A concave–convex lens of refractive index $1.5$ has radii of curvature of its surfaces $30\ \text{cm}$ (concave surface) and $20\ \text{cm}$ (convex surface). The concave surface is upwards and is filled with a liquid of refractive index $1.3$. The focal length of the liquid–glass combination will be:
Answer: (D) $\dfrac{600}{11}\ \text{cm}$
Treat the system as two thin lenses in contact, with light travelling downwards. Both surfaces of the glass have their centres of curvature above the lens.
**Liquid lens** (flat top, bottom curved with $R = 30\ \text{cm}$, centre above): plano-convex,
$$\frac1{f_\ell} = (1.3 - 1)\frac1{30} = \frac1{100}$$
**Glass meniscus** (upper surface $R_1 = -30$, lower surface $R_2 = -20$ in the sign convention):
$$\frac1{f_g} = (1.5 - 1)\left(\frac1{-30} - \frac1{-20}\right) = 0.5\times\frac1{60} = \frac1{120}$$
$$\frac1F = \frac1{100} + \frac1{120} = \frac{11}{600} \Rightarrow F = \frac{600}{11}\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics