Q. The slab of a material of refractive index 2 shown in figure has curved surface $APB$ of radius of curvature $10\ cm$ and a plane surface $CD$. On the left of $APB$ is air and on the right of $CD$ is water with refractive indices as given in figure. An object $O$ is placed at a distance of $15\ cm$ from pole $P$ as shown. The distance of the final image of $O$ from $P$, as viewed from the left is
A) $20\ cm$
B) 30 $cm$
C) 40 $cm$
D) 50 $cm$
Explanation
(b) In case of refraction from a curved surface, we have
$\frac{\mu_{2}}{v} - \frac{\mu_{1}}{u} = \frac{\mu_{2} - \mu_{1}}{R} \Rightarrow \frac{1}{v} - \frac{2}{( - 15)} = \frac{(1 - 2)}{- 10} \Rightarrow v = - 30\ cm$
$i.e.,$ the curved surface will form virtual image $I$ at distance of $30\ cm$ from $P$. Since the image is virtual there will be no refraction at the plane surface $CD$ (as the rays are not actually passing through the boundary), the distance of final image $I$ from $P$ will remain $30\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.