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Equality of the relevant wave impedances across an interface. In the elementary normal-incidence model this makes the reflection coefficient vanish, even when other material properties differ.

Let $\mathcal{R}=|r{|}^2$ be the single-mirror power reflectivity and ${\theta }_g$ the propagation angle inside the gas. Define $\lambda$ as the vacuum wavelength. Successive transmitted complex amplitudes differ by a round-trip factor $\mathcal{R}{e}^{i\delta }$, where, for real dielectric reflection coefficients whose product has positive sign,The cosine arises by comparing successive emerging rays on the same outgoing wavefront; it is not the literal zigzag distance $2d/\cos {\theta }_g$. Equivalently, the round-trip phase is $2k_{z}d$. If the source's $\theta$ is the angle in external air, Snell's law gives $n\sin {\theta }_g=\sin \theta$ andIf $\theta$ denotes the internal angle, simply set ${\theta }_g=\theta$.

For identical lossless mirrors, their power transmission is $1-\mathcal{R}$. In an energy-normalized amplitude convention this is $|t{|}^2=1-\mathcal{R}$; more general interface amplitudes require the corresponding wave impedance factors. The geometric series of transmitted amplitudes givesHenceThis is Fabry–Pérot transmission from multiple reflections.^[ref]

At normal incidence, for nondispersive $n$ and zero net reflection phase modulo $2\pi$, the resonance condition is $\delta =2\pi m$. The resonance frequencies and free spectral range areFor finite $\mathcal{R}<1$ the resonances have finite width, and transmission away from them is nonzero. Thus “only discrete frequencies” is an ideal high-reflectivity description of the selected resonances, not an exact statement about the support of the transmitted spectrum. For example, $\mathcal{R}=0.9$ at $\delta =\pi$ still gives $I_t/I_0=(0.1/1.9{)}^2\simeq 0.00277$.

At half the unit peak, $Q{\sin }^2(\delta /2)=1$. Measuring the phase displacement from a resonance yieldsThe last expression is the high-reflectivity approximation, with finesse $\pi \sqrt{\mathcal{R}}/(1-\mathcal{R})$. At $\mathcal{R}=0.9$, the full phase width is about $0.2109$ radians and $\Delta {\nu }_{\mathrm{FWHM}}/\Delta {\nu }_{\mathrm{FSR}}\simeq 0.03357$. At $\mathcal{R}=0.1$, the minimum transmission is $0.6694$, so the trace never reaches half the peak and this full width at half maximum is not defined. The exact half-maximum condition requires $\mathcal{R}\ge 3-2\sqrt{2}$.

For the absorbing Fabry–Pérot mirror phase and transmission, write $r=\sqrt{\mathcal{R}}{e}^{i{\varphi }_r}$ and let $\mathcal{T}$ be the power transmission of one mirror. Optical absorption makes $\mathcal{R}+\mathcal{T}<1$. The transmitted intensity becomesHere the phase formula is at normal incidence. The peak is reduced below unity, while a change in $\mathcal{R}$ changes the width. The complex reflection phase also moves the resonances: if ${\varphi }_r$ is approximately constant,If either $n$ or ${\varphi }_r$ depends on frequency, this is an implicit condition and the free spectral range need not be constant. No numerical shift is determined without the mirrors' optical response.

At the same tension, $k\propto \sqrt{\rho }$, so $k_2/k_1=3$, $t=1/2$ and $r=-1/2$. The transmitted power fraction is not $t^2$, because the two strings have different wave impedances. Using the energy flux of a transverse string wave,The reflected fraction is $\mathcal{R}=|r{|}^2=1/4$, so $\mathcal{R}+\mathcal{T}=1$. The transmitted displacement amplitude is $2.5\mathrm{mm}$ and its wavelength is $0.5/3\mathrm{m}$. The negative $r$ indicates a reflected phase reversal; it does not mean negative reflected energy.

The intensity of an isotropic radiator at distance $r$ is the radiated power divided by spherical area: $\mathcal{I}=P/(4\pi r^2)$. For a locally plane far-field electromagnetic wave in air, its mean Poynting flux is $\mathcal{I}=E_{\mathrm{rms}}^2/Z_0=E_{\mathrm{peak}}^2/(2Z_0)$, where the vacuum wave impedance is $Z_0\simeq 376.7\Omega$. ThusandBoth conventions are given because “field strength” alone does not specify RMS or peak. This estimate treats the stated transmitter power as radiated power and ignores ground reflection and absorption.

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.

A characteristic wave impedance relates an effort-like quantity to its conjugate flow-like quantity in a single travelling wave: for a mechanical wave these are force and velocity, and their product is power. The sign depends on the propagation and force conventions. Use the transverse force transmitted towards increasing $x$, $F=-T{\Psi }_x$. A rightgoing travelling wave $\Psi =f(x-vt)$ has transverse velocity $u={\Psi }_t=-v{\Psi }_x$, soThis is the positive characteristic impedance of a stretched string. A leftgoing component has $F/u=-Z$ with the same force orientation. Its units differ from pressure-based acoustic impedance because it uses force, rather than force per area.

Use peak-amplitude phasors with time factor $e^{-i\omega t}$ and $\mu ={\mu }_0$. Faraday's law of induction and the Ampere-Maxwell law giveIn the good-conductor limit $\sigma \gg \omega ϵ$, take the root with positive attenuation into the slab:The real physical field is $\mathrm{Re}[{\tilde{E}}_y{e}^{-i\omega t}]$. A thick slab excludes a returning wave from its rear face. The skin depth is the electric-field amplitude decay length; the intensity decay length is $\delta /2$.

Faraday's law of induction fixes ${\tilde{H}}_z=k{\tilde{E}}_y/({\mu }_0\omega )$. The surface impedance of a good conductor and the complete transmitted magnetic field are consequentlyThe sign of the imaginary part belongs to the stated time convention. Average the Poynting vector using complex conjugation and the factor one half appropriate to peak phasors:As an independent energy check, $-{\partial }_x⟨N_x⟩=\sigma |E_0{|}^2{e}^{-2x/\delta }/2$, the mean Joule heating per unit volume.

Ohm's law gives ${\tilde{J}}_y=\sigma {\tilde{E}}_y$ and ${\tilde{B}}_z={\mu }_0{\tilde{H}}_z$. Hence the averaged Lorentz force per unit volume isIntegrating through the illuminated half-space gives the positive inward radiation pressureFinally let $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$ be the vacuum wave impedance. Tangential-field boundary conditions at normal incidence give the electric-amplitude reflection coefficient $r=(Z-Z_0)/(Z+Z_0)$. Since $|Z|\ll Z_0$,Here $N_0=(1-R)I_{\mathrm{inc}}$ by surface energy balance. Combining the leading formulas gives $p\simeq 2I_{\mathrm{inc}}/c$, the nearly perfect-reflector limit. The displayed reflection expansion also assumes a good conductor relative to the incident vacuum wave, $\omega {ϵ}_0/\sigma \ll 1$.

The characteristic impedance of a lossless transmission line is the ratio of the voltage to the current amplitude of a single forward-travelling wave. If $\mathcal{L}$ and $\mathcal{C}$ denote inductance and capacitance per unit length, the lossless telegrapher equations giveA backward wave has $I_{-}=-V_{-}/Z$ when current is positive along $+z$. Thus characteristic impedance is not a resistor's electrical resistance: it describes travelling-wave amplitudes and determines impedance matching, power transfer and reflections. A semi-infinite line, or a line terminated by a matching load, presents that impedance at its input.

At a junction with no lumped branch, voltage and electric current are continuous:Eliminating $V_t$ gives the voltage reflection coefficientFor real positive $Z_1$, incident and reflected waves occupy the same line, so their power ratio is $|r{|}^2$.

For the slightly mismatched coaxial cables, the change in characteristic impedance isConsequentlyThe requested power limit $0.01\%=10^{-4}$ implies $|\Delta b|/b<0.02\ln (b/a)$. Since $\ln (b/a)=2\pi nZ/Z_{\mathrm{vac}}=1.918$ for the stated line, the adjacent cables' outer radii must match within approximately $3.84\%$. If a tolerance is quoted separately as a symmetric manufacturing range $\pm \delta b$ and arbitrary worst-case pairs can differ by $2\delta b$, the corresponding individual tolerance is approximately $\pm 1.92\%$. The leading calculation is for small mismatches; the exact logarithmic expression gives slightly asymmetric positive and negative limits. The vacuum wave impedance used here is approximately $376.73\Omega$.^[ref]

A needle is a shunt electrical resistance $R_n$, so the load seen at its location is the continuation of the line in parallel with $R_n$: $Z_L=Z{R}_n/(Z+R_n)$. The local voltage reflection coefficient and transmission amplitude areFor periodic resistive loading of a transmission line, the reflection from the next needle travels an additional round trip $2L$ and crosses each previous needle twice. Taking the first needle as phase reference, the single-return approximation for periodic line reflections therefore sums toHere $v_p$ is the propagation phase velocity in the cable, $c/n$ for the stated nondispersive nonmagnetic filling. Squaring the amplitude gives the requested expressionFor a finite number $N$ of needles the amplitude has the additional numerator $1-[t^2{e}^{i\varphi }{]}^N$. The displayed semi-infinite result assumes sufficiently many needles, no terminal reflection, and neglects intrinsic propagation loss.

For real $r<0$, write $q=t^2$. The denominator of the power coefficient is $1+q^2-2q\cos \varphi$. It has constructive peaks at $\varphi =2\pi j$ and minima at $\varphi =(2j+1)\pi$, repeating with frequency spacing $\Delta \nu =v_p/(2L)$. Within this approximation,Weak individual shunts thus give narrow interference peaks, with phase width of order $|r|$.

The requested sum excludes paths that reflect repeatedly between different needles. It should not be read as an exact multiple-scattering result, particularly at the coherent peaks. For example the exact semi-infinite amplitude satisfiesAt $\varphi =2\pi j$ this equation reduces to $(R+1{)}^2=0$: the exact ideal infinite array has unit power reflection there, rather than the first-return approximation's value near $1/4$. This does not change the requested geometric-series derivation, but identifies its physical limitation.

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]

In a homogeneous linear medium, write $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$. Without free charge or current, Maxwell's equations becomeTaking the curl of Faraday's law, using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, givesTheir plane-wave solutions have ${\omega }^2=k^2/(\mu ϵ)$ and speed $v=1/\sqrt{\mu ϵ}$; in vacuum this is $c=1/\sqrt{{\mu }_0{ϵ}_0}$. For a wave $\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, the divergence equations give $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$, while Faraday's law gives ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$. Thus the fields and propagation direction form an orthogonal triad, with $B_0=E_0/v$.

The two independent transverse electric components allow linear, circular and elliptical polarization of an electromagnetic wave. Their relative amplitudes and phase determine the polarization state. In a lossless nondispersive medium the energy density and energy flux areA travelling plane wave has equal electric and magnetic energy contributions and time-averaged intensity $⟨S⟩=E_0^2/(2Z)$, where $Z=\sqrt{\mu /ϵ}$ is its wave impedance and $E_0$ denotes peak electric-field amplitude. Reflection and refraction follow by imposing the electromagnetic boundary conditions. Frequency-dependent constitutive response introduces dispersion; conductivity introduces attenuation, so the simple real constant-speed result has a specified domain of validity.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.

The Poynting theorem identifies the loss as Joule heating. The cycle-averaged heating density for a complex conductivity isThe wave impedance is $Z=\mu \omega /k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the skin depth.

At high enough frequency that $\omega \gg \nu$, a metal's free-electron response also approaches the collisionless plasma form. The low-frequency formula $\delta \propto {\omega }^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the electronic band structure and transitions between bands. Both plasmas and metals are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.

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 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 $R+T=1$.

The energy flux carried into a transmitted wave, integrated over the relevant cross section. The power transmission coefficient compares this with incident power. A displacement or field-amplitude transmission coefficient must be converted with the appropriate wave impedances; its squared modulus alone need not be the power fraction. This distinction applies both to electromagnetic waves and transverse string waves.

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.