A spherical surface separates two media of refractive indices $1$ and $1.5$ as shown in the figure. The object O is $0.2\ \text{m}$ from the surface in medium 1 and the centre of curvature C is in medium 2, with $R = 0.4\ \text{m}$. The distance of the image of the object O is:
Answer: (B) $0.4\ \text{m}$ left to the spherical surface
Light goes from medium 1 ($\mu_1 = 1$) to medium 2 ($\mu_2 = 1.5$). With the surface as origin and light travelling to the right: $u = -0.2\ \text{m}$, $R = +0.4\ \text{m}$ (C is on the right).
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$
$$\frac{1.5}{v} + \frac{1}{0.2} = \frac{0.5}{0.4} \Rightarrow \frac{1.5}{v} = 1.25 - 5 = -3.75 \Rightarrow v = -0.4\ \text{m}$$
The image is virtual, $0.4\ \text{m}$ to the left of the surface.
Solution by Sreeraj P, M.Sc Physics