DP024 Module Mutual Physics - Perak Matriculation College

1

TABLE OF CONTENTS 1
5
1.0 SIMPLE HARMONIC MOTION 8
2.0 MECHANICAL WAVES 12
3.0 ELECTRIC CURRENT AND DIRECT CURRENT CIRCUITS 20
4.0 CAPACITOR 29
5.0 MAGNETISM & ELECTROMAGNETIC INDUCTION 35
6.0 GEOMETRICAL OPTICS
7.0 PHYSICAL OPTICS

2

Hakimah
Simple Harmonic Motion

CHAPTER 1.0: SIMPLE HARMONIC MOTION

1.1 Simple Harmonic Motion

L/O 1.1 c): Interpret a-x graphs for SHM

a = - 2 x where,
a : acceleration
Compares : angular velocity
x : displacement
y = m x+ c m : gradient

∴ m = - 2

Example 1:

Figure 1.1
FIGURE 1.1 shows a graph of acceleration, a against displacement, x of an oscillating object.

(i) Calculate the angular frequency of the motion

(ii) Calculate the time taken for the object for one oscillation

Solution: , = − 2 (ii) 2
(i) −8 − 8 =
= 1.26
= 5 − (−5) 2
= 1.26
= −1.6
− 2 = −1.6 = 4.99
= 1.26 −1
Exercise1-2:
Exercise 1-1: FIGURE 1.2 below represent a graph of
acceleration against displacement of an object.

Figure 1.2 Figure 1.3
FIGURE 1.2 shows a graph of acceleration, a against If the object takes 3 s to complete 1 oscillation,
displacement, x of an oscillating object. Determine the determine the value of x.
angular frequency of the motion and the time taken for
the object for one oscillation. [-4.58 cm]

[1.26 rad s-1, 4.99 s]

1

Hakimah
Simple Harmonic Motion

L/O 1.1 f) : Use velocity

= where,
v : velocity
dx : change in displacement
dt : change in time

Example 2:
The displacement x of a particle undergoes linear SHM is given by

= 5 sin(5 )

where x and t are measured in cm and second respectively. Deduce the expression of the particle’s velocity.

Solution:

=



= [ = 5 sin(5 )]

= 5(5 ) cos (5 )

= 25 cos (5 )

Exercise 2-1: Exercise 2-2:
The displacement x of a particle undergoes linear Given an equation of velocity as function of time in
SHM is given by SHM,

= 20 sin(10 ) = 3 sin (2 )

where x and t are measured in cm and second where x and t are measured in cm and second
respectively. Deduce the expression of the particle’s respectively. Deduce the expression of the particle’s
velocity.
displacement.
[ ( )]
[− ( )]



2

Hakimah
Simple Harmonic Motion

L/O 1.1 f) : Use acceleration

= where,
a : acceleration
dx : change in velocity
dt : change in time

Example 3:
The velocity v of a particle undergoes linear SHM is given by

3
= 5 sin(5 )
where v and t are measured in cm s-1 and s respectively. Deduce the expression of the particle’s acceleration.

Solution:
=



= [ = 5 sin(3 )]

5

= 5 (3 ) cos (3 )

55

= 3 cos (3 )

5

Exercise 3-1: Exercise 3-2:
The velocity v of a particle undergoes linear SHM is Given an equation of acceleration as function of time
given by in SHM,

= 5 cos(3 ) = 1.5 cos (2 )

where v and t are measured in m s-1 and s respectively. where a and t are measured in m s-2 and s respectively.
Deduce the expression of the particle’s acceleration. Deduce the expression of the particle’s velocity.

[− ( )] [ ( )]



3

Hakimah
Simple Harmonic Motion

L/O 1.1 g) : Solve problems related to SHM

Example 4:
A mass is hung at the end of the spring. The mass is pulled down and released so that it oscillates vertically.
If the mass takes 0.75 s to travel up between upper and lower point. Calculate the period of the motion and
the angular frequency.

Solution: (ii)
(i) Angular frequency,
Period,

= 2 = 2
= 2(0.75)

= 1.5
= 2

1.5

= 1.33 −1

Exercise 4-1: Exercise 4-2:
A mass is hung at the end of the spring. The mass is
pulled down and released so that it oscillates An object undergoes simple harmonic motion with
vertically. If the mass takes 0.74 s to travel up angular frequency 0.8π rads-1. Calculate the time
between upper and lower point. Calculate the period
of the motion and the angular frequency. taken for five complete vibrations.
[12.5 s]
[1.48 s, 1.35π rads-1]

4

Hakimah
Mechanical Waves

CHAPTER 2.0: MECHANICAL WAVES
2.1 Mechanical Waves
L/O 2.1 (e) : Use equation for progressive wave,

( , ) = sin( ± ) where;
y : distance moved by a particle from its
equilibrium position
x : distance of particle from the origin
A: amplitude of the wave
k : wave number
: angular velocity
t : time

Example 5:
A progressive wave is given by the equation

y(x,t) = 2 sin π(0.2t – 4x)

where y in cm, x in m and t in second.

i) Determine the direction of the wave propagation
ii) Calculate the wavelength and period of the wave

Solution:
(ii) ℎ

2 2
(ii) = =

2 2
= 4 = 0.2
= 0.5 = 10

Exercise 5-1: Exercise 5-2:
A progressive wave is given by the equation
A progressive wave travelling with a velocity of
y(x,t) = 2 sin π(4t + 0.2x) 100 m s-1 and frequency of 200 Hz and amplitude

where y in cm, x in m and t in second. 20 cm is propagating from left to right. Write a

i) Determine the direction of the wave propagation wave equation for the progressive wave.
ii) Calculate the wavelength and period of the wave [ y(x,t) = 20 sin (400πt+4πx)]

[to the right, 10 m, 0.5 s]

5

Hakimah
Mechanical Waves

L/O 2.1 (g) : Use particle vibrational velocity as

where;
= v = velocity
dy = change in velocity
dt = change in time

Example 6:
Given the expression for speed of the particle by

= 600 cos(40 − 0.2 )

where vy in cm s-1 and x in cm and t in s.
Calculate the speed for the particle at x = 5.0 cm and t = 0.5 s

Solution:
= 600 cos(40 90.5) − 0.2 (5))
= −600 −1

Exercise 6-1: Exercise 6-2:
Given the expression for speed of the particle by
Given the expression for speed of the particle by
= 500 cos(30 − 0.5 ) ( , ) = 5 sin(2 − 3 )

where vy in cms-1 and x in cm and t in s. where y and x in m and t in s.
Calculate the speed for the particle at x = 5.0 cm Write the expression for vibrational speed, vy..
and t = 0.5 s
[10π cos (2πt - 3πx)]
[387.1π cm s-1]

6

Hakimah
Mechanical Waves

L/O 2.1 (h) : Use the wave propagation velocity,

= where ;

v : wave propagation velocity

f : frequency
λ : wavelength

Example 7:
A wave travelling along a string is described by

y(x,t) = 0.327 sin (2.72t – 72.1x)

where y in cm, x in m and t in second. Determine the velocity which the wave moves along the string.

Solution: = 2 =
= (0.433)(0.0871)
= 2 = 0.037 −1
= 2
2.72

72.1 = 2 = 2
= 0.433


= 0.0871

Exercise 7-1: Exercise 7-2:
A wave travelling along a string is described by A progressive wave travelling with a velocity of
200 m s-1 and frequency of 400 Hz. Calculate the
y(x,t) = 0.03 sin (2πt – 200πx) wavelength of the wave.

where y in cm, x in m and t in seconds. [0.5 m]
Determine the velocity which the wave moves
along the string.

[0.01 m s-1]

7

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

CHAPTER 3 : ELECTRIC CURRENT AND DIRECT-CURRENT CIRCUITS
3.0 Electric Current and Direct-Current Circuits
L/O 3.0 (b) : Use electric current

I = dQ where,
dt I = magnitude of current

dQ = net charge that moving passes
through an area
dt = time interval

Example 8:
In an electrical circuit, 8 x 10-5 C of charges moving through a cross section of a wire in 0.04s. Calculate the

magnitude of the current.

Solution:

I = dQ
dt

= 8 x 10 −5
0.04

= 2 mA

Exercise 8-1: Exercise 8-2:
In an electrical circuit, 1.6 x 10-4 C of charges A current of 5 A flows in the circuit for 3 seconds.
Calculate the total charges flowing during this time.
moving through a cross section of a wire in 0.032s.
[15 C]
Calculate the magnitude of the current.

[5 mA]

8

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.0 (d) : Use resistivity where,
ρ = resistivity
 = RA
l R = electrical resistance

A = cross sectional area of conductor
l = length of conductor

Example 9:

A copper extension cord has diameter of 0.912 mm and length of 30 m. Calculate the resistance of the copper

extension at 20 ºC.
(ρcopper = 1.67 x 10-8 Ω m)

Solution:

( )A = d 2 =  0.912 x10−3 2 = 6.53 x10−7 m2
44
( )→
 = RA R= l = 1.67 x10−8 (30 ) = 0.767 
l A 6.53 x10
−7

Exercise 9-1: Exercise 9-2:
A tungsten light bulb filament extension cord has Determine the resistivity of a metal wire if the
diameter of 0.02 mm and length of 4 cm. Calculate resistance is 6 Ω, uniform cross sectional area of the
the resistance of the tungsten light bulb filament wire is 8 mm2and length is 0.5 m.
extension cord at 20 ºC.
(ρtungsten = 5.4 x 10-8 Ω m) [9.60 x 10-5 Ω m]

[6.88 Ω]

9

L/O 3.0 (f) : Use Ohm’s Law Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits
V = IR
where,
I = magnitude of current
V = potential difference or voltage drop
R = electrical resistance

Example 10: A light bulb uses a single 3 V battery to provide a current of 0.4 A in the filament. Calculate
the resistance of the glowing filament.

Solution:

R = V = 3 = 7.5
I 0.4

Exercise 10-1: Exercise 10-2:
A potential difference of 15 V is applied to a 12-Ω
A light bulb uses two 1.5 V batteries to provide a
resistor . Calculate the current flowing through the
current of 0.3 A in the filament. Calculate the
resistor.
resistance of the glowing filament.
[1.25 A]
[10 Ω]

10

Mohd Khairul Azmi & Noraida
Electric Current and Direct-Current Circuits

L/O 3.0 (h) : Calculate the effective resistance of resistors in series and parallel. R1
R2
Resistors in Series: Resistors in Parallel: R3
R1
R2 R3 R3 = 1 Ω
B
Reff = R1 + R2 + R3 1 =1+1+1
Reff R1 R2 R3 R3 = 1 Ω

Example 11: R1 = 2 Ω
Calculate the equivalent resistance between point A and B (R123).
R2 = 2 Ω
Solution: A R12 = 1 Ω

1 = 1 + 1 = 1 + 1 =1 
R12 R1 R2 2 2
R12 = 1 
R123 = R12 + R3 = 1 + 1 = 2 

Exercise 11-1: Exercise 11-2:

Calculate the equivalent resistance between point A Calculate the equivalent resistance between point A

and B (R123). and B.
[4 Ω]
[7.5 Ω]

R1 = 5 Ω R3 = 5 Ω R1 = 6 Ω R3 = 6 Ω
A
AB B
R2 = 6 Ω
R2 = 5 Ω

11

Siti Farahiyah & Wan Idayu Farhana
Capacitor

CHAPTER 4.0 : CAPACITOR
4.1 Capacitance and Capacitors in Series and Parallel
L/O 4.1b): Use capacitance,

C=Q where,
V C= capacitance
Q= magnitude of charge on each plate
V= potential difference across two plates

Example 12:
When the potential difference between the plates of a capacitor is increased by 3.25 V, the magnitude of the
charge on each plate increases by 13.5 C. Calculate the capacitance of this capacitor

Solution:

C = Q = 13.510−6 = 4.15 µF
V 3.25

Exercise 12-1: Exercise 12-2:
A capacitor stores 100 pC of charge when it is The capacitance of this capacitor 1000 F. The
connected across a potential difference of 20 V. magnitude of the charge on each plate increases by 80
Calculate the capacitance of the capacitor. C. Calculate the potential difference between the
plates of a capacitor.
[5 pF]
[0.08 V]

12

Siti Farahiyah & Wan Idayu Farhana
Capacitor

L/O 4.1 c): Calculate the effective capacitance of various arrangements by using the following formulae:

i. Series: where,
Ceff = effective capacitance
1 = 1 + 1 + 1 + .... + 1 C1, C2, C3, Cn = capacitance
Ceff C1 C2 C3 Cn

Example 13: Calculate the effective capacitance between A and B

C1=15 F C2=3 F Solution:
A
B 1 = 1 + 1 = 1 +1= 4
Ceff C1 C2 15 3 15

C eff = 15 μF
4

=2.5

Exercise 13-1: Exercise 13-2:
Calculate the effective capacitance between A and B The effective capacitance between A and B is 1.2 µF
Calculate the value of C1
C1=12 F C2=4 F
A C1 C2=4 F
AB

[3 µF] C3=2 F

B
[12 µF]

13

Siti Farahiyah & Wan Idayu Farhana
Capacitor

L/O 4.1 c): Calculate the effective capacitance of various arrangements by using the following formulae:
i. Series:

1 = 1 + 1 + 1 + .... + 1 where,
Ceff C1 C2 C3 Cn C = capacitance
Q = magnitude of charge on each plate
V = potential difference across two plates

Q = Q1 = Q2 = Q3 V = V1 +V2 +V3

Example 14: Calculate the charge on each capacitor if the circuit is connected to 24 V.

C1=15 F C2=3 F
A
Solution: B

= 15
4
15
= = 4 (24) = 60

1 = 2 = = 60

Exercise 14-1: Exercise 14-2:
Calculate the charge on each capacitor if the circuit Calculate the charge on each capacitor if the circuit
is connected to 12 V. is connected to 6 V.

[36 µC] [72 µC]

C1=12 F C2=4 F C1 C2=4 F
A A
B
C3=2 F

B

14

Siti Farahiyah & Wan Idayu Farhana
Capacitor

4.1 c): Calculate the effective capacitance of various arrangements by using the following formulae:
i. Parallel:

Ce ff = C1 + C2 + C3 + ...Cn where,
Ceff = effective capacitance
C1, C2, C3, Cn = capacitance

Example 15: Calculate the effective capacitance between A and B.

C1= 15 F

A B

Solution: C2=6 F

Ce ff = C1 + C2
= 15 + 6
= 21μ F

Exercise 15-1: Exercise 15-2:
Calculate the effective capacitance between A and B FIGURE shows an arrangement of capacitors. If the
equivalent capacitance is 1 F, calculate the value of
C1= 5 F C.

C

AB

C2= 12 [17 µF] 0.6 F
F

[0.4 µF]

15

Siti Farahiyah & Wan Idayu Farhana
Capacitor

4.1 c): Calculate the effective capacitance of various arrangements by using the following formulae:
i. Parallel:

Ce ff = C1 + C2 + C3 + ...Cn where,
C = capacitance
V = V1 = V2 = V3 Q = Q1 + Q2 + Q3 Q = magnitude of charge on each plate
V = potential difference across two plates

Example 16: Calculate the charge on each capacitor if the circuit is connected to 24 V.

C1= 15 F

A B

C2=6 F

Solution:

V = V1 = V2 = 24V

Q1 = C1V = (15)(24) = 3.610−4 C = 360 μC

Q1 = C2V = (6)(24) = 144 μC

Exercise 16-1: Exercise 16-2:
Calculate the charge on each capacitor if the circuit Calculate the charge on each capacitor if the circuit
is connected to 15 V between A and B is connected to 9 V between A and B

C1= 5 F C1=0.4
µF

AB C2=0.6
C2= 12 F
F
[ 3.6 µC, 5.4 µC]
[ 75 µC, 180 µC]

16

Siti Farahiyah & Wan Idayu Farhana
Capacitor

4.2 : Charging and Discharging of Capacitors

L/O 4.2a) Use

τ = RC where,
τ = time constant

R = resistance

C = Capacitance

Example 17: A 40 µF capacitor has a charge of 250 µC is connect to a 30 kΩ resistor. Calculate the time
constant.

Solution:

τ = RC

( )( )= 30 103 40 10−6

= 1.2 s

Exercise 17-1: Exercise 17-2:
A 20 µF capacitor has a charge of 240 µC is connect The time constant of RC circuit is 100 s. The
to a 50 kΩ resistor. Calculate the time constant. capacitor has a charge is connecting to a 50 kΩ
resistor. Calculate the capacitance.
[1 s]
[2 mF]

17

Siti Farahiyah & Wan Idayu Farhana
Capacitor

L/O 4.2 d) Use: where,
i. for discharging Qo = maximum charge
R = resistance of the resistor
−t C = capacitance of the capacitor

Q = Q0e RC

Example 18: The 1000 F capacitor has a maximum charge 6.0 mC charge on each plate. The capacitor is
then discharged through a 100 k resistor. Calculate the charge stored in capacitor at t = 2 s

Solution:

−t

Q = Q0e RC

( )= −2
6  10 −3
e ( )( )100103 100010−6

= 5.88 10 −3 C

= 5.58 mC

Exercise 18-1: Exercise 18-2:
A fully charged 12.0 F capacitor
The 100 pF capacitor has a maximum charge 4.0 mC has 6.0 mC charge on each plate. The capacitor is then
discharged through a 100  resistor. Calculate the
charge on each plate. The capacitor is then discharged time taken if just 20% of charge is left in the capacitor.

through a 100  resistor. Calculate the charge stored [1.93 ms]

in capacitor at t = 0 s [4 mC]

18

L/O 4.2 c) Use Siti Farahiyah & Wan Idayu Farhana
ii. for charging Capacitor

Q = Q0 1 − − t  where,
RC Qo = maximum charge
e R = resistance of the resistor
C = capacitance of the capacitor

Example 19: A 20 µF capacitor is connected to a 50 kΩ resistor. If the maximum charge id 240 µC, calculate
the charge induced in capacitor at time, t = 0.1 s

Solution:

Q = Q0 1 − − t 
RC
e

−6 1 − (50103 0.1 ) 

= 240  10 − e )(2010−6

= 2.28 10 −5 C

Exercise 19-1: Exercise 19-2:
50 µF are connected in series with a 15 kΩ resistor. If

the maximum charge is 360 µC, calculate the charge

induced in capacitor at time, t = 0.5 s

[1.7510−4 C ]

Figure shows a circuit which is used to charge a
capacitor of 50 µF which is initially without charge.
Calculate the charge stored in capacitor at t = 1 s.

[1.4810−5 C ]

19

Muhammad Dzarfan
Magnetism and Electromagnetic Induction

CHAPTER 5.0 MAGNETISM AND ELECTROMAGNETIC INDUCTION

5.2 Magnetic field produced by current-carrying conductor

L/O 5.2 (b): Calculate the magnetic field by using :

i. for a long straight wire

where

B = oI B : magnetic field strength at a point near the wire
2r µ0 : permeability of free space
I : current

r : perpendicular distance from the wire to a point

Example 20:
Calculate the value of magnetic field in air at a point 5 cm from a long straight wire carrying a current of
25 A.

Solution:

B =  I
2r

= (4 10−7 )(25)

(2 ) 510−2
( )B

B =1.010−4 T

Exercise 20-1: Exercise 20-2:

Calculate the flux density in air at a point 16 cm A straight wire is placed in a magnetic field of

from a long straight wire carrying a current of 2.5x10-6 T horizontally. Calculate the current
5 A. [6.25x10-6 T]
flowing in the straight wire if the straight wire is 15

m from it. [187.5 A]

20

Muhammad Dzarfan
Magnetism and Electromagnetic Induction

L/O 5.2 (b): Calculate the magnetic field by using :
ii. at the centre of a circular coil

B = oNI where
2r B : magnetic field strength at a point near the wire
µ0 : permeability of free space
I : current
r : perpendicular distance from the wire to a point
N : Total numbers of loops

Example 21:

A circular coil has 30 turns and a radius of 22.5 cm. If the current flowing in the coil is 5.96 A calculate the
magnetic field strength at the centre of the coil is 5.0×10-4 T.

Solution:

B =  nI

B =  NI
2r

= (4 10−7 )(30)(5.96)

(2) 22.510−2
( )B

B = 5.010−4 T

Exercise 21-2: Exercise 21-3:

Calculate the magnetic field at the center of a A flat circular coil with 50 loops of wire has a
circular loop of wire of radius,r 4.0 cm when a
diameter of 32 cm. Determine current must flow in
current,I of 2.0 A flows in the wire.
[3.14x10-5 T] its wires to produce a field,B of 5.0 x 10-4 Wb m-2 at

its center. [2.55 A]

21

Muhammad Dzarfan
Magnetism and Electromagnetic Induction

L/O 5.2 (b): Calculate the magnetic field by using : where
iii. at the centre of solenoid B : magnetic field strength at a point near the wire
µ0 : permeability of free space
B = oNI I : current
l r : perpendicular distance from the wire to a point
N : Total numbers of loops

Example 22:
A 300 turns solenoid of length 25 cm. A current of 8 A flows through the solenoid. Calculate the magnetic
field at centre of solenoid.

Solution:

B =  nI

B =  NI
l

B = (4 10−7 )(300)(8)
25 10−2

B = 0.012T

Exercise 22-2: Exercise 22-3:
A solenoid 30 cm long has 840 turns.The
current in the solenoid is 5.0 A. Calculate the A 30.0 cm long solenoid and 1.25 cm in diameter is
magnetic field at the centre of the solenoid.
to produce a field,B of 0.385 T at its center.
[0.0176 T]
Determine current should the solenoid carry if it has

975 turns of the wire. [94.3 A]

22

Ummi Atiah & Siti Farahiyah
Magnetism and Electromagnetic Induction

5.3 Magnetic flux
L/O 5.3 (a) Use Magnetic flux,

 = B.Acos where,

B = magnitude field

A = surface area
θ = angle between normal A and

B

Example 23:
A magnetic field of strength 0.3 T is directed perpendicular to a plane circular loop of area 0.196 cm2.
Calculate the magnetic flux through the area enclosed by this loop.

Solution:

 = 0

 = BAcos

= (0.3)(0.196 10−4 )(cos0)

= 5.88 10−6 Wb

Exercise 23-1:
A magnetic field of strength 0.5 T is directed perpendicular to a plane circular loop of area 2 cm2. Calculate
the magnetic flux through the area enclosed by this loop.

[110−4 Wb ]

Exercise 23-2: Exercise 23-3:
A magnetic field is directed 30o to a plane circular
A rectangular loop of area 0.126 m2 is placed in a loop. The magnetic flux produced in the circular loop
of radius 25 cm is 5.910 −2 Tm2 . Calculate the
uniform magnetic field of magnitude 0.85 T. If the strength of the magnetic field.

magnetic flux through the loop is 9.310−2 Tm2 , [0.60 T]

calculate the angle between the normal to the loop and

the magnetic field. [29.7⁰]

23

L/O 5.3 (b): Use Magnetic flux linkage, Ummi Atiah & Siti Farahiyah
Magnetism and Electromagnetic Induction
 = N = NBAcos
where,
N = number of turns

 = magnetic flux

Example 24:
Calculate the flux linkage in a coil of 15 turns and area 25 cm2 in a field of strength 5 T perpendicular to the

plane.

Solution:
Φ = NBA

= (15) (5) (25 x 10-4 )

= 0.1875 Wb

Exercise 24-1: Exercise 24-2:
Calculate the flux linkage in a coil of 20 turns and area A circular loop is placed in a uniform magnetic field
4 cm2 in a field of strength 8 T perpendicular to the perpendicularly. If the flux linkage through the loop is
plane.
9.3 10−2 T m2 and the flux for each loop is
[0.64 Wb] 4.65 10−4 Wb . Calculate number of turns of the

loop.
[200]

24

5.4 Induced emf Ummi Atiah & Siti Farahiyah
L/O 5.4 (b) Use Faraday’s Law. Magnetism and Electromagnetic Induction

ε = −N d where,
dt N = number of turns

d

= rate of change of magnetic flux
dt

Example 25:
A 50-turn coil has a rate of change of magnetic flux of 0.5 Wb s-1 through a uniform magnetic field. Calculate

the emf induced in the coil?

Solution:

ε = −N d
dt

= −(50)(0.5)

= −25 V

Exercise 25-1: Exercise 25-2:
A 60-turn coil has a rate of change of magnetic flux
of 0.4 Wb s-1 through a uniform magnetic field. An 150V emf is produced by a coil of 100 turns that
Calculate the emf induced in the coil. passes through a uniform electric field. Calculate the

[-24 V] rate of change of magnetic flux of the coil.
[-1.5 Wb s-1]

Exercise 25-3:
The magnetic flux through one turn of a 60-turn coil of wire is reduced from 35 Wb to 5 Wb in 0.8 s, which
results in an induced current of 360 A. Calculate the total resistance of the wire in the coil.

[6.25  ]

25

Ummi Atiah & Siti Farahiyah
Magnetism and Electromagnetic Induction

L/O 5.4 (c)(i) Use induced emf in a straight conductor,

 = Blvsin where,
B = magnetic field strength

l = length of conductor

v = velocity of conductor
θ = angle between B and v

Example 26:

The rod shown above moves to the right on essentially zero-resistance rails at a speed of 3 m s-1. If the
magnetic 0.75 T everywhere in the region, Calculate the induced emf in a straight conductor?

Solution:

 = Blvsin
= (0.75)(4 10−2 )(3) sin 90

= 0.09 V

Exercise 26-1:
The rod of 5 cm moves to the right on essentially zero-resistance rails at a speed of 5 m s-1. If the magnetic
0.9 T everywhere in the region, calculate the induced emf in a straight conductor?

[0.225 V]

Exercise 26-2: Exercise 26-3:
A conducting rod of 10 cm is sliding on a metal rail.
The apparatus is in a uniform magnetic field of A 4 cm rod is in a uniform magnetic field which is
strength 0.25 T, which is directly into the page. The
conducting rod is moving to the right by a force. If the directly into the page as shown below. The rod is
induced emf produced is 0.125 V, calculate the pulled to the right at a constant speed of 5 m s-1 by a
velocity of the moving rod.
force . The only significant resistance in the circuit
[5 m s-1] comes from the 2.0 Ω resistor with the 0.025 A

current flowing. Calculate the magnetic field strength.

[0.25 T]

26

Ummi Atiah & Siti Farahiyah
Magnetism and Electromagnetic Induction

L/O 5.4 (c)(ii) Use induced emf in a coil,

Changing Magnetic Field,  = −NA dB where,
dt dB : change in magnetic field strength
A : surface Area
N : number of turns
dt :change in time

Example 27:
A coil of 1000 turns encloses an area of 25cm2. The coil is placed in an external magnetic to a magnetic field
of 6 ×10-5T. The coil is then pulled out of the field in 0.3 s, calculate the average emf induced during this

interval?

Solution:

 = −NA dB
dt

= −(1000)(25 10−4 ) (0 − 6 10−5 )
0.3

= 5 10−4 V

Exercise 27-1: Exercise 27-2:
A coil of 900 turns encloses an area of 40cm2. The A 5 ×10-4 V emf is induced in a 2000 turns coil of an
coil is placed in an external magnetic to a magnetic area of 25cm2. This coil is initially placed in a region
field of 9 ×10-5T. The coil is then pulled out of the
field in 0.5s, calculate the average emf induced of magnetic field and later it is moved out of the field
during this interval.
in 0.3 s. Calculate the strength of the magnetic field.
[ 6.4810−4 V ]
[ 310−5 T ]

27

L/O 5.4 (e)(ii) Use induced emf in a coil, Ummi Atiah & Siti Farahiyah
Magnetism and Electromagnetic Induction
Changing area  = −NB dA
dt Where,
B = magnetic field strength
dA = change in surface Area
N = number of turns
dt = change in time

Example 28:
A loop of wire of area 0.5 m2 is moving into a 01.5 T magnetic field. If it takes 4 seconds for the loop to be
stretched until its area becomes 0.1 m2 what emf is produced?

Solution:

 = −NB dA
dt

= −(1)(0.15) (0.1− 0.5)
4

= 0.015 V

Exercise 28-1: Exercise 28-2:
A loop of wire of area 0.7 m2 is moving into a 0.9 T
magnetic field. If it takes 5 seconds for the loop to be A 100-turn loop of wire is placed in a 2 T magnetic
stretched until its area becomes 0.08 m2 . Calculate the field and an induced emf of 4 V is produced. If the
area of the loop is being decreased to 0.02 m2 over a
emf that is produced.
period of 3 seconds, calculate the initial area of the
[0.11 V]
loop.
[0.08 m2]

28

Nageswary & Noorsuraya
Geometrical Optics

CHAPTER 6.0: GEOMETRICAL OPTICS

6.1 Reflection at Spherical Surface

L/O 6.1 c) Use mirror equation,

1 =1+1 where,
f uv f = focal length of mirror
u = object distance from the pole

for real object only. v = image distance from the pole
*Sign convention for focal length, f:

(i) Positive f for concave mirror

(ii) Negative f for convex mirror

Example 29:
A dentist uses a concave mirror attached to a thin rod to examine one of your teeth. When the tooth is 1.20
cm in front of the mirror, the image it forms is 9.25 cm behind the mirror.
Calculate the focal length of the mirror.

Solution:

1 =1+1
f uv

1 = 1 + 1
f 1.20
(− 9.25)

f = +1.38 cm

Exercise 29-1: Exercise 29-2:

A mirror forms a real image of a light bulb on a wall Olivia is inspecting the concave primary mirror of the

of a bathroom that is 3.5 m from the mirror. If the bulb Hubble telescope before its launched into space. She

is placed 7.0 cm from the mirror, Calculate the focal stands 87.0 m in front of the mirror which has a focal
length of 58.0 m. Calculate Olivia’s image distance.
length of the mirror. [6.87 cm]
[174.0 m]

29

L/O 6.1 c) Use mirror equation, Nageswary & Noorsuraya
1 =1+1 Geometrical Optics
f uv
where,
for real object only. f = focal length of mirror
u = object distance from the pole
v = image distance from the pole

*Sign convention for focal length, f:
(i) Positive f for concave mirror
(ii) Negative f for convex mirror

Example 30:
A Barbie doll is placed 10.0 cm in front of a convex mirror and an image is formed 6.0 cm behind the mirror.
Calculate the focal length of the mirror.

Solution:

1 =1+1
f uv

1 = 1 + 1
f 10
(− 6)

f = −15 cm

Exercise 30-1: Exercise 30-2:

A Barbie doll is placed 6.0 cm in front of a convex An external side mirror of a car is convex with a radius

mirror with a focal length of 12.0 cm. Calculate the of curvature 18 m. Calculate the location of the image

position of the image formed. [- 4.0 cm] for an object 10 m from the mirror. [ - 4.74 m]

30

L/O 6.1 e) Use magnification Nageswary & Noorsuraya
Geometrical Optics
m = hi = − v
ho u where,
m = magnification
u = object distance from the pole
v = image distance from the pole
hi = height of image
ho = height of object

Example 31:
A dentist uses a small mirror attached to a thin rod to examine one of your teeth. When the tooth is 1.20 cm
in front of the mirror, the image it forms is 9.25 cm behind the mirror.
Calculate the magnification of the image.

Solution:

m=−v
u

= − (−9.25)
1.20

= 7.71

Exercise 31-1: Exercise 31-2:
A 4.0 cm tall light bulb is placed at a distance of 35.5
A dentist uses a small mirror attached to a thin rod to cm from a convex mirror. If the height of the image
formed is 1.38 cm, Calculate the image distance.
examine one of your teeth. When the tooth is 24.0 mm
[- 12.25 cm]
in front of the mirror, the image it forms is 9.0 mm

behind the mirror. Calculate the magnification of the

image. [0.375]

31

Nageswary & Noorsuraya
Geometrical Optics

6.2 Thin lenses 1 =1+1 where,
L/O 6.2 b) Use thin lens equation, f uv f = focal length of lens
u = object distance from the pole
for real object only. v = image distance from the pole

*Sign convention for focal length, f:
(i) Positive f for convex lens
(ii) Negative f for concave lens

Example 32:
An object of height 10 cm is placed at 30 cm from a converging lens. The real image formed 60 cm from the
lens. Calculate focal length of the lens.
Solution:

1 =1+1
f uv

=1+1
30 60

f = 20 cm

Exercise 32-1: Exercise 32-2:
An object of height 10 cm is placed at 30 cm from a An object is placed in front of a converging lens with
converging lens. The virtual image formed 60 cm focal length 10 cm. The virtual image formed 30 cm
from the lens. Calculate focal length of the lens. from the lens. Calculate the object distance.

[60 cm] [7.5 cm]

32

L/O 6.2 b) Use thin lens equation, Nageswary & Noorsuraya
Geometrical Optics
1 =1+1
f uv where,
f = focal length of lens
for real object only. u = object distance from the pole
v = image distance from the pole

*Sign convention for focal length, f:
(i) Positive f for convex lens
(ii) Negative f for concave lens

Example 33:
An object of height 10 cm is placed at 30 cm from a diverging lens. The virtual image formed 20 cm from the
lens. Calculate focal length of the lens.
Solution:

1 =1+1
f uv

=1+ 1
30 − 20

f = −60 cm

Exercise 33-1: Exercise 33-2:
A thumbtack is placed at 20 cm from a concave lens. An object is placed 30 cm in front of a concave lens
The virtual image of the thumbtack formed 5 cm from with focal length 10 cm. Calculate the image distance.
the lens. Calculate focal length of the lens.
[-7.5 cm]
[-6.67 cm]

33

Nageswary & Noorsuraya
Geometrical Optics

L/O 6.2 d) Use magnification

m = hi = − v where,
ho u m = magnification
u = object distance from the pole
v = image distance from the pole
hi = height of image
ho = height of object

Example 34:
An object of height 10 cm is placed at 30 cm from a diverging lens. The virtual image formed 20 cm from the
lens. Calculate the magnification of the image.
Solution:

m = − v = − − 20 = 2 , Upright, diminished, virtual
u 30 3

Exercise 34-1: Exercise 34-2:
An object of height 10 cm is placed at 30 cm from a A 4.0 cm Marvel figurine is placed at a distance of 40
converging lens. The real image formed 20 cm from cm from a convex lens. If the height of the image
the lens. Calculate the magnification of the image. formed is 3.0 cm, Calculate the image distance.

[-2/3] [- 30 cm]

34

Nik Norhasrina & Salmi
Physical Optics

CHAPTER 7.0 PHYSICAL OPTICS

7.3 INTERFERENCE OF TRANSMITTED LIGHT THROUGH DOUBLE-SLITS

L/O 7.3 : Interference of transmitted light through double-slits

a) Use: where,
i. for bright fringes (maxima)

ym = mD ym = separation between central bright and mth bright

d fringes
m = order
where m =0, +1, +2, +3, … λ = wavelength
D = distance between double slits and screen

d = separation between double slit

Example 35:
In a Young’s double-slits experiment, distance between the screen and the double-slits is 2 m. Distance

between the two slits is 0.03 mm. If distance between the central bright fringe and the fourth bright fringe is

16.4 cm. Calculate the wavelength of the light.

Solution:

ym = mD
d

16.4 10 −2 = 4 (2)
0.03 10 −3

 = 6.15 10 −7 m

 = 615 nm

Exercise 35-1: Exercise 35-2:
In a Young’s double-slits experiment, distance In Double slit experiment, the distance between the
between the screen and the double-slits is 5 m. centre of the fringes to the fringes is 5 Mm. The
Distance between the two slits is 0.15 mm. If distance separation between the slits is 0.2 mm. If the distance
between the central bright fringe and the seventh from the slits to the fringes is a meter away and a
bright fringe is 10 cm. Calculate the wavelength of wavelength is 500 nm, what is the number of order of
the light. the bright fringes.

[429 nm] [2]

35

Nik Norhasrina & Salmi
Physical Optics

L/O 7.3 : Interference of transmitted light through double-slits

a) Use: where,
i. for bright fringes (maxima)

ym = (m + 1 )D ym = separation between central bright and mth bright
2
fringes
d m = order
λ = wavelength
where m =0, +1, +2, +3, … D = distance between double slits and screen

d = separation between double slit

Example 36:
The distance between the screen and the double-slits is 2m. Distance between the two slits is 0.03mm. If the
distance between the central bright and the fourth dark fringe is 16.4 cm, calculate the wavelength of the light
Solution:

(m + 1)D
2
y3 = d

(3 + 1)(2)
2
1.64 10−2 = 0.03 10 −3

 = 7.0310−7 m

Exercise 36-1: Exercise 36-2:
The distance between the screen and the double-slits The separation between the slits and its wavelength
is 4m. Distance between the two slits is 0.6 mm. If the are 0.500 mm and 0.2 mm in a Young Double slits
distance between the central bright and the fifth dark experiment. However, the interference pattern on a
fringe is 15.3 mm, calculate the wavelength of the screen is 3.30 m away. The distance from the center
light. to the minimum of order for dark fringes is 4.62 m.
What is the number of order of the dark fringes?
[ 5.110−7 m ]
[3]

36

Nik Norhasrina & Salmi
Physical Optics

L/O 7.3 (b) Interference of transmitted light through double-slits

Use: where,

y = D y = separation between consecutive bright or dark fringes

d λ = wavelength

D = distance between double slits and screen

d = separation between double slit

and explain the effect of changing any of the variables

Example 37:
In a Young’s double-slit experiment, the split separation is 0.20mm and a screen is 0.50m from the double slit.
A light source that emits red and violet light of wavelengths 700nm is used. Calculate the separation between
red fringes.

Solution:

y red = red D
d

= (700 10−9 )(0.5)
0.2 10−3

= 1.75 10−3 m

Exercise 37-1: Exercise 37-2:
In a Young’s double-slit experiment, the split separation The slit separation in Young Double Slit
experiment is 0.20mm with the screen distance is
is 0.45 mm and a screen is 0.450m from the double slit. A 0.40m. A light source that have separation between
violet fringes on the screen is 3.22 mm. Find the
light source that emits red and violet light of wavelengths wavelength used.

555nm is used. Calculate the separation between red   = 1.6110−6 m

fringes.  y = 5.55 10−4 m

37