Download interference of light_Module 1 (engineering physics) second semester and more Lecture notes Physics in PDF only on Docsity!
Introduction
The wave theory of light was first put forward by Christian Huygens in 1678. According to him,
light produces periodic disturbance in the medium which travels in the form of a wave. He assumed
that light waves are longitudinal in nature. On the basis of wave nature of light we can easily
explained interference, diffraction and polarization of light.
Interference of Light
When two or more coherent light waves of same frequency and almost same amplitude superpose
on each other, then at some points on the screen, intensity is found to be maximum and at some
other points intensity is found to be minimum. This redistribution of intensities is called
interference of light. There are two types of interference:
(i) Constructive Interference
If phase difference between two waves is zero or even multiple of i.e. waves are in same phase
then the amplitude and intensity of resultant light wave will be maximum, this type of interference
is called constructive interference.
(ii) Destructive Interference
If phase difference between two waves is 180
o or odd multiple of i.e. waves are in opposite phase
then the amplitude and intensity of resultant light wave will be minimum, this type of interference
is called destructive interference.
Principle of Superposition of Waves
According to the principle of superposition of wave, “ the resultant displacement at a particular
point produced by a number of waves is the vector sum of the displacements produced by the
individual waves and depends on the phase difference between them ”. This can be seen in the figure
given below (for waves in same phase):
According to the principle of superposition the resultant displacement
y = y 1 + y 2 + y 3 + y 4 +........
Interference due to Thin Film
Thin films, such as soap bubbles, oil spread over the surface of water or a glass plate, show
colourful patterns, this is known as thin film interference, because it is the interference of light
waves reflecting from the top surface of a film with the waves reflecting from the bottom surface.
To obtain a nice coloured pattern, the thickness of the film has to be similar to the wavelength of
light.
According to above figure, let a ray of light (SA) from a monochromatic source (S) is incident on
surface PQ at a point A at an angle i. This incident ray is partially reflected as AR 1 and partially
refracted and transmitted rays are produced.
(A) Interference in Reflected Rays
According to above figure, it is clear that there will be a definite path difference between the
reflected rays AR 1 , CR 2 , ER 3 , ……. Let us consider two reflected rays AR 1 and CR 2. Now to
determine the path difference between the reflected rays AR 1 and ABCR 2 , we draw a perpendicular
S M
M
i (^) i Gi
K
R 1 R 2 R 3
I
P Q
C E
t
R
r r
B H D J F
S
L
T 1 T 2 T 3
A
sinr
sini ………. (12)
From equation (11) and (12), we have
sini
sinisin r 1 cosr
2 μt
2
cosr
2 μt [1 – sin
2 r]
cosr
2 μt cos
2 r
= 2t cos r ………. (13)
But According to Stoke’s theorem, when a light ray suffers reflection at denser medium, then a
phase change of or path difference of 2
λ occurs.
Hence, actual path difference between the reflected ray AR 1 and ABCR 2 will be
= 2t cos r - 2
λ ………. (14)
Case I:Constructive Interference OR Maxima
For constructive interference the path difference between waves will be zero or even multiple of 2
λ .
i.e. = 2n 2
λ
Now form equation (14), we have
2 t cos r - 2
λ = 2n 2
λ
2 t cos r = ( 2 n + 1) 2
λ ………. (1 5 )
where n = 0, 1, 2, ……..
In this case whole film will appear bright.
Case II:Destructive Interference OR Minima
For destructive interference the path difference between waves will be odd multiple of 2
λ .
i.e. = (2n-1) 2
λ
Now from equation (14), we have
2 t cos r - 2
λ = (2n-1) 2
λ
2 t cos r - 2
λ = n- 2
λ
2 t cos r = n ………. ( 16 )
where n = 0, 1, 2, 3, ……….
In this case film will appear dark but not completely dark.
MICHELSON’S INTERFEROMETER
Principle
In this experiment coherent waves are produced by method of division of amplitude. The produced
waves propagate in perpendicular directions and incident normally on the mirrors. After reflection
from these mirrors, when they superimpose on each other, they produce interference fringes.
Construction and working
A schematic diagram of the Michelson Interferometer experiment is shown in figure. Light
emerging from monochromatic source of light(S) is made to fall on convex lens (L) to get parallel
beam of light and then it is allowed to fall on half silvered plate (beam splitter), P which is inclined
at an angle of 45o^ to the incident parallel beam of monochromatic light. Since the rear surface of
plate P is half silvered, hence beam is partially reflected and partially transmitted from half silvered
and precede towards the mirror M 1 and M 2 respectively. These two rays get reflected back from
mirror M 1 and M 2 respectively and meet again at O on half silvered portion of plate P and then
move along OT and superimpose on each other to produce interference fringes. These fringes can
be seen by telescope T.Fringes may be circular, hyperbolic and linear depending on the position of
M 1 and M 2.
Action of CompensationPlate
In Michelson experiment, after division of ray at point O, in absence of plate C the reflected ray
passes the plate P twice where as the refracted ray does not pass even once through the plate P.
therefore the optical path of the reflected and refracted rays are not equal when mirror M 1 and M 2
P
M 1 (Mirror)
M 2 (Mirror)
S
M 2 (Image of Mirror M 2 )
C
Convex Lens
LightSource
(^45) o o
Telescope T
P = Half silvered plate
C = Compensation plate d = Thickness of imaginary air
film
d
Case I:Bright Fringes or Maxima
For bright fringes the path difference must be
= 2n( 2
λ ) ………. (2)
Now from equation (1) & (2), we have
2d cos- 2
λ = 2n( 2
λ )
2d cos = (2n+1) 2
λ ………. (3)
Case: II: Dark Fringes or Minima
For dark fringes the path difference must be
= (2n-1) 2
λ ………. (4)
Now from equation (1) & (4), we have
2d cos- 2
λ = (2n-1) 2
λ
2d cos = n ………. (5)
From equation (3) & (5), we see that bright and dark circular fringes (rings) will form due to
constructive and destructive interference respectively.
(ii) Localized of fringes
When the mirrors M 1 and M 2 are not perpendicular to each other, the mirror M 1 and image of M 2 ,
i.e. M 2
' encloses a virtual wedge shaped film whose thickness increase uniformly. The shape of the
fringes depends on the orientation of the mirrors as shown in figure.
APPLICATIONS OF MICHELSON’S INTERFEROMETER:
(1) Determination of wavelength of monochromatic light
To determine the wavelength of monochromatic light, firstly we have to arrange Michelson
interferometer experiment in such a way that circular fringes are seen through the telescope.
The movable mirror M 1 is moved back ward or forward by the micrometer screw, so that a path
difference is introduced in the path of rays.
Now if the mirror M 1 is displaced through a distance d, then the path difference introduced in path
of rays will be 2d.
So for dark fringe
2d = n ……. (1)
From equation (1) it is clear that when the separation between M 1 and M 2 is d, then no. of fringe
displaced will be n as seen through the telescope.
Now due to displacement of mirror M 1 fringes cross the field of view. Let on moving the mirror M 1
by a distance x, the N fringe cross the field of view.
2 (d + x) = (n + N)
2d + 2x = n + N
2x = N (2d = n from eq
n
N
2x ………. (3)
So, from equation (3), wavelength of monochromatic light can be determined easily.
(2) Determination of wavelength separation of D lines of sodium light
To determine the wavelength difference Michelson Interferometer is adjusted in such a way that the
circular fringes are observed. As we know that there are two spectral lines of sodium light D 1 and
D 2 these two spectral lines are very close to each other. Let the wavelength of D 1 line is 1 and the
wavelength of D 2 line is 2. These wavelengths are nearly equal. These two lines will produce its
own interference pattern but as the difference between two wave lengths is very small
(approximately of the order of 6 A
o ), these interference patterns will overlap. Let for 1 , n
th fringe
coincides with (n + 1)
th fringe of 2.
Let the mirror M 1 is moved by a distance x and next pattern is observed.
2x = n 1 = (n + 1) 2 ………. (1)
n = λ 1
2x ………. (2)
and n + 1 = λ 2
2x …….……(3)
Subtracting equation (2) from equation (3), we have
(n + 1) – n = λ 2
2x
2x
1 = 2x
2 λ 1
λ
1 2 = 2x ( 1 – 2 )
(a) Phase Condition
This condition requires that the wave reflected from the top and bottom surfaces of thin film. The
optical path difference between the reflected waves.
2 cos (^) 2 2
f t^ r
Where t is the thickness of the film and f is the refractive index of the film material. There are two
additional path differences first 2
corresponds to phase change at the surface (air to film
boundary) and second 2
corresponds to phase change at the bottom surface (film to glass
boundary).
Assuming normal incident of light cos ( + r) = 1, we get
= 2ft –
But waves reflected from the top and bottom of the film are in opposite phase, i.e. destructive
interference so path difference will be odd multiple of 2
i.e.
2 1 2
n
(^2) 2 1 2
f t^ n
(^2 2) 2 1 2 2 2
f t^ n^ n
(^2) 2 1 2
f t^ n
The thickness will be minimum when n = 0, so
(^2) min 2
f t
min (^4) f
t
min
1
(^4) f
t
^ (^) ^
It means that the optical thickness of the AR coating should be of one quarter wavelength.
(b) Amplitude Condition
The amplitude condition requires that the amplitudes of reflected waves are equal, i.e. E 1 = E 2
The above condition requires
f a g f
f a g f
Where a, g and f are the refractive indices of air, glass and thin film respectively.
We get as a = 1
f g f
f g f
By cross produce
f g – g + f^2 – f = f g – f^2 + g – f 2 f^2 = 2 g
1/
f g
Thus, refractive index of the coating should be equal to the square root of the refractive index of
glass.
(ii) Multilayer Antireflection (AR) Coating: A single layer AR coating is effective only at one
particular wave length and cannot totally reduce the reflection at that wavelength. Multilayer
coatings are more effective and useful. In two layer coating, each layer is quarter wave (/4) thick.
In practice, three-layer coatings are widely used and are highly effective over most of the visible
spectrum.
The centre layer is of high refractive index such as zirconium dioxide ZrO 2 ( = 2.10) and is half
wave (/2) thick. The outside layer is of magnesium fluoride MgF 2 ( = 1.38) having (/4)
thickness and the inside layer is again a (/4) thick cesium fluoride CeF 3 ( = 1.63). Such coating
are often referred to as quarter-half-quarter coatings. In important applications AR coatings are
made using upto 100 layers of alternating high and low refractive index materials.
d
a 1 g a 1 2 g a 1 2 3 g
MgF 2 ZrO 2 CeF 3
IMPORTANT QUESTIONS
- Describe construction and working with diagram of Michelson Interferometer. How is it used to
find the wavelength of light? What is the order of the central fringe?
- Explain the working of Michelson Interferometer. How circular fringes be produced with it.
- Show that the Michelson’s interferometer is used to find the wavelength of light.
- What are the conditions for obtaining
(i) Circular fringes and
(ii) St. line fringes in Michelson’s interferometer.
- Draw a labeled diagram of Michelson’s Interferometer. How shall we use this device to
determine the wavelength of monochromatic light.
- Describe working of Michelson Interferometer. How it is used to measure the difference in
wavelength between the D lines of sodium light?
- Explain the working principle of Antireflection coating and derive an expression for the
minimum thickness of the coating.
TUTE SHEET- 1
- In Michelson’s interferometer, the readings of movable mirror for a pair of consecutive
distinctness of fringes are 0.7 mm and 0.99 mm respectively. The source of light used has a
doublet of wavelengths whose mean value is taken as 5893Ao. Find the difference between
wavelength components.
- Michelson interferometer experiment is performed with a source which has two wavelengths
4882A
o and 4886A
o
. By what distance does the mirror have to be moved between positions of
disappearance of fringes?
- A Michelson interferometer is set to form circular fringes with light of wavelength 5000Ao. By
changing path length of movable mirror slowly. 50 fringes cross the centre of view. How much
path length has been changed?
- In Laser Michelson’s interferometer when a transparent thin glass plate of refractive index 1.
is introduced in the path of one of the beams, 100 fringes cross the field of view at given point.
If wavelength of laser used is 6328A
o , find the thickness of the plate.
5. A single layer of coating of thickness /4 is deposited on a convex lens of g = 1.9 to reduce its
reflectivity minimum. What is the refractive index of coating?
