# A thin convex lens L ( refractive index = $1.5$  ) is placed on a plane mirror M. When a pin is placed at A , such that $OA=18\:cm$ , its real inverted image is formed at A itself , as shown in figure . When a liquid of refractive index $\mu _{l}$ is put between the lens and the mirror , the pin has to be moved to A' such that $OA'=27\:\:cm$ , to get its inverted real image at A' itself . The value of $\mu _{l}$ will be : Option 1) $\frac{4}{3}$ Option 2) $\frac{3}{2}$ Option 3) $\sqrt{3}$ Option 4) $\sqrt{2}$

S solutionqc

$P_{initial} = 2P_{L} = \frac{1}{9} \Rightarrow P_{L} = \frac{1}{18}$

$P_{final}=2P_{L}+2P^{'} = \frac{2}{27}$

$2P^{'}=\frac{2}{27}-\frac{1}{9} \Rightarrow P^{'}=-\frac{1}{54}$

$-\frac{1}{54}=(\mu-1)\left ( \frac{1}{-18}-\frac{1}{\infty } \right )$

$\Rightarrow \mu=\frac{4}{3}$

Option 1)

$\frac{4}{3}$

Option 2)

$\frac{3}{2}$

Option 3)

$\sqrt{3}$

Option 4)

$\sqrt{2}$

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