 Solution

Use peak-amplitude phasors with time factor $e^{-i\omega t}$ and $\mu ={\mu }_0$. Faraday's law of induction and the Ampere-Maxwell law giveIn the good-conductor limit $\sigma \gg \omega ϵ$, take the root with positive attenuation into the slab:The real physical field is $\mathrm{Re}[{\tilde{E}}_y{e}^{-i\omega t}]$. A thick slab excludes a returning wave from its rear face. The skin depth is the electric-field amplitude decay length; the intensity decay length is $\delta /2$.

Faraday's law of induction fixes ${\tilde{H}}_z=k{\tilde{E}}_y/({\mu }_0\omega )$. The surface impedance of a good conductor and the complete transmitted magnetic field are consequentlyThe sign of the imaginary part belongs to the stated time convention. Average the Poynting vector using complex conjugation and the factor one half appropriate to peak phasors:As an independent energy check, $-{\partial }_x⟨N_x⟩=\sigma |E_0{|}^2{e}^{-2x/\delta }/2$, the mean Joule heating per unit volume.

Ohm's law gives ${\tilde{J}}_y=\sigma {\tilde{E}}_y$ and ${\tilde{B}}_z={\mu }_0{\tilde{H}}_z$. Hence the averaged Lorentz force per unit volume isIntegrating through the illuminated half-space gives the positive inward radiation pressureFinally let $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$ be the vacuum wave impedance. Tangential-field boundary conditions at normal incidence give the electric-amplitude reflection coefficient $r=(Z-Z_0)/(Z+Z_0)$. Since $|Z|\ll Z_0$,Here $N_0=(1-R)I_{\mathrm{inc}}$ by surface energy balance. Combining the leading formulas gives $p\simeq 2I_{\mathrm{inc}}/c$, the nearly perfect-reflector limit. The displayed reflection expansion also assumes a good conductor relative to the incident vacuum wave, $\omega {ϵ}_0/\sigma \ll 1$.