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At normal incidence from vacuum, the Fresnel equations give the reflectanceA good metal has $R$ close to one below its plasma frequency. At very low frequency the electromagnetic field penetrates only a skin depth; the local Drude limit gives $1-R\simeq \sqrt{8{ϵ}_0\omega /{\sigma }_0}$. In the collisionless negative-permittivity range the transmitted field is evanescent and reflection remains strong. Near the plasma edge the reflectance drops sharply; above it ${ϵ}_r^{\prime }>0$ and propagating waves become possible. In the ideal free-electron high-frequency limit,Finite damping rounds the plasma edge, and interband absorption changes the detailed response of actual metals.

Let the interface be $y=0$, with glass at $y<0$ and water at $y>0$, and choose $x$ parallel to the incident beam's tangential component. Define $k=\omega /c$ as the vacuum wavenumber, with glass and water refractive indices $n_g$ and $n_w$. Continuity of phase along the interface fixes $k_x=k{n}_g\sin \theta$. In water, the dispersion relation is $k_x^2+k_y^2=k^2{n}_w^2$, so above the critical angle,Choose the positive imaginary part because a field growing exponentially into the water would be unphysical. The evanescent field in total internal reflection is thereforeIt propagates in phase parallel to the interface but decays normally. Here $E_0$ is the transmitted field amplitude at the interface; relating it to the incident amplitude requires the appropriate polarization-dependent Fresnel equations. If the symbol $k$ instead denotes the incident glass wavenumber, replace every vacuum $k$ in the formula by $k/n_g$; equivalently $k_x=k\sin \theta$ and $\kappa =k\sqrt{{\sin }^2\theta -(n_w/n_g{)}^2}$. This states the wavenumber convention left implicit in the source.

Steven J. Byrnes developed a detailed treatment of multilayer optical calculations together with the tmm Python implementation. His primary preprint derives how reflected, transmitted and absorbed fields follow from interface relations and propagation through planar layers, including absorbing media and incoherent thick layers.^[ref]

The calculation retains optical phase as well as amplitude. Reflections from two surfaces can therefore cancel even though either surface alone reflects light. Treating the two reflected powers as independent contributions would miss this interference.

For an original normal-incidence antireflection coating example, take lossless refractive indices $n_0=1$, $n_1=1.5$ and $n_s=2.25$. At a vacuum wavelength of $600\mathrm{nm}$, a coating thickness $d=\lambda /(4n_1)=100\mathrm{nm}$ gives one-way phase $\delta =2\pi n_{1}d/\lambda =\pi /2$. The Fresnel equations give amplitude coefficients $r_{01}=r_{1s}=-0.2$, and the single-layer result isThus the ideal reflected power is zero at the stipulated wavelength, whereas the uncoated air–substrate interface reflects $|(1-2.25)/(1+2.25){|}^2\approx 14.8\%$. These are lossless model numbers, not a fabricated coating's performance.

Byrnes's analysis also explains why absorbing incident or final media require careful definitions of power flow: an amplitude coefficient and a transmitted-power fraction are not interchangeable. Sign conventions, complex wavevectors and interference assumptions belong in a reproducible calculation, alongside the layer thicknesses and refractive indices.^[ref]

For a plane electromagnetic wave in a homogeneous, linear, lossless, isotropic medium with permittivity $ϵ$ and magnetic permeability $\mu$, the wave impedance relates the field amplitudes by $Z=E/H=\sqrt{\mu /ϵ}$. Its units are ohms. Impedance differences determine the normal-incidence amplitude Fresnel equations, $r=(Z_2-Z_1)/(Z_2+Z_1)$ and $t=2Z_2/(Z_2+Z_1)$, with a common tangential electric-field axis.