A perfectly reflecting solid sphere of radius $R$ is placed in the path of a uniform beam of large crossiectional area and intensity $I$. Calculate the force exerted on this sphere due to the light beam. Also calculate the force on the sphere if it would have been perfectly absorbing. Support the result obtained by an analytical argument.

Let $O$ be the centre of the sphere and $QP$ be the light incident on the sphere of radius $R$. Let $I$ be the intensity of the incident light. Consider a circular element of radius $r$, thickness $dr$, subtending an angle to at the centre of the sphere as shown in figure.

Then we observe that $r=R\sin \theta$ and $dr=Rd\theta$. If $dA$ be the area of this circular element then

$dA=\left(2\pi r\right)dr=2\pi \left(R\sin \theta \right)\left(Rd\theta \right)=2\pi R^2\sin \theta d\theta$

Since intensity $I$ is defined as the energy incident per second per unit area of a surface held normal to the direction of propagation of light, hence

$I=\frac{dE}{\left(d{A}_n\right)\left(dt\right)}$

where, $d{A}_n=dA\cos \theta$ is the area of the element normal to the direction of propagation of light.

$\Rightarrow I=\frac{dE}{\left(dA\cos \theta \right)\left(dt\right)}$

So, the energy of the light falling on this element of area $dA$ in time $dt$ is$\Rightarrow dF=f\cos \theta =2\left(\frac{IdA}{c}\right){\cos }^3\theta$

$dE=Idt\left(dA\cos \theta \right)$

The momentum $\left(dp\right)$ of light falling on this element of area $dA$ is $\frac{dE}{c}$ along $QP$ and since light incident at $P$ on this element is reflected by the sphere along $PM$ So, the change in momentum of the photon beam incident on the element is directed along $PN$ or $OP$and is given by

$dp=2\left(\frac{dE}{c}\right)\cos \theta =\frac{2}{c}\left(Idt\right)\left(dA{\cos }^2\theta \right)$

The force $\left(f\right)$ on $dA$ due to the light falling on it is directed along $PQ$ is given by

$f=\frac{dp}{dt}=2\left(\frac{IdA}{c}\right){\cos }^2\theta$

It is observed that due to symmetry, the net force on the ring element as well as the sphere is directed along the line $XO$. The component of this force (directed along the line $XO$) acting on the element of area $dA$ is

$dF=f\cos \theta =\left(2\left(\frac{IdA}{c}\right){\cos }^2\theta \right)\cos \theta$ 

$\Rightarrow dF=f\cos \theta =2\left(\frac{IdA}{c}\right){\cos }^3\theta$

$\Rightarrow dF=\left(\frac{2I}{c}\right)\left(2\pi R^2\sin \theta d\theta \right){\cos }^3\theta$

The force on the entire sphere is $F=\int dF$

$\Rightarrow F={\int }_0^{\frac{\pi }{2}}\left(\frac{2I}{c}\right)\left(2\pi R^2\sin \theta d\theta \right){\cos }^3\theta$

$\Rightarrow F=\left(\frac{4\pi R^{2}I}{c}\right){\int }_0^{\frac{\pi }{2}}{\cos }^3\theta \sin \theta d\theta$

$\Rightarrow F=-\left(\frac{4\pi R^{2}I}{c}\right){\int }_0^{\frac{\pi }{2}}{\cos }^3\theta d\left(\cos \theta \right)$

$\Rightarrow F=-{\left.\left(\frac{4\pi R^{2}I}{c}\right)\left(\frac{{\cos }^4\theta }{4}\right)\right|}_0^{\frac{\pi }{2}}=-\frac{4\pi R^{2}I}{c}\left(0-\frac{1}{4}\right)$

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