A planoconvex lens,when silvered at its plane | vedclass

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(B) Let $f_l$ be the focal length of the lens and $R$ be the radius of curvature of the curved surface. For a planoconvex lens,the focal length is giv

Vedclass · Accepted Answer

$14/9$

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(B) Let $f_l$ be the focal length of the lens and $R$ be the radius of curvature of the curved surface. For a planoconvex lens,the focal length is given by $\frac{1}{f_l} = (\mu - 1) \frac{1}{R}$.Case $(i)$: When the plane surface is silvered,the system acts as a mirror with power $P = 2P_l + P_m$. Since the plane surface is a mirror of infinite focal length,$P_m = 0$. The effective focal length $F_1 = -28 \ cm$ (concave mirror).$\frac{1}{F_1} = -(\frac{2}{f_l} + 0) \implies \frac{1}{-28} = -\frac{2}{f_l} \implies f_l = 56 \ cm$.Case $(ii)$: When the curved surface is silvered,the system acts as a mirror with power $P = 2P_l + P_m$. Here,$P_m = \frac{1}{f_m} = \frac{2}{R}$ (since $f_m = R/2$). The effective focal length $F_2 = -10 \ cm$.$\frac{1}{F_2} = -(\frac{2}{f_l} + \frac{1}{f_m}) \implies \frac{1}{-10} = -(\frac{2}{56} + \frac{2}{R})$.Since $\frac{1}{f_l} = (\mu - 1) \frac{1}{R} = \frac{1}{56}$,we have $\frac{1}{R} = \frac{1}{56(\mu - 1)}$.Substituting this: $\frac{1}{10} = \frac{2}{56} + \frac{2}{56(\mu - 1)} \cdot 2 = \frac{1}{28} + \frac{1}{28(\mu - 1)}$.$\frac{1}{10} - \frac{1}{28} = \frac{1}{28(\mu - 1)} \implies \frac{18}{280} = \frac{1}{28(\mu - 1)}$.$\frac{9}{140} = \frac{1}{28(\mu - 1)} \implies \mu - 1 = \frac{140}{9 	imes 28} = \frac{5}{9}$.$\mu = 1 + \frac{5}{9} = \frac{14}{9}$.

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