= 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> give
$$
\nabla\times\mathbf E=i\omega\mu_0\mathbf H,\qquad
\nabla\times\mathbf H=(\sigma-i\omega\epsilon)\mathbf E,
\qquad k^2=\mu_0\epsilon\omega^2+i\mu_0\sigma\omega.
$$
In the good-conductor limit $\sigma\gg\omega\epsilon$, take the root with positive attenuation into the slab:
$$
k\simeq\frac{1+i}{\delta},\qquad \delta=\sqrt{\frac2{\mu_0\sigma\omega}},
\qquad
\boxed{\widetilde E_y(x)=E_0 e^{(i-1)x/\delta}.}
$$
The real physical field is $\operatorname{Re}[\widetilde E_y e^{-i\omega t}]$. 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 $\widetilde H_z=k\widetilde E_y/(\mu_0\omega)$. The <surface impedance of a good conductor> and the complete transmitted <magnetic field> are consequently
$$
\boxed{Z=\frac{\widetilde E_y}{\widetilde H_z}
=\frac{\mu_0\omega}{k}\simeq(1-i)\sqrt{\frac{\mu_0\omega}{2\sigma}}
=\frac{1-i}{\sigma\delta},\qquad
\widetilde{\mathbf H}=\frac{1+i}{\mu_0\omega\delta}E_0e^{(i-1)x/\delta}\widehat{\mathbf z}.}
$$
The 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>:
$$
\langle\mathbf N\rangle=\frac12\operatorname{Re}(\widetilde{\mathbf E}\times\widetilde{\mathbf H}^{\,*})
=N_0e^{-2x/\delta}\widehat{\mathbf x},\qquad
\boxed{N_0=\frac{|E_0|^2}{2\mu_0\omega\delta}=\frac{\sigma\delta|E_0|^2}{4}.}
$$
As an independent <energy> check, $-\partial_x\langle N_x\rangle=\sigma|E_0|^2e^{-2x/\delta}/2$, the mean <Joule heating> per unit <volume>.
<Ohm's law> gives $\widetilde J_y=\sigma\widetilde E_y$ and $\widetilde B_z=\mu_0\widetilde H_z$. Hence the averaged <Lorentz force> per unit <volume> is
$$
\langle f_x\rangle=\frac12\operatorname{Re}(\widetilde J_y\widetilde B_z^*)
=\sigma\mu_0 N_0 e^{-2x/\delta}.
$$
Integrating through the illuminated half-space gives the positive inward <radiation pressure>
$$
\boxed{p=\int_0^\infty\langle f_x\rangle dx=\frac{\sigma\delta\mu_0}{2}N_0.}
$$
Finally let $Z_0=\sqrt{\mu_0/\epsilon_0}$ 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 $|Z|\ll Z_0$,
$$
R=|r|^2=1-\frac{4Z_0\operatorname{Re}Z}{|Z+Z_0|^2}
\simeq1-\frac{4\operatorname{Re}Z}{Z_0},\qquad
\boxed{R\simeq1-2\sqrt{\frac{2\omega\epsilon_0}{\sigma}}.}
$$
Here $N_0=(1-R)I_{\rm inc}$ by surface <energy> balance. Combining the leading formulas gives $p\simeq2I_{\rm inc}/c$, the nearly perfect-reflector limit. The displayed reflection expansion also assumes a good conductor relative to the incident <vacuum> wave, $\omega\epsilon_0/\sigma\ll1$.
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