Question

The radiation force experienced by body exposed to radiation of intensity I, assuming surface of body to be perfectly absorbing is:

A. $${\displaystyle \frac{\pi R^{2}I}{c}}$$
B. $${\displaystyle \frac{\pi RHI}{c}}$$
C. $${\displaystyle \frac{IRH}{2c}}$$
D. $${\displaystyle \frac{IRH}{c}}$$

Answer 1

Hint: In this question we have been asked to calculate the radiation force experienced by a body. It is given that the body is exposed to radiation of intensity I and the surface of the body is perfectly absorbing. We know that radiation pressure is given as the ratio of intensity of radiation and speed of light. Therefore, calculating the area from the given diagram. The radiation force shall be calculated.

Formula used: $$P={\displaystyle \frac{I}{C}}$$
Where,
P is the radiation pressure
I is the intensity of radiation
C is the speed of light

Complete step by step solution:
From the diagram, we can say that the radius of the circle is given as R, the height of the cone is H.
Therefore, from the diagram we can say that area affected by the radiation will be given as,
$$A={\displaystyle \frac{2RH}{2}}$$ …………….. (1)
We also know that radiation pressure applied by a wave on a perfectly absorbing surface is equal to the energy density of the wave. The radiation pressure is given as,
$$P={\displaystyle \frac{I}{C}}$$ …………….. (2)
Now, we know that pressure is also given as force per unit area
Therefore,
$$P={\displaystyle \frac{F}{A}}$$
Solving for force,
$$F=P\times A$$ ……… (3)
Now, from (1), (2) and (3)
We get,
$$F={\displaystyle \frac{IRH}{c}}$$

Therefore, the correct answer is option D.

Note:
The mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field is known as radiation pressure. The radiation pressure is doubled when the radiation is reflected and not completely absorbed. The force associated due to radiation pressure is known as radiation pressure force.