91Ó°ÊÓ

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

$\mathbf{E}\left(z,t\right)\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\left[\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)+\cos \left(\mathrm{kz}+\mathrm{Ó\neg ³Ù}\right)\right]\hat{\mathbf{x}}$

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

(a)

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

$\mathrm{B}=\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)+\cos \left(\mathrm{kz}+\mathrm{Ó\neg ³Ù}\right)\right]\hat{\mathrm{y}}$

Therefore, the magnetic field is$\mathrm{B}=\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)+\cos \left(\mathrm{kz}+\mathrm{Ó\neg ³Ù}\right)\right]\hat{\mathrm{y}}$.

(b)

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

Substitute $\mathrm{B}=\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)+\cos \left(\mathrm{kz}+\mathrm{Ó\neg ³Ù}\right)\right]\hat{\mathrm{y}}$in the above equation.

$\begin{array}{rcl}\mathrm{K}×\left(-\hat{\mathrm{z}}\right)& =& \frac{1}{{\mathrm{μ}}_0}\frac{{\mathrm{E}}_0}{\mathrm{c}}\left[\cos \left(\mathrm{kz}-\mathrm{Ó\neg ³Ù}\right)+\cos \left(\mathrm{kz}+\mathrm{Ó\neg ³Ù}\right)\right]\hat{\mathrm{y}}\\ & =& \frac{{\mathrm{E}}_0}{{\mathrm{μ}}_0\mathrm{c}}\left[2\cos \left(\mathrm{Ó\neg ³Ù}\right)\right]\hat{\mathrm{y}}\\ \mathrm{K}& =& \frac{2{\mathrm{E}}_0}{{\mathrm{μ}}_0\mathrm{c}}\cos \left(\mathrm{Ó\neg ³Ù}\right)\hat{\mathrm{y}}\end{array}$

Therefore, the current K on the surface is$\mathrm{K}=\frac{2{\mathrm{E}}_0}{{\mathrm{μ}}_0\mathrm{c}}\cos \left(\mathrm{Ó\neg ³Ù}\right)\hat{\mathrm{y}}$.

(c)

Write the expression for the force per unit area

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

Substitute $\mathrm{K}=\frac{2{\mathrm{E}}_0}{{\mathrm{μ}}_0\mathrm{c}}\cos \left(\mathrm{Ó\neg ³Ù}\right)\hat{\mathrm{x}}$expression and ${\mathrm{B}}_{\mathrm{avg}}=\left(\mathrm{cosÓ\neg ³Ù}\right)\hat{\mathrm{y}}$in the above expression.

$$\begin{gathered} \mathrm{f}=\left(\frac{2{{\mathrm{E}}_0}^2}{{{\mathrm{μ}}_0}^2\mathrm{c}}\cos \left(\mathrm{Ó\neg ³Ù}\right)\hat{\mathrm{x}}\right)×\left[\left(\mathrm{cosÓ\neg ³Ù}\right)\hat{\mathrm{y}}\right] \\ \mathrm{f}=2{{\mathrm{Î\mu }}_0{\mathrm{E}}_0}^2{\cos }^2\left(\mathrm{Ó\neg ³Ù}\right)\hat{\mathrm{z}} \end{gathered}$$

It is known that the time average of ${\cos }^2\left(\mathrm{Ó\neg ³Ù}\right)$is $\frac{1}{2}$. Hence, the above eqution becomes,

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

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