Question

The incident intensity on a horizontal surface at sea level from the sun is about $1kW{m}^{-2}$. Find the ratio of this pressure to atmospheric pressure $p_0$ (about $1\times {10}^{5}Pa)$ at sea level.
[Assuming that $50$ % of this intensity is reflected and $50$ % is absorbed]

• A. $5\times {10}^{-11}$
• B. $4\times {10}^{-8}$
• C. $6\times {10}^{-12}$
• D. $8\times {10}^{-11}$

Answer 1

The correct option is A $5\times {10}^{-11}$

Given,

$I=1kW/m^2$

$c=3\times {10}^{8}m/s$

$P_0=1\times {10}^{5}Pa$
Pressure exerted due to absorbed light, $P_a=\frac{I}{2c}$

Pressure exerted due to reflected light, $P_r=\frac{2I}{2c}$
Hence, the total radiation pressure on the surface can be given as:
$P_t=P_a+P_r=\frac{3I}{2c}$

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

So, the ratio of the radiation and atmospheric pressure will be:
$\frac{P}{P_0}=\frac{5\times {10}^{-6}}{1\times {10}^5}$

$\Rightarrow \frac{P}{P_0}=5\times {10}^{-11}$
Hence, the correct option is $\left(A\right)$