Given a thin convex lens (refractive index $\mu_2$), kept in a liquid (refractive index $\mu_1$, $\mu_1 < \mu_2$) having radii of curvature $|R_1|$ and $|R_2|$. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?
Answer: (A) $\dfrac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_2|}$
The silvered lens behaves as a mirror whose power (in the liquid) is
$$P = 2P_L + P_M$$
since light crosses the lens twice and reflects once.
Lens in the liquid: $P_L = \dfrac{1}{f_L} = \left(\dfrac{\mu_2}{\mu_1}-1\right)\left(\dfrac{1}{|R_1|}+\dfrac{1}{|R_2|}\right)$. Silvered second surface (concave towards the lens): $P_M = \dfrac{2}{|R_2|}$.
An image forms on the object (real, inverted) when the object is at the centre of curvature of the equivalent mirror, at distance $x = 2F = \dfrac{2}{P}$:
$$\frac{1}{x} = \frac{P}{2} = \frac{\mu_2-\mu_1}{\mu_1}\left(\frac{1}{|R_1|}+\frac{1}{|R_2|}\right) + \frac{1}{|R_2|}$$
$$= \frac{(\mu_2-\mu_1)(|R_1|+|R_2|) + \mu_1|R_1|}{\mu_1|R_1||R_2|} = \frac{\mu_2(|R_1|+|R_2|) - \mu_1|R_2|}{\mu_1|R_1||R_2|}$$
$$x = \frac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_2|}$$
Solution by Sreeraj P, M.Sc Physics