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The focal length of a Plano convex lens when plane surface is silvered and object is in front of curved surface is

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

1 Answer

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.


answered Feb 13, 2014 by meena.p
edited Jul 22, 2014 by thagee.vedartham
Why is the power of comvex surface added twice?

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