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