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In time reversal acoustics, the array records the incoming signal, reverses each recorded time trace, and re-emits it through the same medium. This is phase conjugation in the frequency domain. For the time-harmonic wave convention $e^{-i\omega t}$, reversing a real time trace replaces its positive-frequency wave amplitude by its complex conjugate. The medium must remain unchanged between recording and re-emission.

Let $f(\mathbf{z})={\Psi }_0(0,\mathbf{z})$ and $G_L(\mathbf{\zeta },\mathbf{z})=G(L,\mathbf{\zeta };0,\mathbf{z})$, with the chosen source and array normalizations incorporated into the Green function. Let $a(\mathbf{\zeta })\ge 0$ be the array's aperture weight, equal to the indicator of its receiving region for an ideal uniform array. The recorded field isBy wave reciprocity, back-propagation has the same Green function with the source and receiver exchanged. Therefore the physically re-emitted, back-propagated wave amplitude isIf $Hf=R$ and $P_A$ multiplies by $a$, then ${\Psi }_B=H^T{P}_A\overline{Hf}$, where $T$ denotes the transpose without conjugation. Taking a final complex conjugate instead defines the adjoint reconstruction $\overline{{\Psi }_B}=H^{\ast }{P}_{A}Hf$. This distinction prevents an erroneous conjugation in the time reversal operator.

For a localized Gaussian beam or acoustic point source in a homogeneous medium, a finite aperture admits a limited range of angles. The focal width is of order $\lambda L/A$ when $A$ denotes the aperture diameter. In a random medium, multiple scattering creates paths with a larger angular spread. Each reversed path retraces its route, and the paths interfere constructively at the source. This can produce a larger effective aperture in time reversal and a narrower focus, even though the unreversed field has a complicated speckle pattern.

This comparison concerns a homogeneous reference medium; a deterministic heterogeneous medium can also provide useful multipath propagation. Suitable scale limits or frequency and spatial averaging can make refocusing self-averaging. Such self-averaging is not automatic for every monochromatic source and every random realization. Wave absorption, changing medium parameters, unresolved paths or poor array coverage can spoil refocusing. With complete capture of the propagating modes and a lossless unitary operator $H$, ideal adjoint reconstruction is already exact in either medium. Random scattering can improve finite-aperture wave focusing through angular diversity. See the regime-dependent analysis in Statistical stability in time reversal.

Birefringence gives two orthogonal linear eigenpolarizations with different refractive indices, $n_f$ and $n_s$. They accumulate different wave phases through the same physical thickness $d$. Their relative phase retardation isHere ${\lambda }_0$ is the vacuum wavelength. For the thinnest positive-retardance plate, $d={\lambda }_0/[4(n_s-n_f)]$. The plate uses a difference in phase velocity, rather than differential wave absorption. Equal wave amplitudes along the two axes, obtained from input linear polarization at $45^{\circ }$ to them, acquire phase quadrature and become circular polarization. Reversing that conversion gives linear polarization from the corresponding circular input. A different input angle generally gives elliptical polarization on one pass.

For the prescribed displacement along $z$, the source acceleration is $\mathbf{a}(t_r)=-a{\omega }^2\cos (\omega t_r)\hat{\mathbf{z}}$. If $\theta$ is the angle from the $z$ axis, $|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2=a^2{\omega }^4{\cos }^2(\omega t_r){\sin }^2\theta$. Averaging ${\cos }^2$ over one wave period givesThe angular distribution of electric dipole radiation has zero intensity on the oscillation axis and maximum intensity in the equatorial plane, with no azimuthal dependence. Its total power isThe condition $a\omega \ll c$ is the required nonrelativistic condition. The plotted radial coordinate below is angular power, rather than field wave amplitude.

Work in the dilute, single-scattering limit and add the diffuse scattered light intensities of randomly positioned particles. The incident unpolarized light has equal, mutually uncorrelated $x$ and $y$ electric components. Each particle therefore has an induced electric dipole moment along its permitted axis, with the same mean squared magnitude proportional to $|\alpha {|}^2$.

For an observer in direction $\hat{\mathbf{n}}=(\sin \psi ,0,\cos \psi )$, choose transverse unit directions ${\mathbf{e}}_{\parallel }=(\cos \psi ,0,-\sin \psi )$ in the scattering plane and ${\mathbf{e}}_{\perp }=\hat{\mathbf{y}}$ perpendicular to it. The electric dipole radiation field is proportional to the transverse projection of the electric dipole. Hence an $x$ electric dipole contributes a wave amplitude factor $\cos \psi$ along ${\mathbf{e}}_{\parallel }$, while a $y$ electric dipole contributes its full wave amplitude along ${\mathbf{e}}_{\perp }$.

Let $C>0$ collect the common incident light intensity, electric polarizability, distance and frequency factors. ThenThe two components are uncorrelated, so the coherency matrix is diagonal in this basis. The degree of polarization isA signed degree of linear polarization, defined positive perpendicular to the scattering plane, has the same expression without the absolute value. The dominant direction is $\hat{\mathbf{y}}$ when $N_2>N_1{\cos }^2\psi$, and ${\mathbf{e}}_{\parallel }$ when the inequality reverses. If the denominator vanishes there is no scattered signal and its degree of polarization is undefined. A coherent exactly forward component, including the transmitted incident beam, is not part of this diffuse-intensity calculation; random particle positions do not by themselves destroy forward optical coherence.

Now take $\hat{\mathbf{n}}=(0,\sin \psi ,\cos \psi )$, with the same angle to the incident $z$ direction. The perpendicular direction is $\hat{\mathbf{x}}$ and the in-plane transverse direction is $(0,\cos \psi ,-\sin \psi )$. The roles of the two electric dipole populations interchange: $x$ electric dipoles radiate at full strength, whereas the observed wave amplitude from a $y$ electric dipole is reduced by $\cos \psi$.

Thus $I_{\perp }=C{N}_1$ and $I_{\parallel }=C{N}_2{\cos }^2\psi$, givingThis is also obtained by the symmetry $x\leftrightarrow y$, $N_1\leftrightarrow N_2$ in part (a). For $N_1=N_2$, both planes give the familiar Rayleigh scattering degree of polarization ${\sin }^2\psi /(1+{\cos }^2\psi )$: the diffuse signal is unpolarized forward and fully linearly polarized at right angles. At a dark direction with zero denominator, no degree of polarization can be measured.

Distinguish reflection at the input terminal, reflection at the receiving load, and a returned signal that traverses the free-space link twice. The input antenna is initially matched, so its direct terminal reflection is zero.

In the usual weak-coupling, matched-port two-port network model for the antenna pair, let the forward and reverse wave amplitude transfer coefficients be $t_{21}$ and $t_{12}$. Electromagnetic reciprocity gives equal power transfer factors $|t_{21}{|}^2=|t_{12}{|}^2=\mathcal{F}$, where $\mathcal{F}=(G_a\lambda /(4\pi d){)}^2$ from part (a). The receiving-load reflection returns wave amplitude $t_{12}{\Gamma }_L{t}_{21}$ to the original input port. The load-mismatch contribution to the returned power is thereforeFor fixed gains and impedances this returned power scales as ${\lambda }^4$: it contains two link traversals. By contrast, the reflected power at the receiving load before that return traversal is $P_i\mathcal{F}|{\Gamma }_L{|}^2\propto {\lambda }^2$. Calling that local reflected power the power already back in the input line would omit one propagation loss.

The original wording does not specify an antenna scattering model. The boxed result is the load-dependent echo in an ideal matched-port link, with structural scattering and residual input coupling neglected. The actual total input return can include coherent scattering from the receiving antenna even when ${\Gamma }_L=0$; gains and effective areas alone do not specify that background. If “reflected back” means only the direct reflection at the stated matched transmitting antenna, its value is $0$. These meanings should not be conflated; the ${\lambda }^4$ result requires the two-pass echo interpretation and its stated model.

Take an incident field along $\hat{\mathbf{x}}$, without loss of generality because this component has the same circular eigenmodes for every direction of input linear polarization. Resolve its Jones vector into equal circular components:Using the complex amplitude convention $\mathrm{Re}[\mathbf{E}{e}^{-i\omega t}]$, the real circular contributions areThey counter-rotate and add toIn general the tip traces an ellipse with semiaxis lengths $|a|E_0$ and $|b|E_0$. Equal transmitted circular wave amplitudes produce a line; removal of one circular mode, ${\lambda }_{+}=0$ or ${\lambda }_{-}=0$, produces a circle. The special zero-output case has no defined polarization of waves.

The normalized circular eigenvectors from part (a) define the orthogonal projection matricesThus the spectral decomposition of the Jones matrix isThis simply substitutes $a=({\lambda }_{+}+{\lambda }_{-})/2$ and $b=({\lambda }_{+}-{\lambda }_{-})/2$ and makes each eigenmode's wave amplitude transmission explicit.

When ${\gamma }^{\pm }={\alpha }^{\pm }$ are real, $\gamma$ and $\delta$ are real. The two opposite circular polarizations retain their wave phases but are attenuated at different wave amplitude rates. This circular dichroism changes the ellipticity without rotating the principal axis of the ellipse.

For input $E_0\hat{\mathbf{x}}$, the corrected Jones matrix givesThe common factor $e^{-\gamma z}$ describes the mea wave amplitude attenuation. The $\cosh (\delta z)$ factor reflects the sum of the two transmitted modal wave amplitudes; $\tanh (\delta z)$ fixes their imbalance and hence the signed ellipticity. The real field has components proportional to $\cosh (\delta z)\cos \omega t$ and $\sinh (\delta z)\sin \omega t$, soThe ellipse's principal axis stays along the original input direction. For large differential optical depth it approaches the less attenuated circular eigenmode. A passive medium has ${\alpha }^{\pm }\ge 0$, or $\gamma \ge |\delta |$, so the growing hyperbolic function factor does not imply net optical gain. Using the printed $\mathrm{sech}$ prefactor would give a spurious extra attenuation inconsistent with the modal laws.

The given spatial and carrier factors combine into retarded copies of the same random field:On the $y$ axis, $t_1=ky/(\sqrt{2}\omega )$ and $t_2=-ky/(\sqrt{2}\omega )$. Expand the squared electric-field wave amplitude. Each self term contributes $I_0/2$, and the two cross terms together contribute the real part of the full-field correlation. Stationarity and $\Gamma (-\tau )=\Gamma (\tau {)}^{\ast }$ giveIn vacuum, $k=\omega /c$ and $\tau (y)=\sqrt{2}y/c$. Reversing the delay convention replaces $\tau$ by $-\tau$ without changing this real-part result. This $I$ is the electric-field light intensity convention of the question. For two nonparallel waves it should not automatically be identified with the magnitude of the net Poynting vector; part (d) gives that geometric conversion explicitly.

A line with lifetime broadening has a Lorentzian line shape. Write its half-width at half-maximum in angular-frequency units as $\kappa >0$. The Wiener-Khinchin theorem, or the Fourier transform of the Lorentzian, givesTherefore in vacuumThe sketch is a cosine fringe pattern about baseline $I_0$, bounded by $I_0[1\pm e^{-\sqrt{2}\kappa |y|/c}]$. Its central value is $2I_0$, its spatial interference fringe spacing is $\lambda /\sqrt{2}$, and its visibility decays on length $c/(\sqrt{2}\kappa )$. For an isolated transition whose population lifetime is $T_1$ and which has no extra dephasing, $\kappa =1/(2T_1)$, so the field coherence time is $2T_1$, not $T_1$. Introducing $\kappa$ avoids confusing population, wave amplitude and spectral-width conventions when no numerical linewidth is given.

On the $x$ axis the two geometric delays are identical:Their field wave amplitudes therefore add as equal copies at zero relative delay. Stationarity removes dependence on the common delay, givingindependent of $x$ in the ideal overlapping plane-wave region. The corresponding mean normal energy flux through a transverse plane is $\sqrt{2}{I}_0$. There is no increasing path difference along the bisector, so finite temporal optical coherence does not produce a longitudinal visibility envelope on this axis.

A uniform Lorentz boost preserves the plane wave form, while changing both its frequency and its vector wave amplitude. Decompose vectors parallel and perpendicular to $\mathbf{v}$, so $\mathbf{k}={\mathbf{k}}_{\parallel }+{\mathbf{k}}_{\perp }$. In the observer's primed coordinates,The transformed wave amplitude is specified in part (d). The unprimed “rest frame” in this question is the original coordinate frame, not a rest frame of a vacuum light wave, which has no inertial rest frame.

Use the inverse coordinate transformation,Substitution into the wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ and comparison of the coefficients of ${\mathbf{r}}^{\prime }$ and $t^{\prime }$ givesThe first is the relativistic Doppler effect; the changed direction of the wavevector describes relativistic aberration. Equivalently,The electric-field wave amplitude follows from part (b), after rotating the boost axis to $\mathbf{v}$:For a vacuum plane wave, Maxwell equations give ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$ and $\mathbf{k}\cdot {\mathbf{E}}_0=0$. Thus an expression using only the given electric wave amplitude and wavevector isEquivalently, ${\mathbf{E}}_0^{\prime }=\gamma ({\mathbf{E}}_0+\mathbf{v}\times {\mathbf{B}}_0)-[{\gamma }^2/(\gamma +1)]\mathbf{v}(\mathbf{v}\cdot {\mathbf{E}}_0)/c^2$. The wave amplitude is not a scalar carried unchanged by the wave phase transformation; its direction must remain transverse to the transformed wavevector.

Use Fraunhofer diffraction, appropriate to observation in the far field or in the focal plane of a lens, and uniform illumination across the slit. At observation angle $\theta$, points separated by transverse coordinate $u$ acquire a phase difference $ku\sin \theta$, where $k=2\pi /\lambda$ is the wavenumber. The superposition principle for the coherent wave amplitudes givesThe intensity is proportional to the squared modulus of the wave amplitude, soThe limit at $\beta =0$ is one. The noncentral zeros satisfy $w\sin \theta =m\lambda$ for nonzero integers $m$ for which the angle is physically allowed. In the Rayleigh criterion, two mutually incoherent point images are just resolved when the central maximum of one diffraction pattern lies at the first minimum of the other. For single-slit diffraction this givesThe exact statement here takes one image on the slit normal; for nearby paraxial images the angular resolution is $\lambda /w$. The first propagating minimum exists only for $w\ge \lambda$. The numerical factor $1.22$ belongs to a circular aperture and does not apply to this slit.

The voltage divider relation for the capacitor output gives the transfer functionConsequently the relative wave amplitude isFor input $V_0\cos \omega t$, the output is $V_0|H|\cos (\omega t+\varphi )$, where the continuous phase shift isFor positive frequency and $R>0$, this phase shift goes from $0$ towards $-\pi$, passing through $-\pi /2$ at ${\omega }_0$. A one-argument arctangent without a quadrant correction would wrongly predict a positive phase shift above ${\omega }_0$. The capacitor follows a slowly varying input voltage, while its output falls as $1/({\omega }^{2}LC)$ at high frequency.

Put $x=\omega /{\omega }_0$. Maximizing the output wave amplitude is equivalent to minimizingIts derivative is $D^{\prime }(x)=2x[2x^2-2+Q^{-2}]$. An interior minimum at positive frequency therefore requires $Q>1/\sqrt{2}$, and occurs atIndeed $D^{\prime \prime }(x_{\mathrm{res}})=8x_{\mathrm{res}}^2>0$, and the peak value is $|H{|}_{\max }=Q/\sqrt{1-1/(4Q^2)}$. For $0<Q\le 1/\sqrt{2}$, $D$ increases for $x>0$ and the largest output is the DC value $|H(0)|=1$. At equality the putative peak has reached zero frequency; there is no distinct nonzero-frequency resonance. This threshold concerns capacitor voltage peaking: the series RLC circuit's electric current still has its maximum at ${\omega }_0$ for every positive $R$.

The single secondary turn links the primary magnetic flux ${\mu }_{0}n\pi a^2{I}_p$, so its mutual inductance with the primary isChoose positive winding directions so that positive secondary electric current would reinforce the primary magnetic flux. Faraday's law and Kirchhoff's voltage law then giveAfter the transient proportional to $e^{-Rt/L_s}$ has decayed, put $\alpha ={\tan }^{-1}(\omega L_s/R)$. The steady electric current isIts wave amplitude is $\omega M{I}_0/\sqrt{R^2+(\omega L_s{)}^2}$ and it lags the induced electromotive force by $\alpha$. Relative to the primary $I_0\cos \omega t$, its phase shift is $-\pi /2-\alpha$, modulo $2\pi$. Reversing the chosen secondary winding direction adds $\pi$ to this signed phase shift and leaves all power predictions unchanged.

The lamp's mean Joule heating isFor $\omega L_s\ll R$ the response is nearly in phase with the induced electromotive force; for $\omega L_s\gg R$ the secondary electric current nearly opposes the primary electric current in the stated winding convention.

A suitable electrical circuit has $R$ in series with the input and a shunt branch consisting of $C$ in parallel with the lossy inductor $R_L+i\omega L$; take the output voltage across that shunt branch. This is the middle arrangement in the circuit figure. In terms of its complex impedance,At direct current the capacitor is an open circuit and the inductor contributes only $R_L$, so $H(0)=R_L/(R+R_L)=0.01$. ThereforeWhen $R_L$ is neglected, $Z_p=i\omega L/(1-{\omega }^{2}LC)$. At the antiresonance ${\omega }_0^{2}LC=1$ its electrical admittance vanishes, so $|H|$ reaches its maximum value of one. ConsequentlyThis value uses precisely the requested lossless approximation; keeping $R_L$ changes both the peak position and its height. A transfer function peak and one DC measurement do not uniquely identify every possible two-port network, but the displayed arrangement meets both observations to the stipulated approximation. Also, the angle brackets in the source must denote AC wave amplitudes or root mean square values: the literal time averages of sinusoidal voltages are zero, whose ratio would be undefined.

For a uniformly illuminated single slit of width $b$, summing the wave amplitudes across the slit gives the Fraunhofer diffraction profileThe first minimum occurs at $\sin \theta =\lambda /b$. The Rayleigh criterion places one source's central maximum at the other source's first minimum; hence the slit gives $\Delta \theta \simeq \lambda /b$. A circular aperture instead produces an Airy disk, whose first minimum is at $1.22\lambda /D$. For the camera's circular lens,The factor $1.22$ belongs to circular-aperture diffraction, rather than the single-slit profile.