A planoconvex lens becomes an optical system | vedclass

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(B) Let the focal length of the lens be $f$ and the radius of curvature of the curved surface be $R$. The focal length of the plane surface is $\infty

Vedclass · Accepted Answer

$1.55$

Vedclass · Answer

(B) Let the focal length of the lens be $f$ and the radius of curvature of the curved surface be $R$. The focal length of the plane surface is $\infty$. For the silvered lens, the effective power is $P = 2P_L + P_M$, where $P_L$ is the power of the lens and $P_M$ is the power of the mirror. In Fig. $-A$, the plane surface is silvered. The mirror formed is a plane mirror $(R_M = \infty)$, so $P_M = 0$. The effective focal length $F_1 = -28 \, cm$ (as it acts as a concave mirror). $\frac{1}{F_1} = -\frac{2}{f} - 0 \implies \frac{1}{-28} = -\frac{2}{f} \implies f = 56 \, cm$. Using the lens maker's formula, $\frac{1}{f} = (\mu - 1)(\frac{1}{R})$. In Fig. $-B$, the curved surface is silvered. The mirror formed is a concave mirror with radius $R$, so $P_M = -\frac{1}{f_M} = -\frac{2}{R}$. The effective focal length $F_2 = -10 \, cm$. $\frac{1}{F_2} = -\frac{2}{f} - \frac{2}{R} \implies \frac{1}{-10} = -\frac{2}{56} - \frac{2}{R}$. $\frac{2}{R} = \frac{1}{10} - \frac{1}{28} = \frac{14 - 5}{140} = \frac{9}{140} \implies R = \frac{280}{9} \, cm$. Substituting $f$ and $R$ in the lens maker's formula: $\frac{1}{56} = (\mu - 1)(\frac{9}{280}) \implies \mu - 1 = \frac{280}{56 	imes 9} = \frac{5}{9} \approx 0.555$. Thus, $\mu = 1.555 \approx 1.55$.

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