Ray Optics

JEE Main level, single correct, medium

129 of 295

Question 129 of 295

The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index =1.7=1.7) are same. If the curvature of planoconcave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ____\_\_\_\_.

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Answer: (C) 5:25: 2

Let R1R_1 and R2R_2 be the radii of the front and back surfaces of the biconvex lens. The power of the biconvex lens is P1=(1.5−1)(1R1+1R2)=0.5(1R1+1R2)P_1 = (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).
The plano-concave lens has a flat surface and a concave surface of radius R2R_2 (matching the back surface of the biconvex lens). Its power is P2=(1.7−1)(1R2)=0.7R2P_2 = (1.7 - 1)\left(\frac{1}{R_2}\right) = \frac{0.7}{R_2}.
Since the magnitudes of power are equal: 0.5(1R1+1R2)=0.7R20.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}
Simplifying: 0.5R1+0.5R2=0.7R2\frac{0.5}{R_1} + \frac{0.5}{R_2} = \frac{0.7}{R_2}, which gives 0.5R1=0.2R2\frac{0.5}{R_1} = \frac{0.2}{R_2}
Therefore: R1R2=0.50.2=52\frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac{5}{2}

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