The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (

Question

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 $\_\_\_\_$

Professor PC Thomas Classes · Answer

Using the Lens Maker's Formula, the power of the biconvex lens ($P_1$) is $\frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = 0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$. For the plano-concave lens, its curved surface matches the back of the biconvex lens, so its radius is $R_2$. Its power ($P_2$) magnitude is $|-(1.7 - 1)\left(\frac{1}{\infty} - \frac{1}{R_2}\right)| = 0.7\left(\frac{1}{R_2}\right)$. Equating the magnitudes $|P_1| = |P_2|$ yields $0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2} \implies \frac{0.5}{R_1} = \frac{0.7}{R_2} - \frac{0.5}{R_2} = \frac{0.2}{R_2}$. Therefore, the ratio of the radii is $\frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac{5}{2}$.