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Question

In the ideal double slit experiment, when a glass plate (μ=1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where central maxima occurred previously remains unchanged. The minimum thickness of glass plate is kλ, then the value of k is (Integer only)

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Solution

In order to obtain a maxima at the previous central maxima position, the path differences between interfering waves,

Δx=nλ

The path difference induced by glass slab is,

Δxslab=(μ1)t

According to question,
(μ1)t=nλ

For minimum thickness of slab, the permissible value of n is 1.

t=λμ1=λ1.51

t=2λ

Given, t=kλ k=2

Correct answer: 2
Why this question?
Caution: n=0 can't be taken for minimum thickness as it yields t=0.

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