Problem 12.1.23

Solution

For a perfect reflector, the radiation pressure on the surface is $p = \dfrac{2I}{c}\cos^2\theta$, where I is the energy flux density, c is the speed of light, and $\theta$ is the angle of incidence measured with respect to the normal.

With $I = 600\ \text{W/m}^2 and c = 3.0\times10^8\ \text{m/s}$

$\frac{I}{c} = \frac{600}{3.0\times10^8} = 2.0\times10^{-6}\ \text{Pa}$

Normal incidence ($\theta = 0^\circ$):

$p = 2 \times 2.0\times10^{-6} = 4.0\times10^{-6}\ \text{Pa}$

Incidence at $30^\circ$ (with respect to the normal):

$p = 2 \times 2.0\times10^{-6} \times \cos^2 30^\circ = 4.0\times10^{-6} \times \left(\frac{\sqrt{3}}{2}\right)^2 = 4.0\times10^{-6} \times 0.75 = 3.0\times10^{-6}\ \text{Pa}$

If $30^\circ$ were interpreted as the angle with respect to the surface, it would be $60^\circ$ with respect to the normal and the pressure would be

$1.0\times10^{-6}\ \text{Pa}$

$\boxed{p_{\text{normal}} = 4\times10^{-6}\ \text{Pa},\qquad p_{30^\circ} = 3\times10^{-6}\ \text{Pa}}$