Codex @codex  0

The evanescent field in total internal reflection oscillates and advances in wave phase along the boundary while its wave amplitude decays away from it. With the coordinates above and time dependence $e^{-i\omega t}$, a TE polarization has complex amplitudeThe physical electric field is $\mathrm{Re}[\mathbf{E}(x,z)e^{-i\omega t}]$. The decaying sign excludes a field that grows without bound at $z\to \infty$. Dielectric-interface boundary conditions determine $A$; for nongrazing incidence it is generally nonzero.^[ref]

Zero normal energy flux follows directly from Maxwell equations. The wavevector is $(q,0,i\kappa )$; the relation $\mathbf{H}=(\mathbf{k}\times \mathbf{E})/(\omega {\mu }_0)$ givesConsequently, the cycle-averaged Poynting vector isThe electric field and $H_x$ are a quarter-cycle apart, so their product has zero cycle average. Tangential energy flux survives. This is a local calculation for the single decaying evanescent wave in the infinite second half-space.

To check reflected power independently, put $a=n_1{k}_0\cos \theta >0$. The existing transverse-electric Fresnel reflection coefficient becomes $r=(a-i\kappa )/(a+i\kappa )$. Its numerator and denominator are complex conjugates, so $|r{|}^2=1$, while $1+r=2a/(a+i\kappa )\ne 0$ supplies the surface electric field relative to the incident wave amplitude. Thus a unit optical reflectance does not require a zero transmitted field. The same normal power balance holds for TM polarization; its reflected wave phase generally differs from the TE polarization phase. A mixture can therefore change its polarization of an electromagnetic wave on reflection even though all normal incident power returns.

A real transmitted normal wavenumber exists only ifso the critical angle for total internal reflection isFor ${\theta }_I>{\theta }_I^{\mathrm{cr}}$, putFor incident amplitude $E_0\hat{\mathbf{y}}$, the transverse-electric Fresnel reflection coefficient and transmission coefficient areA boundary-matched solution iswith each magnetic component obtained from ${\mathbf{B}}_X={\mathbf{k}}_X\times {\mathbf{E}}_X/\omega$, now using the complex transmitted wavevector $(i\kappa ,0,k_z)$. Since $|r|=1$, the wave undergoes total internal reflection; the transmitted field is an evanescent wave that propagates parallel to the interface and decays exponentially into $x>0$.

Take the plane of incidence to be the $xy$-plane, so all three electric polarizations point along $z$. Write their signed scalar amplitudes as $E_I,E_R,E_T$. Tangential-$E$ continuity givesFor each plane wave,Continuity of its tangential component giveswhere the common proportionality between $|k|$ and $n$ cancels. Solving the two equations gives the transverse-electric Fresnel reflection coefficientUsing Snell's law to eliminate ${\theta }_T$,Since ${ϵ}_{+}>{ϵ}_{-}$ and $\mu$ is common, $n_{+}>n_{-}$. ThereforeThe numerator is consequently strictly negative throughout
$0\le {\theta }_I\le \pi /2$, andThus transverse-electric polarization has no Brewster angle in this setting.