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Q.

In the Ideal double slit experiment, when a glass – plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ ) the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass – plate is

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a

2λ

b

2λ3

c

λ3

d

λ

answer is A.

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Detailed Solution

Path difference due to slab should be integral multiple of λ  or  Δx=nλ  or  μ−1t=nλ,n=1,2,3.............  or  t=nλμ−1  For minimum value of t,   n = 1  ∴t=λμ−1=λ1.5−1=2λ
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In the Ideal double slit experiment, when a glass – plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ ) the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass – plate is