Q. A spherical surface of radius of curvature $R$ separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $O$, and $PO = OQ$. The distance $PO$ is equal to
A) 5 $R$
B) 3 $R$
C) 2 $R$
D) 1.5 $R$
Explanation
(a) $\frac{\mu_{2}}{v} - \frac{\mu_{1}}{u} = \frac{\mu_{2} - \mu_{1}}{R} \Rightarrow \frac{1.5}{+ OQ} - \frac{1}{( - OP)} = \frac{(1.5 - 1)}{+ R}$
On putting $OQ = OP,\ OP = 5R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.