The intensity of sunlight hitting the earth is about $\mathbf{1300}\frac{\mathbf{W}}{{\mathbf{m}}^{\mathbf{2}}}$ . If sunlight strikes a perfect absorber, what pressure does it exert? How about a perfect reflector? What fraction of atmospheric pressure does this amount to?

Write the expression for the pressure in the case of a perfect absorber.

$\mathbf{P}\mathbf{=}\frac{\mathbf{l}}{\mathbf{C}}$ ….. (1)

Here, Iis the intensity, and c is the speed of light.

Write the expression for the pressure in the case of a perfect reflector.

$\mathbf{P}\mathbf{=}\frac{\mathbf{2}\mathbf{l}}{\mathbf{C}}$ ….. (2)

Substitute the values in equation (1).

$$\begin{gathered} P=\frac{\left(1300W/m^2\right)}{\left(3\times {10}^{8}m/s\right)} \\ P=4.3\times {10}^{-6}N/m^2 \end{gathered}$$

Substitute the values in equation (2).

$$\begin{gathered} P^{\text{'}}=\frac{2\left(1300W/m^2\right)}{\left(3\times {10}^{8}m/s\right)} \\ {P}^{\text{'}}=8.6\times {10}^{-6}N/m^2 \end{gathered}$$

The atmospheric pressure is equal to $1.03\times {10}^5\mathrm{N}/{\mathrm{m}}^2$
.Calculate the fraction of atmospheric pressure for a perfect absorber.

$$\begin{gathered} F_{absorber}=\frac{4.3\times {10}^{-6}N/m^2}{1.03\times {10}^{5}N/m^2} \\ {F}_{absorber}=4.17\times {10}^{-11} \end{gathered}$$

Calculate the fraction of atmospheric pressure for a perfect reflector.

$$\begin{gathered} F_{absorber}=\frac{8.6\times {10}^{-6}N/m^2}{1.03\times {10}^{5}N/m^2} \\ {F}_{absorber}=8.3\times {10}^{-11} \end{gathered}$$

Therefore, the pressure in the case of a perfect absorber and reflector isrole="math" localid="1657366852692" $4.3\times {10}^{-6}\mathrm{N}/{\mathrm{m}}^2$ and$8.6\times {10}^{-6}\mathrm{N}/{\mathrm{m}}^2$ respectively, and the fraction of atmospheric pressure for a perfect absorber and the perfect reflector is$4.17\times {10}^{-11}$ and $8.3\times {10}^{-11}$respectively.