A plano-convex lens,when silvered at its plan | vedclass

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(D) The power of a silvered lens system is given by $P = 2P_L + P_M$,where $P_L$ is the power of the lens and $P_M$ is the power of the mirror. The fo

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(D) The power of a silvered lens system is given by $P = 2P_L + P_M$,where $P_L$ is the power of the lens and $P_M$ is the power of the mirror. The focal length $F$ is given by $\frac{1}{F} = -(\frac{2}{f_L} + \frac{1}{f_M})$.Case $1$: Plane surface is silvered. The lens acts as a lens of focal length $f_L$ and a plane mirror $(f_M = \infty)$.$\frac{1}{f_L} = (\mu - 1)(\frac{1}{R} - \frac{1}{\infty}) = \frac{\mu - 1}{R}$.The equivalent focal length $F_1 = -28 \, cm$ (concave mirror).$\frac{1}{F_1} = -(\frac{2}{f_L} + 0) = -\frac{2(\mu - 1)}{R} = -\frac{1}{28} \implies \frac{R}{\mu - 1} = 56 \quad \dots(1)$Case $2$: Curved surface is silvered. The lens acts as a lens of focal length $f_L$ and a concave mirror of radius $R$ $(f_M = -R/2)$.$\frac{1}{F_2} = -(\frac{2}{f_L} + \frac{1}{f_M}) = -(\frac{2(\mu - 1)}{R} - \frac{2}{R}) = -\frac{2(\mu - 2)}{R} = -\frac{1}{10} \implies \frac{R}{\mu - 2} = 20 \quad \dots(2)$From $(1)$,$R = 56(\mu - 1)$. Substituting into $(2)$:$\frac{56(\mu - 1)}{\mu - 2} = 20 \implies 56\mu - 56 = 20\mu - 40$$36\mu = 16 \implies \mu = \frac{16}{36} = \frac{4}{9}$.Since $\frac{4}{9}$ is not among the options,the correct answer is $D$.

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