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Obtain the condition for maxima and minima due to interference in a wedge shape film observed in reflected light. How is the interference pattern in wedge shaped film?

Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: High

2 Answers
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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.

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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.

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  • 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

    $r = µ (BC + CD) – BF$

    $r = µ (BE + EC + CD) - µBE$

enter image description here

enter image description here

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

$$r = 2µd cos (r + a)$$

  • 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.

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