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

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2 × 10^–3mm

4 × 10^–3mm

6× 10^–3mm

7.2 × 10^–3mm

answer is D.

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Detailed Solution

$\begin{array}{l} n = \frac{(μ - 1) t}{λ} \\ \Rightarrow 8 = \frac{(1 .6 - 1) t}{540 \times 10^{- 9}} \Rightarrow t = 7 .2 \times 10^{- 3} mm \end{array}$

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