A plano-convex lens becomes an optical system of $28\ \text{cm}$ focal length when its plane surface is silvered and illuminated from left to right (light falls on the curved surface first). If the same lens is instead silvered on the curved surface and illuminated from the other side (light falls on the plane surface first), it acts as an optical system of focal length $10\ \text{cm}$. The refractive index of the material of the lens is:
Answer: (A) $1.55$
Let $R$ be the radius of the curved surface. The lens has $\dfrac1f = \dfrac{\mu-1}{R}$.
A silvered lens acts as a mirror of power $P = 2P_\text{lens} + P_\text{mirror}$, with focal length $F = \dfrac1P$.
**Plane surface silvered** (plane mirror, $P_\text{mirror} = 0$):
$$\frac{1}{F_1} = \frac{2(\mu-1)}{R} \quad\Rightarrow\quad F_1 = \frac{R}{2(\mu-1)} = 28\ \text{cm}$$
**Curved surface silvered** (concave mirror of focal length $\frac R2$, $P_\text{mirror} = \frac 2R$):
$$\frac{1}{F_2} = \frac{2(\mu-1)}{R} + \frac{2}{R} = \frac{2\mu}{R} \quad\Rightarrow\quad F_2 = \frac{R}{2\mu} = 10\ \text{cm}$$
So $R = 20\mu$, and substituting:
$$\frac{20\mu}{2(\mu - 1)} = 28 \;\Rightarrow\; 10\mu = 28\mu - 28 \;\Rightarrow\; \mu = \frac{28}{18} \approx 1.55$$
Solution by Sreeraj P, M.Sc Physics