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Q.

A flake of glass of index of refraction 1.6  is placed over one of the openings of double slit separates. There is a displacement of the interference pattern through four successive maxima toward the side where the flake was placed. If the wavelength of the light used is λ=540 nm , calculate the thickness of the flake?

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a

3.6 μm

b

7.2  μm

c

6 μm

d

2 μm

answer is A.

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Detailed Solution

Refractive index μ=1.6 Number of maxima n=4 Wavelength λ=540 nm=540×10−9 m Thickness t=?                t=nλμ−1                  =4×540×10−91.6−1                 =4×54×10−76                =3.6×10−6m          ∴   t=3.6 μm
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