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The source-free Maxwell equations in a linear medium areHere the constitutive relations are $\mathbf{D}=\epsilon \mathbf{E}$ 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,Faraday's law and the Ampère-Maxwell equation give, respectively,Taking $\mathbf{k}\times$ the first equation and using transversality yieldsThus a nonzero plane electromagnetic wave in a linear medium must satisfyIn particular, ${\mathbf{E}}_0$, ${\mathbf{B}}_0$, and $\mathbf{k}$ are mutually orthogonal, with $|{\mathbf{B}}_0|=\sqrt{\mu \epsilon }|{\mathbf{E}}_0|$.

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 writtenThe phase matching at a planar wave interface requires one common angular frequency and one common tangential wavevector:Since $k_{\pm }=\omega \sqrt{{\epsilon }_{\pm }\mu }=n_{\pm }\omega /c$, this gives the Snell law for electromagnetic waves,

For the final polarization, take the signed electric amplitudes along $\hat{\mathbf{z}}$. The dielectric-interface boundary conditions require continuity of tangential $\mathbf{E}$ and $\mathbf{H}$. Sincetheir $z$- and $y$-components giveandEliminating $E_I$ yieldsUsing ${\theta }_R={\theta }_I$ and Snell's law,Therefore the reflected-to-transmitted transverse-electric amplitude ratio iswith the polarization directions chosen so that the displayed signed factor is nonnegative; otherwise its absolute value gives the magnitude ratio.

For a time-independent field, Faraday's law gives $\nabla \times \mathbf{E}=0$. Integrating around a thin rectangular loop crossing $x=0$ and shrinking its height gives continuity of the tangential component:A thin Gaussian pillbox and $\nabla \cdot \mathbf{D}=0$ give continuity of the normal electric displacement field:These are the electrostatic cases of the dielectric-interface boundary conditions.

Let $\hat{n}$ be the unit normal pointing from the minus medium to the plus medium. Integrating the Maxwell equations across an infinitesimal pillbox or loop gives the dielectric-interface boundary conditionsandThus tangential $E$ and $H$ are continuous, as are normal $D$ and $B$.

Because $D=ϵE$, the normal electric field generally jumps according toThe tangential displacement field generally jumps because ${ϵ}_{+}\ne {ϵ}_{-}$. Since the permeability is common, continuity of tangential $H$ is continuity of tangential $B$; together with normal-$B$ continuity, every component of $B$ is continuous. The same common $\mu$ also makes the normal component of $H$ continuous.