A bi-convex lens has radius of curvature of both surfaces the same, $\dfrac16\ \text{cm}$. If this lens is to be replaced by another convex lens of the same material having different radii of curvature on both sides ($R_1 \ne R_2$), without any change in lens power, then a possible combination of $R_1$ and $R_2$ is:
Answer: (B) $\dfrac15\ \text{cm}$ and $\dfrac17\ \text{cm}$
By the lens maker's formula, for a biconvex lens $P = (\mu - 1)\left(\dfrac1{R_1} + \dfrac1{R_2}\right)$ with both radii taken as magnitudes. The power is unchanged if
$$\frac1{R_1} + \frac1{R_2} = \frac{2}{R} = 2\times6 = 12\ \text{cm}^{-1}$$
Check the options: $3 + 3 = 6$; $5 + 7 = 12$ ✔; $3 + 7 = 10$; $6 + 9 = 15$. Option (1) also has equal radii, which is not allowed anyway.
So $R_1 = \frac15\ \text{cm}$, $R_2 = \frac17\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics