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Use optical photolithography for the two masks and a resist profile compatible with the selected process. A controlled wet mesa etch or a calibrated reactive-ion etching process must pass completely through the p–n interface, with limited sidewall damage. Use profilometry or cross-section controls to verify depth and a test device to check leakage. For the front lift-off process, directional evaporation and an undercut resist profile discourage a continuous metal fence bridging the mask sidewall.

The cap-removal step requires a separately qualified selective etchant: its selectivity between gallium arsenide and the particular aluminium gallium arsenide alloy depends on alloy composition, chemistry and processing conditions. Establish etch rate and endpoint on a witness stack; a universal etch time cannot be inferred from a diagram without thicknesses. The front metal must resist that etchant. Back metallization and activation must establish an ohmic contact rather than an unintended rectifying junction. Verify specific contact resistance on test structures and avoid an anneal that degrades the p–n profile. The final dielectric can be deposited by a low-damage evaporation or other compatible thin-film method.

Mesa definition comes first so that the two front contacts can be aligned to an already isolated active area and can avoid damaged mesa edges. Keeping the cap until after metallization provides the intended contact surface and protects the window during the contact process. lift-off process must precede selective cap removal if the retained metal is to act as that etch mask. The cap etch must stop above the emitter rather than destroy the junction.

Rear metallization after the wet-processing stages avoids unnecessarily exposing the rear contact to their etchants. Contact activation precedes the optical coating where possible, because the anneal can alter dielectric properties and optical thickness. The antireflection coating goes last and leaves the pads uncoated for electrical access. The key ordering constraints are clean semiconductor before contacts, electrical isolation without severing the intended current path, selective aperture opening, and a thermal budget compatible with every retained layer. If a different metal or etchant requires a changed order, it must preserve these same functions.

One application is a semiconducting nanotube transistor array. Removing metallic tubes prevents conducting bridges that would bypass gate control; a narrow chirality distribution also narrows the spread of band gaps and switching behaviour. High charge carrier mobility and thin flexible channels are attractive, but contact resistance, tube alignment, residual metallic contamination, dielectric traps and the cost/yield of sorting still limit uniform large-area circuits.

A second is a wavelength-selective near-infrared photodetector or emitter. A selected chirality concentrates absorption and emission around its characteristic nanotube optical transitions, avoiding the broad mixed-tube response. Optical selectivity can be tailored by choosing the species. Limitations include low absolute emission yield, dark exciton states, contact-related recombination, environmental spectral shifts and quenching by bundles or metallic contaminants. Optical and electrical characterization after device processing is needed; the sorting label alone does not guarantee device performance.

Sunlight is approximately unpolarized before its interaction with the environment. For Rayleigh scattering by small isotropic scatterers, the induced electric dipoles radiate only their components transverse to the viewing direction. For unpolarized incident light scattered through angle $\chi$, the perpendicular and parallel intensities are proportional to 1 and ${\cos }^2\chi$. The degree of linear polarization is thereforeAt $90^{\circ }$ single scattering is fully linearly polarized perpendicular to the scattering plane. This explains the polarized sky away from the Sun; multiple scattering and aerosols reduce the ideal contrast.

Reflection at a dielectric surface treats the two linear polarizations differently. At Brewster incidence, $\tan {\theta }_B=n_2/n_1$, the reflected p component vanishes for ideal transparent nonmagnetic media. The remaining reflected glare is s polarized, perpendicular to the incidence plane, approximately horizontal for a horizontal water surface. Away from this angle the reflection is generally partially polarized. The transmitted beam also has unequal s/p transmission.

Refraction and internal reflection in water droplets likewise have polarization-dependent Fresnel coefficients, producing strong linear polarization in rainbow light. These are generic scattering and boundary-condition mechanisms; no biological detection mechanism is required.

Photonic structures control light through the spatial arrangement of dielectric material. Waveguides confine transverse modes; periodic photonic crystals combine interference and scattering to produce a photonic band structure ${\omega }_n(\mathbf{k})$. A photonic band gap excludes propagating modes in specified directions and polarizations. A one-dimensional stop band is not automatically a complete three-dimensional gap.

A simple example is a distributed Bragg reflector with alternating high- and low-index layers. At normal incidence, the quarter-wave choicelets successive interface reflections reinforce near the design wavelength. A departure from periodicity can create a localized defect mode inside the stop band. An isolated defect forms a cavity; a line defect can guide light through otherwise forbidden bulk propagation. Finite structures leak and real materials absorb, so ideal band arguments do not ensure infinite confinement.

Changing the photonic density of states can suppress or enhance spontaneous emission and control how emitters couple to selected modes. This is distinct from electronic band-gap engineering: it is the electromagnetic modes that are being structured. A complete electromagnetic gap overlapping an emitter's transition is the ideal mechanism proposed for suppressing otherwise allowed spontaneous emission.^[ref]

Both Thomson scattering and Rayleigh scattering can be understood as radiation from a dipole driven by the incident electric field. Their different frequency dependences come from the response of a free charge compared with a bound charge.

For a free particle of charge $q$ and mass $m$, a harmonic field gives $m\ddot{\mathbf{x}}=q\mathbf{E}$, hence $\mathbf{p}=q\mathbf{x}=-q^2\mathbf{E}/(m{\omega }^2)$. The radiated field contains two time derivatives of $\mathbf{p}$, cancelling the response's ${\omega }^{-2}$. The Thomson cross section is consequently independent of frequency in the classical low-recoil regime:For incident linear polarization with unit polarization $\mathbf{ϵ}$, summing over outgoing polarizations gives $d\sigma /d\Omega =r_q^2[1-|\hat{\mathbf{n}}\cdot \mathbf{ϵ}{|}^2]$. Radiation vanishes along the acceleration and is strongest perpendicular to it. Averaging an incident unpolarized light beam giveswhere $\Theta$ is the scattering angle. Scattering through $90^{\circ }$ is fully linearly polarized perpendicular to the scattering plane for this ideal isotropic dipole model. The electron case dominates over equally charged heavier particles because the cross section scales as $m^{-2}$. recoil corrections become necessary when photon energy is no longer small compared with $m{c}^2$.

For a bound charge with natural frequency ${\omega }_0$ and damping rate ${\gamma }_d$,Far below resonance, ${\alpha }_{\mathrm{e}}$ is approximately constant, soThis is Rayleigh scattering. It requires a scatterer small compared with the wavelength and, for the simple expression, an isotropic linear dipole response. Atoms, molecules and small dielectric particles provide examples. The stronger scattering of shorter visible wavelengths explains blue skylight; absorption and the nonuniform solar spectrum prevent treating that simple scaling as a complete color calculation.^[ref]

For isotropic electric scatterers the normalized angular and polarization patterns are the same as in Thomson scattering. The frequency scaling distinguishes their limits: a free charge has a displacement proportional to ${\omega }^{-2}$, whereas a tightly bound charge has an approximately frequency-independent displacement. Near resonance, far outside the small-particle limit, or with anisotropic polarizability, the simple Rayleigh formula must be modified. In particular, the electric-only pattern must not silently be applied to a perfectly conducting sphere whose magnetic dipole response is also retained.

A scattering measurement should subtract stray-light and detector backgrounds, control scatterer density and optical thickness, and calibrate detector response and incident power when comparing wavelengths. A polarization analyzer at $90^{\circ }$ tests the dipole pattern; multiple scattering or unresolved nonisotropic contributions can reduce its polarization.

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.

Birefringence is polarization-dependent propagation in an optically anisotropic medium. Its permittivity tensor relates $\mathbf{D}$ to $\mathbf{E}$. In a lossless uniaxial dielectric, take the optic axis along $z$ and the principal relative permittivities as $(n_o^2,n_o^2,n_e^2)$. For a wave vector making angle $\vartheta$ with that axis, the ordinary mode has index $n_o$, while the extraordinary phase-index angular dependence isThe ordinary electric field is perpendicular to the plane containing the wave vector and optic axis. The extraordinary electric displacement field lies in that plane and is perpendicular to the wave vector, but its electric field need not be. Its energy-flow direction can therefore differ from the wave-normal direction: birefringent walk-off. Along the optic axis the two indices coincide. Refraction generally splits an incident beam into two differently polarized rays, producing double images, as in calcite.

A waveplate instead uses the relative phase of two eigenpolarizations. With the axes chosen for the propagation direction, the retardance is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate gives $|\Delta |=\pi /2$ modulo $2\pi$ and converts equal-amplitude orthogonal components between linear and circular polarization. A half-wave plate gives $|\Delta |=\pi$ and rotates a linear polarization. This is a phase effect; unequal absorption is a separate property.^[ref]

A photonic crystal has a spatially periodic dielectric response on the scale of the optical wavelength. For a nonmagnetic, lossless, nondispersive dielectric, the source-free Maxwell equations can be written as the eigenproblemThe Bloch theorem gives ${\mathbf{H}}_{n\mathbf{k}}=e^{i\mathbf{k}\cdot \mathbf{r}}{\mathbf{u}}_{n\mathbf{k}}(\mathbf{r})$ with periodic $\mathbf{u}$. The allowed frequencies form photonic band structures. Coupling of counterpropagating waves near a Brillouin zone boundary opens a stop band. A one-dimensional dielectric multilayer is a simple example; two-dimensional hole arrays and three-dimensional dielectric networks allow more extensive control.

A complete photonic band gap excludes bulk propagating modes for every direction and polarization in a frequency interval. Periodicity alone does not guarantee such a gap: dimensionality, symmetry and refractive-index contrast matter. A missing or altered element can support a localized defect state in the gap, making a cavity; a line defect can guide light. Finite crystals have leakage and wavelength-dependent transmission rather than an infinitely sharp exclusion. Applications include selective mirrors, optical filters, small resonators and integrated waveguides.^[ref]

Spherical-particle frequency scaling. For a dielectric sphere of radius $a\ll \lambda$, the quasistatic electric polarizability isThe electric dipole radiation formula and $I_0={ϵ}_{0}c|E_0{|}^2/2$ giveThus Rayleigh scattering scales as ${\omega }^4$, or ${\lambda }^{-4}$, when the polarizability varies slowly with frequency. Material resonances and absorption can change this simple scaling. For the incident $z$ polarization, the dipole is along $z$: intensity is proportional to ${\sin }^2\chi$, where $\chi$ is the angle from $z$. There is no radiation on the dipole axis and a maximum in the $xy$ plane. In a direction $\hat{\mathbf{n}}$ the scattered electric field is along $\hat{\mathbf{z}}-(\hat{\mathbf{z}}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}$, hence linearly polarized wherever nonzero.^[ref]

Needle model used in parts (a)–(d). Let its axis be $\mathbf{u}=(u_x,u_y,u_z)=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. A very slender, high-permittivity needle has dominant longitudinal response, so use the rank-one needle polarizability $\mathbf{\alpha }\simeq {\alpha }_{\parallel }\mathbf{u}{\mathbf{u}}^{\mathsf{T}}$. Its transverse response is neglected at leading order. For observation along $y$, define $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, $s=|\mathbf{v}{|}^2=1-u_y^2$ andFor a unit incident Jones vector $\mathbf{e}$, the common far-field relation is ${\mathbf{E}}_s\propto \mathbf{v}(\mathbf{u}\cdot \mathbf{e})$, and $I_s=K{I}_{0}s|\mathbf{u}\cdot \mathbf{e}{|}^2$. Here $I_s$ is irradiance at distance $R$; multiplying by $R^2$ gives power per solid angle. No absolute value of ${\alpha }_{\parallel }$ is given or needed for the requested relative intensities.

Diffuse random cloud, with the incident polarization of (a). The single-particle amplitudes through $x$ and $z$ analysers are proportional to $u_x{u}_z$ and $u_z^2$. The isotropic orientation fourth moment gives the needle-scattering analyzer momentsFor example, $⟨u_z^4⟩=\frac{1}{2}{\int }_{-1}^1{\mu }^{4}d\mu =1/5$; symmetry and $u_x^2+u_y^2+u_z^2=1$ give $2⟨u_x^2{u}_z^2⟩=1/3-1/5$. Since the particles scatter mutually incoherently, intensities add. For $N$ illuminated particles,The cross correlation vanishes by reflection symmetry. The degree of linear polarization is $(I_z-I_x)/(I_z+I_x)=\boxed{1/2}$: random anisotropic scatterers produce partially depolarized light, even though every individual needle produces a single linear polarization.

Continuous homogeneous dielectric. One must sum amplitudes coherently and include their spatial phases, rather than retain the preceding sum of intensities. Random orientations give $⟨\mathbf{u}{\mathbf{u}}^{\mathsf{T}}⟩=\mathbf{I}/3$, so the averaged response is isotropic and its induced polarization is along $z$. Its coherent radiation toward $y$, when allowed by the sample geometry, is therefore purely $z$ polarized, with $I_x=0$. The independent-needle prefactor is no longer a quantitative dense-medium model; local-field and collective effects determine the effective permittivity.

There is also coherent continuum phase matching: a uniform bulk driven by a plane wave has no arbitrary outgoing $90^{\circ }$ plane-wave component. In an ideal infinite homogeneous medium there is no diffuse side scattering at all; propagation, refraction and reflection are determined by the medium and its interfaces. A finite sample's coherent side signal depends on its shape and boundaries. Thus the old $1:3$ diffuse ratio is replaced by a coherent isotropic-medium response, not by a universal nonzero $N^2$ side-scattering intensity.

State a time convention before using complex response: take fields proportional to $e^{i(kz-\omega t)}$, with $k=k_r+i\kappa$ and $\kappa \ge 0$ for a passive wave entering the material. In a homogeneous Ohmic medium, Maxwell equations with $\mathbf{J}=\sigma \mathbf{E}$ giveThe ohmic-conductor complex wavenumber describes both phase advance and amplitude decay. Real constant conductivity is only a low-frequency approximation. In the Drude model, $m\dot{\mathbf{v}}+m\gamma \mathbf{v}=-e\mathbf{E}$ giveswith a fixed positive ionic background and no additional bound-electron contribution in this simple model. The plasma frequency marks the inertial electron-response scale. More elaborate band structure, interband transitions and nonlocal transport limit this scalar treatment.^[ref]

In the collisionless nonmagnetic limit, the dispersion relation is ${\omega }^2={\omega }_p^2+c^2{k}^2$. For $\omega >{\omega }_p$,A crest can move faster than $c$ without carrying a new signal faster than light. Below cutoff the ideal wave is evanescent, with penetration length $c/\sqrt{{\omega }_p^2-{\omega }^2}$; it does not propagate as an ordinary real-$k$ bulk wave. Evanescence in the lossless limit is not Joule dissipation. Nonzero collision rate gives a dissipative conductivity and smooths this ideal cutoff picture.

For a good conductor with approximately real $\sigma \gg \omega ϵ$, the spatial branch becomesA field amplitude falls as $e^{-z/\delta }$, while intensity falls as $e^{-2z/\delta }$. The phase velocity is $\omega \delta$ in this limit. Magnetic and electric amplitudes are not related by the vacuum ratio: $Z_s=E/H=(1-i)\sqrt{\mu \omega /(2\sigma )}$ with the chosen time convention. Its positive real part is the surface resistance. The local mean heating is $\frac{1}{2}\mathrm{Re}\sigma |E{|}^2$, consistent with electromagnetic energy flowing into the metal. skin depth is an amplitude length; interpreting it as an intensity length gives a factor-of-two error.

A transmission-line mode propagates chiefly through the dielectric between conductors, even though finite metal conductivity causes loss at its boundaries. With per-length parameters $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$, the telegrapher's equations give, for the same $e^{-i\omega t}$ convention,Writing a forward wave as $e^{-\Gamma z}$, $\Gamma =\alpha -i\beta$, gives ${\Gamma }^2=(R^{\prime }-i\omega L^{\prime })(G^{\prime }-i\omega C^{\prime })$. In the low-loss regime $R^{\prime }\ll \omega L^{\prime }$ and $G^{\prime }\ll \omega C^{\prime }$, set $Z_0\simeq \sqrt{L^{\prime }/C^{\prime }}$ to obtainThis low-loss transmission-line attenuation separates series conductor loss from dielectric leakage loss. Average forward power falls as $e^{-2\alpha z}$. In the normal skin-effect regime, reduced conducting area makes the AC conductor resistance increase roughly as $\sqrt{\omega }$; dielectric loss can supply a separate frequency dependence. Reflection from a mismatched termination changes delivered power without being, by itself, dissipation in the line.^[ref]

To test loss, compare several lengths of otherwise identical line with matched terminations, calibrate connectors and incident/reflected power, and monitor temperature. Separate amplitude attenuation from phase delay and from standing-wave minima. Bulk penetration, line attenuation and dielectric propagation speed are related electromagnetic quantities but describe different experimental geometries.

Gauss's law connects electric flux through a closed surface to its enclosed charge:The differential form follows from the divergence theorem. The law is exact; extracting a field magnitude by simple division additionally needs symmetry. For spherical symmetry, $E_r(r)=Q(r)/(4\pi {ϵ}_0{r}^2)$; a uniformly charged ball therefore has a field linear in $r$ inside and an inverse-square field outside. An infinite line of linear charge density $\Lambda$ gives $E_r=\Lambda /(2\pi {ϵ}_{0}r)$ by a cylindrical surface. A uniformly charged infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side by a pillbox.

An electrostatic conductor has zero interior field and constant potential; excess charge occupies its surface. A pillbox gives $E_{n,\mathrm{out}}=\sigma /{ϵ}_0$ when the inside is conducting. More generally, $\hat{\mathbf{n}}\cdot ({\mathbf{E}}_2-{\mathbf{E}}_1)=\sigma /{ϵ}_0$. For dielectric matter, $\nabla \cdot \mathbf{D}={\rho }_{\mathrm{free}}$ separates bound polarization charge from free charge. Gauss's law alone does not establish a uniform field for an arbitrary finite plate or recover a detailed nonsymmetric charge distribution from one flux measurement.

For the dielectric half-space the two formal sources coincide at $(a,0,0)$, outside the region being represented. Their sum is $Q+Q^{\prime }=2{ϵ}_{r}Q/(1+{ϵ}_r)$ in the dielectric Coulomb kernel, henceThere is no charge singularity inside this half-space. The expression satisfies Laplace's equation there, decays at infinity, and matches the vacuum solution's tangential electric field and normal electric displacement. The coincident image is a mathematical representation, not an extra physical charge in the vacuum region.

Apply Faraday's law to a shrinking rectangular loop crossing the interface. For bounded fields the magnetic-flux term vanishes with its area, giving ${\mathbf{E}}_{t,+}={\mathbf{E}}_{t,-}$. A shrinking Gaussian pillbox similarly gives $D_{n,+}-D_{n,-}={\sigma }_{\mathrm{free}}$. Thus normal $D$ is continuous when no free surface charge is present, even though a bound polarization charge may remain. These are the assumptions needed for the stated dielectric-interface boundary conditions.

Write $ϵ={ϵ}_r{ϵ}_0$, $r_{+}=\sqrt{(x-a{)}^2+y^2+z^2}$ and $r_{-}=\sqrt{(x+a{)}^2+y^2+z^2}$. Use each region's own permittivity in its Coulomb potential:At $x=0$, continuity of potential and normal displacement givesSolving,This permittivity convention for transmitted image charges matters: if the vacuum Coulomb prefactor is used even for ${\varphi }_{-}$, its combined effective source is $2Q/(1+{ϵ}_r)$ instead. The physical potential is the same. The two numbered leaves identify how these sources satisfy the equations in their respective regions.

The real charge feels only the image contribution after removing its own singular self-field. The image at $(-a,0,0)$ is separated from it by $2a$, soFor the bound charge, the outward normal of the dielectric is $+\hat{x}$ and $\mathbf{P}={ϵ}_0({ϵ}_r-1){\mathbf{E}}_{-}$. Differentiating ${\varphi }_{-}$ at $x=0$ gives $E_{-,x}=-a(Q+Q^{\prime })/[4\pi ϵ(a^2+y^2+z^2{)}^{3/2}]$. ThereforeIts integral over the plane is $Q^{\prime \prime }$, consistent with the external image representation. For ${ϵ}_r=1$ both the image force and induced surface charge vanish.

In a homogeneous, source-free, nonmagnetic dielectric the electric wave equation isSet $\theta =\omega t-kx$ and $E_y=f(z)\cos \theta$. Substitution gives $f^{\prime \prime }+({ϵ}_r{\omega }^2/c^2-k^2)f=0$. For the labelled field form (a), $f=E_0\cos Kz$, so $K^2={ϵ}_r{\omega }^2/c^2-k^2>0$. For form (b), $f=E_0^{\prime }{e}^{s{K}^{\prime }z}$ with $s=\pm 1$, so $K^{\prime 2}=k^2-{ϵ}_r{\omega }^2/c^2>0$. These give the two stated inequalities, with equality a zero-transverse-wavenumber limiting case.

Faraday's law gives the associated wave magnetic field, without an added static field,Thus in case (a), $H_x=-K{E}_0\sin (Kz)\sin \theta /({\mu }_0\omega )$ and $H_z=k{E}_0\cos (Kz)\cos \theta /({\mu }_0\omega )$. In case (b), $H_x=s{K}^{\prime }{E}_0^{\prime }{e}^{s{K}^{\prime }z}\sin \theta /({\mu }_0\omega )$ and $H_z=k{E}_0^{\prime }{e}^{s{K}^{\prime }z}\cos \theta /({\mu }_0\omega )$. In both cases $H_y=0$.

For the planar dielectric waveguide the core is dielectric but the exterior is vacuum. Consequently the exterior decay constant has $K^{\prime 2}=k^2-{\omega }^2/c^2$, while the interior oscillation has $K^2={ϵ}_r{\omega }^2/c^2-k^2$. Adding gives the relation labelled (c):For an even bound mode the complete profile can be writtenContinuity of tangential $E_y$ fixes the exterior amplitude; continuity of tangential $H_x$ makes $f^{\prime }$ continuous because both media have magnetic permeability ${\mu }_0$. At $z=d$,which is the relation labelled (d). The same even profile also matches at $z=-d$; continuity of $H_z$ follows from that of $E_y$.

In the requested $K,K^{\prime }$ plot, (c) is a first-quadrant circle and (d) rises from the origin to its first pole at $Kd=\pi /2$. Equivalently put $u=Kd$, $w=K^{\prime }d$, and $\mathcal{V}=(\omega d/c)\sqrt{{ϵ}_r-1}$. At $u=0$ the circle has $w=\mathcal{V}>0$ whereas $u\tan u=0$. At the other end of $0<u<\min (\mathcal{V},\pi /2)$, the rising tangent curve exceeds the falling circle. Monotonicity therefore gives one intersection on this first branch for every $\mathcal{V}>0$.

The fundamental even mode of a symmetric dielectric slab has no positive cutoff in index contrast or thickness. Its electric profile is a nodeless cosine maximum at the mid-plane, smoothly joined to decaying exterior tails. The qualifier ${ϵ}_r>1$ is essential: for the printed limiting value ${ϵ}_r=1$, the circle collapses to $K=K^{\prime }=0$ and there is no transversely confined bound mode. A uniform unconstrained wave is not a guided mode. This even-mode argument does not assert that the specified cosine describes every possible odd guide mode.

The original diagram uses the illustrative normalized radius $\mathcal{V}=1.4$; it is a scaled $K{K}^{\prime }$ plot with $u=Kd,w=K^{\prime }d$, followed by the corresponding field profile.

For inner radius $a$ and outer conductor's inner radius $b$, Gauss's law gives $E(s)=\lambda /(2\pi {ϵ}_0{ϵ}_{r}s)$. Integrating across the dielectric gives voltage $V=\lambda \ln (b/a)/(2\pi {ϵ}_0{ϵ}_r)$. Thus the capacitance per unit length isThe stated outer radius is interpreted as the radius of the dielectric-facing surface. Magnetic permeability affects inductance, but does not enter this electrostatic capacitance.

A transmission line stores electric and magnetic energy along its length, so voltage and current depend on both position and time. A distributed circuit uses inductance $L^{\prime }$ and capacitance $C^{\prime }$ per length, together with resistance $R^{\prime }$ and leakage conductance $G^{\prime }$. Applying voltage and charge balance to a short segment gives the telegrapher's equations,For a lossless line, both variables obey wave equations with speed $v=1/\sqrt{L^{\prime }{C}^{\prime }}$ and characteristic impedance $Z_0=\sqrt{L^{\prime }/C^{\prime }}$. Forward and backward waves obey $I^{+}=V^{+}/Z_0$ and $I^{-}=-V^{-}/Z_0$. Matching a termination $Z_L$ gives the voltage reflection coefficientA matched load absorbs without reflection; an open circuit has ${\Gamma }_L=1$, and a short has ${\Gamma }_L=-1$. Reflections can produce standing waves. A coaxial line confines the fields between conductors and, for a uniform dielectric TEM wave, has $v=1/\sqrt{\mu ϵ}$. Finite loss attenuates waves; frequency-dependent losses or material properties can also distort pulses. A long line is a wave-carrying distributed system, so it cannot generally be replaced by one lumped capacitor.

Polarization density $\mathbf{P}$ is electric dipole moment per unit volume. Displacing positive and negative bound charges leaves the bulk nearly neutral locally but can produce uncompensated charge where polarization varies or terminates. For a dielectric occupying a volume with outward unit normal $\mathbf{n}$,The divergence theorem makes their total sum zero for a body described entirely by bound dipoles. Uniform polarization gives no interior volume charge but generally gives opposite surface charges on different faces.

In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}=ϵ\mathbf{E}$. Gauss's law becomes $\nabla \cdot \mathbf{D}={\rho }_f$, separating free from bound charge. Across a static interface, tangential electric field is continuous, while the normal displacement jumps by the free surface charge. Polarization charges alter the internal field, as in the dielectric cylinder calculation. Molecular distortion and orientation contribute differently, and response can depend on temperature and frequency. Bound charge is a redistribution of material charge, not an additional free-charge reservoir.

Use a cold, unmagnetized, collisionless electron plasma with immobile ions. For an oscillating field proportional to $e^{-i\omega t}$, the electron displacement follows $-m_e{\omega }^2\mathbf{x}=-e\mathbf{E}$. Hence $\mathbf{P}=-n_{e}e\mathbf{x}=-n_e{e}^2\mathbf{E}/(m_e{\omega }^2)$ andThe induced polarization opposes the field because electron inertia changes the phase of its displacement. Maxwell's equations give transverse electromagnetic dispersionThus transverse waves propagate above the plasma frequency, while below it the wavenumber is imaginary and a semi-infinite lossless plasma reflects an incident wave with an evanescent interior field. In the cold electrostatic model, a longitudinal density oscillation instead has frequency ${\omega }_p$.

Collisions at rate $\nu$ give ${ϵ}_r=1-{\omega }_p^2/[\omega (\omega +i\nu )]$ for this time convention, introducing absorption. A background magnetic field makes the response anisotropic, and thermal motion supplies additional dispersion and static screening. The divergent zero-frequency cold collisionless formula should not be mistaken for a complete description of static Debye screening. The simple dielectric function is a dynamical electron-inertia response under stated assumptions.

Put line charges $+\lambda$ and $-\lambda$ on the two axes. For $d\gg a_1,a_2$, the surface potentials differ byThus the capacitance per unit length in vacuum isThe two logarithms belong to one potential difference, not two independent capacitances in parallel. A homogeneous dielectric replaces ${ϵ}_0$ by its absolute permittivity; finite radii and proximity corrections are neglected.