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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 a weakly absorbing doped semiconductor with one dominant carrier species and approximately constant background dielectric function ${ϵ}_b>1$, the Drude model gives a normal-incidence reflectivity minimum near ${ϵ}_r=1$. Thus ${\omega }_{pc}^2=({ϵ}_b-1){\omega }_{\min }^2$, with ${\omega }_{pc}^2=n{e}^2/({ϵ}_0{m}^{\ast })$ and ${\omega }_{\min }=2\pi c/{\lambda }_{\min }$. Eliminating the plasma frequency proves the displayed mass estimate. The real-dielectric zero instead has ${\omega }_0\simeq {\omega }_{pc}/\sqrt{{ϵ}_b}$ and is the nearby screened plasma edge, so its wavelength requires a different denominator. With appreciable loss, fit the full complex Fresnel equations rather than identifying the minimum with an exactly real dielectric value.

A doped semiconductor at frequencies far below its bound-electron resonances can haveFor a weakly absorbing nonmagnetic half-space adjacent to vacuum, the normal-incidence optical reflectance is minimized near refractive-index matching, ${ϵ}_r\simeq 1$. When ${ϵ}_b>1$ and ${\gamma }_c$ is small this gives ${\omega }_{\min }\simeq {\omega }_{pc}/\sqrt{{ϵ}_b-1}$. The zero of the real part of ${ϵ}_r$, near ${\omega }_{pc}/\sqrt{{ϵ}_b}$, is a distinct frequency. Increasing the carrier density shifts the minimum to shorter wavelength if the carrier effective mass and background response remain fixed. With appreciable loss, the minimum must be found from the full complex Fresnel equations.

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.

For $N$ ion pairs per unit volume, a relative displacement $x$ produces ionic polarization $P=Nqx$. Solving the driven Lorentz oscillator model with the source's $e^{-i\omega t}$ convention givesThe constant ${ϵ}_{\infty }$ includes the fast electronic response: infinity here means well above the phonon band but below electronic resonances. Writing $\Delta ϵ={ϵ}_s-{ϵ}_{\infty }$ makes the numerator $\Delta ϵ{\omega }_{\mathrm{TO}}^2$. In an undamped longitudinal mode the bulk displacement field vanishes without an external free charge, so a nonzero electric field requires ${ϵ}_r({\omega }_{\mathrm{LO}})=0$. ThereforeThis is the Lyddane-Sachs-Teller relation for the single polar mode.

At normal incidence from vacuum the Fresnel equations giveBelow the transverse resonance the dielectric response increases, raising reflectivity. Between the transverse and longitudinal frequencies the undamped dielectric function is negative, giving an evanescent transmitted wave and nearly total reflection: the Reststrahlen band. Above the longitudinal edge it becomes small and positive, passing through ${ϵ}_r\simeq 1$ at the reflectivity minimum and eventually approaching its electronic background.

Reading the original graph to its limited precision gives $\hslash {\omega }_{\mathrm{TO}}\simeq 30\mathrm{meV}$ and $\hslash {\omega }_{\mathrm{LO}}\simeq 37\mathrm{meV}$, hence ${\omega }_{\mathrm{TO}}\simeq 4.6\times 10^{13}{\mathrm{s}}^{-1}$ and ${\omega }_{\mathrm{LO}}\simeq 5.6\times 10^{13}{\mathrm{s}}^{-1}$. Extrapolating the low-energy reflectivity to roughly $R(0)\simeq 0.25$ givesValues around $9$–$10$ and $5.5$–$6.5$ are appropriate to this graphical estimate. Neither the dip near $39\mathrm{meV}$ nor the finite-energy right endpoint represents the high-frequency asymptote; treating that endpoint as $R(\infty )$ would underestimate ${ϵ}_{\infty }$.

At $\omega ={\omega }_{\mathrm{TO}}$, ${ϵ}_r={ϵ}_{\infty }+iA$ with $A=\Delta ϵ{\omega }_{\mathrm{TO}}/\gamma$. Weak damping gives $A\gg {ϵ}_{\infty }$, so for the complex refractive index $n+i\kappa$,Squaring and substituting $A$ provesUse the reflectivity at the estimated transverse resonance, not automatically the plateau maximum. A representative reading $R_{\mathrm{TO}}\simeq 0.9$, with $\Delta ϵ\simeq 3.1$, gives $\gamma /{\omega }_{\mathrm{TO}}\simeq 0.0043$, $\gamma \simeq 2.0\times 10^{11}{\mathrm{s}}^{-1}$ andGraphical uncertainty in the transverse edge makes this only a few-picosecond estimate: readings $R_{\mathrm{TO}}\simeq 0.85$–$0.93$ give approximately $2$–$11\mathrm{ps}$. The displacement amplitude decays as $e^{-\gamma t/2}$ and has lifetime $2/\gamma$; phonon occupation/energy decays as $e^{-\gamma t}$ and has lifetime $1/\gamma$. Anharmonic forces allow an optical phonon to decay into two lower-energy phonons when energy and crystal momentum are conserved. Spontaneous decay can persist at low temperature; defects and electron–phonon scattering can add further linewidth.

For a nonmagnetic sample bordering air, the normal-incidence Fresnel equations giveAt short infrared wavelengths the carrier response becomes small and $R$ approaches the background value $R_{\infty }$. Extrapolating the short-wavelength trend gives about $0.29$, henceThe leftmost point at $4\mu \mathrm{m}$ is about $0.27$ and has not quite reached that asymptote. As a consistency check, using the minimum below and $R(4\mu \mathrm{m})\simeq 0.27$ gives ${ϵ}_{\infty }\simeq 11$ without treating that point as the limiting value.

The minimum is near ${\lambda }_{\min }\simeq 13\mu \mathrm{m}$. For weak damping, zero optical reflectance means refractive-index matching, ${ϵ}_r=1$, rather than ${ϵ}_r=0$. Thususing $N=5.4\times 10^{24}{\mathrm{m}}^{-3}$. Given graphical uncertainty, about $0.08m_e$ is an appropriate estimate. This is a carrier effective mass from a reflectivity minimum, not an independent determination of the bare electron mass.

Just beyond the minimum, $\mathrm{Re}{ϵ}_r$ passes through zero near ${\lambda }_p={\lambda }_{\min }/\sqrt{1-1/{ϵ}_{\infty }}\simeq 13.6\mu \mathrm{m}$. At longer wavelengths it becomes negative and reflection rises sharply toward the conducting regime. Finite damping rounds the edge, prevents a strictly exact zero minimum, and keeps the plotted long-wavelength reflectivity below unity.

A forward-biased p-n junction injects electrons from its n-type semiconductor and electron holes from its p-type semiconductor into an active region. A direct band gap allows radiative transitions without a compensating phonon, so stimulated emission can compete effectively with nonradiative processes. At thermal equilibrium there is one Fermi level; forward injection instead establishes separate electron and valence-electron quasi-Fermi levels $F_c$ and $F_v$. Sufficient injection makes conduction states more occupied than the corresponding valence states, producing optical gain.

Two parallel cleaved facets supply an optical resonator. At normal incidence their approximate intensity reflectance follows the Fresnel equations:A resonant normal mode grows when its round-trip amplification exceeds its losses. For length $L$, material optical gain $g$, overlap factor $\Gamma$, internal intensity loss ${\alpha }_i$ and facet reflectances $R_1,R_2$,The pump supplies the emitted energy; the facets provide feedback and let some output escape. Positive optical gain alone does not establish laser threshold.

Polarization under a boost. Lorentz invariance of normalized plane-wave polarization preserves the polarization ellipse's shape, handedness and degree of polarization when its transverse basis is carried to the aberrated ray. The propagation direction changes; equality of a fixed laboratory Cartesian pair of field components is not the appropriate comparison.

Rotate the $yz$ axes so that a general incident vacuum plane wave has unit direction $\hat{\mathbf{n}}=(\cos \vartheta ,\sin \vartheta ,0)$. Choose the transverse basis ${\mathbf{e}}_p=(-\sin \vartheta ,\cos \vartheta ,0)$, ${\mathbf{e}}_s=(0,0,1)$, and arbitrary complex amplitudes $E_p,E_s$. ThenPutting $\beta =v/c$, the given transformation givesThe four-wavevector transforms with Doppler factor $\mathcal{D}=\gamma (1-\beta \cos \vartheta )>0$, givingIn the corresponding basis ${\mathbf{e}}_p^{\prime }=(-\sin {\vartheta }^{\prime },\cos {\vartheta }^{\prime },0)$, ${\mathbf{e}}_s^{\prime }={\mathbf{e}}_s$, the transformed field isBoth components of the Jones vector acquire the same positive real multiplier. Their amplitude ratio and relative wave phase are unchanged, proving preservation of arbitrary linear, circular or elliptical polarization. All Stokes parameters scale by ${\mathcal{D}}^2$, so their normalized values and the degree of polarization are unchanged. An unpolarized light ensemble remains unpolarized under this single-ray boost. A mixture of different ray directions requires separate transformations and is outside this statement.

Reflection from the moving slab. Take the upper interface as $z=0$, the material below it, and the incident ray along $-\hat{\mathbf{z}}$. This fixes the sign left implicit by “along the $z$-axis”; choosing the opposite surface convention changes no result below. Treat $n$ as the real refractive index in the slab's rest frame, with unit relative magnetic permeability. In the laboratory,Transform to the slab rest frame:The incidence is now oblique, in the $xz$ plane, at angle $i$ to the surface normal:The incoming ensemble remains unpolarized light. In this frame the ordinary stationary-interface Fresnel equations give the amplitude coefficients:For generic oblique incidence, $R_s=|r_s{|}^2$ and $R_p=|r_p{|}^2$ differ. Equal incident $s$ and $p$ intensities therefore produce unequal reflected intensities: the reflection is partially polarized. Its degree of linear polarization isThe inverse boost preserves this degree, so the same conclusion holds in the laboratory even though the laboratory incident ray is normal to the surface. The moving material cannot be treated as an isotropic stationary dielectric in that frame.

The reflected rest-frame wavevector is $(\omega /c)(\gamma ,-\gamma \beta ,0,+1)$. Boosting back gives $(\omega /c)(1,0,0,+1)$: the reflected laboratory ray is normal to the surface and has the incident frequency. There is no contradiction with the oblique rest-frame geometry. Translation tangential to a uniform interface does not itself introduce a moving normal mirror.

Complete linear polarization. For Brewster reflection from a tangentially moving dielectric, at Brewster's angle, $r_p=0$ and $\tan i_B=n$. Combining this with the aberration relation givesOnly the $s$ component is reflected. The plane of incidence in the slab frame is $xz$, so the reflected electric field is along $y$, perpendicular to the slab velocity. Boosting this $s$ field back keeps its electric direction along $y$. For example, the reflected rest-frame $s$ field has $B_z^{\prime }=-\beta E_y^{\prime }/c$; hence $E_y=\gamma (E_y^{\prime }+v{B}_z^{\prime })=E_y^{\prime }/\gamma$ and $E_x=E_z=0$.

This exact extinction assumes a transparent nonmagnetic dielectric with $n\ne 1$. At $v=0$ the reflected intensities are equal and the reflected beam is unpolarized. If $n=1$, both reflection coefficients vanish and there is no reflected beam whose polarization can be assigned. With dispersion, evaluate $n$ at the shifted frequency ${\omega }^{\prime }=\gamma \omega$; the Brewster-speed equation is then implicit. absorption generally prevents a real incidence angle from making the complex $r_p$ exactly zero.

At an air–glass interface, resolve the incident unpolarized light into components perpendicular ($s$) and parallel ($p$) to the plane of incidence. The Fresnel equations give the parallel reflection coefficientAt Brewster's angle, $r_p=0$. Together with Snell's law, this gives ${\theta }_i+{\theta }_t=\pi /2$ and $\tan {\theta }_i=n_2/n_1$. Thus, taking the incident medium to be air,The remaining reflected component has linear polarization perpendicular to the plane of incidence. This assumes a transparent, nonmagnetic glass with negligible absorption.

Assume normal dispersion, so shorter visible wavelengths have larger refractive indices. At fixed external angle $\theta$, the relevant criticality parameter iswhich increases with $n$ on this route. Thus shorter-wavelength blue light is totally reflected at B and reaches C, while longer-wavelength red light can transmit through B; green lies at the critical threshold. The prism separates the routes by wavelength near the green cutoff.

The totally reflected rays emerging from C have the same external deviation $180^{\circ }-2\alpha +2\theta$ despite their different refractive indices; their exit positions can differ. Red light also has a partially reflected component by the Fresnel equations, so C is not a mathematically perfect colour cutoff. The observation concerns total versus partial reflection, with the transmitted B rays spread by wavelength. Anomalous dispersion, if present, would require the actual $n(\lambda )$ instead of the normal ordering assumed here.

The wave impedance is the travelling-wave field ratio $Z=E/H$; for the stated nonmagnetic media it is $Z_0/n$. A reflection coefficient is a reflected-to-incident complex amplitude ratio with an explicitly chosen field orientation. With this reflection-amplitude sign convention, using the question's formula for incidence inside silicon givesWith a common tangential electric-field direction, the usual electric-field Fresnel equations instead give $r_E=(Z_2-Z_1)/(Z_2+Z_1)=+5/9$ for this incidence. The printed formula corresponds to the opposite reflected-field sign convention, for example the magnetic amplitude convention. It gives the same reflectance and the same round-trip factor $r^2$.

Each extra internal round trip multiplies the transmitted amplitude by $r^2{e}^{i\delta }$, with $\delta =4\pi nd/\lambda$. A geometric series therefore givesThis is Fabry–Pérot transmission from multiple reflections. Assuming lossless silicon with wavelength-independent $n$, the full intensity transmittance between air regions isMaxima have $2nd/\lambda =j$, minima $2nd/\lambda =j+\frac{1}{2}$. Here $2nd=7000\mathrm{nm}$, so within the specified tuning range the only maximum is $1400\mathrm{nm}$, and the two minima are $1555.6\mathrm{nm}$ and $1272.7\mathrm{nm}$. The contrast is

Identical highly reflecting coatings increase the finesse: transmission maxima become narrower and the intensity between them falls. Ideal symmetric lossless coatings retain $T_{\max }=1$; absorption or unequal mirrors reduce it. The illustrated maxima stay fixed only because additional reflection wave phases are held unchanged. Real coating wave phases can also shift the resonances.

For a lossless single interface at normal incidence, $R=1-T=0.04$ and the Fresnel equations give $R=[(n-1)/(n+1){]}^2$, taking air's refractive index as one. Selecting the glass branch $n>1$ givesThe reciprocal root $2/3$ is the mathematical lower-index branch, not ordinary glass.

In thin-film interference, coherent beams reflected at the two faces of a transparent film overlap. Let the film have refractive index $n_f$, thickness $t$, and internal ray angle ${\theta }_f$, determined by Snell's law. For vacuum wavelength ${\lambda }_0$, the propagation part of the relative phase isAlthough the internal round-trip ray length is $2t/\cos {\theta }_f$, the two emergent rays must be compared on the same outgoing wavefront. Subtracting their external lateral path contribution gives the effective optical path difference $2n_{f}t\cos {\theta }_f$, explaining the cosine rather than its reciprocal.

Reflection also changes phase. At normal incidence in lossless positive-index media, the electric-field Fresnel equations give $r_{ij}=(n_i-n_j)/(n_i+n_j)$: reflection towards higher index reverses sign, whereas reflection towards lower index does not. If the two interfering reflections differ by one such phase reversal, their total relative phase is ${\delta }_{\mathrm{prop}}+\pi$ modulo $2\pi$. Then reflected maxima and minima obeyIf there are zero or two reversals, the conditions interchange. These bright/dark conditions assume overlapping beams with adequate temporal optical coherence; complete cancellation further requires equal amplitudes. In transmitted light, a lossless complete film has complementary intensity variations because reflected plus transmitted power is conserved.

For multiple internal reflections at normal incidence, summing the geometric series gives the exact film reflection amplitudeThis includes both the amplitude mismatch and repeated-beam interference. A normal-incidence antireflection coating on substrate index $n_s>n_0$ can cancel reflection at one wavelength when $n_f=\sqrt{n_0{n}_s}$ and $t={\lambda }_0/(4n_f)$: both interface reflections reverse sign and the film's round-trip propagation adds $\pi$, with equal interface amplitudes. A varying thickness or angle selects different maxima for different wavelengths, producing the colours of transparent films in white light. Broad bandwidth and large incidence angles generally prevent perfect cancellation across all wavelengths.

For a nonabsorbing interface at normal incidence, the Fresnel equations give the intensity reflectanceThe two air–glass interfaces each transmit a fraction $1-\mathcal{R}$ of their incident power. Omitting all internally reflected paths,The power-transmission factors, rather than electric-field amplitudes alone, are multiplied. Absorption, coatings and interference are excluded by this approximation.

Reflected and transmitted waves arise because the fields must satisfy the electromagnetic boundary conditions. With no free surface charge or current, tangential $E,H$ and normal $D,B$ are continuous. At normal incidence between lossless media, define the real wave impedance $Z_j=\sqrt{{\mu }_j/{ϵ}_j}$. Writing $r=E_r/E_i$ and $t=E_t/E_i$ in a common transverse electric-field direction givesFor the power normalization of Fresnel amplitudes, power fractions are $R=|r{|}^2$ and $T=(Z_1/Z_2)|t{|}^2$, so $R+T=1$. An electric-field amplitude coefficient alone is not a transmitted power fraction.

At oblique incidence, phase matching along the interface gives Snell's law, $n_1\sin {\theta }_i=n_2\sin {\theta }_t$. Applying the same boundary conditions separately to the two linear polarizations gives the Fresnel equations. For nonmagnetic dielectrics the parallel-polarized reflection vanishes at $\tan {\theta }_B=n_2/n_1$. Above the critical angle into a lower-index lossless medium, the transmitted field is evanescent and reflection is total. A perfect conductor instead has zero tangential electric field and unit reflected power; finite conductivity introduces penetration and absorption.^[ref]

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.

Reflection and central spot. At a boundary from lower to higher refractive index, the near-normal electric-field reflection coefficient is negative, $r=(n_1-n_2)/(n_1+n_2)$. This gives a $\pi$ wave phase reversal on reflection from the glass plate through the air film. Reflection at the lens's glass-to-air underside has no such reversal. These are consequences of the Fresnel equations for the reflected electric field.

At ideal contact the film thickness is zero, so the two reflected contributions have zero geometrical optical path difference but a relative $\pi$ phase reversal. They produce destructive interference, and the central spot is dark. This conclusion concerns the film reflection; other lens surfaces can add a weak background.

Rings and their radii. At radial distance $r$, let the air thickness be $t$. The two relevant rays reflect at the upper and lower film interfaces. The lower reflection accumulates an extra round-trip optical path length $2t$, plus the relative $\pi$ reflection shift. As $t$ increases away from contact, alternating constructive and destructive thin-film interference forms Newton's rings. At near-normal incidence, a bright ring satisfiesThe extended source can illuminate these localized film fringes: each incident contribution interferes with its own two reflected paths; unrelated source points need not be mutually coherent.

For the spherical lens surface, geometry gives $r^2+(R-t{)}^2=R^2$, so $r^2=2Rt-t^2$. Since $t\ll R$, $t\simeq r^2/(2R)$. Combining the equations yieldsThe $m=0$ bright ring is the first bright ring around the dark contact point.

Curvature from ring diameters. Writing $D_m=2r_m$, subtract the squared diameters to eliminate the unknown initial order:Using SI units,

Liquid-filled film. With film refractive index $n=1.4$, still smaller than both glass indices, there remains exactly one relative reflection phase reversal. The round-trip optical path length is now $2nt$, soHere $\lambda =500\mathrm{nm}$ is the illuminating vacuum wavelength, approximately its air value, rather than its shorter value inside the liquid. The ring labelled $m=2$ therefore has diameterFilling the gap contracts each ring radius by $1/\sqrt{n}$ at fixed order and curvature.

In a direct optical transition, a photon promotes an electron between the valence band and conduction band at essentially the same crystal wavevector: optical photon momentum is small on the scale of the Brillouin zone. An indirect optical transition also needs a phonon or another momentum source. Near an indirect gap, phonon emission requires photon energy approximately $E_g+\hslash \Omega$, while phonon absorption permits $E_g-\hslash \Omega$ with a thermally occupied phonon.

Strong direct interband absorption needs both many allowed transitions and a nonzero optical matrix element. The joint density of states counts pairs with $E_c(\mathbf{k})-E_v(\mathbf{k})=\hslash \omega$; nearly parallel bands and critical points of this energy difference can enhance it. Dipole selection rules and polarization weight that count, so a large joint density of states alone does not guarantee a strong absorption peak.

The supplied infrared spectrum lies below the interband threshold and instead probes free-carrier response. In the Drude model, $ϵ(\omega )={ϵ}_{\infty }-{\omega }_p^2/[\omega (\omega +i/\tau )]$, with ${\omega }_p^2=n{e}^2/({ϵ}_0{m}^{\ast })$. At long wavelengths the mobile electrons screen the field and the reflectivity is high. As wavelength decreases through the screened plasma edge, the real dielectric function becomes positive and propagating light can enter. The reflectivity reaches its minimum near index matching, $ϵ\simeq 1$, then rises toward the bound-electron background reflectance.

For background ${ϵ}_{\infty }=16$, its refractive index is four and the normal-incidence Fresnel equations give $R_{\infty }=[(4-1)/(4+1){]}^2=9/25=0.36$. The short-wavelength side of the figure is approaching a reflectance of this order, consistent with that background; its shortest plotted wavelength still has a carrier correction. Here “infinity” means high frequency relative to the free-carrier scale while remaining below the remote interband resonances.

The Drude-Lorentz reflectivity minimum is near $ϵ=1$, not the plasma-edge zero $ϵ=0$. Reading the deep minimum as ${\lambda }_{\min }\simeq 22\mu \mathrm{m}$ and neglecting damping there gives ${\omega }_p^2=({ϵ}_{\infty }-1)(2\pi c/{\lambda }_{\min }{)}^2$. The carrier effective mass from a reflectivity minimum is thereforeThe graph supports only a rough read-off: shifting the minimum by $0.5\mu \mathrm{m}$ changes this mass by about five percent. The deep narrow minimum supports the weak-loss assumption $\omega \tau \gg 1$ in this region, although finite scattering explains why the long-wavelength reflectivity is below unity. The estimate also assumes one dominant carrier species, a nonmagnetic bulk sample and a nearly constant background dielectric function. If the plasma-edge zero is used instead, its denominator is ${ϵ}_{\infty }$ and its wavelength is slightly longer; the two wavelengths must not be interchanged.

In a simple metal, conduction electrons can move collectively in response to an incident electromagnetic wave. The Drude model gives, with time dependence $e^{-i\omega t}$,In the low-loss range $\gamma \ll \omega <{\omega }_p$, the relative permittivity is approximately $1-{\omega }_p^2/{\omega }^2<0$. The wavevector inside is predominantly imaginary, so the field is evanescent and penetrates only a short distance. Surface currents reradiate the field; little energy enters a good conductor. The normal-incidence Fresnel equations giveMany ordinary metals have their plasma frequency in the ultraviolet, placing visible light below this reflection edge. For $\omega >{\omega }_p$ the free-carrier contribution permits propagation; at $\omega \gg {\omega }_p$, ${ϵ}_r\to 1$ and $R\sim {\omega }_p^4/(16{\omega }^4)\to 0$. The electrons' inertial response becomes too small to screen the rapidly oscillating field.

Ultraviolet transparency is a simple-metal model prediction, not a universal property of real metals. Electronic transitions between bands and bound-electron absorption can remain strong, shift the plasma edge and prevent transparency; transmission also depends on thickness and attenuation. The robust explanation is the change from a reflective negative-permittivity range to a possible propagating range above the plasma edge.