Question

The incident intensity of light from the sun on a horizontal surface at sea level is about $1{\text{kW/m}}^2$. Assuming that $50\%$ of this light is reflected and $50\%$ is absorbed, the radiation pressure on this horizontal surface will be -

• A. $5\times {10}^{-6}\text{Pa}$
• B. $6\times {10}^{-6}\text{Pa}$
• C. $7\times {10}^{-6}\text{Pa}$
• D. $8\times {10}^{-6}\text{Pa}$

Answer 1

The correct option is A $5\times {10}^{-6}\text{Pa}$

Radiation pressure,

$P=50\%\times \frac{2I}{c}+50\%\times \frac{I}{c}$

$[\because P=P_{\text{reflection}}+P_{\text{absorption}}]$

$\Rightarrow P=\frac{I}{c}+\frac{I}{2c}=\frac{3I}{2c}$

$\Rightarrow P=\frac{3\times 1\times {10}^3}{2\times 3\times {10}^8}=5\times {10}^{-6}\text{Pa}$

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