$(a)\;\frac{R}{\mu} \\ (b)\;\frac{R}{2\mu} \\ (c)\;\frac{R}{3\mu} \\ (d)\;\frac{3R}{\mu} $

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In this case

$\large\frac{1}{f_L}=(\mu-1) \bigg[ \large\frac{1}{\infty}-\frac{1}{-R} \bigg]$

$\qquad= \large\frac{\mu-1}{R} $ and $f_m =\large\frac{(-R)}{2}$

So, $P_L= \large\frac{1}{f_L}=\frac{(\mu-1)}{R}$

and $P_M=\large\frac{-1}{f_m} =\frac{2}{R}$

and hence power of system

$P=P_L+P_m+P_L$

$\quad=2P_L +P_M$

$P= \large\frac{2( \mu-1)}{R} +\frac{2}{R}$

$\quad= \large\frac{2 \mu}{R}$

$\therefore F= \large\frac{-1}{P}=\large\frac{-R}{2\mu}$

ie the lens will be equivalent to a converging mirror of focal length $\bigg( \large\frac{R}{2y}\bigg)$

Hence b is the correct answer.

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