Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: High

Consider a film of non-uniform thickness as shown in Fig. It is bound by two surfaces OX and OX′ inclined at an angle θ. The thickness of the film gradually increases from O to X. Such a film of non-uniform thickness is known as wedge shaped film. The point O at which the thickness is zero is known as the edge of the wedge. The angle θ between the surfaces OX and OX′ is known as the angle of wedge. Let μ be the refractive index of the material of the film. Let a beam AB of monochromatic light of wavelength λ be incident at an angle ‘i’ on the upper surface of the film. It is reflected along BF and is transmitted along BC. At C also the beam suffers partial reflection and refraction and finally we have the ray DR2 in the reflected system. Thus as a result of partial reflection and refraction at the upper and lower surfaces of the film, we have two coherent rays BR1and DR2 in the reflected system.

To find the path difference between these two rays, draw DF perpendicular to BR1. The optical path difference between the rays BR1and DR2 is

Δ = μ (BC + CD ) – BF. ………………………………..(1)

Extend BC further to appoint P such that CP = CD. Draw a perpendicular DE to BC such that BE = EC is also perpendicular DP to OX’ at I.

Therefore Optical Path Difference Δ = μ (BE + EC + CD ) – BF.

Δ = μ (BE + EC + CP ) – BF……………………………….(2)

In triangle BDE and BFD

sin⁡i/sin⁡r =BF/BE

But sin⁡i / sin⁡r =μ

⸫ μ =BF/BE

⸫ BF = μ BE

Equation 2 becomes Δ = μ (BE + EC + CP ) – μ BE

Δ = μ (EC + CP) i.e Δ =μ (EP)

In triangle DPE

Cos (r+θ) = EP/DP

EP = DP Cos (r+θ)

DP = 2DI

DP = 2t

Therefore, Δ=2μtcos(r+θ )

Due to reflection at B, an additional path change of λ/2 .

Δ=2μtcos(r+θ ) ± λ/2

Condition for constructive interference (or maxima or brightness)]

If the OPD is an integral multiple of λ, then the waves interfere constructively.

Δ =nλ

From equation (2)

2μtcos(r+θ)±λ/2 =nλ

2μtcos(r+θ) =(2n±1)λ/2…………………………………(3)

Condition for destructive interference (or minima or darkness)

If OPD is odd multiple of λ/2, then the rays interfere destructively,

Δ =(2n±1)λ/2

From equation (2)

2μtcos(r+θ) ±λ/2 =(2n± 1)λ/2

2μtcos(r+θ) =nλ-----------------------------------------------------(4)

The interference pattern in wedge shaped film consists of alternate dark and bright bands which are parallel to each other and they are equally spaced.

• The shape of the fringe depends on how the thickness of the air film enclosed varies.

• In wedge shaped film the thickness of the air is constant over a straight line along the width of the wedge.

• Hence the fringes in wedge shaped film is straight.

• In a newton’s ring set up the air film is enclosed below the convex lens. The thickness of the film is constant over a circle (or concentric circles) having center at the center of the lens. Hence the fringes are circular.

Conditions for maxima and minima for interference in wedge shaped film

• Let us consider two plane surfaces GH and G1H1 inclined at an angle α and enclosing a wedge shaped film. The thickness of the film increases from G to H as shown in the figure.

• Let µ be the refractive index of the material of the film.

• When this film is illuminated there is interference between two systems of rays, one reflected from the front surface and the other obtained by internal reflection at the back surface.

• The path difference Ґ is given by

  $r = µ (BC + CD) – BF$

  $r = µ (BE + EC + CD) - µBE$

$$r = µ (EC + CD) = µ (EC + CP) = µ EP$$

$$r = 2µd cos (r + a)$$

• Due to reflection an additional phase difference of λ/2 is introduced.

  Hence, r = 2µd cos (r + a) + λ/2

• For constructive interface

  2µd cos (r + a) + λ/2 = nλ

  Or 2µd cos (r + a) = (2n - 1) λ/2 where n = 1, 2, 3,….

• For destructive interface

  2µd cos (r + a) + λ/2 = (2n - 1) λ/2

  Or 2µd cos (r + a) = nλ Where n = 0, 1, 2, 3,…..

• Spacing between two consecutive bright bands is obtained as follows.

• For nth maxima

  2µd cos (r + a) = (2n - 1) λ/2

• Let this band be obtained at a distance $X_n$ from thin edge as shon in figure. For near normal incident, r=0. Assuming, µ=1, From the figure, $d = X_n tanα$

  2$X_n$ tan a cos a = (2n - 1) λ/2

  2$X_n$ sin a = (2n - 1) λ/2

• For (n+1)th maxima

  2$X_{n+1}$ sin a = (2n - 1) λ/2

  2 ($X_{n+1} - X_n$) sin a = λ

  Or fringe spacing, $β = X_{n+1} - X_n$

  Where α is small and measured in radius.