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Use sinusoidal phasors with physical values $v(t)=\mathrm{Re}(\hat{V}{e}^{j\omega t})$, $i(t)=\mathrm{Re}(\hat{I}{e}^{j\omega t})$, and $j^2=-1$. The complex impedance is $Z=\hat{V}/\hat{I}=X+jY$: its modulus is the voltage-to-current amplitude ratio, and its complex argument is the voltage-minus-current phase difference. Here $X$ is the electrical resistance part and $Y$ the signed electrical reactance; the source's $X,Y$ notation differs from the common $R,X$ notation.

The component laws $v_R=Ri$, $v_L=Ldi/dt$ and $i_C=Cd{v}_C/dt$ give the complex impedances of an ideal resistor, inductor and capacitor:In a series circuit the same electric current flows through both components, so Kirchhoff's voltage law adds their voltage phasors. The resulting series RC impedance isFor positive $R,C,\omega$, the negative phase difference means that electric current leads the applied electric potential difference.

In the parallel LC circuit both components have the same voltage phasor. Adding their current phasors by Kirchhoff's current law gives the total electrical admittance:ThusAt the ideal parallel resonance, ${\omega }_0=1/\sqrt{LC}$, the two branch current phasors cancel. The net electrical admittance is zero and the magnitude of the complex impedance tends to infinity: the parallel combination behaves as an open circuit at that frequency. Its complex argument is undefined exactly at the pole. Finite component losses replace the infinite peak by a large finite electrical impedance. This conclusion concerns a voltage-driven sinusoidal response; an ideal resonant current drive does not have a bounded sinusoidal steady-state voltage.

Use the $e^{i\omega t}$ phasor convention and interpret impedance matching as conjugate matching for the maximum power transfer theorem. The load electrical impedance of the series circuit isFor $Z=A+i\omega B$, the condition $Z_{RC}=Z^{\ast }$ equates the real parts and imaginary parts:These formulas require $B>0$, $C>0$ and a passive realization with $A\ge 0$, evaluated at the matching angular frequency; if $A,B$ depend on $\omega$, the equation for $\omega$ is implicit.

There is a convention ambiguity in the word “matched”. Literal equality $Z_{RC}=Z$, as required for equal bridge arms in part (b), instead gives $R=A$ and ${\omega }^{2}BC=-1$, requiring $B<0$. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive angular frequency because its electrical reactance is positive while that of the series RC impedance is negative.

Let the source's top terminal have voltage phasor $V_0$ and its bottom terminal zero. Denote the left and right junctions by $L$ and $R$. The two left resistors are equal, so at balance $V_L=V_0/2$. Zero electric current through the ammeter requires $V_R=V_L$, and therefore equal upper and lower right-arm electrical impedances:Write $x=\omega R_x{C}_x$. Separating real parts and imaginary parts givesThe second equation implies $1+x^2={\omega }^2{R}_x^{2}C{C}_x$. Subtract $x^2={\omega }^2{R}_x^2{C}_x^2$ to obtainA finite balance with positive components requires $0<C_x<C$, and then $0<R_v<R_x$. This equal-ratio RC bridge uses equality of arm electrical impedances, independent of the conjugate-matching convention in part (a).

Detector convention. The stated negligible-electrical impedance ammeter is an ideal short circuit, whose actual terminal voltage is zero at every angular frequency. The nontrivial bridge output usually intended is the open-circuit junction difference, obtained with the detector removed or replaced by a high-input impedance voltmeter. Define its polarity byFor a detector of finite electrical impedance $Z_A$, its terminal voltage is insteadHere $Z_{\mathrm{th}}$ is the Thévenin equivalent electrical impedance, found by shorting the ideal source. This exhibits why the ideal-ammeter limit $Z_A\to 0$ has zero terminal voltage even when the bridge is unbalanced.

An ideal operational amplifier has infinite input impedance and open-loop gain, zero output impedance and no offsets or response limits. Its input currents vanish. Stable operational-amplifier negative feedback, a finite output and unsaturated linear operation then imply a virtual short, $v_{+}=v_{-}$. Real amplifiers have finite frequency-dependent gain, input bias currents, offset voltage, output loading, finite bandwidth, slew rate and supply-rail limits. Equal input voltages cannot be assumed in an open-loop or saturated circuit.

Here the noninverting input is grounded. In the ideal-feedback limit the inverting node is a virtual ground, so the resistor current charges the feedback capacitor:The initial capacitor charge fixes the integration constant. This is an op amp integrator, with $H(i\omega )=-1/(i\omega RC)$.

A tiny direct-current input, input bias current or offset can drive the output towards saturation. A parallel feedback resistor $R_f$ provides a direct-current return path, giving $H(s)=-(R_f/R)/(1+s{R}_{f}C)$; choose $\omega R_{f}C\gg 1$ over the intended integration band. A reset switch sets initial charge; low-offset components and suitable signal amplitude reduce drift and saturation. The useful band must also remain below the amplifier's bandwidth and slew-rate limits.

For the stipulated finite but constant real gain $A=100$, infinite input impedance and negligible output impedance, write the inverting-node voltage as $v=-V_{\mathrm{out}}/A$. With phasors proportional to $e^{i\omega t}$,ThereforeThe low-frequency gain is $-100$. The corner is ${\omega }_c=1/(101RC)$; above it the amplitude falls as $100/(101\omega RC)$ and the phase approaches $+\pi /2$. phase $+\pi$ and $-\pi$ describe the same low-frequency inversion. This finite-gain circuit is a low-pass response with an integrating high-frequency regime, not an ideal integrator down to zero frequency. A genuinely frequency-dependent $A(i\omega )$ would introduce additional high-frequency limitations beyond this specified model.

An ideal operational amplifier draws no current at either input. With stable operational-amplifier negative feedback in its unsaturated linear range, infinite open-loop gain also gives a virtual short, $V_{+}=V_{-}$. The amplifier adjusts its output to enforce that equality; it is not a conducting connection between the inputs. Here $V_{+}=0$, so the ideal summing node is a virtual ground. Kirchhoff's current law gives $V_{\mathrm{in}}/R_1=-V_{\mathrm{out}}/R_2$, hence the ideal gain is $-R_2/R_1$.

For the finite-input-resistance inverting amplifier let the inverting-node voltage be $v$, and continue to neglect output impedance. The controlled output and node-current balance areEliminating $v$ givesDividing the denominator by $R_2$ gives the requested resistance-sum form. The ideal result follows whenever $|A|\gg 1+R_2/R_1+R_2/R_{\mathrm{in}}$; taking $R_{\mathrm{in}}\to \infty$ alone is insufficient if $A$ is small. At finite frequency $A$ and the gain are phasor ratios, subject to feedback stability.

With equal resistors and sufficiently large loop gain, $V_{\mathrm{out}}\simeq -V_{\mathrm{in}}$. This gives a controlled sign reversal or 180-degree phase inversion, useful for subtraction and opposite-phase signal paths, with amplitude set by a resistor ratio rather than uncertain open-loop gain. The source still sees approximately $R_1$, so this is not a high-input-impedance voltage follower. For the ideal-input, single-pole model, the noise gain of an operational amplifier is two and the closed-loop bandwidth is approximately half the gain-bandwidth product. Output swing, slew rate and load constraints must also permit linear operation.

Take the aperture in $z=0$ and the observation point $P=(0,0,b)$, with $b>0$. Under scalar diffraction theory, the Huygens-Fresnel principle sums complex amplitudes from area elements, rather than summing their intensities. Apart from a common normalization,Here $h$ is the complex aperture transmission, $b$ is the axial distance, and $\theta$ is the angle of the secondary ray to the aperture normal. The obliquity factor $K$ describes the direction dependence of the secondary contribution; one Kirchhoff convention is $K=(1+\cos \theta )/2$ for normal illumination. Other scalar boundary approximations have slightly different angular weights, all tending to unity near the axis.

Put $r^2=x^2+y^2$. The paraxial approximation gives $R\simeq b+r^2/(2b)$ in the phase and $1/R\simeq 1/b$ in the amplitude. Removing common $e^{ikb}/b$ yieldsAssumptions are uniform normally incident plane-wave illumination, coherent monochromatic propagation, aperture dimensions large enough for the scalar description, $r/b\ll 1$, and phase accuracy $k{r}^4/(8b^3)\ll 1$ over the relevant aperture. The last condition justifies dropping the quartic distance term; geometric smallness alone does not guarantee phase accuracy.

For a circular aperture, $h=1$ for $r\le a$ and zero elsewhere. Rotational symmetry and $s=r^2$ give $dxdy=rdrd\phi$, $rdr=ds/2$, andThe constant $\pi$ can be absorbed into the proportionality factor, producing the required one-dimensional integral.

At a focus a distance $f$ away, half-period-zone boundaries obeyWith slowly varying obliquity and illumination, each zone has almost the same area and its integrated phasor has the opposite sign to its neighbour. A zone plate blocks alternate zones, so the remaining phasors add instead of nearly cancelling. Choosing an open central zone, the first three open regions have outer boundaries $n=1,3,5$:These are outer radii, not zone widths; the second open region is the annulus between $r_2$ and $r_3$.

For the specified outer radius, $R^2/(\lambda f)=49$: there are approximately 49 half-period zones, of which 25 odd zones are open. Normalize to the field ${\psi }_0$ of the unobstructed plane wave. The normalized Fresnel aperture integral isFor any odd open annulus this gives $2$, so the whole plate gives $\psi /{\psi }_0\simeq 50$ andThe axial intensity gain of an amplitude zone plate grows as the square of the number of open zones. Treating the number of zones continuously gives $49^2\simeq 2.40\times 10^3$, the same accuracy as an estimate. This large local intensity gain is consistent with conservation of energy because light is concentrated into a small focal spot. It is not a power efficiency; the ideal first-order power efficiency of a binary amplitude zone plate is only $1/{\pi }^2$.^[ref]

“Without the zone plate” has here meant removing the finite plate and its outer stop, giving the unobstructed incident wave. If instead a clear circular aperture of radius $R$ is retained, its axial field is ${\psi }_{\mathrm{clear}}/{\psi }_0=1-e^{i49\pi }=2$, so the ratio to that finite-aperture reference is $2500/4=625$. The two reference experiments differ; neither ratio should be confused with total transmitted power.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.

Write the real displacement as $U_n(t)=\mathrm{Re}[u{e}^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store energy in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing $a$, with longitudinal displacements along the chain.

The two spring forces on site $n$ giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, $\omega \simeq a\sqrt{\alpha /M}|q|$, and the first Brillouin zone ends at $|q|=\pi /a$. The maximum angular frequency is ${\omega }_m=2\sqrt{\alpha /M}$.

Time averaging uses $⟨[\mathrm{Re}(z{e}^{i\omega t}){]}^2⟩=|z{|}^2/2$. ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.

Use periodic boundary conditions on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi )$. Using folded wave-number counting for a monatomic chain combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.

For positive $q$, $\omega ={\omega }_m\sin (qa/2)$ andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:

A non-ideal amplifier is not specified by infinite input impedance alone: its open-loop gain and output impedance must also be modeled. Take its internal controlled output to be $u=-A(s)v$, where $v$ is the inverting-input potential, and place a resistance $r_o$ between $u$ and the actual output $y=V_{\mathrm{out}}$. The non-inverting input remains grounded; assume no external load. This gives an explicit finite-gain operational-amplifier integrator model.

Writing Kirchhoff's current law at the input and output nodes givesThe second equation gives $y=(sC{r}_o-A)v/(1+sC{r}_o)$. Substitution in the first gives the general transfer functionFor sinusoidal phasors, set $s=i\omega$. The numerator and denominator are complex, so the magnitude gives the amplitude ratio and their arguments give the relative phase.

The simplest model takes $r_o=0$ and a constant finite gain $A_0>0$. It givesThus below ${\omega }_l=[(1+A_0)RC{]}^{-1}$ the amplifier has gain magnitude $A_0$ and inversion. Above that frequency it approaches an integrator with magnitude $A_0/[(1+A_0)\omega RC]\simeq 1/(\omega RC)$ and phase $+90^{\circ }$. Constant finite gain by itself predicts no further high-frequency turnover.

For a more realistic sketch, use the standard dominant-pole open-loop gain $A(s)=A_0/(1+s/{\omega }_a)$, still with $r_o=0$. ThenThe exact frequency response isAssuming $A_0\gg 1$ and $A_0{\omega }_{a}RC\gg 1$, the turnover frequencies are well separated: ${\omega }_l\simeq 1/(A_{0}RC)$ and ${\omega }_h\simeq A_0{\omega }_a$, the approximate unity-gain angular frequency of the amplifier. The regimes arewith magnitude slopes $0,-20,-40\mathrm{dB}$ per decade and smooth phase transitions. This is the solid curve in the figure; $180^{\circ }$ and $-180^{\circ }$ denote the same phase.

If finite $r_o$ is retained and $A(i\omega )\to 0$, the general expression instead approaches $r_o/(R+r_o)$ at very high frequency: the capacitor couples the input to the output through a passive voltage divider. The dashed curve illustrates that eventual feedthrough. Other amplifier poles or an external load would alter the response again. No particular values of $A_0,{\omega }_a,r_o$ are given in the paper, so the numerical sketch is illustrative and the general expression states the model dependence explicitly.

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, $\omega {ϵ}_0/\sigma \ll 1$.

Use peak phasors with time dependence $e^{-i\omega t}$, real positive electrical conductivity $\sigma$, and absolute permeability ${\mu }_a=\mu {\mu }_0$. Ampère-Maxwell equation and Faraday equations giveso ${ϵ}^{\prime }=ϵ+i\sigma /(\omega {ϵ}_0)$. The sign of the imaginary term follows from the stated time convention. For a transverse plane wave in a homogeneous conductor,In the good-conductor limit, choose the root that decays into $z>0$:The electric phasor and physical field are consequently ${\tilde{E}}_x=E_0{e}^{(i-1)z/\delta }$ and $E_x=\mathrm{Re}[E_0{e}^{i(z/\delta -\omega t)-z/\delta }]$, as required. Tangential boundary matching sets $E_0$; it need not equal the incident amplitude.

Faraday's law fixes the magnetic field, including its phase:For real $E_0$ and $\psi =z/\delta -\omega t$, this means $H_y=E_0{e}^{-z/\delta }(\cos \psi -\sin \psi )/(\omega {\mu }_a\delta )$, with other components zero. The time-averaged Poynting vector isThe energy entering a thin slab exceeds that leaving it by $-d⟨N_z⟩/dz$ per unit volume, givingDirectly, $\mathbf{J}=\sigma \mathbf{E}$ and $\mathbf{J}\cdot \mathbf{E}=J^2/\sigma$ at each time. Therefore $⟨J^2/\sigma ⟩=\sigma |{\tilde{E}}_x{|}^2/2$ is exactly this loss. The factors one half use peak rather than root mean square phasors; amplitude decays over $\delta$, while intensity decays over $\delta /2$.

For a cold collisionless unmagnetized plasma, electron inertia gives $\tilde{\sigma }=i{ϵ}_0{\omega }_p^2/\omega$ and real ${ϵ}^{\prime }=1-{\omega }_p^2/{\omega }^2$. Above the plasma frequency the wave propagates without this resistive attenuation. Below it, a semi-infinite plasma supports an evanescent field with electric and magnetic components in quadrature and zero mean inward energy flux: the incident energy is reflected, not absorbed as Joule heating. electron kinetic energy is reversibly stored, so the nondispersive dielectric field-energy formula alone is incomplete. Collisions introduce a real dissipative electrical conductivity and can produce absorption.^[ref]

Field equations. In a homogeneous source-free nonmagnetic medium, $\mathbf{D}=ϵ{\epsilon }_0\mathbf{E}$ and $\mathbf{B}={\mu }_0\mathbf{H}$. Maxwell's equations becomeThe first equation follows from $\nabla \cdot \mathbf{D}=0$ for spatially constant nonzero $ϵ$. When the effective relative permittivity is negative below, its charge response is already included in $\mathbf{D}$; it must not also be counted as an extra free source.

Amplitude relations and bulk dispersion. Work with complex phasors proportional to $e^{ikx-Kz-i\omega t}$. For $\mathbf{H}=(0,H_y,0)$, the Ampere-Maxwell law givesso the amplitudes must obeyThey are not arbitrary independent constants: the dispersion relation alone would not make arbitrary amplitudes satisfy the equations. These relations give $ik{E}_x-K{E}_z=0$, so the electric divergence vanishes; the magnetic divergence vanishes because there is no $y$ dependence. The only nonzero component of Faraday's law of induction isSubstitution then givesThis proves that all four field equations hold for the specified transverse magnetic polarization when both the amplitude relations and the dispersion relation are imposed.

Interface matching. Let region 1 be vacuum above the interface, with ${ϵ}_1=1$, $K={\beta }_1>0$; region 2 has ${ϵ}_2=ϵ$, $K={\beta }_2<0$. These signs make the evanescent fields decay as $z\to \pm \infty$. In the absence of free surface current, tangential $H_y$ is continuous, giving a common interface amplitude. Tangential $E_x$ is also continuous, henceNormal $D_z$ is continuous as well: both sides have $D_z=-k{H}_y/\omega$. This does not require $E_z$ itself to be continuous; the material can have polarization surface charge. The normal magnetic field is zero on both sides.

Existence and surface dispersion. The opposite signs of ${\beta }_1$ and ${\beta }_2$ require $ϵ<0$. The two bulk equations areUse ${\beta }_2^2={ϵ}^2{\beta }_1^2$ to eliminate the decay constants. For the nontrivial interface branch,A positive ${\beta }_1^2$ together with real propagating $k$ gives $ϵ<-1$. A negative permittivity between $-1$ and zero is insufficient. This lossless planar surface-plasmon dispersion describes a surface plasmon polariton; $ϵ=1$ would instead be identical media and supplies no decaying bound mode.

Frequency range and long wavelengths. With $ϵ=1-{\omega }_p^2/{\omega }^2$, the existence condition becomesThe upper value is a limiting frequency approached as $k\to \infty$, not an attained maximum at finite $k$. Substitution into the dispersion relation givesFor $\omega \ll {\omega }_p$, choosing propagation in the positive $x$ direction,It approaches the vacuum light line from larger $k$. The vacuum decay constant is then ${\beta }_1\simeq {\omega }^2/(c{\omega }_p)$, so confinement becomes weak at low frequency. This calculation assumes the stated real, lossless effective permittivity; dissipation would require complex propagation and decay constants.

Effective permittivity. Use peak phasors with time dependence $e^{-i\omega t}$. The Ampere-Maxwell law isThe conductive contribution to complex permittivity is thereforeHere ${ϵ}^{\prime }$ is the source's effective relative permittivity, not just the real part of a complex response. The positive imaginary term is the loss sign appropriate to this time convention.

The penetrating electric wave. Combine this equation with Faraday's law of induction. In a uniform source-free medium the transverse phasor satisfiesFor a good conductor of electrical conductivity $\sigma$, $\sigma \gg \omega ϵ{\epsilon }_0$, so $k^2\simeq i{\mu }_0\sigma \omega$. The root which decays into $z>0$ isChoose the surface phasor amplitude $E_0$ real. Taking the real part of ${\tilde{E}}_x=E_0{e}^{ikz}$ givesThe skin depth $\delta$ is an amplitude-decay distance in metres.

The magnetic wave and phase. The curl of an $x$-polarized field varying in $z$ is in the $y$ direction. Since $\nabla \times \tilde{\mathbf{E}}=i\omega {\mu }_0\tilde{\mathbf{H}}$,ThusThe phase sign follows the specified convention. Equivalently the surface impedance of a good conductor is $E_x/H_y=(1-i)/(\sigma \delta )$.

How the fields differ from those in vacuum. The two metal fields have the same exponential envelope and spatial period $2\pi \delta$, but their phase arguments differ by $\pi /4$.

An individual travelling wave in the lossless exterior has no attenuation, real wave impedance and in-phase electric and magnetic fields. The exterior total field also includes the reflected wave and may show interference, so it need not share that simple phase relation everywhere. At the interface the tangential fields match; $E_0$ is the field just inside the metal, not necessarily the incident amplitude. A good conductor reflects strongly and its wave impedance is much smaller than the vacuum impedance.

Average inward energy flux. For peak phasors, the Poynting vector averaged over a cycle is $⟨\mathbf{S}⟩=\mathrm{Re}(\tilde{\mathbf{E}}\times {\tilde{\mathbf{H}}}^{\ast })/2$. The fields giveConsequently the constant in the question isThe flux is in the $+z$ direction and has units $\mathrm{W}{\mathrm{m}}^{-2}$.