Plane electromagnetic wave ($\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$)

A monochromatic vacuum plane electromagnetic wave has phase $\mathbf{k}\cdot \mathbf{r}-\omega t$, where $\mathbf{k}$ is the wavevector, $\omega >0$ the angular frequency, and $\omega =c|\mathbf{k}|$. Its field amplitudes obey $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$ and ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$.

For the linear polarization and orientation in the visualization, choose a right-handed Cartesian coordinate system and putConstant phase means $z=ct+\text{constant}$, so propagation is along $+z$. The cross product $\hat{\mathbf{x}}\times \hat{\mathbf{y}}=\hat{\mathbf{z}}$ fixes the positive Poynting vector. Both fields reverse together every half-period and are in phase. Reversing the propagation direction with the same electric orientation reverses the magnetic orientation.

These fields are spatially uniform across every transverse plane. An infinite plane electromagnetic wave is an idealization with infinite total electromagnetic energy; it is useful as a local approximation and as a building block for superposition. For frequency $f=100\mathrm{MHz}$, $\omega =2\pi f$ and wavelength $\lambda =c/f\simeq 3.00\mathrm{m}$, independently of $E_0$.

For a source-free plane electromagnetic wave in vacuum, let $n^{\mu }=(1,\mathbf{s})$ be its null propagation Lorentz four-vector. In signature $(+---)$, take $F^{0i}=-E_i/c$, $F^{ij}=-{ϵ}_{ijk}{B}_k$, and define the dual electromagnetic field tensor by ${\tilde{F}}^{0i}=-B_i$, ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. Lower both indices with the Minkowski metric. Direct contraction givesBoth contractions vanish exactly when the transverse plane electromagnetic wave Maxwell equations hold. Being zero covectors, they remain zero in every inertial frame. Thus the transformed spatial wavevector is perpendicular to both transformed fields, even though the separate three-dimensional dot products are not themselves Lorentz scalars for arbitrary fields.

One contraction alone is insufficient without the other Maxwell equations. In particular, adding a magnetic field parallel to $\mathbf{s}$ leaves $n^{\mu }{F}_{\mu \nu }$ unchanged but violates $\mathbf{s}\cdot \mathbf{B}=0$. Conversely the two scalar perpendicularity conditions alone do not impose the required relation $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$ between field strengths and directions.

The polarization of an electromagnetic wave describes the path traced by its transverse electric-field vector at a fixed point. A monochromatic wave may be linearly, circularly, or elliptically polarized; in vacuum it has two independent transverse polarization states.

This specializes polarization of waves to electric-field oscillations.

For a vacuum plane electromagnetic wave, a Lorentz boost multiplies both transverse Jones vector amplitudes by the same positive Doppler factor when the basis is carried to the aberrated ray. Their ratio and phase difference therefore remain unchanged. All Stokes parameters gain the same squared factor, leaving normalized parameters, ellipse shape, handedness and degree of polarization unchanged. This does not mean the propagation direction or the components in an arbitrarily fixed pair of spatial axes remain unchanged.

The Faraday effect is magnetic field-induced circular birefringence: opposite circular polarizations acquire different wave phases, rotating transmitted linear polarization. For a transparent medium with isotropy before applying the magnetic field, in the linear magnetic-response regime at fixed vacuum wavelength ${\lambda }_0$ and temperature $T$,Here $V$ is the Verdet constant, $L$ is the material path length, and $B_{\parallel }$ is the applied magnetic flux density component along the ray. This $\mathbf{B}$ is the external bias, distinct from the oscillating magnetic field of the light itself. Michael Faraday discovered the effect in 1845.^[ref]

Fix the sign before using the formula. On the forward path choose a right-handed Cartesian coordinate system, propagation along $+z$, and measure $\theta$ from $+x$ toward $+y$ for the electric field's oscillation line. Positive $\theta$ is counterclockwise in the $x$-right, $y$-up end view seen from $z>0$, looking toward the source. Then $B_{\parallel }=B_z$ and positive $V$ means positive rotation for positive $B_z$. This convention differs from the clockwise-positive convention in optical rotation. A linear polarization line is unchanged by $180^{\circ }$; a measured orientation determines accumulated rotation only modulo $\pi$, unless continuity selects a branch.

For uniform $B_{\parallel }$, the formula reduces to $\theta =VBL$. Its appeal is that a weak change of refractive index becomes a visible change of orientation. Magnetic reversal measurement of Faraday rotation turns this into an inference of $V$ or of the magnetic field path integral. Faraday rotation on a round trip explains the directional behavior used by an optical isolator. In an astronomical cold plasma, the high-frequency limit instead leads to the rotation measure law $\theta =\mathrm{RM}{\lambda }_0^2$; that wavelength dependence is not a universal law for transparent solids.

The Verdet constant is the coefficient relating Faraday rotation to the magnetic flux density path integral. Its SI units are $\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. It depends on material, wavelength and temperature; a sign requires the stated viewing convention.^[ref]

The name commemorates Émile Verdet. For nonuniform material or temperature, use $\theta =\int V(\ell )B_{\parallel }(\ell )d\ell$ when the local scalar description remains valid. A single central magnetic field reading cannot generally replace this weighted integral. Strong additional linear birefringence or appreciable circular dichroism requires following the full polarization of an electromagnetic wave, rather than treating the output as merely a rotated line.

The factor of one half follows from superposition, without assigning ambiguous right/left names to the circular polarizations. Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}(z)e^{-i\omega t}]$ and fixed transverse unit vectorsFor equal transmitted wave amplitudes, let the accumulated wave phases be ${\varphi }_{+}$ and ${\varphi }_{-}$. Their sum isThe common wave phase $\bar{\varphi }$ sets the timing of the electric field oscillation; the real direction in parentheses sets its linear polarization. Opposite circular polarizations have changed their relative wave phase, yet their sum still oscillates along a line. Unequal transmitted wave amplitudes would generally give elliptical polarization instead.

In a uniform sample, with refractive indices $n_{+},n_{-}$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the phase difference is $(k_{+}-k_{-})L=k_0(n_{+}-n_{-})L$. ThusThe second relation identifies the small signed circular birefringence responsible for the rotation in this basis. Unlike a waveplate, whose separated modes have orthogonal linear polarizations, this ideal rotator has modes with opposite circular polarizations.

Faraday rotation exhibits optical nonreciprocity: with the bias magnetic field held fixed, a retracing beam gains another rotation in the same laboratory sense. Natural optical activity in a reciprocal medium instead unwinds on retracing the path.^[ref]

To reconcile this with the signed $B_{\parallel }$ formula, retain the laboratory $x,y,z$ axes for comparing physical lines. For the backward ray choose a right-handed ray basis $x^{\prime }=x$, $y^{\prime }=-y$, $z^{\prime }=-z$. Its magnetic flux density component is $B_{\parallel }^{\prime }=-B_z$, so its local rotation increment is $-\theta$. But positive $y^{\prime }$ is negative laboratory $y$: the backward increment is therefore $+\theta$ in the laboratory orientation plot. Reversing the ray reverses both the longitudinal component and the basis used for its signed angle. Reversing only one creates the false prediction of cancellation.

In fixed laboratory transverse components, define the rotation matrixFor an ideal mirror at normal incidence that contributes the same complex amplitude factor $r$ to both transverse components, the round-trip map is $rR(\theta )R(\theta )=rR(2\theta )$. The common mirror factor does not change the oscillation line. A reciprocal natural rotator instead gives $rR(-\alpha )R(\alpha )=rI$. This comparison assumes a retraced straight path and no additional polarization-changing elements.

For example, a chosen single-pass $30^{\circ }$ Faraday rotation produces a returned line at $60^{\circ }$ to the original laboratory $x$ axis. A chosen reciprocal $30^{\circ }$ natural rotator returns it to $0^{\circ }$. The same addition makes an ideal optical isolator possible: polarizers at $0^{\circ }$ and $45^{\circ }$ surround a $45^{\circ }$ Faraday rotator. Forward light follows $0^{\circ }\to 45^{\circ }$ and passes. Returning light selected at $45^{\circ }$ reaches $90^{\circ }$ at the entrance polarizer and is rejected by Malus's law. Real transmission loss and incomplete extinction limit the isolation.

Consider an original model at fixed vacuum wavelength $600\mathrm{nm}$ and temperature $20^{\circ }\mathrm{C}$. Stipulate a transparent sample with Verdet constant $V=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$, length $L=0.050\mathrm{m}$, uniform adjustable $B_z$, and a field-independent analyzer-zero offset ${\theta }_0=0.020\mathrm{rad}$. These are chosen values, not measured glass properties. On a continuously tracked small-angle branch,Taking half the difference gives ${\theta }_{\mathrm{odd}}=0.100\mathrm{rad}$; half the sum recovers ${\theta }_0=0.020\mathrm{rad}$. Hence $V={\theta }_{\mathrm{odd}}/(LB)=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. A field-reversal difference of $0.200\mathrm{rad}$ is twice the single-field rotation, not that rotation itself. More generally this subtraction removes every contribution that is even under $B\mapsto -B$, provided the sample and apparatus remain otherwise matched.

An ideal polarimeter analyzer at laboratory angle $a$ reads normalized light intensity $I(a,B)/I_0={\cos }^2[a-{\theta }_{\mathrm{obs}}(B)]$. Here $I_0$ is the light intensity immediately before the analyzer, separately calibrated for each condition. A scan locates the maximum at $a={\theta }_{\mathrm{obs}}$ modulo $\pi$. At $a=\pi /4$, expansion about zero gives $I/I_0=1/2+{\theta }_{\mathrm{obs}}+O({\theta }_{\mathrm{obs}}^3)$, with angles in radians. At crossed extinction, the response instead begins as ${\theta }_{\mathrm{obs}}^2$ and loses the sign near zero. The favorable $45^{\circ }$ slope follows directly by differentiating Malus's law; its sign reverses if the angular convention is reversed.

A William & Mary teaching experiment places a $10\mathrm{cm}$ glass rod inside a solenoid. Align a polarized helium-neon laser through the rod's centre onto a photodiode after a rotatable polarimeter analyzer. First scan the analyzer from maximum to minimum transmission in $5^{\circ }$ steps and record detector voltage, establishing the angular response. A function generator and electronic amplifier then drive sinusoidal electric current through the solenoid; a multimeter measures the electric current and an oscilloscope reads the photodiode. Trigger the oscilloscope from the generator, remove the large constant detector component by AC coupling, and average repeated traces. Convert the small synchronized voltage modulation to rotation using the calibrated angular slope. The manual fits rotation against magnetic flux density, using its stated coil calibration and converting measured root mean square current to peak current. This is a specified laboratory procedure, rather than measured results reproduced here.^[ref]

To test the inference, add matched zero-electric current readings before and after each reversal, hold the analyzer fixed during paired readings, and compare equal positive and negative electric currents. A blocked beam tests electrical pickup in the detector channel. These are proposed controls. Joule heating scales as current squared, so reversing electric current at equal magnitude helps separate an odd rotation from even heating effects; it does not remove drift between readings. Record temperature and allow repeatable settling. Changes in transmission, mechanical motion, stressed-glass birefringence and detector pickup can imitate a signal. Synchronized averaging suppresses unsynchronized noise, but cannot reject an artifact synchronized to the drive.

Quantitative extraction of the Verdet constant also needs the integral of magnetic flux density through the rod: a finite solenoid can have a nonuniform axial magnetic field. Use its calibrated path integral, or justify a uniform-field approximation; do not silently multiply a central field by the whole rod length. A signed reversal or phase-sensitive readout tests the part odd in applied magnetic field. Its agreement with the simple rotation law supports the model under these controls, rather than proving that every change of transmitted light intensity is Faraday rotation.

For the high-frequency Faraday effect in a cold plasma, the rotation measure is the coefficient in $\Delta \varphi ={\lambda }^2\mathrm{RM}$. It is a density-weighted line integral of the line-of-sight magnetic field; the coefficient $a_0$ depends on the units. A sign convention for the line of sight fixes its sign. Correlations across an extended source can constrain a foreground magnetic energy spectrum, subject to statistical assumptions.

For constant electron density, common path length $L$, and a solenoidal isotropic spectral tensor, the rotation-measure correlation contains the window $|{\int }_0^L{e}^{i{k}_{z}z}dz{|}^2$. As $L$ becomes large compared with the correlation length, this window acts as $2\pi L\delta (k_z)$ within the integral. The transverse projector then has $zz$ component one, yielding the displayed two-dimensional Fourier transform. For finite $L$, the windowed integral retains longitudinal modes and endpoint corrections. The method of using such rotation measure correlations to infer magnetic spectra is developed by Enßlin and Vogt.

Invert the two-dimensional Fourier transform in the long-path projection of a magnetic correlation, then perform the angular integration in polar coordinates. The angular integral is $2\pi J_0(kr)$, a Bessel function of the first kind, giving the displayed Hankel transform with the stated Fourier-Hankel normalization for an isotropic spectrum. The result recovers the scalar tensor coefficient $H$, whose shell-integrated magnetic energy spectrum has an additional factor $k^2$.

A linear polarizer transmits one linear polarization and suppresses the orthogonal one. Two ideal crossed polarizers block light unless an intervening medium changes its polarization of an electromagnetic wave. A twisted nematic field effect uses this selection to convert field-driven reorientation into brightness modulation.

A prism polarizer associated with William Nicol. Its birefringent crystal arrangement selects one transmitted polarization; two such prisms can serve as the polarizer and analyser in a polariscope.

A polarizing element used to examine the polarization of incident light. Rotation converts that polarization information into an intensity variation, described by Malus's law for an incident linearly polarized beam and an ideal linear analyser.

A reciprocal optical circular polarizer can be assembled from a linear polarizer followed by a quarter-wave plate whose axes make $45^{\circ }$ to the transmitted linear state. Forward passage converts unpolarized light into circular light of half the ideal incident intensity. Reverse passage encounters the retarder first; it leaves unpolarized light unpolarized until the linear polarizer transmits half as linear light. Thus the one-pass intensity can be equal in both directions even though the outgoing polarization differs. Use a coherency matrix, not a single Jones vector, for unpolarized input.

The Stokes parameters $I,Q,U,V$ describe intensity, two linear-polarization components and circular polarization. $Q$ and $U$ depend on the chosen transverse basis. On the celestial sphere, $Q\pm iU$ transform as spin-two fields rather than as ordinary scalars.

A wave is linearly polarized when its oscillating electric field always lies along one fixed line. For a transverse plane wave, that line is perpendicular to the wavevector. Unlike circular polarization, the field oscillates along the line rather than rotating through directions.

An ideal linear polarizer transmits the fraction ${\cos }^2\theta$ of incident linearly polarized light when its axis makes angle $\theta$ to the polarization. For a single photon, the corresponding conditional transmission probability is $|⟨\theta |0⟩{|}^2={\cos }^2\theta$, by the Born rule; rejection has probability ${\sin }^2\theta$. The classical field calculation projects its electric-field amplitude onto the transmission direction and squares that amplitude to obtain intensity. This is a conditional probability given the incident polarization and is not automatically the joint probability of passing one analyzer and failing another.

An electromagnetic wave has circular polarization when its transverse electric field has constant magnitude and rotates once during each period. In complex-amplitude notation, two orthogonal components have equal magnitude and a relative phase of one quarter-cycle.

At the surface of a perfect conductor, the tangential electric field and normal magnetic field vanish. An incident plane electromagnetic wave is therefore accompanied by a reflected wave whose tangential electric field cancels the incident field at the surface.

The linear carrier acceleration equation $\dot{\mathbf{j}}={ϵ}_0{\omega }_p^2\mathbf{E}$ and Faraday's law imply this conserved local quantity. Setting its initial value to zero gives the second London equation and London screening. An undamped response alone does not select that initial value, so it must not be used to infer expulsion of a pre-existing magnetic field for every magnetic history.^[ref]

If a metal surface is $z=h(x)$ and the incident field is independent of $y$, Maxwell's equations separate into TE polarization and TM polarization. The metal condition that the tangential electric field vanish sets $E_y=0$ in the TE sector. In the TM sector $\mathbf{E}=i\nabla \times (H_y\hat{\mathbf{y}})/(\omega ϵ)$, so the same condition gives $({\partial }_z-h^{\prime }{\partial }_x)H_y=0$. This explains why the TE problem agrees with scalar Dirichlet acoustics while the TM problem has different rough-surface boundary forcing.

At normal incidence, the reflected electric amplitude is the negative of the incident amplitude at the surface, while the reflected magnetic amplitude has the same sign. Their superposition forms a standing wave with an electric node and a magnetic antinode at the conductor.

Across an interface with unit normal $\mathbf{n}$ from medium one to medium two, Maxwell equations imply $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$, $\mathbf{n}\cdot ({\mathbf{B}}_2-{\mathbf{B}}_1)=0$, $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$, and $\mathbf{n}\times ({\mathbf{H}}_2-{\mathbf{H}}_1)={\mathbf{K}}_f$. Here ${\sigma }_f$ and ${\mathbf{K}}_f$ are the free surface electric charge and electric current densities.

Without free surface charge or current, the tangential components of $E$ and $H$ and the normal components of $D$ and $B$ are continuous across an interface.

Equality of the boundary fields for every tangential position and time forces equal frequency and equal tangential wavevector for incident, reflected, and transmitted plane waves.

For refractive indices $n_{-}$ and $n_{+}$,

Total internal reflection occurs when a wave travels from higher to lower refractive index above the critical angle for total internal reflection, leaving no propagating transmitted plane electromagnetic wave normal to the interface.

At an infinite planar interface, total internal reflection returns all incident normal power in the reflected plane electromagnetic wave. Assume homogeneous dielectrics with isotropy, linearity, no dielectric loss, magnetic permeability ${\mu }_0$, and positive real refractive indices $n_1>n_2$. Above the critical angle for total internal reflection, both TE polarization and TM polarization have unit optical reflectance despite the external evanescent wave.^[ref]

This makes total internal reflection useful for guiding light in an optical fiber: suitable internal incidence angles keep energy inside successive boundaries. Its more surprising feature is that perfect optical reflectance coexists with an external electric field. The evanescent field in total internal reflection can interact with a nearby object, as the frustrated total internal reflection experiment demonstrates.

For an original idealized example, choose $n_1=1.5$, $n_2=1$ at vacuum wavelength ${\lambda }_0=633\mathrm{nm}$. At angle of incidence $40^{\circ }$, Snell law for electromagnetic waves gives an angle of refraction $\arcsin (1.5\sin 40^{\circ })\approx 74.6186^{\circ }$, so a transmitted light ray still emerges. At ${\theta }_c\approx 41.8103^{\circ }$, its direction becomes tangent to the boundary. At $50^{\circ }$, the proposed sine of a real angle of refraction would be $1.5\sin 50^{\circ }\approx 1.14907>1$: the transmitted wavevector must instead have an imaginary normal component. These are predictions for the stipulated media, not measured properties of a particular block.

A real semicircular-block experiment isolates this threshold. The University of Maryland laboratory instructions place a laser, motor-driven rotation stage and photodetector together; an aperture limits the detector voltage. The semicircular block is centered on the rotation axis, with the photodetector aligned along its flat face. The instructions identify a strong grazing-beam signal and a weaker signal just before the critical angle for total internal reflection. The readout is detector voltage versus stage angle. Retracing the incident laser beam identifies normal incidence for alignment.^[ref]

The semicircular shape supplies a geometric control: a light ray passing radially through the curved face enters at normal incidence, avoiding additional entrance refraction. The weaker signal near the flat face locates a limiting angle of refraction; a vanishing remote-detector signal alone would not establish that the external electric field vanishes. Beam width and alignment limit angular resolution; no measured threshold or decay depth is supplied.^[ref]

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 ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. 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 $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, 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 $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

The evanescent field in total internal reflection oscillates and advances in wave phase along the boundary while its wave amplitude decays away from it. With the coordinates above and time dependence $e^{-i\omega t}$, a TE polarization has complex amplitudeThe physical electric field is $\mathrm{Re}[\mathbf{E}(x,z)e^{-i\omega t}]$. The decaying sign excludes a field that grows without bound at $z\to \infty$. Dielectric-interface boundary conditions determine $A$; for nongrazing incidence it is generally nonzero.^[ref]

Zero normal energy flux follows directly from Maxwell equations. The wavevector is $(q,0,i\kappa )$; the relation $\mathbf{H}=(\mathbf{k}\times \mathbf{E})/(\omega {\mu }_0)$ givesConsequently, the cycle-averaged Poynting vector isThe electric field and $H_x$ are a quarter-cycle apart, so their product has zero cycle average. Tangential energy flux survives. This is a local calculation for the single decaying evanescent wave in the infinite second half-space.

To check reflected power independently, put $a=n_1{k}_0\cos \theta >0$. The existing transverse-electric Fresnel reflection coefficient becomes $r=(a-i\kappa )/(a+i\kappa )$. Its numerator and denominator are complex conjugates, so $|r{|}^2=1$, while $1+r=2a/(a+i\kappa )\ne 0$ supplies the surface electric field relative to the incident wave amplitude. Thus a unit optical reflectance does not require a zero transmitted field. The same normal power balance holds for TM polarization; its reflected wave phase generally differs from the TE polarization phase. A mixture can therefore change its polarization of an electromagnetic wave on reflection even though all normal incident power returns.

Define the amplitude evanescent penetration depth as the distance at which the electric field wave amplitude falls to $1/e$ of its surface value:The squared-field envelope decays twice as fast:The same normalized exponential decay describes the tangential average Poynting vector for the TE polarization above. An intensity penetration depth convention often means $d_{1/e}$; it must not be confused with amplitude depth $\delta$ or interpreted as outgoing normal light intensity.

For the stipulated $n_1=1.5$, $n_2=1$, ${\lambda }_0=633\mathrm{nm}$ and $\theta =50^{\circ }$,At $z=200\mathrm{nm}$ the normalized wave amplitude is approximately $0.325099$, and its squared envelope is approximately $0.105689$. At $z=\delta$, these become $e^{-1}$ and $e^{-2}$, respectively. The quantity inside the square root is dimensionless, so $\kappa$ has inverse-length units and both depths have length units.

The sharp angular dependence has a simple explanation. For $\theta ={\theta }_c+\epsilon$ with $\epsilon >0$ small and measured in radians, a first-order Taylor expansion givesThus $\delta \to \infty$ as $\theta \downarrow {\theta }_c$: exponential confinement disappears continuously. With fixed refractive indices, $\delta$ decreases as the angle of incidence increases and approaches ${\lambda }_0/(2\pi \sqrt{n_1^2-n_2^2})\approx 90.1091\mathrm{nm}$ as $\theta \to 90^{\circ }$. This is a limit of the normalized depth; exactly grazing incidence supplies zero incident normal power. Real beam widths, angular spreads and optical dispersion matter when applying these ideal plane electromagnetic wave predictions near the threshold.

Frustrated total internal reflection occurs when a nearby second high-index dielectric couples to the evanescent field in total internal reflection, allowing power to cross a finite low-index gap. The extra boundary changes the dielectric-interface boundary conditions; the gap field need not be a single decaying evanescent wave. Consequently, the zero normal Poynting vector calculation for two infinite half-spaces does not forbid transmission in this three-medium geometry.

Harvard’s actual demonstration uses two right-angle glass prisms, a helium-neon laser, a small clamp and two perpendicular screens as detectors. The contact faces are cleaned and placed opposite each other, with the clamp protecting the corners. Increasing pressure narrows the intervening air film; the observed beam routing changes gradually from total internal reflection toward onward transmission. The broad-gap reflecting configuration supplies the comparison for the narrow-gap transmitting configuration. This is a qualitative readout: the page explicitly says the gap is not measured.^[ref]

The inference is that the lower-index region can mediate electromagnetic energy transfer when another boundary is nearby. The demonstration does not determine a numerical evanescent penetration depth, resolve gap uniformity, or establish a three-layer transmission coefficient. For the constructed $50^{\circ }$ example, $\delta \approx 178\mathrm{nm}$ identifies the scale over which the isolated-interface field remains appreciable; substituting $e^{-2g/\delta }$ for a gap transmission fraction would omit the changed boundary fields and their interference. A quantitative gap calculation must solve the full Maxwell equations with both boundaries.

The Fresnel equations give reflection and transmission amplitudes for both transverse-electric and transverse-magnetic polarization of an electromagnetic wave at a planar interface. They follow by applying interface conditions for electromagnetic fields to incident, reflected and refracted plane electromagnetic waves.

The specular reflection of a wave at a planar dielectric interface with amplitudes described by the Fresnel equations. Below the critical angle, the reflected fraction is generally nonzero even when transmission is allowed; above the critical threshold, ideal lossless total internal reflection has unit reflected power. The reflected electric-field sign depends on polarization and the chosen basis conventions.

With one fixed tangential electric-field direction, normal-incidence reflection from impedance $Z_1$ to $Z_2$ gives $r_E=(Z_2-Z_1)/(Z_2+Z_1)$. Using one fixed tangential magnetic-field direction reverses the reflected wave's sign, giving $r_H=-r_E$. Reversing the chosen reflected basis can produce the same change. The reflectance $|r{|}^2$ is invariant, but coherent amplitudes require a consistent convention throughout. Two internal reflections at identical opposite interfaces give the same round-trip product in either convention.

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.

For reflection or refraction at a flat isotropic interface, p-polarization has its electric field in the incidence plane and perpendicular to its wavevector. At a lossless nonmagnetic dielectric interface, its reflected amplitude vanishes at Brewster's angle. This does not mean that its field is generally parallel to the interface surface.

For reflection or refraction at a flat isotropic interface, s-polarization has its electric field perpendicular to the plane containing the incident wavevector and interface normal. At an isotropic dielectric boundary it is an eigenpolarization of the Fresnel reflection problem. The label is relative to the specified interface, not an absolute direction in space.

For real wave impedances at normal incidence, the electric-field transmission coefficient $t$ gives transmitted power fraction $T=(Z_1/Z_2)|t{|}^2$. Reflection gives $R=|r{|}^2$. Enforcing both tangential electric and magnetic boundary conditions gives $r=(Z_2-Z_1)/(Z_2+Z_1)$ and $t=2Z_2/(Z_1+Z_2)$, which satisfy $R+T=1$.

For an interface between two lossless, isotropic, nonmagnetic media with unequal refractive index values $n_1,n_2$, Brewster's angle is the incidence angle at which the reflected amplitude in transverse magnetic polarization vanishes. The Fresnel equations and Snell's law give $\tan {\theta }_B=n_2/n_1$; the reflected and transmitted rays are perpendicular there. The usual formula assumes equal magnetic permeabilities.

For sunlight reflected from a horizontal flat dielectric interface, incidence angle is measured from the vertical while solar elevation is measured from the horizontal. The Brewster condition $\tan {\theta }_B=n_2/n_1$ therefore gives $h_B=\pi /2-{\theta }_B={\tan }^{-1}(n_1/n_2)$. Reflected light has its electric field perpendicular to the vertical incidence plane, so its surviving linear polarization is horizontal. This ideal result assumes real refractive indices, negligible absorption, nonmagnetic media and a smooth interface.

At a transparent nonmagnetic dielectric interface, the Fresnel reflection coefficients differ for s and p linear polarization. At Brewster incidence the reflected p amplitude vanishes and reflected unpolarized light becomes s polarized. For a horizontal reflecting surface this is horizontal glare. Away from Brewster incidence it is generally partial polarization. Refraction and internal reflection in droplets also select the two components differently, explaining polarization of rainbow light without a biological mechanism.

A vacuum ray incident normally on a dielectric moving parallel to its surface becomes oblique in the material rest frame, with $\sin i=|\beta |$. For a transparent nonmagnetic dielectric, the rest-frame Brewster angle satisfies $\tan i_B=n$, giving $|{\beta }_B|=n/\sqrt{1+n^2}$. The reflected electric field is perpendicular to the plane containing the ray and the velocity. A real refractive index $n\ne 1$ is assumed; dispersion makes the condition implicit through the rest-frame frequency, and absorption generally removes exact extinction.

In the sharp planar, nonmagnetic ionosphere model with frequency above the plasma frequency, the refractive index is $n=\sqrt{1-{\omega }_p^2/{\omega }^2}<1$. From air, the Brewster angle satisfies $\tan {\theta }_B=n$. At that angle the reflected p-polarized wave vanishes, while s polarization retains a nonzero reflected component. A ground-to-ground single reflection from height $h$ spans $D=2h\tan {\theta }_B=2hn$. This is below the critical angle for total internal reflection for total internal reflection, which reflects both polarizations in this lossless isotropic model. The distinction diagnoses ordinary Brewster reflection rather than total internal reflection.

When the permeabilities agree, the signed reflected-to-incident electric-field ratio for polarization normal to the plane of incidence isFor incidence from lower to higher refractive index it never vanishes.

For equal permeabilities on the two sides of an interface, tangential-field continuity and Snell law for electromagnetic waves givefor consistently signed electric amplitudes normal to the plane of incidence.

For $D_i={\epsilon }_{ij}{E}_j$, $B=\mu H$, and a plane wave with electric polarization $e$, the source-free Maxwell equations implyIf $e$ is an eigenvector of $\epsilon$, then $e\perp k$, the dispersion relation is $|k{|}^2={\omega }^2\mu {\epsilon }_e$, and the Poynting vector is parallel to $k$.