A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $24\ \text{W}$. The radius of curvature of hemisphere is $10\ \text{cm}$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ______ $\times10^{-8}\ \text{N}$.
Numerical value type. Enter your answer.
Answer: 4
Intensity on the surface: $I=\dfrac{P}{4\pi R^2}$. Light falls normally and is reflected, giving pressure $\dfrac{2I}{c}$.
By symmetry only the component along the axis survives, and $\int\cos\theta\,dA$ over the hemisphere equals its projected area $\pi R^2$:
$$F=\frac{2I}{c}\pi R^2=\frac{2P}{4\pi R^2c}\pi R^2=\frac{P}{2c}=\frac{24}{2\times3\times10^8}=4\times10^{-8}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics