Electromagnetic waves > Properties of emission and reflection of electromagnetic waves > Problem 12.1.26

Statement

$12.1.26.$

The average pressure of a plane sinusoidal wave incident at an angle α on
a metal surface is equal to P. Determine the amplitude of the electric field
strength of this wave.

Solution

The average radiation pressure on a perfect mirror for a sinusoidal plane wave incident at an angle$ \alpha $(measured with respect to the normal) is:

$P = \frac{2I}{c}\cos^2\alpha$

where I is the average intensity of the incident wave. For a plane electromagnetic wave in vacuum

$ I = \dfrac{1}{2}\varepsilon_0 c E_0^2 $

where E_0 is the amplitude of the electric field. Substituting and solving for$ E_0$:

$P = \varepsilon_0 E_0^2 \cos^2\alpha
\quad\Longrightarrow\quad
\boxed{E_0 = \frac{1}{\cos\alpha}\sqrt{\frac{P}{\varepsilon_0}}}$

If the angle were measured with respect to the surface,$ \cos\alpha$ is replaced by$ \sin\alpha$. In the CGS system, the equivalent expression is \displaystyle $E_0 = \frac{\sqrt{4\pi P}}{\cos\alpha}$
(But this is only for knowledge )

Answer

$\boxed{E_0 = \frac{1}{\cos\alpha}\sqrt{\frac{P}{\varepsilon_0}}}$