Spherical Surfaces & Thin Lenses question from JEE Main 2025 (29 Jan, Shift 2)

Answer: (b)

1. Use $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$. Both centres of curvature lie in the air, i.e. on the incident side, so $R = -R$ for each surface. $OB = \frac{R}{2}$, $OA = \frac{3R}{2}$.
2. Surface B: $\frac{1.5}{v} + \frac{2}{R} = -\frac{0.5}{R} \Rightarrow v = -0.6R$ (image $0.6R$ from B, on the air side).
3. Surface A: $\frac{1.5}{v} + \frac{2}{3R} = -\frac{0.5}{R} \Rightarrow v = -\frac{9R}{7} \approx -1.286R$ (image $1.286R$ from A).
4. Taking P as origin: images at $R - 0.6R = 0.4R$ and $-R + 1.286R = 0.286R$.
5. Separation $= 0.4R - 0.286R \approx 0.114R$. So the answer is (b).