Question

A partially reflecting surface ( Solid hemisphere) of radius $R$ placed in the path of a parallel beam of light of large aperture. If the beam carries an intensity $I$ and reflectivity is $0.75,$ The force exerted by the beam on the hemisphere is-

• A. $\frac{I\pi R^2}{c^2}\left(1.75\right)$
• B. $\frac{I\pi R^2}{c}\left(1.75\right)$
• C. Zero
• D. $\frac{I\pi R^2}{c}\left(0.25\right)$

Answer 1

The correct option is B $\frac{I\pi R^2}{c}\left(1.75\right)$

For a partially reflecting surface, if the radiation falls normally on it, the radiation pressure exerted on it is,

$P=\frac{F}{A}=\frac{I}{c}(1+r)$

$\Rightarrow F=PA=\frac{I}{c}(1+r)A.......\left(1\right)$

Where, $I=\text{Intensity}c=\text{Speed of light}r=\text{Reflectivity}=0.75$



The effective area $\perp$to the energy flux = $\pi R^2$

From (1) we get,

$F=\frac{I}{c}(1+0.75)\times A=\frac{I\pi R^2}{c}\left(1.75\right)$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.

Why this question ? Key concept : $\text{Radiation force}=\text{Radiation pressure(due to absorbtion)}\times \text{Effective area}\perp \text{to the flow of energy}$