Question

A plane wave of intensity $I=0.70{\text{W cm}}^{–2}$ illuminates a sphere with ideal mirror surface. The radius of sphere is $R=5.0\text{cm}$. From the standpoint of photon theory, find the force that light exerts on the sphere (in $\mu N$).

Answer 1

Imagine the sphere to be made of thin circular rings of radius $r$, thickness $ds=Rd\theta$ and subtending an angle of $\theta$ at the center.




Since, surface of mirror is considered to be ideal, i.e., reflection coefficient is unity, photons suffer momentum change in normal direction only.
Along normal momentum per second of incident photons,
${\left(\frac{dP}{dt}\right)}_{incident}=\frac{I}{c}dAco{s}^2\theta$
Due to collision rate of change of momentum of photons is
${\left(\frac{dP}{dt}\right)}_{photon}=\frac{2I}{c}dAco{s}^2\theta$
Force experienced by the ring along normal is
$d{F}_n={\left(\frac{dP}{dt}\right)}_{ball}=+\frac{2I}{c}dAco{s}^2\theta$
This force may be resolved into horizontal and vertical components.
The vertical component $d{F}_n\sin \theta$ is cancelled because every element on the upper half has a symmetrically placed element in the lower half.
So, resultant force on the ball is
$F=\int d{F}_{n}cos\theta =\int \frac{2I}{c}dAco{s}^3\theta$
$dA=\left(2\pi R\sin \theta \right)Rd\theta$
$F={\int }_0^{\frac{\pi }{2}}\frac{4I}{c}{R}^2{\cos }^3\theta \sin \theta d\theta$
$F=\frac{4\pi R^{2}I}{c}{\int }_0^{\frac{\pi }{2}}{\cos }^3\theta \sin \theta d\theta =\frac{\pi R^{2}I}{c}$
On substituting values,
$F=\pi \times 25\times {10}^{-4}\left(\frac{0.70\times {10}^4}{3\times {10}^8}\right)N$
$F=0.18\mu N$.