Question

The slab of a material of refractive index $2$ shown in figure has a curved surface $APB$ of radius of curvature $10cm$ and a plane surface $CD$ . On the left of $APB$ is air on the right of $CD$ is water with refractive indices as given in the figure. An object $O$ is placed at a distance of $15cm$ from the pole $P$ as shown. The distance of the final image of $O$ from $P$ , as viewed normally from the left is:

• A. $30cm$
• B. $40cm$
• C. $45cm$
• D. $50cm$

Answer 1

Answer: (A)

From the formula of the refraction from the curve surface for object kept at a distance $u=-15cm$ and radius of curvature $R=-10cm$ , then,
$\frac{\mu _{1}}{v}-\frac{\mu _{2}}{u}=\frac{\mu _{1} - \mu _{2}}{R}$
$\frac{1}{v}-\frac{2}{\left(- 15\right)}=\frac{1 - 2}{- 10}$
$v=-30cm$
It will be virtual image there will be no refraction at the plane surface $CD$ , so the final distance of image will be $30cm$ .