= Critical angle for total internal reflection
{title2=$\theta_c$}
For <refractive indices> $n_->n_+$, the <critical angle for total internal reflection> satisfies $\sin\theta_c=n_+/n_-$. Above it, the transmitted normal <wavenumber> is imaginary and the transmitted field is an <evanescent wave>.
The <critical angle for total internal reflection> is the boundary between outgoing <refraction> and an <evanescent wave>, rather than an <angle> at which the external <electric field> abruptly disappears. Write $n_1=n_-$ and $n_2=n_+$. At the threshold, <Snell law for electromagnetic waves> has $\theta_t=90^\circ$, hence
$$
\theta_c=\arcsin\!\left(\frac{n_2}{n_1}\right),\qquad 0<n_2<n_1.
$$
Here <inverse sine> selects $0<\theta_c<90^\circ$. For $n_1=1.5$ and $n_2=1$, $\sin\theta_c=2/3$ and $\theta_c\approx41.8103149^\circ$. The threshold is independent of <polarization of an electromagnetic wave> under the stated assumptions.
A short proof using <phase matching at a planar wave interface> reveals all three cases at once. Choose a <Cartesian coordinate system> with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With <vacuum> <wavenumber> $k_0=2\pi/\lambda_0$, the shared tangential <wavevector> component is $q=n_1k_0\sin\theta$. The <dispersion relation> in medium two gives
$$
k_{2z}^{\,2}=n_2^2k_0^2-q^2=k_0^2\left(n_2^2-n_1^2\sin^2\theta\right).
$$
Below $\theta_c$, choose the positive real <square root> for an outgoing <plane electromagnetic wave>. At $\theta_c$, $k_{2z}=0$: there is a grazing <wave>, with zero normal average <power> but no exponential confinement in $z$. Above $\theta_c$, choose $k_{2z}=i\kappa$ with $\kappa>0$, so the <wave amplitude> decays into the second <half-space>.
If $n_1\leq n_2$, no nongrazing <angle of incidence> produces this threshold. As $n_2/n_1\to1$ from below, $\theta_c\to90^\circ$; as that ratio tends to zero, $\theta_c\to0$. These <limits> explain why a larger <refractive index> contrast permits confinement over a wider range of internal <angles>.
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