Question

A plank of mass “m” is lying on a smooth surface in situation as shown in figure. Assume that the plank slips over the surface and surface of plank exposed to radiation is black. Find the force on the plank due to radiation.

The light beam is incident on the black body at an angle $\theta$ to the normal.

• A. $\frac{Iab}{c}$
• B. $\frac{Iab}{c}cos\theta$
• C. $\frac{Iab}{c}sin\theta$
• D. $\frac{Iab}{c}tan\theta$

Answer 1

The correct option is B

$\frac{Iab}{c}cos\theta$

So the power of the incident beam is

$P=Iabcos\theta$

The number of photons incident per unit time can be found by

$P=\frac{dn}{dt}\times \frac{hc}{\lambda }$

So

$\frac{dn}{dt}=\frac{Iabcos\theta \lambda }{hc}$ .

Each of the photon will be completely absorbed by the surface. This will produce an impulse on the slab in the horizontal direction alone. The impulse in the normal direction will be absorbed by the surface on which it a kept.
Radiation force experience by surface is

$F=\frac{dn}{dt}\times p$,

Where p is the momentum of the photon incident. By de-Broglie’s hypothesis,

$P=\frac{h}{\lambda }$

$F=\frac{Iabcos\theta }{c}$