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(A) For a lens with one surface silvered,the equivalent power is given by $P_{eq} = 2P_L + P_m$,where $P_L$ is the power of the lens and $P_m$ is the

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$20$

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(A) For a lens with one surface silvered,the equivalent power is given by $P_{eq} = 2P_L + P_m$,where $P_L$ is the power of the lens and $P_m$ is the power of the mirror.First,calculate the focal length of the lens $(f_L)$ using the lens maker's formula:$\frac{1}{f_L} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (1.5 - 1) \left( \frac{1}{40} - \frac{1}{-40} ight) = 0.5 	imes \left( \frac{2}{40} ight) = \frac{1}{40} \implies f_L = 40 \, cm$.The power of the lens is $P_L = \frac{1}{f_L} = \frac{1}{40} \, cm^{-1}$.The silvered surface acts as a concave mirror with radius of curvature $R = 40 \, cm$. Its focal length is $f_m = -\frac{R}{2} = -20 \, cm$. The power of the mirror is $P_m = -\frac{1}{f_m} = -\frac{1}{-20} = \frac{1}{20} \, cm^{-1}$.The equivalent power of the system is $P_{eq} = 2 \left( \frac{1}{40} ight) + \frac{1}{20} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \, cm^{-1}$.The equivalent focal length is $f_{eq} = -\frac{1}{P_{eq}} = -10 \, cm$ (the negative sign indicates it acts as a concave mirror).Under auto-collimation,the object must be placed at the center of curvature of the equivalent mirror,which is at a distance $u = 2|f_{eq}| = 2 	imes 10 = 20 \, cm$.

An object is placed in front of a symmetrical convex lens with refractive index $1.5$ and radius of curvature $40 \, cm$. The surface of the lens further away from the object is silvered. Under auto-collimation condition,the object distance is.......$cm$.

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