CHAPTER 7: OPTICS

100

Waves that reflect from You often see bright bands of
objects at different locations
color when light reflects from a
can interfere with one another soap bubble or from a thin layer
just like the light pass through of oil floating on water.
double slits.

**The colors are a result of
constructive interference
between light reflected from the
two surfaces of the thin film.

101

Reflection and Phase Change

102 7.7 (a) IDENTIFY the occurrence of phase change upon reflection (From lower

to higher refractive index, phase change = rad or path difference = ! ).
"

THE OCCURRENCE OF PHASE CHANGE UPON REFLECTION

Ü Let say the light waves travel from medium 1 to medium 2:

medium 1 ( refractive index, n1)

Incident ray Reflected ray

medium 2 ( refractive index, n2)

Ü The reflection is occurred at the boundary of medium
Ü Reflection from lower to higher refractive index (n2 > n1) :

Reflected ray undergoes a phase change of 180⁰ / π rad / path

difference of λ/2.

Ü Reflection from higher to lower refractive index (n2 < n1 ) :

Reflected ray undergoes no phase change/no path difference.

103

2 TYPES OF INTERFERENCE IN THIN FILM

1. In – phase interference
2. Anti – phase interference

Thin layer of oil on water Soap buble

**The colors are a result of constructive interference
between light reflected from the two surfaces of the thin film.

104

7.7 (b) DESCRIBE with the aid of a diagram the interference of light in thin films at Normal incidence.
(Limited to 3 media)
180° 180° In – phase
IN - PHASE INTERFERENCE (Constructively)
ÖÖ

(n1˂ n ˂ n2 ) +=

Incident ray Ray 1 Ray 2

n1 = 1.00

n = 1.40

n2 = 1.52
Ray 1 : undergoes phase change
Ray 2 : undergoes phase change and travel at extra distance of 2t
Both rays will overlap (interfere) either constructively or destructively.

Ü Interference occur with in-phase sources
when the path difference , 2nt = mλ : interfere constructively

when the path difference ,2nt = æ m + 1 ö λ: interfere destructively
èç 2 ø÷

105

ANTI - PHASE INTERFERENCE 0° 180° Anti – phase
(n1 > n ˂ n2 )
C Ö (Destructively)
Incident ray +=

Ray 1 Ray 2

n1 = 1.52

n = 1.00

n2 = 1.52

Ray 1 : no phase change
Ray 2 : undergoes a phase change and travel at extra distance of 2t
Both rays will overlap (interfere) either constructively or destructively.

Ü Interference occur with anti-phase sources

when the path difference , 2nt = æ m + 1 ö λ : interfere constructively
èç 2 ø÷

when the path difference , 2nt = mλ : interfere destructively

106

ANTI - PHASE INTERFERENCE 180° 0° Anti – phase
(n1˂ n > n2 )
Ö C (Destructively)
Incident ray + =

Ray 1 Ray 2

n1 = 1.00
n = 1.33

n2 = 1.00

Ray 1 : undergoes a phase change
Ray 2 : no phase change and travel at extra distance of 2t
Both rays will overlap (interfere) either constructively or destructively.

Ü Interference occur with anti-phase sources

when the path difference , 2nt = æ m + 1 ö λ : interfere constructively
çè 2 ÷ø

when the path difference , 2nt = mλ : interfere destructively

107

SUMMARY FOR INTERFERENCE IN THIN FILM

Constructive 2nt = mλ m = 1,2,3,...
Destructive
In-phase 1
2
2nt = æ m + ö λ m = 0,1,2,3,...
èç ø÷ m = 0,1,2,3,...

Anti-phase Constructive 2nt = æ m + 1 ö λ
Destructive çè 2 ø÷

2nt = mλ m = 1,2,3,...

where :
n = refractive index of thin film ( the middle

medium)
t = thickness of thin film
m = order
λ = wavelength in vacuum

108

7.7 (e) EXPLAIN the application of thin films (eg: solar panel, glass tint).

Applications of Thin-Film Solar Panels
• Thin-film solar panels require a larger area for

installation so they can be installed in institutional and
commercial buildings with large rooftops/open spaces.
• They can be used in forest areas as well.
• Thin-film solar panels are suitable to be used in solar
farms.
• They can be used to power traffic and street lights.
• These panels can be installed on the rooftop of
buses/RVs to power small appliances, Wi-Fi modems,
fans, etc. They can also help in maintaining the
temperature of a bus.
• Thin-film solar panels can be installed in large-sized
steel water tanks to provide electricity for pumping
water.

109

Applications of Thin-Films in Glass Tint
There are many different grades, shades, colours, and
thicknesses of available window films built to offer solutions to
a variety of challenges. Window films are a retrofit upgrade for
existing glass that can be used to address problems inherent to
glazing, including:
• heat and glare reduction
• thermal insulation
• UV filtration
• safety and security
• Privacy, Automotive styling
• decoration, signage and branding

Window films are an extremely cost-effective method to reduce
heating and cooling costs in existing buildings by reducing the
amount of heat transfer through glazing.

110

EXAMPLE 7.7.1 Solution:

A soap bubble appears green (λ = 540 2nt min = æ m+ 1 ö λ
nm) at the view point on its front èç 2 ÷ø
surface nearest the viewer. What is its
minimum thickness? Assume n = 1.35. æ 1 ö
èç 2 ÷ø
180° 0° 2(1.35)t min = 0+ λ

Ö C Anti – phase tmin =100 nm
+=
Incident ray
(Destructively)

Ray 1 Ray 2
n1 = 1.00

t n = 1.35

n1 = 1.00

111

EXAMPLE 7.7.2 Solution:

A lens appears greenish yellow (λvacuum 2ntmin = mλ
= 570 nm is strongest) when white 2(1.25)tmin = (1)(570×10-9 )
light reflects from it. What minimum tmin = 228 nm
non-zero thickness of coating (n =
1.25) do you think is used on such a
(glass) lens ?

180° 180°

ÖÖ In – phase
+= (Constructively)

Incident ray Ray 1 Ray 2
n1 = 1.00

t n = 1.25

n2 = 1.50

112 Solution:

EXAMPLE 7.7.3

White light is incident normally on a 2nt min = æ m+ 1 ö λ
lens ( n = 1.52) that is coated with a èç 2 ø÷
film of MgF2( n = 1.38). For what
minimum thickness of the film will 2(1.38)t min = æ 0+ 1 ö 550 ´10-9
yellow light (λvacuum = 550 nm) be çè 2 ÷ø
missing in the reflected light ?

180° 180° tmin = 99.64 nm

ÖÖ In – phase
+=

Incident ray Ray 1 Ray 2 (Constructively)

n1 = 1.00

t n = 1.38

n2 = 1.52

113 EXERCISE 7.7

1. A non-reflective coating of MgF2( n = 1.38) covers the glass (n= 1.52) of a

camera lens. Assuming that the coating prevents reflection of yellow-green
clioghattin(λgvaccauunmh=a5v6e5. nm) , determine the minimum nonzero thickness that the
(102
nm)
2. A plastic film with index of refraction 1.80 is put on the surface of a car

window to increase the reflectivity and thereby to keep the interior of the
car cooler. The window has index of refraction 1.60. What minimum
thickness required if light of wavelength 600 nm in air reflected from the two
sides of the film is to interfere constructively. (83 nm)

3. A If the index of refraction of soap is 1.333 and red light’s wavelength is 650
nm, determine

(a) Three smallest thicknesses of the soap bubble that produce constructive
interference for that red light.
(122 nm, 366 nm, 610 nm)

(b) Three smallest thicknesses that give destructive interference?
(0, 244 nm, 488 nm)

114

7.8 DIFFRACTION BY A SINGLE SLIT

Learning Outcomes:

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

(a) DEFINE diffraction

(b) EXPLAIN with the aid of a diagram the diffraction of a single slit

(c) USE

) = for DARK fringes ( minima)


) = / for BRIGHT fringes (maxima)


where n = ±1, ±2, ±3, ...

115

7.8 (a) DEFINE diffraction

DIFFRACTION

• Diffraction is the bending of a wave around the edges of an

opening or other obstacle.

• Diffraction is the result of interference between sources on the

same wave front.

λλ λ

(a) obstacle (b) a > λ (c) a ≈ λ

116 7.8 (b) EXPLAIN with the aid of a diagram the diffraction of a single slit

DIFFRACTION OF A SINGLE SLIT

• Consider parallel light rays strike the slit.
• According to Huygens’s principle, each portion of the slit acts as a

source of new light waves (secondary wavelet) that radiate
towards the screen (diffracted light pattern is observed on the
screen).

• All the wavelets going forward in phase and travel the same
distance to the screen and interfere constructively to produce
central maximum.

117

Central maximum 3rd dark fringe

1st dark fringe 1st bright fringe

118

• Dark fringes are produced when all the rays
interfere destructively with path difference is :

sin =

• Bright fringes are produced when all the rays
interfere constructively when path difference is:

sin = + %
&

119

Diffraction pattern

Central bright 2nd bright fringe Width of central
with 2nd dark fringe bright fringe
alternating
dark and 1st bright fringe width = 2λD
bright fringes a
occurring on 1st dark fringe
either side. Central
maximum

1st dark fringe
1st bright fringe

2nd dark fringe
2nd bright fringe

Most of the light intensity is concentrated in the broad central
bright fringe.
In the forward direction, θ = 0°, all wave interfere constructively
giving max intensity at the center of the diffraction pattern.

120 In general, a minimum / dark fringe is obtained at angle θn
sin 0 = , n = ±1, ±2, ±3,

… wahe–rwe idth of the single slit
θn – angle for the nth minimum measured from the
central axis
λ – wavelength of the monochromatic light used

Since, angle θ is small, thus : The distance of nth order dark
yn fringe (minimum) to the
sin qn = tan qn = D center of the central bright

fringe: =


The distance of nth order
bright fringe (maxima) to the
center of the central bright
fringe:
(
=

121 7.8 c) Use equation bright and dark fringes for single slit
SUMMARY OF DIFFRACTION OF SINGLE SLIT

Constructive Destructive Width of central
bright fringe
sainsin θ== æçè m ++12 ö÷ø12λ
assiinn θ ==mλ width = 2λD
a
m +"12 öø÷ y m== m λ D
= ym æ a λ D a Bright / Dark Order
çè fringe number,
= mn

1st DF 1
1st BF

where: 2nd DF 2
a = width of slit (a < λ ) 2nd BF

mn = order 3rd DF

λ = wavelength of light 3rd BF 3
D = distance between slit and screen
mn th bright/dark fringe to
yθ m==adnigstlaenbceetwbeetewnemnenth bright/dark fringe to the the central bright
central bright

122

EXAMPLE 7.8.1 Solution:

A single slit of width 0.140 mm is ym = nmlD
illuminated by monochromatic
light, and diffraction bands are a
observed on a screen 2.00 m away.
If the second dark band is 16.0 mm 16 ´10-3 = (2)l(2)
from the central bright band, what
is the wavelength of the light ? 0.14 ´10-3

l = 560 nm

123

EXAMPLE 7.8.2 Solution:

Visible light of wavelength of 550 a) The width of the slit
nm falls on a single slit and
produces its second diffraction a sinq = nml
minimum at an angle of 45.0o
relative to the incident direction of ( )a sin 45° = 2 550´10-9
the light.
a) What is the width of the slit? a = 1.56 ´10-6 m
b) At what angle is the first
b) The angle is the first minimum
minimum produced? produced

a sinq = nml

( )1.56´10-6sinq = (1) 550´10-9

q = 20.64°

124

EXAMPLE 7.8.3 Solution:

How wide is a slit if it diffracts 590 width = 2λD
nm light so that its central a
diffraction peak is 2.0 cm wide on
a screen 1.80 m away ? 2.0 ×10-2 = 2(590 ×10-9 )(1.80)
a

a = 0.106 ×10-3m

125

EXERCISE 7.8

1. Monochromatic light falls on a slit 2.2 x 10-3 mm wide. If the angle between the
first dark fringes on both side of the central maximum is 30⁰ , what is the
wavelength of the light used ? (589 nm)

2. Light with a wavelength of 589 nm from a distance source is incident on a slit
0.720 mm wide, and the resulting diffraction pattern is observed on a screen 2.00
m away. What is the distance between two consecutive dark fringes?
(1.64 mm)

3. Monochromatic light from a distant source is incident on a slit 0.800 mm
wide. On a screen 3.00 m away, the distance from the central maximum of the
diffraction pattern to the first minimum is measured to be 2.42 mm. Calculate
the wavelength of the light ? (645 nm)

4. The width of a slit is 2.0 x 10-5 m. Light with a wavelength of 480 nm passes
through this slit and falls on a screen that is located 0.50 m away. In the
diffraction pattern, by using appropriate diagram, find the width of the bright
fringe that is next to the central bright fringe. (0.012 m)

126

EXERCISE 7.8

5. A single slit has a width of 2.1 x 10-6 m and is used to form a
diffraction pattern. Find the angle that locates the second
dark fringe when the wavelength of the light is
(a) 430 nm
(b) 660 nm.
(24.17⁰, 38.94⁰)

6. When 450 nm light falls on a slit, the central diffraction peak
on a screen 2.0 m away is 6.0 cm wide. Calculate the slit
width.
(3 x 10-5 m)

127

7.9 DIFFRACTION GRATING

Learning Outcomes:

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

(a) EXPLAIN with the aid of a diagram the formation of diffraction.

(b) APPLY sin = where d = 1
N

128

DIFFRACTION GRATING

• Diffraction grating is a device consisting of a great number of

narrow, closely spaced, equidistant and parallel slits or lines.

d = 1
N

• d = grating spacing = spacing between adjacent slits
• N = number of lines per unit length

129

line
a dd
(1) (2) (3)
(1) 2 lines , 1 slit (diffraction single slit)
(2) 3 lines , 1 slits (double slit)
(3) many lines , (diffraction grating)

130

7.9 (a) EXPLAIN with the aid of a diagram the formation of diffraction.

THE FORMATION OF DIFFRACTION

Monochromatic Ө1 Ө2 n=2
light n=1

n=0

n=1

n=2
• A monochromatic light strikes a diffraction grating
• A diffraction grating consists of thousands of slits per cm
• According to Huygens’s Principle, each point on the wave-

front striking the slits act as a point source of secondary
wavelet.
• After passing through the diffraction grating, the light is
diffracted.
• The diffraction pattern is observed as shown in figure.
• The bright fringes are observed when all the rays interfere
constructively in phase with path difference = nλ

131

• Figure 1 illustrates light travels to a distant viewing screen from five

slits of the grating.

first order maximum

(n = 1)

incoming plane central or zeroth order
wavefront of light
maximum (n = 0)

diffraction grating first order maximum
Figure 1
(n = -1)
Figure 2
d sin q

dθ
θ

132

7.9 Apply equation of diffraction grating

sin =

d= 1
N

where The highest number
n = 0,1,2,3,...(order number) of order that can be
d = grating spacing observed is limited by
θ = angle of nth bright fringe the fact that θ can
λ = wavelength of light not be more than 90°

133

ORDER NUMBER FOR DIFFRACTION GRATING:

Bright fringe Order number Order number,
(maximum) n
Central BF 0
1st BF 0th order maximum 1
2nd BF 1st order maximum 2
3rd BF 2nd order maximum 3
4th BF 3rd order maximum 4
4th order maximum

134

Compact disc as a diffraction grating

The surface of a compact disc has a spiral grooved track ( with
adjacent grooves having a separation on the order of 1μm ).

The surface of compact disc acts as a diffraction grating.
When compact disc is viewed in
white light, the light reflecting
from the regions between these
closely spaced grooves
interferes constructively in
certain directions forming
colored `lanes` .

135

EXAMPLE 7.9.1 Solution:

When a grating with 300 lines d = 1 = 300 1
per mm is illuminated normally N mm-1
with a parallel beam of
monochromatic light. A second d = 3.33×10-3 mm = 3.33×10-6 m
order principal maximum is
observed at 18.9o. Find the d sinq = mn l
wavelength of the light ? 3.33´10-6 sin18.9° = 2l
l = 5.39 ´10-7 m

136

EXAMPLE 7.9.2 Solution:

The first-order line of 550 nm sin =
light falling on a diffraction
grating is observed at a 12° = sin
angle. How far apart are the
slits ? 1 550×10,-
sin 12°
=

= 2.65×10,/