In a long glass tube, a mixture of two liquids A and B with refractive indices $1.3$ and $1.4$ respectively forms a convex refractive meniscus towards A. If an object placed at $13\ \text{cm}$ from the vertex of the meniscus in A forms an image with a magnification of $-2$, then the radius of curvature of the meniscus is
Answer: (D) $\dfrac{2}{3}\ \text{cm}$
Light goes from A ($\mu_1 = 1.3$) into B ($\mu_2 = 1.4$); $u = -13\ \text{cm}$.
Magnification at a single refracting surface: $m = \dfrac{\mu_1v}{\mu_2u}$.
$$-2 = \frac{1.3\,v}{1.4\times(-13)} \Rightarrow v = 28\ \text{cm}$$
Refraction formula:
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \Rightarrow \frac{1.4}{28} + \frac{1.3}{13} = \frac{0.1}{R}$$
$$0.05 + 0.1 = \frac{0.1}{R} \Rightarrow R = \frac{2}{3}\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics