= Solution
The <source-free Maxwell equations in a linear medium> are
$$
\boxed{
\nabla\mathbin{\cdot}\mathbf D=0,\qquad
\nabla\mathbin{\cdot}\mathbf B=0,\qquad
\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t},
\qquad
\nabla\times\mathbf H=\frac{\partial\mathbf D}{\partial t}.}
$$
Here the <constitutive relations> are $\mathbf D=\varepsilon\mathbf E$ and $\mathbf B=\mu\mathbf H$.
Substitution of the sinusoidal plane waves into the two divergence equations gives the <transversality of an electromagnetic plane wave>,
$$
\mathbf k\mathbin{\cdot}\mathbf E_0=0,
\qquad
\mathbf k\mathbin{\cdot}\mathbf B_0=0.
$$
<Faraday's law> and the <Ampère-Maxwell equation> give, respectively,
$$
\mathbf k\times\mathbf E_0=\omega\mathbf B_0,
\qquad
\mathbf k\times\mathbf B_0=-\mu\varepsilon\omega\mathbf E_0.
$$
Taking $\mathbf k\times$ the first equation and using transversality yields
$$
-k^2\mathbf E_0
=-\mu\varepsilon\omega^2\mathbf E_0.
$$
Thus a nonzero <plane electromagnetic wave in a linear medium> must satisfy
$$
\boxed{
k^2=\mu\varepsilon\omega^2,\qquad
\mathbf B_0=\frac{\mathbf k\times\mathbf E_0}{\omega},\qquad
v_{\rm ph}=\frac{\omega}{k}=\frac1{\sqrt{\mu\varepsilon}}.}
$$
In particular, $\mathbf E_0$, $\mathbf B_0$, and $\mathbf k$ are mutually orthogonal, with $|\mathbf B_0|=\sqrt{\mu\varepsilon}\,|\mathbf E_0|$.
Take the interface to be $x=0$ and the plane of incidence to be the $xy$-plane. The incident, reflected, and transmitted wavevectors may be written
$$
\begin{aligned}
\mathbf k_I&=k_-(\cos\theta_I,\sin\theta_I,0),\\
\mathbf k_R&=k_-(-\cos\theta_R,\sin\theta_R,0),\\
\mathbf k_T&=k_+(\cos\theta_T,\sin\theta_T,0).
\end{aligned}
$$
The <phase matching at a planar wave interface> requires one common angular frequency and one common tangential wavevector:
$$
k_-\sin\theta_I
=k_-\sin\theta_R
=k_+\sin\theta_T.
$$
Since $k_\pm=\omega\sqrt{\varepsilon_\pm\mu}
=n_\pm\omega/c$, this gives the <Snell law for electromagnetic waves>,
$$
\boxed{\theta_R=\theta_I,\qquad
n_-\sin\theta_I=n_+\sin\theta_T.}
$$
For the final polarization, take the signed electric amplitudes along $\widehat{\mathbf z}$. The <dielectric-interface boundary conditions> require continuity of tangential $\mathbf E$ and $\mathbf H$. Since
$$
\mathbf H=\frac{\mathbf k\times\mathbf E}{\mu\omega},
$$
their $z$- and $y$-components give
$$
E_I+E_R=E_T,
$$
and
$$
k_-\cos\theta_I(E_I-E_R)
=k_+\cos\theta_T E_T.
$$
Eliminating $E_I$ yields
$$
\frac{E_R}{E_T}
=\frac12\left(
1-\frac{k_+\cos\theta_T}{k_-\cos\theta_I}
\right).
$$
Using $\theta_R=\theta_I$ and Snell's law,
$$
\frac{k_+\cos\theta_T}{k_-\cos\theta_I}
=\frac{\sin\theta_R\cos\theta_T}
{\sin\theta_T\cos\theta_R}
=\frac{\tan\theta_R}{\tan\theta_T}.
$$
Therefore the <reflected-to-transmitted transverse-electric amplitude ratio> is
$$
\boxed{
\frac{|\mathbf E_R|}{|\mathbf E_T|}
=\frac12\left(
1-\frac{\tan\theta_R}{\tan\theta_T}
\right),}
$$
with the polarization directions chosen so that the displayed signed factor is nonnegative; otherwise its absolute value gives the magnitude ratio.
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