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Young’s double slit experiment is conducted in a liquid of refractive index $μ_{1}$ as shown in figure. A thin transparent slab of refractive index $μ_{2}$ and thickness t is placed in front of the slit $s_{2}$ . The magnitude of optical path difference at ‘O’ is :-

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$| (\frac{μ_{2}}{μ_{1}} - 1) t |$

$| (\frac{μ_{1}}{μ_{2}} - 1) t |$

$| (μ_{2} - 1) t |$

$| (μ_{2} - μ_{1}) t |$

answer is D.

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Detailed Solution

${(S_{1} O)}_{o p t i c a l} = μ_{1} (S_{1} O) {(S_{2} O)}_{o p t i c a l} = μ_{1} (S_{2} O - t) + μ_{2} t$

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