A spherical surface of radius of curvature $R$ separates air from glass (refractive index $= 1.5$). The centre of curvature is in the glass medium. A point object $O$ placed in air on the optic axis of the surface forms its real image at $I$ inside glass. The line $OI$ intersects the spherical surface at $P$ and $PO = PI$. The distance $PO$ equals
Answer: (A) $5R$
Refraction at a single spherical surface:
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2-\mu_1}{R}$$
Light goes from air ($\mu_1 = 1$) into glass ($\mu_2 = 1.5$); the centre of curvature is in the glass, so $R$ is positive. Let $PO = PI = x$: $u = -x$, $v = +x$.
$$\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R} \Rightarrow \frac{2.5}{x} = \frac{0.5}{R} \Rightarrow x = 5R$$
Solution by Sreeraj P, M.Sc Physics