In a modified YDSE, the region between the screen and slits is immersed in a liquid whose refractive index varies with time t as $μ_{l} = (\frac{5}{2}) - (\frac{t}{4})$ until it reaches a steady state value of 5/4 .A glass plate of thickness $36 μ m$ and refractive index $\frac{3}{2}$ is introduced in front of one the slits. Find the instant of time (in s) when central maxima is at point O located on the x-axis.

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Detailed Solution

Consider a point P above O on the screen : ${(S_{1} P)}_{l i q u i d} = {[μ (S_{1} P)]}_{a i r}$

${(S_{2} P - t)}_{l i q u i d} + t_{g l a s s} = {[μ_{1} (S_{2} P - t)]}_{a i r} + {[μ_{g} t]}_{a i r} = {[{μ_{1} (S_{2} P)} + (μ_{g} + μ_{1}) t]}_{a i r} ∴ Δ x = [μ_{s} S_{2} P + (μ g - μ_{1}) t] - μ_{1} S_{1} P = μ_{1} (S_{2} P - S_{1} P) + (μ_{g} - μ_{1}) t = μ_{1} \frac{y d}{D} + (μ_{g} + μ_{1})$

For central maximum $∴ Δ x = 0; ∴ μ_{1} \frac{y d}{D} + (μ_{g} - μ_{1}) t = 0; y - (μ_{g} - μ_{1}) t \frac{D}{d μ_{2}}$

$y = - \frac{\frac{3}{2} - (\frac{5}{2} - \frac{T}{4})}{(\frac{5}{2} - \frac{T}{4})} \frac{t D}{d} = [\frac{- 1 + \frac{T}{4}}{\frac{T}{4}}] = \frac{t D}{4} = (\frac{4 - T}{10 - T}) \frac{t D}{d};$