[The naive explanation for the pressure of light offered in section 9.2.3 has its flaws, as you discovered if you worked Problem 9.11. Here's another account, due originally to Planck.] A plane wave travelling through vaccum in the z direction encounters a perfect conductor occupying the region $\mathbf{z}\mathbf{\ge }\mathbf{0}$, and reflects back:

$\mathbf{E}\left(z,t\right)\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\left[\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)-\cos \left(\mathrm{kz}+\mathrm{\omega t}\right)\right]\hat{\mathbf{x}}\mathbf{,}\mathbf{}\left(z>0\right)$

1. Find the accompanying magnetic field (in the region $\left(z>0\right)$)
2. Assuming $\mathbf{B}\mathbf{=}\mathbf{0}$inside the conductor find the current K on the surface $\mathbf{z}\mathbf{=}\mathbf{0}$, by invoking the appropriate boundary condition.
3. Find the magnetic force per unit area on the surface, and compare its time average with the expected radiation pressure (Eq.9.64).

Write the expression for the electric field for $\mathrm{z}>0$.

$\mathbf{E}\left(z,t\right)\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\left[\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)+\cos \left(\mathrm{kz}+\mathrm{\omega t}\right)\right]\hat{\mathbf{x}}$

Here k is the wave number, $\mathrm{\omega }$is the angular frequency and t is the time.

(a)

Since, $\left(\mathrm{E}\times \mathrm{B}\right)$points in the direction of propagation, write the equation for the magnetic field.

$\mathrm{B}=\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)+\cos \left(\mathrm{kz}+\mathrm{\omega t}\right)\right]\hat{\mathrm{y}}$

Therefore, the magnetic field is$\mathrm{B}=\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)+\cos \left(\mathrm{kz}+\mathrm{\omega t}\right)\right]\hat{\mathrm{y}}$.

(b)

It is known that $\mathrm{K}\times \left(-\hat{\mathrm{z}}\right)=\frac{1}{{\mathrm{\mu }}_0}\mathrm{B}$.

Substitute $\mathrm{B}=\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)+\cos \left(\mathrm{kz}+\mathrm{\omega t}\right)\right]\hat{\mathrm{y}}$in the above equation.

$\begin{array}{rcl}\mathrm{K}\times \left(-\hat{\mathrm{z}}\right)& =& \frac{1}{{\mathrm{\mu }}_0}\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{\omega t}\right)+\cos \left(\mathrm{kz}+\mathrm{\omega t}\right)\right]\hat{\mathrm{y}}\\ & =& \frac{{\mathrm{E}}_0}{{\mathrm{\mu }}_0\mathrm{c}}\left[2\cos \left(\mathrm{\omega t}\right)\right]\hat{\mathrm{y}}\\ \mathrm{K}& =& \frac{2{\mathrm{E}}_0}{{\mathrm{\mu }}_0\mathrm{c}}\cos \left(\mathrm{\omega t}\right)\hat{\mathrm{y}}\end{array}$

Therefore, the current K on the surface is$\mathrm{K}=\frac{2{\mathrm{E}}_0}{{\mathrm{\mu }}_0\mathrm{c}}\cos \left(\mathrm{\omega t}\right)\hat{\mathrm{y}}$.

(c)

Write the expression for the force per unit area

$\mathrm{f}=\mathrm{K}\times {\mathrm{B}}_{\mathrm{avg}}$

Substitute $\mathrm{K}=\frac{2{\mathrm{E}}_0}{{\mathrm{\mu }}_0\mathrm{c}}\cos \left(\mathrm{\omega t}\right)\hat{\mathrm{x}}$expression and ${\mathrm{B}}_{\mathrm{avg}}=\left(\mathrm{cos\omega t}\right)\hat{\mathrm{y}}$in the above expression.

$$\begin{gathered} \mathrm{f}=\left(\frac{2{{\mathrm{E}}_0}^2}{{{\mathrm{\mu }}_0}^2\mathrm{c}}\cos \left(\mathrm{\omega t}\right)\hat{\mathrm{x}}\right)\times \left[\left(\mathrm{cos\omega t}\right)\hat{\mathrm{y}}\right] \\ \mathrm{f}=2{{\mathrm{\epsilon }}_0{\mathrm{E}}_0}^2{\cos }^2\left(\mathrm{\omega t}\right)\hat{\mathrm{z}} \end{gathered}$$

It is known that the time average of ${\cos }^2\left(\mathrm{\omega t}\right)$is $\frac{1}{2}$. Hence, the above eqution becomes,

$$\begin{gathered} \mathrm{f}=2{\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2\left(\frac{1}{2}\right)\hat{\mathrm{z}} \\ \mathrm{f}={\mathrm{\epsilon }}_0{{\mathrm{E}}_0}^2 \end{gathered}$$

This is twice the pressure in Eq. 9.64, but that was for a perfect absorber, whereas this is a perfect reflector.