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A transverse electric polarization with constant permeability and conserved transverse wavenumber $k_0\sin {\theta }_0$ satisfies $e^{\prime \prime }+[{\beta }^2-({\nu }^2-1/4)/(z-z_0{)}^2]e=0$, with ${\beta }^2=k_0^2(a^2-{\sin }^2{\theta }_0)$ and ${\nu }^2=1/4+k_0^2{b}^2$. Substitution $e=\sqrt{|z-z_0|}{Z}_{\nu }(\beta |z-z_0|)$ reduces it to the Bessel equation. The profile is singular at $z_0$, so each side requires its own boundary conditions.

Use time dependence $e^{i\omega t}$ and consider fields independent of $y$, with $\mathbf{E}=\hat{\mathbf{y}}{E}_y(x,z)$. The source-free Maxwell's equations are $\nabla \times \mathbf{E}=-i\omega \mu \mathbf{H}$ and $\nabla \times \mathbf{H}=i\omega \epsilon \mathbf{E}$. The first givesSubstitute these into the $y$ component of the second equation. The TE wave equation in a stratified magnetodielectric isBecause its coefficients depend only on $z$, ${\partial }_x{E}_y$ obeys the same equation. Since it is proportional to $\mu H_z$, substitution and division by $\mu$ giveFor a conserved transverse wavenumber $k_x$, replace ${\partial }_x^2$ by $-k_x^2$ to obtain the corresponding ordinary Helmholtz equations.

If “propagating along the $z$ axis” is read as strict normal incidence with no $x$ dependence, then $H_z=0$ identically. The nonzero magnetic component is $H_x$, and its equation isThe $H_z$ equation above applies to the oblique transverse electric polarization used in the next part and also includes the zero normal-incidence solution. This distinguishes the named component from the usual transverse magnetic field.

Assume a hollow, perfectly conducting rectangular electromagnetic waveguide filled with air approximated by vacuum. Let its wider side be $a=0.040\mathrm{m}$ and its narrower side $b=0.020\mathrm{m}$. The longitudinal propagation constant of a mode isso the waveguide cutoff frequency isFor a mode with transverse electric polarization, one of $m,n$ may vanish, but they cannot both vanish. For a mode with transverse magnetic polarization, both must be positive. A single hollow conducting guide has no transverse electromagnetic mode with zero cutoff.

The lowest cutoff therefore belongs to the ${\mathrm{TE}}_{10}$ mode, varying across the wider side:Strictly, a traveling mode transporting energy down the guide requires $f>f_c$. At exactly $f_c$, $k_z=0$ and the axial group velocity vanishes. Below cutoff the field is evanescent. A different dielectric filling would replace $c$ by the wave speed in that material.

An incident electromagnetic wave generally produces reflected and transmitted waves because the two media have different electromagnetic response. Consider a flat interface between homogeneous, linear, lossless media with permittivities ${ϵ}_j$ and magnetic permeabilities ${\mu }_j$. Their wave speeds and wave impedances areFor each plane electromagnetic wave, $\mathbf{H}=\hat{\mathbf{k}}\times \mathbf{E}/Z_j$. With no free surface charge or current, the interface conditions for electromagnetic fields require continuous tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{D},\mathbf{B}$. Equality of the phase at every point on the boundary gives equal angular frequencies and tangential wavevectors. Hence the reflection angle equals the incidence angle and Snell's law iswhere $n_j=c/v_j$ and each angle is measured from the normal.

At normal incidence use the same tangential axis for all electric amplitudes. The two independent boundary equations areSolving gives the amplitude Fresnel equationsFor nonmagnetic media, these become $r=(n_1-n_2)/(n_1+n_2)$ and $t=2n_1/(n_1+n_2)$. Reflection into a medium of larger refractive index therefore reverses the electric amplitude's sign at normal incidence.

The normal Poynting vector determines the reflected and transmitted fractions of incident power. At normal incidence,It is the power fractions, rather than the squared electric amplitudes alone, that add to unity.

At oblique incidence separate transverse electric polarization ($s$, electric field perpendicular to the incidence plane) from transverse magnetic polarization ($p$, electric field in that plane). For nonmagnetic media, writing $c_i=\cos {\theta }_i$ and $c_t=\cos {\theta }_t$, the Fresnel equations areHere choose a common $s$ direction and, for each ray, the $p$ direction as $\hat{\mathbf{s}}\times \hat{\mathbf{k}}$. This makes the reflected $p$ basis reverse its tangential direction at normal incidence, explaining why the displayed $r_p$ then has the opposite sign to $r_s$. Physical reflectance is independent of this sign convention. For either polarization with propagating transmission,For $p$ polarization, zero reflection occurs at the Brewster angle $\tan {\theta }_B=n_2/n_1$. If $n_1>n_2$, incidence above the critical angle for total internal reflection $\sin {\theta }_c=n_2/n_1$ produces an evanescent field in total internal reflection and $\mathcal{R}=1$ for an isolated lossless interface. A lossy second medium requires complex response coefficients and an absorption term in the energy balance.

For a ray meeting an interface at nonzero incidence angle, the plane spanned by the incoming wavevector and the surface normal. transverse electric polarization has electric field perpendicular to this plane, while transverse magnetic polarization has electric field in it. At exact normal incidence, the surface normal and ray alone do not select a unique plane.

With permittivity and permeability depending only on $z$ and time dependence $e^{i\omega t}$, a transverse electric polarization $\mathbf{E}=\hat{\mathbf{y}}{E}_y(x,z)$ obeys the displayed equation. Maxwell's equations give $H_x=E_{y,z}/(i\omega \mu )$ and $H_z=-E_{y,x}/(i\omega \mu )$. The latter satisfies $H_{z,xx}+H_{z,zz}+({\mu }^{\prime }/\mu )H_{z,z}+[{\omega }^2\mu \epsilon +(\log \mu {)}^{\prime \prime }]H_z=0$. At normal incidence $H_z=0$; the nonzero magnetic component is $H_x$.

Relative to a propagation plane, transverse electric polarization has its electric field perpendicular to that plane. For a conducting corrugation independent of the perpendicular coordinate, the scalar electric component obeys the Helmholtz equation with Dirichlet data at the metal surface.