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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.