A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $$24$$ W. The radius of curvature of hemisphere is $$10$$ cm and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ______ $$\times 10^{-8}$$ N.

Solution :

The source is placed at the centre of curvature of the hemisphere.

Hence, all light rays strike the reflecting surface normally.

For complete reflection, radiation pressure is :

$$p = \frac{2I}{c}$$

Consider a small surface element on hemisphere.

Force on opposite elements have horizontal components that cancel due to symmetry.

Only axial components add.

Net force on hemisphere is :

$$F = \frac{P}{2c}$$

Given :

$$P = 24\text{ W}$$

$$c = 3 \times 10^8\text{ m s}^{-1}$$

Therefore,

$$F = \frac{24}{2 \times 3 \times 10^8}$$

$$= \frac{24}{6 \times 10^8}$$

$$= 4 \times 10^{-8}\text{ N}$$

Final Answer :

$$4$$