INTERFERENCE OF LIGHT

Everyone is familiar with the beautiful colours produced by a thin film of oil on the surface of water and also by the thin film of a soap bubble. These beautiful colour effects arise from interference between light waves
(1)Reflected and
(2)Transmitted from the two surfaces of thin transparent films.
Interference due to reflected light
Consider a transparent film of thickness‘t’ and refractive index ‘µ’. A ray AB is incident on the upper surface of the film. The light partly gets reflected along BP and partly refracted along BC. At C, part of it is reflected along CD and finally emerges out along DQ. The rays BP and DQ interfere and we observe interference pattern.

The path difference between the two rays BP and DQ can be calculated. Draw DM normal to BP and CN normal to BD. It is evident from figure,
Ð BCN = r and Ð BDM = i
Optical path difference between the two waves = µ (BC+CD) - BM
But from figure, BC = CD
\ Optical path difference = 2µBC-BM ----- (1)
From D BCN, cos r = CN/BC = t/BC
Or BC = t/cos r ----- (2)
From D BDM, sin i = BM/BD
Or BM = BD sin i ----- (3)
From D BCN, tan r = BN/CN = BN/t
Or BN = t tan r
But BD = 2 BN
\ BD = 2 t tan r ----- (4)
Substituting the value of BD from equation (4) in equation (3),
BM = 2 t tan r sin i
=2 t sin r/cos r x sin i
=2 t sin r/cos r x sin i/sin r x sinr
=2 µ t sin2r/cos r ----- (5) [µ = sin i/sin r]
Substituting the value of BM from equation (5) and BC from equation (2) in equation (1),
Optical path difference = 2 µ t/cos r - 2 µ t/cos r x sin2r
= 2 µ t/cos r [1 - sin2r]
= 2 µ t/cos r x cos2r
= 2 µ t cos r
By electromagnetic theory, when light is reflected from the surface of an optically denser medium, a phase change ofp, equivalent to a path difference of l/2 occurs.
\ Correct path difference = 2 µ t cos r - l/2
Condition for constructive interference ( Bright bands)
Path difference = nl
\ 2 µ t cos r - l/2 = nl
2 µ t cos r = nl + l/2
2 µ t cos r = (n+1) l/2
Condition for destructive interference (Dark bands)
Path difference = (2n+1) l/2
\ 2 µ t cos r - l/2 = (2n+1) l/2
2 µ t cos r = (2n+1) l/2 + l/2
= (2n+2) l/2
= (n+1) l
Since n is an integer, (n+1) can also be taken as n.
\ 2 µ t cos r = nl
Colours in thin films
When light is incident on a thin film, the reflected light will not include the colour whose wavelength satisfies the equation 2 µ t cos r = nl. Therefore the film will appear coloured and the colour will depend upon thickness and the angle of incidence (refraction). If ‘r’ and‘t’ are constant, the colour will be uniform. In the case of oil on water, different colours are observed because ‘r’ and‘t’ vary.

NEWTON’S RINGS-REFLECTED SYSTEM

When a plano convex lens is placed on a glass plate, a thin film of air is enclosed between the lower surface of the lens and upper surface of the plate. The thickness of the film is very small at the point of contact and gradually increases from the centre outwards. Concentric circular fringes are produced, uniform in thickness with the point of contact as the centre. When viewed with white light, the fringes are colored. With monochromatic light, bright and dark circular fringes are produced.
Here interference is due to the light reflected from the lower surface of the lens and upper surface of the glass plate. The fringes are circular because the air film has a circular symmetry.
Theory
Suppose the radius of curvature of the lens is R. The thickness of the air film be‘t’ at a distance ‘r’ from the centre.
From figure,
AB2 + BC2 = AC2
i.e. (R-t)2 + r2 = R2
R2 + t2- 2Rt + r2 = R2
2Rt = t2 + r2
Since t is very small, t2 can be neglected.
\ 2Rt = r2
Or, t = r2/2R ----- (1)
Dark rings
Condition is 2 µ t cos r = nl
For air film, µ = 1. For normal incidence, cos r = 1
\2t = nl
Substituting‘t’ from equation (1),
2r2/2R = nl
r2 = nlR
rn = ÖnlR
Diameter of nth dark ring, Dn = 2rn = Ö4nlR
Bright rings
Condition is 2 µ t cos r = (2n+1)l/2
For air film, µ = 1. For normal incidence, cos r = 1
\2t = (2n+1)l/2
Substituting‘t’ from equation (1),
2r2/2R = (2n+1)l/2
r2 = (2n+1)lR/2
rn = Ö(2n+1)lR/2
Diameter of nth bright ring, Dn = 2rn = Ö2(2n+1)lR
Determination of wavelength of light using Newton’s rings experiment
With the help of traveling microscope, measure the diameters of mth and nth dark rings.
Dm = Ö4mlR
Dn = Ö4nlR
Dm2 - Dn2 = 4(m-n) lR
\l = Dm2 - Dn2 / 4(m-n)R

AIR WEDGE/WEDGE SHAPED THIN FILM

Let an air wedge be formed between two glass plates, slightly inclined to each other at an angle q as shown. Illuminate the wedge by a parallel beam of monochromatic light. Interference occurs between the rays reflected from the upper and lower surfaces of the air film. Alternate dark and bright bands of equal width are observed.
Let‘t’ be the thickness of the air film at a distance ‘x’ from the edge. From figure,
tanq = t/x
Since q is small, tanq~q
\q = t/x
Or t = qx ----- (1)
Condition for occurrence of a dark band is
2 µ t cos r = nl
For air film, µ = 1. For normal incidence, cos r = 1
\2t = nl
Substituting‘t’ from equation (1),
2qx = nl
Or, xn = nl/2q
\ xn+1 = (n+1)l/2q
Fringe width b = xn+1 - xn
= l/2q
For a liquid wedge,
b = l/2µq

Applications of air wedge
Determination of thickness (diameter) of thin objects
The given object is placed between two glass plates to form an air wedge. On illumination, we can see alternate bright and dark bands with bandwidth b = l/2q

From figure,
q = d/L
\ b = lL/2d
Or, d = lL/2b
Testing of optical planeness of surfaces
A surface is said to be optically flat if it is plane upto 1/10 th of the wavelength of light used. In order to test the planeness, we can use interference fringes obtained from an air film. The surface to be tested is placed in contact with an optically flat glass plate and the fringes are viewed. If the fringes of equal thickness are formed, then the surface is flat. Irregular and distorted fringe pattern is obtained if the surface is not flat.