A double slit experiment is shown in the figure. Each slit has width W. A thin piece of glass of thickness δ refractive index μ, is placed between one of the slits and the screen. The intensity at the central point is measured as a function of thickness δ. For what values of δ is the intensity at C minimum.

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λ2μ

λ2(μ−1)

zero

answer is B.

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Detailed Solution

For point C, without slab path difference is 0In presence of slab path difference is (μ−1)⋅d for point Cfor minimum intensity path difference should be2n-1λ2 so (μ−1)d=(2n−1)λ2⇒d=(2n−1)λ2(μ−1)for minimum intensity n = 1 soδmin=λ2(μ−1)

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