A thin large film having refractive index $μ = \frac{3}{2}$ , is placed in front of slits $S_{1}$ and $S_{2}$ of Young's double slit experiment such that its thickness varies according to the equation $t = t_{0} \cos^{2} (\frac{5 π y}{4 d})$ where y = 0 is taken at $S_{1}$ . Find the value of $t_{0}$ such that there is central maxima at P (refer diagram) $(A s s u m e D > > d)$

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$\frac{d^{2}}{2 D}$

$\frac{d^{2}}{4 D}$

$\frac{2 d^{2}}{D}$

$\frac{d^{2}}{D}$

answer is B.

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Detailed Solution

$\begin{array}{l} t_{1} = t_{0} \cos^{2} 0 = t_{0} \\ t_{2} = t_{0} \cos^{2} [\frac{5 π}{4 d} (- d)] = \frac{t_{0}}{2} \\ For central maxima at P \\ (S_{2} P - S_{1} P) + (μ - 1) (t_{2} - t_{1}) = 0 \\ \Rightarrow \frac{dxd / 2}{D} = (\frac{3}{2} - 1) (t_{0} - \frac{t_{0}}{2}) \\ \Rightarrow t_{0} = \frac{2 d^{2}}{D} \end{array}$

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