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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 $λ = 540 n m$ , calculate the thickness of the flake?

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$2 μ m$

$6 μ m$

$3.6 μ m$

$7.2 μ m$

answer is A.

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Detailed Solution

Refractive index $μ = 1.6$

Number of maxima $n = 4$

Wavelength $λ = 540 n m = 540 \times 10^{- 9} m$

Thickness $t = ?$

$t = \frac{n λ}{μ - 1}$

$= \frac{4 \times 540 \times 10^{- 9}}{1.6 - 1}$

$= \frac{4 \times 54 \times 10^{- 7}}{6}$

$= 3.6 \times 10^{- 6} m$

$∴ t = 3.6 μ m$

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