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 ______
Answer: (B) $5:2$
Let the biconvex lens have radii $R_1$ (front) and $R_2$ (back). The plano-concave lens has one curved surface of radius $R_2$.
$$P_1 = (1.5-1)\left(\frac1{R_1}+\frac1{R_2}\right),\qquad |P_2| = (1.7-1)\frac1{R_2}$$
Setting them equal:
$$\frac{0.5}{R_1} + \frac{0.5}{R_2} = \frac{0.7}{R_2} \Rightarrow \frac{0.5}{R_1} = \frac{0.2}{R_2}$$
$$\frac{R_1}{R_2} = \frac{0.5}{0.2} = \frac52$$
Solution by Sreeraj P, M.Sc Physics