A thin glass plate of thickness $\frac{2500}{3} λ$ ( $λ$ is wavelength of light used) and refractive index µ= 1.5 is inserted between one of the slits and the screen in Young’s double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :

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1:4

4:3

4:1

2:1

answer is C.

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Detailed Solution

${Δx}^{'} = (μ - 1) t = \frac{λ}{2 π} ϕ^{'}; (1 .5 - 1) \frac{2500}{3} λ = \frac{λ}{2 π} ϕ^{'}$
$ϕ = 2 π [\frac{1250}{3}] = 2 π [146 + \frac{2}{3}]$
$\frac{I}{I^{'}} = \frac{\cos^{2} \frac{2 π}{2}}{\cos^{2} \frac{2 π [146 + \frac{2}{3}]}{2}} = \frac{\cos^{2} π}{\cos^{2} \frac{2 π}{3}} = \frac{1}{\frac{1}{4}} = \frac{4}{1}$

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