Question

Radiation force experienced by a body (shown in the figure) exposed to radiation of intensity $I$ assuming the surface of the body to be perfectly reflecting is ($c$ is the velocity of light)

• A. $\frac{\pi R^{2}I}{c}$
• B. $\frac{2\pi R^{2}I}{c}$
• C. $\frac{2R^{4}I}{c(R^2+H^2)}$
• D. $\frac{2\pi R^{4}I}{c(R^2+H^2)}$

Answer 1

The correct option is D $\frac{2\pi R^{4}I}{c(R^2+H^2)}$


The change in momentum of photons incident on the elemental area $dA$, along the perpendicular direction of the reflecting surface, is

$\Delta P=2p\sin \theta =2\left(\frac{h}{\lambda }\right)\sin \theta ...\left(1\right)$

Total number of photon incident per unit of time,

$dn=\frac{I\left(d{A}^{\prime }\right)}{E}=\frac{I\left(d{A}^{\prime }\right)}{\left(\frac{hc}{\lambda }\right)}$

Force on the element :

$\therefore dF=dh\left(\Delta p\right)$

$=\frac{I\left(d{A}^{\prime }\right)}{\left(\frac{hc}{\lambda }\right)}\times \left(2\frac{h}{\lambda }\right)\sin \theta =\frac{2I}{c}\left(d{A}^{\prime }\right)\sin \theta$

Since, the vertical components of $dF$ over the elements cancel each other due to symmetry shape.

Now, horizontal component of the force,

$d{F}^{\prime }=dF\sin \theta$

$\therefore$ Total Force $F=\int d{F}^{\prime }=\int dF\sin \theta$

$=\int \frac{2I}{c}\left(d{A}^{\prime }\right)\sin \theta \times \sin \theta =\frac{2I}{c}{\sin }^2\theta \int d{A}^{\prime }$

$\therefore F=\frac{2I}{c}{\sin }^2\theta \int d{A}^{\prime }$
Where, $\int d{A}^{\prime }=\text{Projected area}=A{R}^2$

$\Rightarrow F=\frac{2I}{c}{\sin }^2\theta \times \pi R^2$

From figure, $\sin \theta =\frac{R}{\sqrt{R^2+H^2}}$

$\Rightarrow F=\frac{2I}{c}\times \left[\frac{R^2}{R^2+H^2}\right]\times \pi R^2$

$\therefore F=\frac{2\pi I{R}^4}{c(R^2+H^2)}$

Hence, $\left(D\right)$ is the correct answer.