Interference in Thin Wedge Films:

Consider the interference pattern produced by the film of varying thickness that is a film which is not plane parallel which is produced by the wedge. It comprises of two non-parallel surfaces inclined at the angle Θ.

Assume the ray of light from the monochromatic source S strikes half silvered mirror and reflected onto an air wedge.

Light from the monochromatic source is partially reflected from mirror onto air wedge. Air wedge is formed by inclining two microscope slides at very small angle Θ as shown in Figure above. If region of reflection at slides is magnified, result is shown in above figure. It can be observe from the figure that some of the light is reflected from lower surface of the top slide and some from top side of the lower slide. Both wave trains R₁ and R₂ then combine together and gives rise to interference patterns when viewed from above half silvered mirror by the eye or microscope.

Two coherent sources are produce by division of amplitude. This is different from Young's experiment in which sources are produced by division of wave-front.

Phase Change in Reflection:

There is a very important fact concerning reflection of waves from surface of higher refractive index μ. Phase change of Π (or 180^o) takes place when light strikes boundary from side of rarer medium.

Therefore, one can say that light reflected by the material of higher refractive index than medium in which rays are traveling goes through a phase change of Π (or 180^o). This phase change is equal to path change of λ/2.

The Air Wedge:

In figure given below Θ is an angle of wedge. S is the distance between bright bands. B and B' are two consecutive bright bands.

As Θ is small in triangle ΔACA',

Therefore tan Θ = (λ/2)/s

Or Θ = λ/2s....................Eq.1

Also, the value of tan Θ can be found as

tan Θ  = t/a

As value of Θ is very small. Thus

Or Θ = t/a.................Eq.2

On comparing Eq.1 and Eq.2, we get

λ/2s = t/a = Θ

Here Θ is measured in radians

Assume that m^th bright fringe is seen above A, one would expect that

2AB = mλ

Generally, band is bright if

2AB = (m - 1/2)λ...............................Eq.4

And for dark band, the condition is

2AB = mλ where m = 0, 1, 2........................Eq.5

If at A', the (m + 1)^th bright fringe is seen then it follows that extra path difference is

2A'B' - 2AB = λ

A'B' - AB = λ/2

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