In Young's double slit experiment fringe width is found to be $β$ and separation between the slits is d. If one of the slits is covered by a transparent sheet of thickness t and refractive index 1.4, the central bright fringe is formed in front of the slit which is covered. Then wave length of light used is

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$\frac{3 βt}{5 d}$

$\frac{5 βt}{4 d}$

$\frac{3 βt}{4 d}$

$\frac{4 βt}{5 d}$

answer is D.

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Detailed Solution

Fringe shift $S = (μ - 1) \frac{Dt}{d} = (μ - 1) (\frac{λD}{d}) \frac{t}{λ}$

$\Rightarrow S = (μ - 1) \frac{βt}{λ}$

Here S = d/2 $∴ \frac{d}{2} = (1.4 - 1) \frac{βt}{λ} \Rightarrow λ = \frac{4 βt}{5 d}$

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