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Question

In the figure shown, a parallel beam of light is incident on the plane of the slits of Young's double slit experiment. Light incident on the slit S1 passes through a medium of variable refractive index μ=1+ax (where x is the distance from the plane of slits as shown), up to a distance ‘𝑙’ before falling on S1. Rest of the space is filled with air. If at O a minima is formed , then the minimum value of the positive constant a (in terms of l and wave length λ in air) is



A
l2λ
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B
λl
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C
None of these
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D
λl2
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Solution

The correct option is D λl2
We know that: Δx=(2n+1)λ2

Optical path shift due to introduction of film
=(μ1)t

Given: μ=1+ax

Total length of slab =l

For a small element dx change in path difference can be calculated as,

Δx=(μ1)dx

Δx=[(1+ax)1]dx=ax dx

For the whole length,
Δx=10ax dx=(al2)2

For a minima to be at O

Δx=(2n+1)λ2

i.e,al22=(2n+1)λ2

For minimum a,n=0

(al2)2=λ2

a=λl2


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