A stationary swimmer is inside a liquid of refractive index $n_{1}$ is at a distance ‘d’ from a fixed point P inside the liquid. A rectangular block of Thickness ‘t’ and refractive index $n_{2} (n_{2} < n_{1})$ is now placed between S and P. Now S will observe P to be at a distance _______

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$d - t (\frac{n_{1}}{n_{2}} - 1)$

$d - t (1 - \frac{n_{2}}{n_{1}})$

$d + t (1 - \frac{n_{2}}{n_{1}})$

$d + t (\frac{n_{1}}{n_{2}} - 1)$

answer is D.

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Detailed Solution

Since $n_{2} {<n}_{1}$ the apparent shift of the final image of P will be $P P^{'}$ ,

$\begin{array}{l} N o w P P^{'} = (\frac{1}{_{1} n_{2}} - 1) t \\ = [\frac{1}{(n_{2} / n_{1})} - 1] t = (\frac{n_{1}}{n_{2}} - 1) t \\ ∴ {New distance =SP+PP}^{'} =d+ (\frac{n_{1}}{n_{2}} -1) t \end{array}$

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