= Solution
<Electromagnetic waves> drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent <refractive index> is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.
Use <phasors> proportional to $e^{i(kz-\omega t)}$, with $k=k'+ik''$ and $k''\geq0$ for a passive medium. In a homogeneous, <isotropic>, linear material, let $\epsilon_b$ represent the <permittivity> associated with bound-charge <electric polarization>, $\mu$ the <magnetic permeability> and $\sigma(\omega)$ the mobile-charge conductivity. The macroscopic <Maxwell equations> give
$$
\nabla\times\mathbf E=i\omega\mathbf B,\qquad
\nabla\times\mathbf H=(\sigma-i\omega\epsilon_b)\mathbf E.
$$
For a transverse plane wave, taking another curl yields
$$
\boxed{k^2=\mu\epsilon_b\omega^2+i\mu\sigma(\omega)\omega
=\mu\omega^2\epsilon_{\rm eff}(\omega),\qquad
\epsilon_{\rm eff}=\epsilon_b+\frac{i\sigma(\omega)}\omega.}
$$
This division of the response avoids counting mobile <Electrons> both in $\epsilon_b$ and in $\sigma$. Define the <complex refractive index> $\widetilde n=ck/\omega=n+i\kappa$. For a nonmagnetic medium, $\widetilde n^2=\epsilon_{\rm eff}/\epsilon_0$. Its real part determines the phase variation, while the positive <optical extinction coefficient> $\kappa$ gives amplitude attenuation $e^{-\omega\kappa z/c}$ and intensity attenuation $e^{-2\omega\kappa z/c}$. The amplitude penetration length is
$$
\delta=\frac1{k''}=\frac c{\omega\kappa}.
$$
A simple microscopic explanation comes from the <Drude model>. Let <Electrons> of charge $-e$, mass $m$ and <number density> $n_e$ move against an effectively stationary positive background. Collisions relax their mean <velocity> at rate $\nu$:
$$
m(\dot{\mathbf v}+\nu\mathbf v)=-e\mathbf E,\qquad
\mathbf J=-n_ee\mathbf v.
$$
Consequently
$$
\boxed{\sigma(\omega)=\frac{n_ee^2}{m(\nu-i\omega)}
=\frac{\epsilon_0\omega_p^2}{\nu-i\omega},\qquad
\omega_p^2=\frac{n_ee^2}{m\epsilon_0}.}
$$
With a vacuum-like bound-charge background, this gives
$$
\frac{\epsilon_{\rm eff}}{\epsilon_0}
=1-\frac{\omega_p^2}{\omega(\omega+i\nu)}.
$$
Its imaginary part is positive for positive $\omega,\nu$, consistent with dissipation under the chosen <phasor> convention. Real materials can require an effective <Electron> mass, a nontrivial bound-charge response, or additional absorption associated with the <electronic band structure>.
A <plasma> is a conducting ionized medium with mobile <Electrons> and <ions>, approximately neutral on large scales because of <plasma screening>. For the electron-frequency transverse waves considered here, the heavy <ions> can often be treated as stationary. In a dilute weakly collisional <plasma>, $\nu\ll\omega$, so
$$
\boxed{n^2=1-\frac{\omega_p^2}{\omega^2},\qquad
\omega^2=\omega_p^2+c^2k^2.}
$$
For $\omega>\omega_p$, $k$ is real in this collisionless model and the wave propagates. Its <phase velocity> is $v_{\rm ph}=c/n>c$, while its <group velocity> is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega<\omega_p$, the refractive-index square is negative and
$$
k=i\frac{\sqrt{\omega_p^2-\omega^2}}c,
\qquad \delta=\frac c{\sqrt{\omega_p^2-\omega^2}}.
$$
The transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless <plasma> reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a <plasma> cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; <plasmas> also support longitudinal collective modes.
In a <metal>, positive <ions> usually form a <crystal> lattice. Mobile conduction <Electrons> occupy states described by the <electronic band structure> and normally form a degenerate <Fermi gas>. The conduction-electron <density> is usually much higher than in a dilute gas <plasma>. Thus its <plasma frequency> is commonly in the optical or <ultraviolet> range, rather than the lower radio-frequency values possible in dilute <plasmas>. <Electron> scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every <plasma> is collisionless and every <metal> collisional: the controlling ratio for either material is $\omega/\nu$, together with its carrier <density> and bound-charge response.
At sufficiently low frequency, $\omega\ll\nu$, the <metal>'s conductivity is approximately real, $\sigma_0=n_ee^2/(m\nu)$. In a good conductor the conduction current dominates the displacement current, $\sigma_0\gg\omega\epsilon_b$. The wave equation then gives
$$
k^2\simeq i\mu\sigma_0\omega,\qquad
\boxed{k\simeq(1+i)\sqrt{\frac{\mu\sigma_0\omega}{2}}=\frac{1+i}{\delta},
\quad\delta=\sqrt{\frac{2}{\mu\sigma_0\omega}}.}
$$
For example, a tangential electric field varies as
$$
E(z,t)=E_0e^{-z/\delta}\cos\left(\frac z\delta-\omega t\right).
$$
The current $\mathbf J=\sigma_0\mathbf E$ is therefore concentrated within a few skin depths of the surface: this is the <skin effect>. As frequency increases within this regime, $\delta\propto\omega^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.
The <Poynting theorem> identifies the loss as <Joule heating>. The cycle-averaged heating <density> for a complex conductivity is
$$
\langle\mathbf J\cdot\mathbf E\rangle=\frac12\operatorname{Re}\sigma\,|\mathbf E|^2.
$$
The <wave impedance> is $Z=\mu\omega/k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{\mu_0/\epsilon_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the <skin depth>.
At high enough frequency that $\omega\gg\nu$, a <metal>'s free-electron response also approaches the collisionless <plasma> form. The low-frequency formula $\delta\propto\omega^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the <electronic band structure> and transitions between bands. Both <plasmas> and <metals> are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.
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