CHAPTER 7_ PHYSICAL OPTICS

PHYSICAL OPTICS

CD

1

INTRODUCTION

• Physical Optics : The study of wave nature of light .
• Focused on : Interference’s and Diffraction’s phenomenon.
• 7 subtopics :

23.1 : Huygen’s Principle
23.2 : Constructive interference and destructive interference
23.3 : Interference of transmitted light through double -slits

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Subtopic 23.1
HUYGEN’S PRINCIPLE

Learning Outcomes:

At the end of this subtopic, the students should be able to:
(a) State Huygen’s Principle
(b) Sketch and explain the wave front of light after passing through

a single slit and obstacle using Huygen’s principle

Remarks : Include spherical and plane wave fronts

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HUYGEN’S PRINCIPLE

Huygen’s Principle state that every point on a wavefront is a
source of wavelets that spread out in the forward direction at
the same speed as the wave itself. The new wavefront is a line
tangent to all of the wavelets.

(a) Plane Wavefront (b) Spherical wavefront

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Step of drawing the wave front

1) Draw a plane / spherical wave front
2) Choose a few point on the wave front ( ~ 5 points )
3) Draw the wavelets for each point – (make sure the

wavelets is forwarded semicircles and all the wavelets
have the same radius)
4) Join all the tangent line of the wavelets to build the new
wavefront

Let’s draw
together!!!

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• After several time, the results of wavefront:

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WAVE FRONT OF LIGHT AFTER PASSING
THROUGH A SINGLE SLIT and OBSTACLE

Explanation:

❖ Based on the Huygen’s Principle, the edges of the

wavefront bend after passing through the opening or slit.
❖ The bending of waves is called as diffraction.
❖ The degree of bending is more extreme for a small opening.

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Same goes to obstacle, the edge of the wavefront bend after
passing through the obstacle.

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INTERFERENCE PHENOMENON

➢Interference is the process of overlapping of two or more waves.
➢To produce the stable interference, the overlapping of the wave

must obey 2 conditions:
i. The source must be coherent
ii. The source must have roughly the same amplitude.
( to obtain total cancellation at minimum and a good
contrast at maximum)
iii. The distance between the coherent sources should be as

small as possible of the light wavelength (  ).

➢What is the ‘coherence sources’ ??? 9
➢Coherent means the source of overlapping wave must have:

i. The same wavelength
ii. Constant phase difference

Path difference, L

• is defined as the difference in distance from each source
to a particular point.

P
x1

S1 x2

S2 L screen

Path difference, L = |S2P − S1P|

= |x2 –x1|

TYPE OF INTERFERENCE

1. Constructive Interference :
Reinforcement of amplitude that will produce a bright
fringes (maximum)
Constructive interference : In- phase sources
Constructive interference for in-phase sources occurs if and
only if the path difference, ∆L between the waves has to
be any whole number of wavelength:
∆L = 0, λ, 2λ, 3λ,…
∆L = mλ where m=0,1,2,3,…

∆L= ІS10 – S20 І = І 3λ - 3λ І = 0

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Interference of two coherent sources in phase
• Path difference for constructive interference

• S1 and S2 are two coherent sources in phase

S1 x1
P (maximum)

S2 x2

+=

Destructive Interference : In- phase sources

Destructive interference for in-phase sources occurs if and
only if the path difference, ∆L between the waves is an
odd multiple of λ/2 wavelength:

ΔL = λ , 3λ , 5λ , 7λ ,...
22 2 2

ΔL =  m + 1  λ where m = 0,1, 2, 3,...
 2 

O

∆L= І S10 – S20 І = І 3λ/2 - 4λ І = λ/2

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• Path difference for destructive interference
• S1 and S2 are two coherent sources in antiphase

S1 x1
Q (minimum)

S2 x2

+=

Figure 23.17

Constructive Interference : Anti- phase sources

Constructive interference for anti-phase sources occurs if and
only if the path difference, ∆L between the waves is an odd
multiple of λ/2 wavelength:

ΔL = λ , 3λ , 5λ , 7λ ,...
22 2 2

ΔL =  m + 1  λ where m = 0,1, 2, 3,...
 2 

∆L= І S10 – S20 І = І 3λ/2 - 4λ І = λ/2

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2. Destructive Interference
Total cancellation of amplitude that will produce a dark
fringes (minimum)
Destructive Interference : Anti- phase sources
Destructive interference for anti-phase sources occurs if
and only if the path difference, ∆L between the waves
has to be any whole number of wavelength:
∆L = 0, λ, 2λ, 3λ,…
∆L = mλ where m=0,1,2,3,….

∆L= ІS10 – S20 І = І 3λ - 3λ І = 0

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Conditions of constructive interference and
destructive interference

Two coherent Interference Path Difference, ∆L
sources

In-Phase Constructive mλ
Anti-Phase Destructive
Constructive  m + 1  λ
Destructive  2 

 m + 1  λ
 2 

mλ

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Subtopic 23.3
INTERFERENCE OF TRANSMITTED LIGHT
THROUGH DOUBLE-SLITS

Learning Outcomes:

At the end of this subtopic, the students should be able to:

(a) Derive and use:
mλD
i) ym = d for bright fringes ( maxima)

ii)  m + 1  λD for dark fringes (minima) where
 2 
ym =
d

m = 0, ±1,±2,±3,...

(b) vUasreiaΔblye=s.λdD and explain the effect of changing any of the

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YOUNG DOUBLE-SLITS EXPERIMENT

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20

YOUNG DOUBLE-SLITS EXPERIMENT

Introduction of Young Double-Slits Experiment:
One of the experiment that shows us the results of interference
phenomenon.

FIGURE (a):
▪
▪ The monocromatic wave of wavelength, λ, entering the slits S1
and S2 which are at a distance, d apart.

After passing the slits, the wave spread out in all direction.
▪ Since these two waves emerged from the same source , they
could be considered coming from two coherent sources.
▪ Light waves from these two sources would interfere.
▪ The interference pattern was then projected onto a screen and we
can see the pattern of alternating bright and dark fringes .

FIGURE (b):
3-Dimension of interference process

FIGURE (c):
Interference pattern on screen

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ym

D

• Consider Young’s Double Slit’s schematic diagram as shown in the
above figure. Let P be a point on the screen.

➢D = the distance between the slits and the screen

➢d = the separation distance between the slits, S1 and S2
➢ym = the distance from the centre of the central maximum to the

mth fringe ( bright or dark fringe) 22

DERIVATION OF DOUBLE SLIT EQUATION

* consider the sources are ‘in phase’.

m

tan θ = ym 23
D

for small angle, sinθ  tan θ

sin θ = ym
D

DERIVATION OF DOUBLE SLIT EQUATION

* considered the cases is ‘in phase’ interference.

Constructive interference Destructive interference
( Bright Fringes / Maximum) (Dark Fringes / Minimum)

ΔL = d sin θ ΔL = d sin θ
ΔL = mλ
d sin θ = mλ ΔL =  m + 1  λ
 2 

d  ym  = mλ d sin θ =  m + 1  λ
 D   2 

mλD d  ym  =  m + 1  λ
ym = d  D   2 

m = 0, ±1,±2,±3,...  m + 1  λD
 2 
where ym =

d

where m = 0, ±1,±2,±3,... 24

Constructive interference Destructive interference
( Bright Fringes / Maximum) (Dark Fringes / Minimum)

mλD  m + 1  λD
ym = d  2 
ym =
d

Meaning of each symbol :
ym = distance of fringes from central maximum
m = order
λ = wavelength
d = distance between both slits
D = distance between the slit and the screen

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HOW TO DETERMINE THE ORDER IN DOUBLE-SLITS
INTERFERENCE

mλD  m +1  λD
ym = d  2 
ym = d

Bright Bright Order, Dark Dark Order,
fringe order m fringe order m
0 0
Central 0th 1 1st 0th 1
1st 2 2nd 1st 2
1st 2nd 3 3rd 2nd 3
2nd 3rd … 4th 3rd
3rd … …
4th … …
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m=3
m=3
m=2
m=2
m=1
m=1
m=0
m=0
m=0
m=1
m=1
m=2
m=2
m=3
m=3

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EXAMPLE 23.3.1

An interference pattern is formed on a screen when light of
wavelength 550 nm is incident on two parallel slits 50 μm
apart. The second order bright fringes is 4.5 cm from the
central maximum. How far from the slits is the screen?

Solution:

ym = mλD
d

4.5× 10-2 = 2 ( 550× 10-9 )D
50× 10-6

D = 2.05 m

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EXAMPLE 23.3.2

Red light of wavelength 600 nm from a point source passes
through two parallel and narrow slits which are 1.5 mm apart.
Determine the distance between the central bright fringe and
the forth dark fringes formed on a screen parallel to the plane
of the slits, if the distance between the slits and the screen is
1.2 m.

 m+ 1  λD
 2 
Solution: ym =

d

 3+ 1  ( 600× 10-9 )(1.2)
 2  1.5× 10-3
=

ym = 1.68×10-3 m 29

EXAMPLE 23.3.3

In a lab experiment, monochromatic light passes through two
narrow slits that are 0.050 mm apart. The interference pattern
is observed on a white wall 1.0 m from the slits, and the second
bright is 2.4 cm from the center of the central maximum.

a) What is the wavelength of the light?
b) What is the distance between third order and fifth order

dark fringes?

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Solution:

(a) mλD
ym = d

2.4 × 10-2 = 2 ( λ)(1.0)
0.050× 10-3

λ = 6×10-7 m

(b)
x = y5 − y3

 m+ 1  λD  m+ 1  λD
 2   2 
= −
dd

 5 + 1  (6 10−7 )(1)  3 + 1  (6 10−7 )(1)
 2   2 
= −
0.05 10−3 0.05 10−3

= 0.024 m 31

THE SEPARATION BETWEEN TWO CONSECUTIVE
(ADJACENT) DARK OR BRIGHT FRINGES, ∆Y

• Equation: Δy = λD
∆y d
∆y where :

screen ∆y = distance between consecutive (adjacent)
bright fringes / dark fringes

λ = wavelength
D = distance between slits and screen
d = separation distance between the slits

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THE EFFECT OF CHANGING λ, D AND d TO THE

INTERFERENCE PATTERN Δy = λD
d
❑ Interference pattern, ∆y is depends on λ, D and d

• If D and d constant,

❖ λ increase : ∆y increase : interference pattern is become wider.

❖λ decrease : ∆y decrease : interference pattern is become closer.

• If λ and d constant,
❖D increase : ∆y increase : interference pattern is become wider.
❖D decrease : ∆y decrease : interference pattern is become closer.

• If λ and D constant,
❖d increase : ∆y decrease : interference pattern is become closer
❖d decrease : ∆y increase : interference pattern is become wider.

• Table below shows the range of wavelength for colours of visible
light.

Colour Range of λ/ nm
Violet
Blue 400 – 450
Green 450 – 520
Yellow 520 – 560
Orange 560 – 600
600 – 625

Red 625 - 700