The magnitudes of power of a biconvex lens (r | vedclass

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(A) Let the radii of curvature of the biconvex lens be $R_1$ and $R_2$. The power of the biconvex lens is given by $P_A = (\mu - 1) \left( \frac{1}{R_

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$5 : 2$

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(A) Let the radii of curvature of the biconvex lens be $R_1$ and $R_2$. The power of the biconvex lens is given by $P_A = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$.For the plano-concave lens, the radius of curvature of the curved surface is $R_2$ and the other surface is flat $(R = \infty)$. Its power is $P_B = -(\mu' - 1) \left( \frac{1}{R_2} + \frac{1}{\infty} \right) = -(1.7 - 1) \left( \frac{1}{R_2} \right) = -0.7 \left( \frac{1}{R_2} \right)$.Given that the magnitudes of power are equal, $|P_A| = |P_B|$.$0.5 \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = 0.7 \left( \frac{1}{R_2} \right)$.$0.5 \left( \frac{1}{R_1} \right) = (0.7 - 0.5) \left( \frac{1}{R_2} \right) = 0.2 \left( \frac{1}{R_2} \right)$.$\frac{0.5}{R_1} = \frac{0.2}{R_2} \implies \frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac{5}{2}$.

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