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(B) Given: Focal length of convex lens $f_l = 20 \,cm$,object distance $u = -10 \,cm$,refractive index $\mu = 1.5$,and radius of curvature of the silv

Vedclass · Accepted Answer

$11$

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(B) Given: Focal length of convex lens $f_l = 20 \,cm$,object distance $u = -10 \,cm$,refractive index $\mu = 1.5$,and radius of curvature of the silvered surface $R = 22 \,cm$.The power of the lens is $P_l = \frac{1}{f_l} = \frac{1}{20} \,cm^{-1}$.The power of the silvered surface (acting as a concave mirror) is $P_m = -\frac{1}{f_m} = -\frac{1}{-R/2} = \frac{2}{R} = \frac{2}{22} = \frac{1}{11} \,cm^{-1}$.The equivalent power of the system is $P_{eq} = 2P_l + P_m = 2(\frac{1}{20}) + \frac{1}{11} = \frac{1}{10} + \frac{1}{11} = \frac{21}{110} \,cm^{-1}$.The equivalent focal length is $F = -\frac{1}{P_{eq}} = -\frac{110}{21} \,cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{F}$:$\frac{1}{v} + \frac{1}{-10} = -\frac{21}{110}$$\frac{1}{v} = \frac{1}{10} - \frac{21}{110} = \frac{11 - 21}{110} = -\frac{10}{110} = -\frac{1}{11}$.Thus,$v = -11 \,cm$.The negative sign indicates the image is formed in front of the silvered surface at a distance of $11 \,cm$.

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