Q155 / 160 · TS EAMCET 2019 (06 May Shift 1) · Physics
A parallel beam of monochromatic light of frequency $v$ is incident on a surface. The intensity of the beam is $I$ and area of the surface is $A$. Find the force exerted by the light of beam on the surface is perfectly reflecting and the light beam is incident at an angle of incidence $\theta$. (The speed of light is denoted as $c$.)
- A$\frac{2 / A \sin ^2 \theta}{\pi C}$
- B$\frac{\mid A \cos ^2 \theta}{C}$
- Correct answerC$\frac{21 A \cos ^2 \theta}{c}$
- D$\frac{\mid A \cos ^2 \theta}{\sqrt{2} C}$
Solution
When a light beam is incident at an angle of
incidence $\theta$, the intensity across the surface is geometrically reduced by the cosine of that angle and the component of the radiation force against the surface will also be reduced by the $\operatorname{cosine}$ of $\theta$, hence resulting pressure,
$p_{\text {incident }}=\frac{I}{c} \cos ^2 \theta$.
Where, $c$ is velocity of light.
Since, light wave reflects also from reflecting surface, hence the recoil due to the reflected wave will further contribute to the radiation pressure.
$\begin{aligned} \therefore \quad p_{\text {enitted }} & =\frac{I}{c} \cos ^2 \theta \\ \therefore \quad p_{\text {net }} & =p_{\text {incident }}+p_{\text {emitted }} \\ & =\frac{I}{c} \cos ^2 \theta+\frac{I}{c} \cos ^2 \theta \\ p_{\text {net }} & =\frac{2 I}{c} \cos ^2 \theta\end{aligned}$
$\therefore$ Force exerted by the light beam,
$F=p_{\text {net }} \times \operatorname{Area}(A)=\frac{2 L A \cos ^2 \theta}{c}$
Hence, the correct answer is option: 'C'.