Question

One end of a cylindrical rod is grounded to a hemispherical surface of radius $R=20mm.$ It is immersed in water of refractive index $\mathrm{1.33.}$ If the refractive index of the hemisphere is $\mathrm{1.5.}$ Find the position of an image of an object placed on the axis of the rod inside water at $10cm$ from the pole.

Answer 1

Given: One end of a cylindrical rod is grounded to a hemispherical surface of radius $R=20mm$. It is immersed in water of refractive index 1.33. If the refractive index of the hemisphere is 1.5.

To find the position of an image of an object placed on the axis of the rod inside water at 10 cm from the pole.

Solution:

As per the given condition,

Refractive index of water, ${\mu }_1=1.33$

Refractive index of the hemisphere, ${\mu }_2=1.5$

Radius of curvature, $R=20mm=2cm$

object distance, $u=-10cm$

Substituting these values in the following formula, we get

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}⟹\frac{1.5}{v}-\frac{1.33}{-10}=\frac{1.5-1.33}{2}⟹\frac{1.5}{v}=\frac{0.17}{2}-\frac{1.33}{10}⟹\frac{1.5}{v}=\frac{(0.17\times 5)-1.33}{10}⟹v=\frac{1.5\times 10}{(-0.48)}⟹v=-31.25cm$

is the required position of an image of an object placed on the axis of the rod.