$(a)\;\frac{R}{2 (y-1)} \\ (b)\;\frac{3R}{2(\mu-1)} \\ (c)\;\frac{R}{(\mu-1)} \\ (d)\;\frac{2R}{(\mu-1)} $

In this situation

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

and $F_m =\large\frac{\infty}{2}$$=\infty$

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

and $P_M=\large\frac{-1}{f_M}$

$\qquad= \large\frac{1}{\infty}$

$\qquad= 0$

and hence power of system

$P= P_l+P_M+P_L$

$\quad= 2PL+P_M$

ie $p=2 \large\frac{(\mu-1)}{R}$$+0$

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

$\therefore F=\large\frac{1}{P}$

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

ie lens will behave as a concave mirror of focal length $\large\frac{R}{2 (\mu-1)}$

Hence a is the correct answer.

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