An equiconvex lens $(μ = 3 / 2)$ is placed over a plane mirror as indicated in diagram. An object placed 12 cm from system such that image and object coincide. Now a liquid is poured between mirror and lens and for the same object the image now formed at 24 cm from mirror. The refractive index of the liquid poured will be

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$\frac{5}{3}$

$\frac{5}{4}$

$\frac{4}{3}$

$\frac{3}{2}$

answer is A.

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Detailed Solution

For initial condition : $\frac{1}{f} = [μ - 1] [\frac{1}{R_{1}} - \frac{1}{R_{2}}] \Rightarrow \frac{1}{12} = [\frac{3}{2} - 1] [\frac{2}{R}]$
R = 12 cm
When liquid is poured

$u = - 12, V = - 24, f_{e q} = ?$
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ [Mirror Formula] $\Rightarrow f_{e q} = - 8$

$P_{e q} = 2 P_{L_{1}} + P_{L_{2}} + P_{M}, \frac{1}{8} = 2 (\frac{1}{12}) + 2 P_{L_{2}} + 0$

$\Rightarrow P_{L_{2}} = \frac{1}{48} = [μ - 1] [\frac{1}{R}], P_{L_{2}} = \frac{1}{48} = [μ - 1] [\frac{1}{R}] \Rightarrow μ = \frac{5}{4}$

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