SP025 Module Mutual - Perak Matriculation College

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.2 (d)(ii) Use induced emf in a coil,

Changing area, where,
B = magnetic field strength
 = −NB dA dA = change in surface Area
dt N = number of turns
dt = change in time

Example 49:

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 49-1: Exercise 49-2:
A loop of wire of area 0.7 m2 is moving into a 0.9 T
A 100-turn loop of wire is placed in a 2 T magnetic
magnetic field. If it takes 5 seconds for the loop to be
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]

49

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.2 (d)(iii) Use induced emf in a rotating coil,

 = NBA sin t where,

t = 

B = magnetic field strength
A= surface area

N = number of turns

 = angular velocity of the rotating coil

Example 50:
A square coil of side 5 cm lies perpendicular to a magnetic field of flux density 4 T. The coil consists of 200
turns of wire. The coil is rotated with angular velocity of 2 rad s-1 in 0.2 seconds. Calculate the magnitude of
the average e.m.f. induced in the coil while it is being rotated.

Solution:

 = NBA sin t
= (200)(4)(5 10−2 )2 (2) sin(2  0.2)
= 1.56 V

Exercise 50-1: Exercise 50-2:
A circular coil with 36 cm diameter lies perpendicular A rotating circular coil generates a maximum emf of
to a magnetic field of 8 T. The coil that consists of 100 500 V. Calculate the number of turns if the radius is
turns of wire is rotated with angular velocity of 5 rad 5 cm and it rotates with the frequency of 50 Hz in the
s-1 in 0.2 seconds. Calculate the of the average e.m.f. region of magnetic field of 1 T.
induced in the coil.
[202.6]
[342.65 V]

50

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

5.3 Self-inductance where,
L/O 5.3 (b)(i) Apply self-inductance,
 = induced emf
L=− 
dI  dI
 dt 
= rate of change of current

dt

Example 51:
Induced emf 0f 15 V is developed across a coil when current flowing through it changes at 30 A s-1.

Determine the self inductance of the coil.

Solution:

L=− 
dI 
 dt 

= − 15
30

= 0.5 H

Exercise 51-1: Exercise 51-2:
An inductor carries a steady current of 1.2 A. When A 2 H inductor carries a steady current of 0.5 A. When
the switch in the circuit is thrown open, the current is the switch in the circuit is thrown open, the current is
effectively zero after 10 ms. If the average induced effectively zero after 10 ms. Calculate the average
emf in the inductor is 600 V, calculate the self induced emf in the inductor during this time?
inductance.
[250 V]
[5 H]

51

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.3 (b)(i) Apply self-inductance,

Apply self-inductance, where,

L = −  = N  = induced emf
dI  I
 dt  dI

= rate of change of current

dt

N= number of turns

 =magnetic flux

I = current

I

Example 52: :

An emf of 24 mV is induced in a 500-turn coil when the current is changing at a rate of 10 As-1.Calculate thec
magnetic flux through each turn of the coil at an instant when the current is 4 A?

Solution: u

L = −  = N r
dI  I r

 dt  e
n
24 10−3 = (500) t
10 4

 = 1.92 10−5 Wb N

Exercise 52-1: Exercise 52-2: :
An emf of 30 mV is induced in a 800-turn coil when
the current is changing at a rate of 12 A s-1.Calculate A current of 4 A flows in a 200-turn coil and thne
the magnetic flux through each turn of the coil at an
instant when the current is 8 A? magnetic flux in each coil is 2.2510−4 Wbu.

[ 2.510−5 Wb ] Calculate the induced emf of the coil if the current

supply is switched off with the current reduced to zermo

in 5  s. b

[ 9103 Ve]

r

o

f

t

u

r

n

s

: magnetic flux I
:
52 c
u
r
r
e
n
t

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.3 (b)(ii) Apply self-inductance for coil,

. where,
L = self-inductance
L = 0N 2 A N = number of turns

2r 0 = permeability of free space (4π × 10–7 H m-1)

A = cross sectional area of the coil
r = radius of the coil

Example 53:
A 400-turn coil has a radius of 310−5 m and a cross sectional area of 0.8 mm2 respectively. A current of 4 A
flows in the coil. Calculate the self inductance of the coil.

Solution:

L = 0N 2 A
2r

= (4 10−7 )(400)2 (0.810−6 )
2(310−5 )

= 2.68 10−3 H

Exercise 53-1: Exercise 53-2:
A 200-turn coil has a radius of 1.510−5 m and a cross The self inductance of the coil with 200-turn is
sectional area of 0.7 mm2 respectively. A current of 6 7.18 10−3 H . If the cross sectional area of the coil
A flows in the coil. Calculate the self inductance of is 0.8 mm2 , calculate the radius of the coil.
the coil.
[ 2.8 10−6 m ]
[1.17 10−3 H ]

53

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

L/O 5.3 (b)(iii) Apply self-inductance for solenoid,.

L = 0N 2 A where

l L = self-inductance
N = number of turns
μo = permeability of free space (4π × 10–7 H m–1)
A = cross sectional area of the solenoid

l = length of the solenoid

Example 54:
A solenoid of area 2 × 10–3 m2 has 400 turns and a length of 20 cm. Calculate its inductance.

Solution:

L = 0N 2 A
l

= (4 10−7 )(400)2 (2 10−3 )
20 10−2

= 2.0110−3 H

Exercise 54-1:
A solenoid of area 4 × 10–3 m2 has 200 turns and a length of 15 cm. Calculate its inductance.

[1.34 10−3 H ]

Exercise 54-2: Exercise 54-3:
A solenoid of length 30 cm has a diameter of 4.5 cm.
If the self inductance of the solenoid is 1.5 × 10–4 H, A 200 turns solenoid of and a length of 20 cm.
calculate the number of turns of the solenoid.
Calculate its cross sectional area if the self inductance
[150] of the solenoid is 1.8 × 10–4 H.

[7.16 x 10-4 m2]

54

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

5.4 Energy stored in inductor where,
L/O 5.4 (a) Use energy stored in an inductor, L : self inductance
. I : current flowing in the inductor

U = 1 LI 2
2

Example 55:

A coil of inductance 350  H carries current of 1.2 A. Calculate the energy stored in the coil?

Solution:

U = 1 LI 2
2

= 1 (350 10−6 )(1.2)2
2

= 2.52 10−4 J

Exercise 55-1: Exercise 55-2:

A coil of inductance 200  H carries current of 1.5 A. If the energy stored in an 350  H inductor is
1.4 × 10–6 J, calculate the current that is carried by
Calculate the energy stored in the coil.
[ 2.25 10−4 J ] this inductor.

[0.09 A]

55

Ummi Atiah & Siti Farahiyah
Electromagnetic Induction

5.5 Mutual Inductance

L/O 5.5 (b) Use mutual inductance, M = 0 N1N 2 A between two coaxial solenoids.
l

M = 0 N1N2 A where,
μo = permeability of the free space
l N1 = number of turns in solenoid (coil 1)
N2 = number of turns in coil (coil 2)
l = length of coil 2

A = cross sectional area of coil 1

Example 56:
Two coaxial solenoids, P and Q have 400 and 700 turns respectively. Solenoid P has a cross sectional area of
4 × 10–3 m2 and the length of solenoid Q is 30 cm. Calculate the mutual inductance of the two solenoids.

Solution:

M = 0 N1N2 A
l

= (4  10 −7 )(400)(700)(4 10−3 )
30 10−2

= 4.69 10−3 H

Exercise 56-1: Exercise 56-2:
Two coaxial solenoids, A and B have 200 and 500 A mutual inductance of 150-turn solenoid is 350 mH
turns respectively. Solenoid A has a cross sectional when another 210-turn coil is wound around it.
area of 10 × 10–3 m2 and the length of solenoid B is Calculate the length of the solenoid if the cross
20 cm. Calculate the mutual inductance of the two sectional area of the coil is 10 × 10–3 m2.
solenoids.
[1.1310−3 m ]
[ 6.28 10−3 H ]

56

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

CHAPTER 6.0 ALTERNATING CURRENT
6.1 Alternating current
L/O 6.1 (c) : Use sinusoidal voltage = 0

= 0 where,

V = magnitude of voltage

V0 = maximum magnitude of voltage
ω = angular frequency

t = time

Example 57:
A sinusoidal voltage of peak value 3 V is applied with a frequency of 10 Hz. Write the expression for a
sinusoidal alternating voltage.

Solution:
0 = 3
= 2 (10) = 20 −1

Insert into equation = 0 ,
= 3 2 0 , where voltage in volt and time in second

Exercise 57-1: Exercise 57-2:
A sinusoidal voltage of peak value 5 V is applied A sinusoidal voltage is given by = 25 30
with a frequency of 50 Hz. Write the expression for a where voltage in volt and time in second. Calculate
sinusoidal alternating voltage. the maximum voltage and frequency.

[ = [25 V, 4.77 Hz]
where voltage in volt and time in second]

57

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.1 (c) : Use sinusoidal current = 0 ,

= 0 where,

I = magnitude of current

I0 = maximum magnitude of current
ω = angular frequency

t = time

Example 58:
A sinusoidal current of peak value 1.5 A is applied with a frequency of 4 Hz. Write the expression for a
sinusoidal alternating current.

Solution:
0 = 1.5
= 2 (4) = 8 −1

Insert into equation = 0 ,
= 1.5 8 , where current in ampere and time in second

Exercise 58-1: Exercise 58-2:
A sinusoidal current of peak value 2.4 A is applied A sinusoidal current is given by = 1.2 35
with a frequency of 16 Hz. Write the expression for a where current in ampere and time in second
sinusoidal alternating current. Calculate the maximum current and frequency.

[ = . [1.2A, 5.57 Hz]
where current is in ampere and time in second]

58

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

6.2 Root mean square (rms)

L/O 6.2 (b) : Use I rm s= I0
2

I rms= I0 where,
2 Irms = root mean square current
I0 = peak current

Example 59: The alternating current is given by = 0.5 10 where current in ampere and time in
second. Calculate the rms current.

Solution:

= 0 = 0.5 = 0.35
√2 √2

Exercise 59-1: Exercise 59-2:
The alternating current is given by = 2.1 15.5 A sinusoidal current has rms current of 2.5 A.
where current in ampere and time in second. Calculate the maximum current flows.
Calculate the rms current.
[3.54 A]
[1.48 A]

59

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.2 (b) : Use V rm s= V0
2

V rm s= V0 where,
2 Vrms = root mean square voltage
V0 = peak voltage

Example 60: A sinusoidal alternating voltage is given by = 5 10 where current in voltage and time in
second. Calculate the rms voltage.

Solution:

= 0 = 5.0 = 3.54
√2 √2

Exercise 60-1: Exercise 60-2:
A sinusoidal alternating voltage is given by A sinusoidal voltage has rms voltage of 12 V.
= 3 20 where voltage in volt and time in Calculate maximum voltage flows.
second. Calculate the rms voltage.
[16.97 V]
[2.12 V]

60

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

6.3 Resistance, reactance and impedance
L/O 6.3 (a) : Use phasor diagram and sinusoidal waveform to show the phase relationship between current
and voltage for a single component circuit of :

i. resistor, R

Example 61:
An AC current flows through a resistor of resistance 100 Ω. The variation of the current with time and
voltage with time are shown in the graph of FIGURE 6.1. Sketch a phasor diagram for the pure resistor
circuit.

I (A)

0.2 t (s)
0
12 34
-0.2

FIGURE 6.1(a)

V(V) t (s)

20 2 34
FIGURE 6.1(b)
01
-20

Solution:
Phase difference between current and voltage is zero.

IV

Exercise 61-1: Exercise 61-2:

Figure 6.2 IV
Figure 6.2 shows variation for voltage and current
with time. Sketch a phasor diagram for the resistor Figure 6.3
circuit.
Figure 6.3 shows the phasor diagram of parallel
voltage and current in pure circuit. An alternating
current with frequency of 60 Hz is applied on the
circuit. Sketch a sinusoidal graph of current with
time.

61

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (a) : Use phasor diagram and sinusoidal waveform to show the phase relationship between current
and voltage for a single component circuit of :

ii. capacitor, C

Example 62:

Figure 6.4

Figure 6.4 shows variation for voltage and current with time. Sketch a phasor diagram for the capacitor

circuit. The phase difference between the current and the voltage is rad.
2

Solution: I
Voltage lags current by rad.

2

V

Exercise 62-1: Exercise 62-2:

Figure 6.5 Figure 6.6

Figure 6.5 shows a graph of voltage against time of The graph of current against time of a pure

a pure capacitor connected to a sinusoidal AC capacitor connected to a sinusoidal AC supply is

supply. Sketch a phasor diagram that has a phase shown in Figure 6.6. Sketch a phasor diagram that
difference of rad between the current and the the current leads the voltage by rad.

2 2

voltage.

62

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (a) : Use phasor diagram and sinusoidal waveform to show the phase relationship between current
and voltage for a single component circuit of :

iii. inductor, L

Example 63: )

Given a sinusoidal voltage across a pure inductor is = 0sin ( + 2 and the alternating current is

= 0 sin . Sketch a phasor diagram of the voltage and current.

Solution:
Voltage leads current by rad.

2

Exercise 63-1: Exercise 63-2:
The voltage leads the current by rad in pure
Given a sinusoidal voltage across a pure inductor is
= 0cos and the alternating current is 2
= 0 sin . Sketch a phasor diagram of the
voltage and current. inductor circuit. The maximum voltage is 6.5 V and

the maximum current is 0.2 A. The frequency of the

AC is 50 Hz. Write down the equation of voltage and

current for the inductor circuit.
[ = . (  + ) ,



= .  ]

63

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (b) : Use phasor diagram to analyse voltage, current, and impedance of series circuit of RL

Example 64:

In a RL circuit, the rms voltage and rms current are 240 V and 1.5 A respectively. The current lags the
voltage by 600. Draw a phasor diagram for the RL circuit.

Solution: V

600
I

Exercise 64-1: Exercise 64-2: I
Sketch a phasor diagram of resistance and inductive V 710

reactance for the RL circuit when the voltage leads the Figure 6.7
current by 550.

Figure 6.7 shows a phasor diagram of voltage and
current in RL circuit. Determine the phase
difference of the voltage and current in RL circuit.

64

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (b) : Use phasor diagram to analyse voltage, current, and impedance of series circuit of RC

Example 65:

An alternating voltage,V is applied to a RC circuit with the rms voltage and rms current are 250 V and 2.2 A
respectively. The current leads the voltage by 760. Draw a phasor diagram for the RC circuit.

Solution: I
760

V

Exercise 65-1: Exercise 65-2:
Sketch a phasor diagram of resistance and capacitive
I
reactance for the RC circuit when the voltage lags the
current by 350. V
Figure 6.8

Figure 6.8 shows a phasor diagram of voltage and
current in RC circuit. Determine the phase
difference of the voltage and current in RC circuit.

65

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (b) : Use phasor diagram to analyse voltage, current, and impedance of series circuit of RLC

Example 66:
A RLC circuit has an impedance of a phase angle, = 54.50. Draw a phasor diagram of the RLC circuit.

Solution: XL - XC Z

54.50
R

Exercise 66-1: Exercise 66-2:
If in a circuit consists of bulb, inductor and capacitor, If in a phasor diagram of RLC circuit, the voltage lags
then the phase angle is 600 with the resistance. the current by 450. Sketch a phasor diagram of the
Sketch a phasor diagram of the RLC circuit. RLC circuit.

66

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (c) : Use
i. capacitive reactance,

1 where,
= 2 XC = capacitive reactance
f = frequency
C = capacitance

Example 67:
An alternating voltage V with frequency of 120 Hz is applied on a capacitor of 2.0 µF. Calculate the
capacitive reactance, XC.

Solution:
1

= 2
1

= 2 (120)(2 10−6)
= 663 

Exercise 67-1: Exercise 67-2:
The reactance of a capacitor is given by XC = 150 
A capacitor of 1.5 µF is applied to an alternating in an AC circuit which consists of 50 Hz frequency.
Calculate the capacitance of the capacitor.
current of frequency 100 Hz. Calculate the reactance
[21.22 µF]
of capacitor, XC.

[1061 Ω]

67

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (c) : Use where,
ii. inductive reactance, XL= inductive reactance
f = frequency
= 2 L = inductance

Example 68:
An alternating voltage V with frequency of 130 Hz is applied on an inductor of 50 mH. Calculate the
inductive reactance, XL.

Solution:
= 2

= 2 (130)(50 10−3)
= 40.84 

Exercise 68-1: Exercise 68-2:
In a pure inductor circuit, it has an alternating voltage Given the inductive reactance is 55  with 180 Hz
V with frequency of 160 Hz. It is applied on an frequency applied in an inductor circuit. Calculate the
inductor of 50 mH. Calculate the inductive reactance, inductance.
XL.
[48.63 mH]
[50.27 ]

68

L/O 6.3 (c) : Use Nur ‘Ainiza & Mohd Khairul Azmi
iii. impedance, Alternating Current

= √ 2 + ( − )2 where,
Z = impedance
R = resistance
XL = inductive reactance
XC = capacitive reactance

Example 69:
A 200  resistor, a 100  inductive reactance and a capacitive reactance of 22  are connected in series to
an alternating source 250 V. Calculate the impedance of the circuit.

Solution:
= √ 2 + ( − )2

= √2002 + (100 − 22)2
= 214.67

Exercise 69-1: Exercise 69-2:
A RLC circuit has a resistance, R = 330 , capacitive Given impedance of RLC circuit is 44.5 . Calculate
reactance, XC = 70  and inductive reactance, XL = 55 the resistance if the difference between inductive and
. Calculate the impedance of the circuit. capacitive reactance is 32 .

[330.34 ] [30.92 ]

69

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.3 (c) : Use
iv. phase angle,

= −1 ( − ) where,
= phase angle
R = resistance
XL= inductive reactance
XC= capacitive reactance

Example 70:
In an AC circuit, the resistor, inductor and capacitor are connected in series. The resistance is 200 ,
capacitive reactance is 100  and inductive reactance is 150 . Calculate the phase angle.

Solution:

= −1 ( − )

(150 − 100)
= −1 200

= 14.040

Exercise 70-1: Exercise 70-2:
In an AC circuit, the resistor, inductor and capacitor In an RLC series circuit with R = 60 Ω and L=30 mH
are connected in series. The resistance is 22 ,
capacitive reactance is 12  and inductive reactance is driven by an ac source whose frequency and voltage
is 45 . Calculate the phase angle.
amplitude are 500 Hz and 50 V respectively. Given the
[56.310]
phase angle, =27.00 , calculate the capacitance of the

circuit.

[5 μF]

70

6.4 Power and power factor Nur ‘Ainiza & Mohd Khairul Azmi
L/O 6.4 (a) : Apply Alternating Current

i. average power, where,
= average power
= = rms current
= rms voltage
ϕ = phase angle

Example 71:
The rms voltage of an AC source is given by 55.5V and the phase angle is 22.00. Calculate the average power
output of the source if the rms current is 1.02 A.

Solution:
=

= 1.02(55.5) 22
= 52.49

Exercise 71-1: Exercise 71-2:
In an AC circuit, the rms voltage is given by 70.0V In an RLC series circuit, the average power is 44.0 W
and the phase angle is 36.00. Calculate the average and the rms current is 1.22 A. Given that the phase
power output of the source if the rms current is angle, =56.70 . Calculate the rms voltage of the
0.77 A. circuit.

[43.61 W] [65.69 V]

71

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.4 (a) : Apply
ii. instantaneous power,

= where,
P = instantaneous power
= current
= voltage

Example 72:
In a pure resistor circuit the voltage is given by = 115 sin 20 where voltage in volt and time in second.
Calculate the instantaneous power if the current is I = 5.5 sin 20 where current in ampere and time in
second.

Solution:
=

= ( 0 sin  )( 0 sin  )
= 0 0 2
= 5.5(115) 220
= 632.5 220
where power in watt and time in second.

Exercise 72-1: Exercise 72-2:
Given an instantaneous power in resistance circuit is
In a pure resistor circuit, the voltage is given by 555 W. The instantaneous current is 311 mA.
= 100 sin 50 where voltage in volt and time in Calculate the instantaneous voltage of the circuit.

second. Calculate the instantaneous power if the [1784.57 V]
current is I= 3.3 sin 50 where current in ampere

and time in second.
[ = where power in watt and

time in second.]

72

Nur ‘Ainiza & Mohd Khairul Azmi
Alternating Current

L/O 6.4 (a) : Apply
iii. power factor,

 = = where,

 = power factor

= real power
= apparent power
= average power
= rms current
= rms voltage

Example 73:
An RLC series circuit has 1.2 kW of average power. The rms current and rms voltage given by 12 A and
240V respectively. Calculate the power factor of the circuit.

Solution:

 =

1.2 × 103
= 12(240)

= 0.417

Exercise 73-1: Exercise 73-2:
An RLC series circuit has 15 W of average power. The power factor of the circuit in an RLC series circuit
The rms current and rms voltage given by 1.1 A is 0.152. The real power is given by 25 W. Calculate the
and 150 V respectively. Calculate the power factor apparent power.
of the circuit.
[164.47 W]
[0.091]

73

Nageswary & Noorsuraya
Geometrical Optics

CHAPTER 7.0: GEOMETRICAL OPTICS
7.1 Reflection at Spherical Surface
L/O 7.1 c) Use mirror equation,

1 =1+1 *Sign convention for focal length, f:
f uv (i) Positive f for concave mirror
(ii) Negative f for convex mirror

for real object only.
where,
f = focal length of mirror
u = object distance from the pole
v = image distance from the pole

Example 74:
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 74-1: Exercise 74-2:
A mirror forms a real image of a light bulb on a wall Olivia is inspecting the concave primary mirror of
of a bathroom that is 3.5 m from the mirror. If the the Hubble telescope before its launched into space.
bulb is placed 7.0 cm from the mirror, Calculate the She stands 87.0 m in front of the mirror which has a
focal length of the mirror. focal length of 58.0 m. Calculate Olivia’s image
distance.
[6.86 cm]
[174 m]

74

Nageswary & Noorsuraya
Geometrical Optics

7.1 Reflection at Spherical Surface 1 =1+1 *Sign convention for focal length, f:
f uv (i) Positive f for concave mirror
L/O 7.1 c) Use mirror equation, (ii) Negative f for convex mirror

for real object only.
where,
f = focal length of mirror
u = object distance from the pole
v = image distance from the pole

Example 75:
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 75-1: Exercise 75-2:
A Barbie doll is placed 6.0 cm in front of a convex An external side mirror of a car is convex with a
mirror with a focal length of 12.0 cm. Calculate the radius of curvature 18 m. Calculate the location of
position of the image formed. the image for an object 10 m from the mirror.

[ - 4.0 cm] [- 4.74 m]

75

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Geometrical Optics

7.1 Reflection at Spherical Surface where,
L/O 7.1 d) Use magnification m = magnification
u = object distance from the pole
m = hi = − v v = image distance from the pole
ho u hi = height of image
ho = height of object

Example 76:
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: Exercise 76-2:
A 4.0 cm tall light bulb is placed at a distance of 35.5
m=−v cm from a convex mirror. If the height of the image
u formed is 1.38 cm, Calculate the image distance.

= − (−9.25) [- 12.25 cm]
1.20

= 7.71

Exercise 76-1:
A dentist uses a small mirror attached to a thin rod to
examine one of your teeth. When the tooth is 24.0
mm in front of the mirror, the image it forms is 9.0
mm behind the mirror. Calculate the magnification of
the image.

[0.375]

76

7.2 Refraction at a spherical surface Nageswary & Noorsuraya
L/O 7.2 a) Use Geometrical Optics

n1 + n2 = n2 − n1 *Sign convention for radius of curvature, R:
uv R (i) Positive R for convex surface
(ii) Negative R for concave surface

for spherical surface.
where,
u = object distance from the pole
v = image distance from the pole
n1 = refracting index of medium 1 (incident ray)
n2 = refracting index of medium 2 (refracted ray)

Example 77:

Figure below shows an object O placed at a distance 20.0 cm from the surface P of a glass sphere of radius

5.0 cm and refractive index of 1.63. Calculate the position of the image formed by the surface P of the glass

sphere.

Solution:

na + ng = ng − na a P
uv R
O 5.0 cm
1.00 + 1.63 = 1.63 −1.00
20.0 v + 5.0 20.0 cm

v = +21.5 cm (The image is 21.5 cm at the back of the first surface P or opposite side of the object)

Exercise 77-1: Exercise 77-2:
Figure below shows an object O placed at a distance An object is placed 20 cm from the surface of the
25.0 cm from the surface P of a glass sphere of glass sphere of radius 10 cm and refractive index of
radius 6.0 cm and refractive index of 1.55. Calculate 1.55. An image is formed due to refraction at the
the position of the image formed by the surface P of surface of the sphere. Calculate the distance of this
the glass sphere. image from the center of the sphere.

[30 cm] [320 cm from the center of sphere]

a P

O

77

Nageswary & Noorsuraya
Geometrical Optics

CHAPTER 7.0: GEOMETRICAL OPTICS

7.2 Refraction at a spherical surface

L/O 7.2 a) Use

n1 + n2 = n2 − n1 Sign convention for radius of curvature, R:
uv R (i) Positive R for convex surface
(ii) Negative R for concave surface

for spherical surface.
where,
u = object distance from the pole
v = image distance from the pole
n1 = refracting index of medium 1 (incident ray)
n2 = refracting index of medium 2 (refracted ray)

Example 78:

A glass sphere with radius 10 cm has refractive index of 1.52. An object is placed inside the sphere. If the

distance of the object is 5 cm from the inner surface and refractive index for air is 1.00. Calculate the image

distance.

Solution:

n1 + n2 = n2 − n1
uv R

1.52 + 1.00 = 1.00 −1.52
5v −10

v = −3.97 cm

Exercise 78-1: Exercise 78-2:
A glass sphere with radius 20 cm has refractive index A coin is embedded in a solid glass sphere of radius
of 1.55. An object is placed inside the sphere. If the 30 cm as shown in FIGURE 1. The refractive index
distance of the object is 7 cm from the inner surface of the glass sphere is 1.5 and the coin is 20 cm from
and refractive index for air is 1.00. Calculate the the surface. Calculate the image distance.
image distance.
[- 17.14 cm]
[- 5.16 cm]

r = 30 cm Solid
20 sphere
cm

FIGURE 1

78

Nageswary & Noorsuraya
Geometrical Optics

7.3 Thin lenses 1 =1+1 *Sign convention for focal length, f:
L/O 7.3 b) Use thin lens equation, f uv (i) Positive f for convex lens
(ii) Negative f for concave lens
for real object only.
where,
f = focal length of lens
u = object distance from the pole
v = image distance from the pole

Example 79:
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 79-1: Exercise 79-2:
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]

79

Nageswary & Noorsuraya
Geometrical Optics

7.3 Thin lenses 1 =1+1 *Sign convention for focal length, f:
L/O 7.3 b) Use thin lens equation, f uv (i) Positive f for convex lens
(ii) Negative f for concave lens
for real object only.
where,
f = focal length of lens
u = object distance from the pole
v = image distance from the pole

Example 80:
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 80-1: Exercise 80-2:
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 with focal length 10 cm. Calculate the image
from the lens. Calculate focal length of the lens. distance.

[- 6.67 cm] [-7.5 cm]

80

Nageswary & Noorsuraya
Geometrical Optics

7.3 Thin lenses In air, n medium = 1
L/O 7.3 d) Use lens maker’s equation,
R is positive if the surface encountered by
1 =  nm aterial − 1 1 − 1  the light beam is convex
f nm edium R1 R2 R is negative if the surface encountered by
the light beam is concave

where,
nmaterial – index of refraction of the lens material.

n medium --index of refraction of the surrounding medium.
R1 – radius of curvature of the 1st surface.
R2 – radius of curvature of the 2nd surface.

Example 81:
A meniscus convex lens of refractive index 1.50 with radii of curvature of 20 cm and 50 cm. The lens is then
immersed in water with refractive index of 1.33. Calculate the focal length of the lens.
Solution:

1 = ( n2 − 1) 1 − 1 
f n1 r1 r2

1 = ( 1.5 −1) 1 − 1 
f 1.33  20 50 

f = 260.8 cm

Exercise 81-1: Exercise 81-2:
A meniscus convex lens of refractive index 1.50 with A plano-convex lens has refractive index 1.66. The
radii of curvature of 30 cm and 45 cm. The lens is radius of curvature of the convex surface is 5.28 cm.
then immersed in oil with refractive index of 1.6. Calculate the focal length of the lens.
Calculate the focal length of the lens.
[8.00 cm]
[- 1440 cm]

81

Nageswary & Noorsuraya
Geometrical Optics

7.3 Thin lenses where,
L/O 7.3 e) Use magnification m = magnification
u = object distance from the pole
m = hi = − v v = image distance from the pole
ho u hi = height of image
ho = height of object

Example 82:
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 82-1: Exercise 82-2:
A 4.0 cm Marvel figurine is placed at a distance of
An object of height 10 cm is placed at 30 cm from a 40 cm from a convex lens. If the height of the image
formed is 3.0 cm, Calculate the image distance.
converging lens. The real image formed 20 cm from
the lens. Calculate the magnification of the image. [- 30 cm]

[m= - ]



82

Nik Norhasrina & Salmi
Physical Optics

CHAPTER 8.0 PHYSICAL OPTICS
8.3 Interference Of Transmitted Light Through Double-Slits
L/O 8.3 : Interference of transmitted light through double-slits

a) Use: ym = mD where,
i)
d ym = separation between central bright and mth bright fringes

for bright fringes (maxima) m = order
λ = wavelength
D = distance between double slits and screen
d = separation between double slit

Example 83:
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 83-1: Exercise 83-2:
In a Young’s double-slits experiment, distance In Double slit experiment, the distance between the
centre of the fringes to the fringes is 5 Mm. The
between the screen and the double-slits is 5 m. separation between the slits is 0.2 mm. If the distance
from the slits to the fringes is a meter away and a
Distance between the two slits is 0.15 mm. If wavelength is 500 nm, what is the number of order of
the bright fringes.
distance between the central bright fringe and the
[m= 2]
seventh bright fringe is 10 cm. Calculate the

wavelength of the light.

[λ: 429 nm]

83

Nik Norhasrina & Salmi
Physical Optics

8.3 Interference Of Transmitted Light Through Double-Slits
L/O 8.3 : Interference of transmitted light through double-slits

a) Use: ym = (m + 1 )D
ii) 2

d for dark fringes (minima), where m =0, +1, +2, +3, …

where,

ym = separation between central bright and mth bright fringes

m = order
λ = wavelength
D = distance between double slits and screen
d = separation between double slit

Example 84:
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 84-1: Exercise 84-2:
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 are 0.500 mm and 0.2 mm in a Young Double slits
the distance between the central bright and the fifth experiment. However, the interference pattern on a
dark fringe is 15.3 mm, calculate the wavelength of screen is 3.30 m away. The distance from the center to
the light. 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 ]

[m=3]

84

Nik Norhasrina & Salmi
Physical Optics

8.3 Interference Of Transmitted Light Through Double-Slits
L/O 8.3 (b) Interference of transmitted light through double-slits

Use: and explain the effect of changing any of the variables
y = D

d

where,

y = separation between consecutive bright or dark fringes

λ = wavelength
D = distance between double slits and screen
d = separation between double slit

Example 85:

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 85-1: Exercise 85-2:

In a Young’s double-slit experiment, the split The slit separation in Young Double Slit experiment
separation is 0.45 mm and a screen is 0.450m from is 0.20mm with the screen distance is 0.40m. A light
the double slit. A light source that emits red and source that have separation between violet fringes on
violet light of wavelengths 555nm is used. the screen is 3.22 mm. Find the wavelength used.
Calculate the separation between red fringes.
  = 1.6110−6 m
 y = 5.55 10−4 m

85

Nik Norhasrina & Salmi
Physical Optics

8.4 Interference Of Reflected Light In Thin Films
L/O 8.4 (c) : Interference of reflected light in thin films

c) Use the following equations for reflected light with no phase difference (non-reflective coating):

(i) Constructive interference: 2nt = m where m = ±1, ±2, ±3, …

where,

n = refractive index

m = order
λ= wavelength

t = thickness of the thin film

Example 86:

The magnesium fluoride of refractive index 1.38 covers the camera lens of refractive index 1.52. The coating
reflects a blue light of wavelength 400 nm. Determine the minimum non zero thickness of the magnesium
fluoride.

Solution:
In phase - coating prevents reflections (D.I)

2nt = m

2(1.38)tmin = (1)(400 10−9 )

tmin = 1.02 10 −8 m

Exercise 86-1: Exercise 86-2:
The magnesium fluoride of refractive index 1.52 The surface of a lens is coated with a film of magnesium
covers the camera lens of refractive index 2.43. floride which has a refractive index of 1.38. The
The coating reflects a yellow-green light of refractive index of the material of the lens is 1.67.
wavelength 600 nm. Determine the minimum non Calculate the minimum thickness of the magnesium
zero thickness of the magnesium fluoride. floride film that would result in constructive interference
for reflected light of wavelength 450nm.
 tmin = 1.05 10−7 m
[t=1.63x10-7 m]

86

Nik Norhasrina & Salmi
Physical Optics

8.4 Interference Of Reflected Light In Thin Films
L/O 8.4 (d) : Interference of reflected light in thin films

c) Use the following equations for reflected light with no phase difference (non reflective coating):

(ii) Destructive interference : 2nt = (m + ½ )  where m = 0, ±1, ±2, ±3, …

where,

n = refractive index

m = order
λ= wavelength

t = thickness of the thin film

Example 87:

A non-reflective coating of magnesium fluoride of refractive index 1.38 covers the camera lens of refractive
index 1.52. The coating enhance reflection of yellow-green light of wavelength in vacuum 565 nm.
Determine the minimum non zero thickness of the magnesium fluoride.

Solution:
In phase - coating prevents reflections (D.I)

2ntmin =  m + 1 
 2 

2(1.38)tmin =  0 + 1 (565 10−9 )
 2 

tmin = 1.02 10−8 m

Exercise 87-1: Exercise 87-2:
A non-reflective coating of magnesium fluoride of An oil film floats on water. The refractive
refractive index 1.52 covers the camera lens of refractive indices of the oil and water are 1.30 and 1.33
index 2.43. The coating enhance reflection of yellow- respectively. If the oil film has uniform thickness
green light of wavelength in vacuum 638 nm. Determine of 200 nm, calculated the maximum wavelength
the minimum non zero thickness of the magnesium of reflected light for destructive interference.
fluoride.
[λ=1.04x10-6 m]
 tmin = 1.05 10−7 m

87

Nik Norhasrina & Salmi
Physical Optics

8.4 Interference Of Reflected Light In Thin Films
L/O 8.4 d : Interference of reflected light in thin films

d) Use the following equations for reflected light of phase difference (reflective coating):

(i) Constructive interference: 2nt = (m + ½)  where m = 0, ±1, ±2, ±3, …

where,

n = refractive index

m = order
λ= wavelength

t = thickness of the thin film

Example 88:
One face of the glass prism used in a video camera coated with a film of titanium dioxide of refractive index,
n=1.4. This is to ensure that red light of wavelength 550 nm which is at normal incident is completely reflected
from that face. The refractive index of the prism is 1.32. Calculate the minimum thickness of the film.
Solution:
Anti-phase completely reflected (C.I)

2nt =  m + 1 
 2

( )2(1.4)tmin 0 1 
=  + 2  550 10−9

= 9.82 10−8 m

Exercise 88-1: Exercise 88-2:
One face of the glass prism used in a video camera A thin film of oil (n = 1.25) is located on a smooth
coated with a film of titanium dioxide of refractive wet pavement. When viewed perpendicular to the
index, n=2.4. This is to ensure that red light of pavement, the film reflects most strongly red light at
wavelength 680 nm which is at normal incident is 640 nm. Calculate the minimum thickness of the oil.
completely reflected from that face. The refractive
index of the prism is 1.50. Calculate the minimum [ t = 128 nm ]
thickness of the film.

[ t = 71nm]

88

Nik Norhasrina & Salmi
Physical Optics

8.4 Interference Of Reflected Light In Thin Films
L/O 8.4 d : Interference of reflected light in thin films

d) Use the following equations for reflected light of phase difference (reflective coating):

(ii) Destructive interference: 2nt = m where m = ±1, ±2, ±3, …

where,
n = refractive index
m = order
λ= wavelength
t = thickness of the thin film

Example 89:
The wall of soup bubble have about the same index of refraction as that of water n=1.33. Calculated the
wavelength of the light source when no light is found to be reflected from a point on the soap bubble where
its wall is 260 nm thick.

2nt = m

 = 2nt
m

( )= 2(1.33) 260 10−7
1
= 6.92 10−7 m

Exercise 89-1: Exercise 89-2:
The wall of soup bubble have about the same index A glass lens of refractive index 1.53 is coated with a
of refraction as that of water n=1.33. Calculated the thin film or refractive index 1.43. No ligth is found to
wavelength of the light source when no light is found be reflected when the wavelength of the ligth is 600
to be reflected from a point on the soap bubble where nm is used. Calculated the thickness of the film.
its wall is 150 nm thick.
[ t = 196nm ]
[  = 399nm]

89

Nik Norhasrina & Salmi
Physical Optics

8.5 Diffraction By A Single Slit

L/O 8.5 (c) :

Use:

i) yn = nD for dark fringes (minima), where n = ±1, ±2, ±3, ...
a

where,

yn = separation between central bright and nth dark fringes

n = order
λ = wavelength
D = distance between double slits and screen
a = width of the slit

Example 90:

A slit is illuminated with light of wavelength 600 nm. The first minimum for the interference pattern appears
at 8o. Determine the distance from central maximum to the third minimum when different light of wavelength

650 nm is used. The distance from slit to the screen is 1.0 m.

Solution:

a sin m = m yn = nD
a
a = m y3
sin  y3 = (3)(650 10−9 )(1.0)
4.31  10 −6
a = (1)(600 10−9 )
= 0.452m
sin 8

a = 4.3110−6 m

Exercise 90-1: Exercise 90-2:
A slit is illuminated with light of wavelength 445 A screen is located 80 cm from single slit which is
nm. The first minimum for the interference pattern illuminated with a beam of light of wavelength 600
appears at 10o. Determine the distance from central nm. The distance between the first minima and second
maximum to the fifth minimum when different light minima measured from the diffraction pattern is 2.8
of wavelength 750 nm is used. The distance from slit cm. What is the width of the slit?
to the screen is 1.5 m.
[ a = 1.7110−5 m ]

[ a = 2.197m ]

90

Nik Norhasrina & Salmi
Physical Optics

8.5 Diffraction By A Single Slit

L/O 8.5 (c) :
Use:

 n + 1 D
ii) yn =  2
a for bright fringes (maxima), where n = ±1, ±2, ±3, ...

where,

yn = separation between central bright and nth dark fringes

n = order
λ = wavelength
D = distance between double slits and screen
a = width of the slit

Example 91:

A beam of monochromatic light is incident normally on a single slit of width
0.3 mm. The distance between the screen to the slit is 100 cm and the wavelength of the light used is 550
nm. Calculate the distance of forth bright fringe?

 n + 1 D
yn =  2
a

 4 + 1   550 10−9 100 10−2
yn =  2  0.3 10−3

yn = 8.25 10−3 m

Exercise 91-1: Exercise 91-2:
A beam of monochromatic light is incident In single slit experiment, the distance between the
normally on a single slit of width 0.4 mm. The screen and the slits is 2.0 m and the wavelength of the
distance between the screen to the slit is 120 cm light is 5.0 x 10-7 m. If the distance between the center
and the wavelength of the light used is 670 nm. of the interference pattern and the 10th bright fringe is
Calculate the distance of fifth bright fringe? 3.4 cm. What is the width of the single slit?

[ yn = 0.011m ] [ a = 3.09 10−4 m ]

91

Nik Norhasrina & Salmi
Physical Optics

8.6 Diffraction Grating

L/O 8.6 (b) Apply , d sin = n where d = 1 where n = ±1, ±2, ±3, ...
N

where,
d = slit separation

n = order
λ = wavelength

N = number of lines per unit length

Example 92:

Monochromatic light from laser is (  = 632.8nm) is incident normally on a diffraction grating containing

6000 lines per centimeter. Find the angles at which the first order and second order are observed.

Solution:

Given:  = 632.8 10−9 m

Diffraction grating 6000 lines per cm

Grating spacing,

d = L = 110 −2
N 6000

= 1.67 10 −6 m

M

For the first order maximum, n=1

d sin 1 = n

sin 1 = (1)(632.8  10 −9 ) = 0.3789
1.67  10 −6

1 = 22.27

For second order maximum, n=2

d sin  2 = n

sin  2 = (2)(632.8 10 −9 ) = 0.7578
1.67 10 −6

 2 = 49.27

Exercise 92-1: Exercise 92-2:

Monochromatic light from laser is (  = 550.3nm A diffraction grating consists of 1.26 104 uniformly

) is incident normally on a diffraction grating distributed parallel lines in an area of width 25.4 mm.
containing 4724 lines per centimeter. Find the angles The grating is illuminated normally with yellow light
at which the first order and second order are from a sodium vapor lamp. This light consists of two
observed. wavelengths which are almost the same, 589 nm and
589.59 nm. What is the angular separation between the
1 = 15.04  1st order maxima for the two wavelengths?
 2 = 31.28
 = 0.017

92

Abdul Mansor & Mimi Erlieza
Quantization of Light

CHAPTER 9.0 – QUANTIZATION OF LIGHT
9.1 Planck’s Quantum Theory
L/O 9.1 (c) : Use Einstein’s equation for a photon energy

E = hf = hc where,

 E = photon energy (J)

f = frequency of light (Hz)
h = Plank’s constant = 6.63 X 10-34 Js
c = speed of light (ms-1)
λ = wavelength of light (m)

Example 93:
Calculate the energy of the photon from an electromagnetic wave of frequency 2.65 X 10 14 Hz.

Solution:

E = hf = 6.63 10−34  2.65 1014 = 1.76 10−19 J

Exercise 93-1: Exercise 93-2:
Calculate the energy of the photon from an
If the frequency of the electromagnetic wave is 1.4 x
1014 Hz. Calculate the energy of the photon. electromagnetic wave of wavelength 700 nm
[ . × − ]
[ . × − ]

93

Abdul Mansor & Mimi Erlieza
Quantization of Light

9.2 Photoelectric Effect where,
L/O 9.2 (f) : Use Einstein’s photoelectric equation
Kmax = maximum kinetic Energy (J)
K max = eV = hf − Wo e = Charge of electron = 1.60 X 10-19 C

V = stopping potential (V)

f = frequency of light (Hz)
h = Plank’s constant = 6.63 X 10-34 J s

Wo = Work Function (J)

Example 94:
If the stopping potential in a photoelectric effect experiment is 2.65 V, calculate the maximum kinetic energy
of the electron.

Solution

Kmax = eV = 1.60 10−19  2.65 = 4.24 10−19 J

Exercise 94-1: Exercise 94-2:
While conducting a photoelectric effect An electromagnet waves of frequency of 6.0 x 1014 Hz
experiment with a light of a certain frequency. It is strike a metal with a work function of 0.5 eV. Calculate
found that reverse potential difference of 1.25 volt
is required to reduce the current to zero. Calculate the maximum kinetic energy of the photoelectron emitted
the maximum kinetic energy.
from metal.
[2.0 X 10-19J] [3.18 X 10-19J]

94

Abdul Mansor & Mimi Erlieza
Wave Properties of Particle

CHAPTER 10.0 – WAVE PROPERTIES OF PARTICLE

10.1 de Broglie Wavelength

L/O 10.1 (b) : Use de Broglie wavelength

=h = h where,
λ = de Broglie wavelength (m)
P mv h = Plank’s constant = 6.63 X 10-34 J s
P = momentum (kg m s-1)

m = mass

v = velocity

Example 95:

A photon has momentum of magnitude 8.24 x 10-28 kg m s-1. Calculate the de Broglie wavelength of this

photon.

 = h = 6.63 10−34 = 8.05 10−7 m
P 8.24 10−28

Exercise 95-1: Exercise 95-2:
A proton has momentum of magnitude at 5.04 x10-22 Calculate de Broglie wavelength of an electron
kg m s-1. Calculate de Broglie wavelength of the which is moving at 2.0 x 107 m s-1

proton. [3.64 x 10-11 m]
[1.32 x 10-12 m]

95

Abdul Mansor & Mimi Erlieza
Nuclear and Particle Physics

CHAPTER 11.0 NUCLEAR AND PARTICLE PHYSICS
11.1 Binding energy and mass defect
L/O 11.1 (a): Use mass defect

m = (Zmp + Nmn ) − mnucleus = (ZmH + Nmn ) − matom

where,
mp= mass of proton
mn= mass of neutron
mH= mass of hidrogen
mnucleus= mass of nuclues
Z= the number of proton in a nuclues
N= the number of neutron in a nuclues

Example 96:

A mass of nucleus 3851Br is 80.8971 u. Calculate the mass defect of the bromine nucleus.

[Given mass of proton = 1.0073 u, mass of neutron = 1.0087 u]

m = (Zmp + Nmn ) − mnucleus= (35 X 1.0073 + 46 X 1.0087) – 80.8971 = 0.7586 u

Exercise 96-1: Exercise 96-2:

The nuclide of 60 Ni has a mass 59.9154 u. Calculate the mass defect for 6 Li .
28 3

Calculate the mass defect of the nuclide. [mass of nucleus 6 Li = 6.0151, mass of 1 H =
3 1
[Given mass of proton = 1.0073 u, mass of neutron =

1.0087 u] 1.0078 u and mass of neutron = 1.0087 u]

[0.5674 u] [0.0344 u]

96

Abdul Mansor & Mimi Erlieza
Nuclear and Particle Physics

L/O 11.1 (b): Use binding energy where,

EB = mc2 m = mass defect

c = speed of light = 3.0 x 108 m s-1

Example 97:

A mass of nucleus 3851Br is 80.8971 u. Calculate binding energy of the bromine nucleus.

[Given mass of proton = 1.0073 u, mass of neutron = 1.0087 u]

m = (Zmp + Nmn ) − mnucleus= (35 x 1.0073 + 46 x 1.0087) – 80.8971 = 0.7586 u

( )( )EB = mc2 = 0.7586u 1.66 10−27 3.0 108 2 = 1.133 10−10 J

Exercise 97-1: Exercise 97-2:

The nuclide of 60 Ni has a mass 59.9154 u. Calculate binding energy for 6 Li .
28 3

Calculate binding energy of the nuclide. [mass of atom 6 Li = 6.0151, mass of 1 H = 1.0078
3 1
[Given mass of proton = 1.0073 u, mass of neutron =

1.0087 u] u and mass of neutron = 1.0087 u]
[ . × − ]
[ . × − ]

97

Nageswary & Mimi Erlieza
Nuclear and Particle Physics

L/O 11.1 (c): Determine binding energy per nucleon

EB where,
A EB = binding energy
A = nucleon number

Example 98:

A mass of nucleus 3851Br is 80.8971 u. Calculate the mass defect of the bromine nucleus.

[Given mass of proton = 1.0073 u, mass of neutron = 1.0087 u]

m = (Zmp + Nmn ) − mnucleus= (35 X 1.0073 + 46 X 1.0087) – 80.8971 = 0.7586 u

( )( )EB = mc2 = 0.7586u 1.66 10−27 3.0 108 2 = 1.133 10−10 J

EB = 1.133 10−10 = 1.40 10−12 J / nucleon
A 81

Exercise 98-1: Exercise 98-2:

The nuclide of 60 Ni has a mass 59.9154 u. Calculate binding energy per nucleon for 6 Li .
28 3

Calculate binding energy per nucleon of the nuclide. [mass of atom 6 Li = 6.0151, mass of 1 H = 1.0078
[Given mass of proton = 1.0073 u, mass of neutron = 3 1

1.0087 u] u and mass of neutron = 1.0087 u]

[ . × − / ] [ . × − / ]

98