Q58 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index $=1.7$) 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 $\_\_\_\_$.
- A$5: 12$
- B$12: 5$
- Correct answerC$5: 2$
- D$2: 5$
Solution
Let $R_1$ and $R_2$ be the radii of the front and back surfaces of the biconvex lens. The power of the biconvex lens is $P_1 = (1.5 - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = 0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$.
The plano-concave lens has a flat surface and a concave surface of radius $R_2$ (matching the back surface of the biconvex lens). Its power is $P_2 = (1.7 - 1)\left(\frac{1}{R_2}\right) = \frac{0.7}{R_2}$.
Since the magnitudes of power are equal: $0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$
Simplifying: $\frac{0.5}{R_1} + \frac{0.5}{R_2} = \frac{0.7}{R_2}$, which gives $\frac{0.5}{R_1} = \frac{0.2}{R_2}$
Therefore: $\frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac{5}{2}$
The plano-concave lens has a flat surface and a concave surface of radius $R_2$ (matching the back surface of the biconvex lens). Its power is $P_2 = (1.7 - 1)\left(\frac{1}{R_2}\right) = \frac{0.7}{R_2}$.
Since the magnitudes of power are equal: $0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$
Simplifying: $\frac{0.5}{R_1} + \frac{0.5}{R_2} = \frac{0.7}{R_2}$, which gives $\frac{0.5}{R_1} = \frac{0.2}{R_2}$
Therefore: $\frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac{5}{2}$
Hence, the correct answer is option: 'C'.