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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 $c$.)
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The correct answer is:
$\frac{21 A \cos ^2 \theta}{c}$
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,
$F=p_{\text {net }} \times \operatorname{Area}(A)=\frac{2 L A \cos ^2 \theta}{c}$
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,
$F=p_{\text {net }} \times \operatorname{Area}(A)=\frac{2 L A \cos ^2 \theta}{c}$
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