square speeds. Find the ratio of the speed of
the isotopes of chlorine, 37cl and 35cl
A 7L vessel contains 3.5mol of an ideal gas at a pressure of 1.6 x 10 6 Pa. find the temperature and average kinetic energy of the
molecules
Ans PV = nRT , T= PV/nR ,KE = KB T
In a period of 1s, 5 x 10 23 molecules of nitrogen strike an area of 8cm2. If the molecules move with a speed of 300m/s and strike
the walls of the vessel head on in elastic collision, find the pressure on the wall. (mass of N2 = 4.68 x 10-26 Kg)
A 125kg block of an unknown substance with temperature of 90 0c is placed in a cup containing 0.326kg of water at 20 0c. the
system reaches an equilibrium temperature of 22.4 0c. if the heat capacity of the cup is negligible, determine the specific heat
capacity of the unknown substance.
Ans. Cx = Mw Cw ( 22.4 – 20 ) / Mx (90 -22.4) = 390J/Kg
What mass of ice at -10 0c is needed to cool a whale’s water tank containing 1.2 x 10 3m3 from 20 0c to 10 0c
Ans = 1.27 x 10 3 Kg
A 5kg block of ice at 0 0c is added to an insulated container partially filled with 10 kg of water at 15 0c. find the final
temperature of the mixture. Take latent heat of fusion of water to be 3.33 x 10 5 J/kg
Ans = 0 0c
State Newton’s law of cooling and give a mathematical relation for your answer
State the Stefan law of radiation and state a mathematical relation for your answer
An ideal gas absorbs 5 x 10 3 J of energy whiles 2 x 10 3 J of work is done on the environment by the expansion of the gas whiles
maintaining a constant pressure. Name the type of thermal process and determine the internal energy of the gas at the end of
the process.

Ans. Isochric or isobaric process, U = W + Q = but W is – 2 x 10 3J, U = 3 x 10 6J
Suppose the internal energy of an ideal gas rises by 3 x 10 3 J at a constant pressure of 1 x 10 5 Pa whiles it gains energy
through heating of 4.2 x 10 3 J. Find the volume change.
Volume change = 1.2 x 10 -2 m3
During one cycle an engine extracts 2 x 10 3 J of energy from a warm reservoir and transfers heat of 1.5 x 19 3 J to a cold
reservoir. Find the thermal efficiency
Ans. E = 1 – ( Qc/Qh) 25%
For the heat engine as above, determine the work done per cycle
Ans, W = Qh – Qc = 5 x 10 2 J
A steam engine has a boiler operating 5 x 10 2 K. the heat from the boiling changes water to steam that drives turbines to
produce kinetic energy with exhaust temperature of 3 x 10 2 K. what is the efficiency of this engine if mass of steam per cycle is
100kb and there 20 turbines.
E = 1 – (Tc/Th) = 40%
A piece of rock of mass 1 x 10 3Kg at 293K falls from a cliff of height 125m into a large lake which was still at 293K. Find the
change in entropy of the lake assuming all the kinetic energy of the rock is turned into heat on impact with the water
Ans. Total kinetic energy on impact = mgh = 1.23J
Entropy S == Q/T = 4.2 x 10 3 J/K
ELECTROMAGNETISM
A proton moves with the speed of 1 x10 5m/s through a magnetic field of 55uT at a particular location. When the proton moves
eastward, it is deflected upwards by the force due to the field but when the proton moves directly upwards, there is no
deflection. Calculate the magnitude and determine the direction of the magnetic field.
Ans , F = Bqv sin@, F = 8.8 x 10 -19N and points northward
A wire carries a current of 22A from west to east direction. The earth’s magnetic field in the vicinity of the wire has strength of
0.5 x 19 -4 T. Find the magnitude and direction of the force on the 36m long wire in this field if the field is directed from south to
north
Ans , F = BIL sin @ . F= 3.96 x 10 -2 N and directed upwards
A wire loop of radius 1m is placed in a field of magnitude 0.5T. The normal to the plane of the loops makes 300
Angle with the magnetic field and the loops carry a 2A current. Determine the magnetic moment and the torque on the loop
Ans . U = IAN = 6.28Am2, T = UB sin @ = 1.57Nm
A wire rectangular coil of sides 2m and 3m have a current of 2A passing through it. The plane of the coil is 300 angle with the
magnetic field of magnitude 0.5T. Find the torque on the coil
Ans. T = BIAN sin @ == 9.00 Nm
A charged particle enters a magnetic field of a mass spectrometer with a speed of 1.79 x 10 6 m/s and moves along a circular
path of radius 16cm in a uniform magnetic field of 0.3T which is directed perpendicular to the velocity of the particle. Find the
mass to charge ratio of the particle.

Ans, mv2/r = Bqv m/q = Br/v = 3.13 x 10 -8 kg/c
State in sentence and mathematically Amperes’ law
A solenoid consists of 100 turns of wire and has a length of 10cm. If it carries a current of 0.5T, find the magnitude of the
magnetic field around this solenoid.
Ans B = U0 n I where n= N/L B = 6.28 x 10 -4 T
A proton is orbiting inside a magnetic field of 6.28 x 10 -4 T and it is moves in a circular path with radius of 0.02m. Find its
momentum.
Ans . Mv = Bqr , Mv = 2.10 x 10 -24 kgm/s
Approximately what length of a straight wire will be required to create a solenoid with 100 turns if it is to have a radius of
0.05m.
Ans, L = N x 2pi R = 31.4m
Suppose you have a 32m length of copper wire. If the wire is wrapped into a 0.24m long solenoid and the solenoid has a radius
of o.o4m, how strong will be the magnetic field if a current of 12 A flows through the solenoid?
Ans, find N = Length of wire / 2pi R then using B= U0 N I/ L = 8 x 10 -3T
A proton enters a constant magnetic field of magnitude 0.05T and traverses a semicircular path of radius 1mm before exiting
the field. What is the exit speed of the particle?
Ans, mv = Bqr, v = Bqr/m = 4.8 x 10 3m/s
Estimate the magnitude of the magnetic field per unit length between 2 parallel wires 2m apart if they both carry a 3A current.
Ans , F = U0 I1 I2 / 2pi d but since it is same charge , F/l = U0 I2 / 2pid = 9 x 10 -7 N/m
A rectangular coil of wire consisting of 10 loops each with length 0.2m and width of 0.3m lies in the xy plane. If the coil carries a
2A current, what torque is exerted on it from a 0.01T field at 300 to the plane of the coil?
Ans , F = BIAN sin @ = 6 x 10 -3 N
A cardiac pacemaker can be affected by a minimum static magnetic field of 1.7T. How close can a person with a pacemaker
come to a long straight wire carrying a 2A current.
Ans , B = U 0 I / 2pi d , d = U0I / 2pi B = 235cm
2 parallel wires 4cm apart repel each other with a force per unit length of 2 x 10 -4 N/m. The current on one is 5A. Find the
current on the other.
State faradays law and give a mathematical relation for your statement
Ans , The induced EMF created due to a changing magnetic field is directly proportional to the negative rate of change of the
field.
State Lenz law and explain what the statement means
Ans , The induced current flows in a direction such that it creates a magnetic field whose flux opposes the flux of the field
inducing it
State 3 scenarios or processes by which an induced current can be generated
A coil of 25 turns forms a 2quare loop with a side of 1.8cm. If an applied field perpendicular to the loop changes from 0.00 T to
0.50 T 0ver 0.8s, determine the induced emf

Ans. E = -N change in flux/ change in time. Change in flux = Area x change in magnetic field, so flux change = 1.62 x 10 -4wb,
e = -5.06 x 10 -3 V
An airplane with wing span of 30m flies northwards over a location where the downward component of the earth’s magnetic
field is 0.6 x 10 -4T. Find the potential difference between the wings if the plane flies at 2.5 x 10 2m/s
Ans. E = BLV = 0.45V
A sliding metal bar of length 0.5m moves through a 0.2T field at 2m/s. Find the current on the bar if a circuit resistance of 0.5
ohms is present and determine the power delivered to the bar
Ans. I = BLV/R == 0.5A, P = I2R = 0.125w
An AC generator consist of 8 turns wire forming a loop with area 0.09m2. The coil rotates in a magnetic field of 0.5T at a
constant frequency of 30/piHz. Find the maximum induced emf
Ans. W = 2pif = 60rads/s, e = NBAw = 21.6V
A motor has coils with a 10 ohm resistance and supplied by a 1.2 x 10 2 V battery. When the motor is running at maximum
speed, the back emf is 70V. Find the current produced by the motor immediately on turning on and after motor attains
maximum speed.
Initially , I = E/R =12A, at maximum speed, the voltage is reduced by the back emf, I = V – Back E / R = 5A
What is an inductor and name one application of inductors in circuit gadgets?
A solenoid has a current of 50A passing through it each second. If the inductance is given to be 0.18H and the solenoid has 300
turns, determine the back emf in the circuit.
Ans, |E== - N x change current per unit time, e = - 9. 05V
A 12V battery is connected in series to a 25 ohms resistor and a5H inductor. Determine the maximum current and find the
energy stored in the inductor.
Ans , I = E/R = 0.48A, PEL = ½ L I2 = 0.576J
A current in a 5H inductor decreases uniformly at a rate of 2A/s. What is the voltage drop across the circuit?
Ans, E = -10V
A generator contains 100 turns that rotates 10times per second. If each coil has an area of 0.1m2 and a magnetic field through
the coils is 0.05T, what is the maximum emf induced.
Ans, e=NBAw, w = 10rev/s x 2pi = 20pi rad/s , e = 10 pi V or 31.4 V
A uniform magnetic field of magnitude 0.5T is directed perpendicular to the plane of a rectangular loop of dimensions12cm and
8cm. Find the magnetic flux through the loop.
Ans, flux = B.A = 4.8 x 10 -3 wb
State 2 differences between Ac generators and DC generators
Distinguish between a generator and a motor
He2+ ion travels at right angles in a field of 0.8T with a speed of 1 x 10 3 m/s. Find the force it experiences and the effect of this
force if the field is directed away from you and the helium ion is moving in the positive X direction.
Ans F = Bqv, q= 2e+ F = 2.56 x 10 -14N
An ion of charge +2e enters a field whose flux density is 1.2wb/m2 at a velocity of 2.5 x 10 5 m/s perpendicular to the flux lines.
Determine the force on it.

Ans. 9.6 x 10 -14 N
An electron is accelerated from rest through a potential difference of 3750V. It enters a magnetic field of strength 4mT
perpendicular to the velocity of the electron. Calculate the velocity of the electron in the field.
Ans, ½ mv2 = Vq, v = sqrt( 2 Vq/m) = 3.63 x 10 7 m/s
For the same electron described above, determine the radius of the circular path it takes in the magnetic field.
Ans, r = Mv/ Bq others formulars; r= 1/B x sqrt( 2Vm/q) ans, r= 52mm
Define the curie temperature of a magnetic material
It is the temperature above which a ferromagnetic begins to behave as a paramagnet.
State the curie law
The magnetization is directly proportional to the external magnetic field so far as the temperature is below a threshold above
which the magnetization has an inverse dependence on the temperature change.
A coil of radius 200mm is to produce a field of 0.4G in its centre. With a current of 0.25A passing through it, how many turns
should the wire have
Ans. B = N Uo I / 2r , N = 2Br/U0I = 51 turns
5 long straight wires are bound closely to form a cable. Current carried by the individual wires are ( -6a,12A, -7A, 18A and 20A.
Find the magnetic field 10cm from the cable
Ans. B = UoI/ 2pir , I = sum of all the currents =37A, B = 74uT
2 long straight wires 10m apart in air carry currents of 20A and 40A respectively in opposite directions. Determine the resultant
flux density between them
Ans, B12 = B1 –B2, B12 = U0 ( I1 –I2)/2pid, because the currents are in opposite direction, it becomes plus due to double
negative. 2.4 x 10 -4 T
Relate the magnetic field to the magnetization
Ans B = Uo( H + M)
Define the pole strength of a bar magnet
Distinguish between a paramagnet and ferromagnetic using domain theory
Distinguish between diamagnetic, paramagnet and ferromagnetic materials using their magnetic susceptibilities.
Diamagnets have a very small negative susceptibility, paramagnets have a small positive susceptibility and ferromagnets have a
large positive susceptibility.

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1 PHYSICS BY TWO TEACHERS.

Written by Dr. Danso Kumi and Bright Forkuo.

PHYSICS BY TWO TEACHERS

LIFE IS EVER REFRESHING AND REJUVINATING WHENEVER I GIVE A

BY DR. DENNIS DANSO KUMI

AND BRIGHT FORKUO.

physics by two teachers © 2019

ELECTROSTATICS, MAGNETISM, ELECTROMAGNETIC INDUCTION

Maxwell calculated the Speed of

Electromagnetic Wave to be;V= 1/√𝜺˳ ∪ ˳

Where𝜀˳ = Permittivity of free space

In Vacuum, V= C= 3 x 108 m/s

Maxwell’s predictions were first confirmed by Heinrich Rudolf Hertz in 1887.

Resonant frequency of the electromagnetic wave:

Concept of Conductors and Insulators Using Band Theory

a. For conductors, these are free electrons in conduction band.

physics by two teachers © 2019

Step 1, polarization of sphere final sphere has same charge as rod

i. You bring a charged rod close to an uncharged metal sphere.

after inducing polarization, you earth the opposite side of sphere.

F𝛼Q1 Q2, Constant of proportionality is the coulombs constant.

physics by two teachers © 2019

Some Constants to Remember

SOME STRATEGIC APPROACHES TO ELECTROSTATIC PROBLEM SOLVING

(a) Where must a test charge be placed between 2 other charges?

Solve for the Quadratic equation to obtain the value of x.

Ans. Let the distance between Q1 AND Q2 BE A, THEN Q3 IS X FROM Q1,

Q1/Q2= x2/(a-x)2 solve the quadratic to get x = 0.77m or 1.23m

physics by two teachers © 2019

 Work done (electrical potential energy) = (kq1*q2)/r

 Work done per charge (voltage or electric potential) = (kq)/r

Consider this equation

ELECTRIC FIELDS AND ELECTRIC FIELD STRENGTH (ELECTRIC FLUX DENSITY)

The idea of field was introduced by Michael Faraday.

physics by two teachers © 2019

PROPERTIES OF ELECTRIC FIELD

NB: Force experiment by a test charge

MILLIKAN’S OIL DROP EXPERIMENT

Atomized oil drop

Field force = Weight

Ans = 6.4 x 10-19C

ELECTRIC FIELD LINES

A device used to measure electric fields is called the Field Mill.

physics by two teachers © 2019

VAN DE GRAAF GENERATOR

LOOKING AT ELECTRIC FIELDS IN TERM OF FLUX

NB: Electric flux Density = Electric field strength

E = -V/d, and the expression V/d is called the potential gradient.

physics by two teachers © 2019

ELECTRIC POTENTIAL ENERGY

Some Facts about Electric Potentials

WORK ENERGY THEOREM IN ELECTROSTATIC FIELDS

½ mv2 = electric force (Eq) x distance travel (d)

Remember the electric potential energy PE = Vq

Ans. Δ PE = -Eq x ΔX = -1.68 x 10 -17J

physics by two teachers © 2019

CAPACITORS AND DIELECTRICS

Unit for capacitance is the Farad.

b. For series connection, it is same charge but different Voltages

physics by two teachers © 2019