Question

Solar radiation of intensity $1.5{\text{kW/m}}^2$ is incident on a perfectly black surface, making an angle of ${30}^{\circ }$ with the surface. What is the pressure experienced by the surface?

• A. $0.25\times {10}^{-6}{\text{N/m}}^2$
• B. $0.25\times {10}^{-8}{\text{N/m}}^2$
• C. $1.25\times {10}^{-8}{\text{N/m}}^2$
• D. $1.25\times {10}^{-6}{\text{N/m}}^2$

Answer 1

The correct option is D $1.25\times {10}^{-6}{\text{N/m}}^2$

As it is a perfectly black surface, all the incident radiation will be absorbed.
Angle made by the incident radiation with the vertical is, $\theta ={90}^{\circ }-{30}^{\circ }={60}^{\circ }$

The pressure experienced by the surface is,
$P=\frac{F\cos \theta }{A}=\frac{I}{c}{\cos }^2\theta$

$=\frac{1.5\times {10}^3}{3\times {10}^8}\times {\cos }^2{60}^{\circ }$

$=1.25\times {10}^{-6}{\text{N/m}}^2$

Hence, $\left(C\right)$ is the correct answer.