A thin biconvex lens is prepared from glass ( | vedclass

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(A) For a biconvex lens,$R_1 = 20 	ext{ cm}$ and $R_2 = -20 	ext{ cm}$.The focal length of the lens $f_l$ is given by $\frac{1}{f_l} = (\mu - 1) \le

Vedclass · Accepted Answer

$10$

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(A) For a biconvex lens,$R_1 = 20 	ext{ cm}$ and $R_2 = -20 	ext{ cm}$.The focal length of the lens $f_l$ is given by $\frac{1}{f_l} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} ight) = 0.5 	imes \frac{2}{20} = \frac{1}{20}$. Thus,$f_l = 20 	ext{ cm}$.The system acts as a mirror with power $P = 2P_l + P_m$,where $P_l = \frac{1}{f_l}$ and $P_m = -\frac{1}{f_m}$.For the silvered surface (convex),the focal length $f_m = \frac{R}{2} = \frac{20}{2} = 10 	ext{ cm}$. Since it acts as a concave mirror in the system,we use the sign convention $P_m = -\frac{1}{f_m} = -\frac{1}{10}$.However,the effective power of the combination is $P = -\left( 2P_l + P_m ight) = -\left( 2 	imes \frac{1}{20} + \frac{1}{10} ight) = -\left( \frac{1}{10} + \frac{1}{10} ight) = -\frac{2}{10} = -\frac{1}{5}$.The effective focal length of the system is $F = \frac{1}{P} = -5 	ext{ cm}$.For the image to form at the same position as the object,the object must be placed at the center of curvature of the equivalent mirror,which is at a distance $u = 2|F| = 2 	imes 5 = 10 	ext{ cm}$.

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The final image of the object forms on the object itself. If the focal length of the lens is $30 \ cm$,then the radius of curvature of the mirror is .... $cm$.

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