EP025 Note 7

CHAPTER 7 : PHYSICAL OPTICS

7.1 Huygens Principle.
7.2 Constructive & Destructive Interferences.
7.3 Interference Of Transmitted Light Through Double-Slits.
7.4 Interference Of Reflected Light In Thin Films.
7.5 Diffraction By A Single Slit.
7.6 Diffraction Grating.

SLT (HOURS)

Lecture :3

Tutorial :8

Practical :2

Ind. Learning : 11

TOTAL : 24

7.0 INTRODUCTION

What is the nature of light ?
Does it travel as a stream of particles
or actually in a form of certain types
of waves ?

7.1 HUYGENS’S PRINCIPLE
LEARNING OUTCOMES (LO) :

7.1 HUYGENS’S PRINCIPLE

Phenomena related to light :
Reflection, Refraction, Interference, Diffraction.

All the phenomena can be explained physically by using a
principle called ‘Huygens Principle’ which is based
on the so-called ‘wave front’.

WAVE FRONTS

Wave fronts Wave front is a line or surface which
AD joins all the adjacent points which have
BE v the same phase in a wave.
CF
It always perpendicular to the direction
λ of wave propagation.

Line joining all point of circular
adjacent wave, e.g. A, B and C wave front
or D,E and F forms wave fronts


Plane ray
Wave front
λ

HUYGEN’S PRINCIPLE

Every point on a wave front is a source of wavelets that
spread out in the forward direction at the same speed as
the wave itself. The new wave front is a line tangent to all

of the wavelets.



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7.2 CONSTRUCTIVE & DESTRUCTIVE INTERFERENCE
LEARNING OUTCOMES (LO) :

7.2 CONSTRUCTIVE & DESTRUCTIVE INTERFERENCE

Interference occurs when  2
coherent waves overlap at a point.

Coherent waves are waves that
have the following properties :
• same frequency
• fixed phase difference.

WAVE PHASE (measured in O or radian)

360o 0o
(2 rad) (0 rad)

180o
( rad)

Condition for interference :
• The sources must be coherent.
• The superposition principle must be applied.

There are two types of interference :
• Constructive interference.
• Destructive interference.

The type of interference
depends on a value called
‘path difference, L’

It refers to the difference of
distance between two waves
at a point from the respective
sources, S1 and S2.

Condition for constructive
interference :

Path difference, L = m
where m = 0, 1, 2,...
(crest-crest or trough-trough).

Condition for destructive
interference :

Path difference, L = (m+ ½)
where m = 0, 1, 2,...
(crest-trough).

7.3 INTERFERENCE OF TRANSMITTED LIGHT THROUGH DOUBLE-SLITS
LEARNING OUTCOMES (LO) :

7.3 INTERFERENCE OF TRANSMITTED LIGHT THROUGH DOUBLE-SLITS

Superposition between light waves from
S1 and S2 (coherent sources) produces
interference patterns in the form of
symmetrical bright and dark stripes
(fringes) with the same thickness.

Bright and dark fringes order, m :

Equation for mth bright fringe :
where m = 0, 1, 2, 3,…

Equation for m’th dark fringe :
where m’ = 0, 1, 2, 3,…

The separation between
successive (consecutive) bright
or dark fringes, y is given by :

Therefore, the change in pattern, y depends on :
• the wavelength of light,  (y  ).
• the distance apart, d of the double slits, (y  d-1).
• distance between slits and the screen, D (y  D).

&

QUESTION 1

In a Young’s double slits experiment, the distance
between the 8th order of bright fringes on the two sides

of the centre is 12.5 cm.
If the separation between the slits and screen is 1.5 m

and the wavelength is 550 nm, calculate the slit
separation.

ANSWERS
d = 1.056x10-4 m

&

QUESTION 2

Monochromatic light of wavelength 550 nm is used in a
double-slits experiment with the slit separation being

0.05 cm.
Calculate the angle between the 5th order bright fringe

to the 3rd dark fringe.

ANSWERS
 = 0.157o

7.4 INTERFERENCE OF REFLECTED LIGHT IN THIN FILMS
LEARNING OUTCOMES (LO) :

7.4 INTERFERENCE OF REFLECTED LIGHT IN THIN FILMS

PHASE CHANGE UPON REFLECTION
Example 1 :
Wave incidents from less dense medium to denser medium.

Effect :
Phase changed =  rad = 180o.
Path difference = ½ 

Example 2 :
Wave incidents from denser medium to less dense medium.

Effect :
Phase changed = 0 rad = 0o = unchanged.
Path difference = 0  = unchanged.

Example 3 :
Wave incidents onto other medium of the same density.

Effect :
Phase changed = 0 rad = 0o = no reflection.
Path difference = 0  = no reflection.

A thin film is a layer of transparent material ranging from fractions of
a nano-metre to several micro-metres in thickness.

Since the thickness of the film is
uniform, then the pattern produced
is either bright (reflective) or dark
(anti-reflective) only and not in
fringes.

REFLECTED LIGHT WITH NO PHASE DIFFERENCE

The formation of interference depends
on their path difference in film, Lfilm,
ACE which is also depends on :
• the thickness, t of the film.
• refractive index, n of the film.

Graphically :

For normal incident,  → 0o, then cos → 1, thus:

…..(i)

Since n3 > n2 > n1, then reflection at point
D and C produces phase difference, 
of  radian respectively (equivalents with
/2 respectively).
n1
The path difference for constructive
n2 interference (reflective coating) :

…..(ii)
n3

where m = 0, 1, 2, 3, …

Then, by substituting (ii) into (i), the
equation of constructive interference
(reflective coating) :
n1

n2 where m = 0, 1, 2, 3, …

n3

Note :

m = 0 refers to the occurrence of

constructive interference where the

n1 thickness of the film is too thin, t → 0 m
(a layer of film atoms).

n2 Normally, we take m = 1 to determine the
minimum thickness of thin film.

n3

The path difference for destructive
interference (anti-reflective coating) :

n1 …..(iii)
n2 where m = 0, 1, 2, 3, …
n3

Then, by substituting (iii) into (i), the
equation of destructive interference
(anti-reflective coating) :

n1

n2 where m = 0, 1, 2, 3, …

n3

REFLECTED LIGHT WITH PHASE DIFFERENCE  radian

Since n2 > n1, then reflection at point D
produces phase changed by  radian
(equivalents /2).
n1 However no phase changed at C since n2 > n3,
produces phase difference,  of  radian
n2 between point C and D.

n3

The path difference for constructive interference
(reflective coating) :

…..(iv)
n1

n2 Then, by substituting (iv) into (i), the equation of
constructive interference (reflective coating) :

n3

where m = 0, 1, 2, 3, …

The path difference for destructive interference
(anti-reflective coating) :

…..(v)

n1
Then, by substituting (v) into (i), the

n2 equation of destructive interference
(anti-reflective coating) :

n3

where m = 0, 1, 2, 3, …

Note :
m = 0 refers to the occurrence of
constructive interference where the
thickness of the film is too thin, t → 0 m
n1 (a layer of film atoms).

n2 Normally, we take m = 1 to determine the
minimum thickness of thin film.

n3

&

QUESTION 1

Light from a sodium lamp with wavelength of 589 nm is
shone normally on a soap film of refractive index 1.40.
If the soap film appeared dark when viewed from the
incident side, determine the minimum thickness of the

soap film.

ANSWERS
tmin = 21.04 m

7.5 DIFFRACTION BY A SINGLE SLIT
LEARNING OUTCOMES (LO) :

7.5 DIFFRACTION BY A SINGLE SLIT

Diffraction is a phenomenon refers to
the spreading or bending of light waves
over the geometrical region as they pass
through an obstacle or an aperture
whose the size of which are comparable
to its wavelength.

Geometrical region



The diffraction pattern consists of a greatest
width & intensity of central bright fringe (or
also known as central maximum)
surrounded by fringes of lower intensity and
thickness.

In the diffraction by a single slit, the
thicknesses of dark fringe (or also known as
minimum) are much shaper and narrower
than that of bright fringes.

Therefore, the positions of minimum can
be determined more accurately.

The condition of minimum
of diffraction by a single slit :

where m’ = 1, 2, 3,…

Note :
• To calculate the greatest number of orders observed, then θ → 90o.

The distance of m’th order minimum
from central maximum :

where m’ = 1, 2, 3,…

Note :
• The width of central maximum is 2y1

DIFFERENCES INTERFERENCE-DIFFRACTION PATTERN

&

QUESTION 1

A single slit diffraction pattern is formed on the screen
which is placed 1.35 m from the slit. The width of the
slit is 0.11 mm and the 3rd dark band is formed at a

distance of 14.5 mm from the central maximum.
Calculate the wavelength of the light used.

ANSWERS
 = 394nm

&

QUESTION 2

A single slit diffraction pattern is obtained on a screen
which is placed at a distance of 30 cm from the slit of

width 7.3 m. The wavelength of the light used is
650 nm.

Calculate the central bright fringe width :
(a) in degree (b) in cm.

ANSWERS
(a)  = 10.2o
(b) (b) y = 5.34cm

7.6 DIFFRACTION GRATING
LEARNING OUTCOMES (LO) :