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Let each slit alone give local light intensity $I_s$, and assume equal coherent illumination. In Fraunhofer diffraction, the two wave amplitudes differ in wave phase by $\delta =2\pi s\sin \theta /\lambda$. Adding the wave amplitudes before squaring givesFor a flat screen, $\sin \theta =x/\sqrt{d^2+x^2}$. In the paraxial region $|x|\ll d$, the double-slit interference profile is thereforeThe given $d\gg s^2/\lambda$ justifies the far-field expansion. Slit widths have been neglected; a finite width multiplies this profile by the single-slit diffraction envelope. At large angles the one-slit light intensity and geometric spreading must also be retained, rather than treating $I_s$ as constant.

Using $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ directly with the specified plane wave givesThe electric field, magnetic flux density and propagation direction form a right-handed triad, and for a nonzero wave amplitudeThe electric field and magnetic field have the same wave phase in vacuum.

For a lossless antireflection coating, cancellation requires both equal reflected wave amplitudes and an odd half-cycle difference in their round-trip wave phase. At normal incidence the interface amplitude coefficients are $r_{01}=(n_0-n)/(n_0+n)$ and $r_{12}=(n-n_s)/(n+n_s)$. Choosing $n_0<n<n_s$ makes both negative: both reflections acquire the same $\pi$ phase reversal. A quarter-wave optical thickness supplies their remaining relative phase $\pi$.

Equal amplitudes require $r_{01}=r_{12}$, which reduces to $n^2=n_0{n}_s$. ThusThe exact multilayer reflection amplitude is zero for these choices, not merely the sum of the first two reflected rays: its numerator is $r_{01}+r_{12}{e}^{2i\beta }=0$, with $\beta =2\pi nd/{\lambda }_0=\pi /2$. Any positive odd multiple of this thickness has the same cancellation at the design wavelength; the smallest is the thin-layer choice. absorption, index dispersion, angular spread and deviations from the selected wavelength limit the practical cancellation.

Model the room's decaying acoustic waves as a weakly damped resonator. The quality factor from ringdown is related to stored energy $E$ bywhere $A$ is the pressure wave amplitude and $\omega =2\pi f$. The measured frequency $f=440\mathrm{Hz}$ gives $\omega \simeq 2.76\times 10^3{\mathrm{s}}^{-1}$. If the stated $5\mathrm{s}$ is interpreted as an approximate wave amplitude exponential relaxation time,an estimate of order $10^4$. Taking $5\mathrm{s}$ as an energy exponential relaxation time instead gives $Q\simeq 1.4\times 10^4$.

There is a real convention ambiguity: “heard to die away” gives neither an initial sound level nor an audibility threshold, so it does not determine either exponential relaxation time. More generally, if the energy falls by a factor $E(0)/E(t_d)$ in $t_d$,In room acoustics, a reverberation time $T_{60}$ usually denotes a decibel drop of $60\mathrm{dB}$, or an energy factor $10^6$. If $5\mathrm{s}$ means $T_{60}$, the corresponding estimate is instead $\boxed{Q\simeq 2\pi (440)(5)/(6\ln 10)\simeq 1.0\times 10^3}$. The simple decay-time estimate and the conventional reverberation time estimate should not be silently equated.

Place the junction at $x=0$ and use the complex amplitude convention $e^{-i\omega t}$. With incident displacement wave amplitude $A_i$, writeHere $r$ and $t$ are displacement wave reflection and transmission coefficients, and $k_j=\omega /c_j$. The junction is assumed massless, with no externally applied transverse force, and both strings have the common equilibrium tension $T$. Continuity of displacement and transverse force givesSince $T{k}_j=\omega Z_j$, these become $1+r=t$ and $Z_L(1-r)=Z_{R}t$. Solving,A larger right-hand mechanical impedance produces a reflected displacement with reversed sign, as expected in the fixed-end limit $Z_R\to \infty$. Equal mechanical impedances give $r=0$ and $t=1$.

The instantaneous rightward energy flux along a string is the transverse force times the transverse velocity,For a rightgoing time-harmonic wave of peak displacement wave amplitude $|A|$, averaging over a wave period givesA leftgoing time-harmonic wave has the opposite signed energy flux. The positive ratios of reflected and transmitted power to incident power are thereforeAll explicit angular frequency factors cancel. In the ideal wave equation on a string, $T$ and the two linear mass densities are constant, so the mechanical impedances have no frequency or wavelength dependence. Consequently $\mathcal{T}$ is independent of wavelength even though the two string wavelengths generally differ.

Finally,This is conservation of energy at a lossless, massless junction. The transmitted displacement wave amplitude can exceed the incident one when $Z_R<Z_L$; the factor $Z_R/Z_L$ is essential when comparing power.

Write $p$ and $q$ for the axial source and observation distances, and put $R_{\mathrm{eff}}=pq/(p+q)$. The aperture is $-d/2<x,y<d/2$. Under the Fresnel diffraction, the spherical incident wave amplitude is effectively constant over this small aperture, while its quadratic wave phase must be retained. Therefore the two transverse integrals separate:where $K$ contains the source strength, the propagation distances and a common phase factor.

Use the dimensionless variable $u=x\sqrt{2/(\lambda R_{\mathrm{eff}})}$ and define the optical normalization of Fresnel integralsThe oddness of $F$ makes the bracket $\sqrt{2\lambda R_{\mathrm{eff}}}F(w)$, soThis square-aperture Fresnel diffraction normalization is especially useful for comparing with no aperture. In the same Fresnel diffraction, the unrestricted one-dimensional Fresnel integral is $\sqrt{\lambda R_{\mathrm{eff}}}{e}^{i\pi /4}$; its square is $i\lambda R_{\mathrm{eff}}$. Thus, writing ${\psi }_0$ for the unobscured field at the same point,The normalization has the correct no-aperture limit because $C(\infty )=S(\infty )=1/2$ in this convention.

The crystal has periodic potential $V(x+a)=V(x)$. By Bloch's theorem, an energy eigenstate may be chosen asExpand the periodic factor in a Fourier series:Multiplication by $e^{ikx}$ gives the stated plane-wave expansion. Here $x$ is position, $a$ is the primitive unit cell length, $G_1$ is the primitive reciprocal lattice wavenumber, $n$ runs over every integer, and $C_{k,n}$ are complex numbers specifying the component wave amplitudes. The wavevector $k$ labels the Bloch state modulo $G_1$; $\hslash k$ is its crystal momentum, rather than generally its measured free-space momentum.

The component $e^{i(k+n{G}_1)x}$ is a plane wave of wavevector $k+n{G}_1$. All these components acquire the same phase factor $e^{ika}$ under $x\mapsto x+a$, since $e^{in{G}_{1}a}=1$. Their mixing therefore respects the crystal's translation symmetry. The coefficients and energy eigenvalue follow from the Schrödinger equation; the expression is not a claim that the coefficients are equal or that only finitely many are nonzero. The energy band index is suppressed in the notation, so different energy bands at a common $k$ generally have different coefficient sets.

The components are in phase quadrature when the axes of the ellipse coincide with the coordinate axes. Since the $x$ wave amplitude is one and the $y$ wave amplitude is two, the Jones vector must have $A=\pm 2i$.

A handedness convention is essential. Adopt the optical convention in which right-handed elliptical polarization rotates clockwise when viewed toward the source, from the positive-$z$ side. With the paper's $e^{+i\omega t}$ phase convention for complex amplitudes, this givesThe tip follows ${\Psi }_x^2+{\Psi }_y^2/4=1$. In particular,

The figure states the viewing direction and shows increasing time. Under the alternative convention that calls rotation from $+x$ toward $+y$ right-handed when looking along propagation, the equally consistent answer is $A=-2i$ and the second vector is $(0,+2)$. The original paper does not state a handedness convention, so its wording alone cannot select between these labels; the real components and viewing direction remove the ambiguity.

Take horizontal interfaces, water depth $h_w=50\mathrm{m}$, and sediment thickness $h_s=20\mathrm{m}$. Both source and receiver are just below the sea surface. Every angle below is measured from the vertical interface normal. Snell's law preserves the horizontal seismic ray parameterA basement seismic head wave of speed $V_h$ has horizontal propagation at the basement interface, so $p=1/V_h$. The two possibilities are:

The water supports only P waves. At oblique incidence the seabed can convert a P wave into an SV-wave in the sediment, and the return SV-wave can convert back to a P wave in the water. Further seismic mode conversion can occur at the basement. Thus the down-going and up-going sediment legs can independently be P or S. The S entries refer to SV-waves; a fluid P wave source in this two-dimensional isotropic geometry does not excite the decoupled SH-wave sector.

Let $L_h\ge 0$ be the distance travelled along the basement interface. For sediment legs $a,b\in \{P,S\}$,The first-appearance distance is the limiting value at $L_h=0$:

For example, the unconverted P branch gives $2(50+20)\tan 36.870^{\circ }=105\mathrm{m}$. The earliest geometrically permitted basement P seismic head wave is the doubly converted path with sediment S legs, at approximately $90\mathrm{m}$. The corresponding basement S family first appears at approximately $297\mathrm{m}$. These limiting distances describe existence of ray branches; a finite detectable wave amplitude also depends on conversion efficiency and acquisition noise.

There is no separate finite-distance P head wave at the water–sediment interface: $V_{p,w}=V_{p,s}=1500\mathrm{m}{\mathrm{s}}^{-1}$, so the nominal critical angle is $90^{\circ }$, and a ray from a source 50 m above the interface cannot reach it at finite horizontal distance. Nor can the slower sediment S speed, $900\mathrm{m}{\mathrm{s}}^{-1}$, support a seismic head wave launched from water: it would require $\sin {\theta }_w=1500/900>1$. A seismic reflection still exists at the seabed because the densities, hence the acoustic impedances, differ.

First appearance and first arrival differ. For each permitted branch the travel time isThe all-P basement P branch has $t_h=x/2500+0.0746667\mathrm{s}$. It overtakes the direct water P wave, $t_d=x/1500$, only at $x=280\mathrm{m}$. The converted P-head branches have larger seismic intercept times, so even where they exist at shorter offsets they are later arrivals. The S-head branches are likewise later than the all-P P-head branch wherever they exist in this model. Reflections and layer reverberations can supply additional later phases.

Phase conjugation creates a returning monochromatic wave whose spatial complex amplitude is proportional to the complex conjugate of the incident field. Fix the time convention throughout:Here $E$ is one scalar component of the electric field, $A\ge 0$ is the wave amplitude, $\omega >0$ is the angular frequency, and $\gamma$ is a spatially constant complex number setting the return strength and reference wave phase. The complex conjugation acts on $U$, including its spatial carrier, while the time factor remains $e^{-i\omega t}$. Conjugating both factors would merely give another representation of the same real field. With $\gamma =1$, the new real field equals $E(\mathbf{r},-t)$ in this ideal monochromatic wave model; that identity does not assert reversal of an arbitrary pulse or of the apparatus producing it.

The spatial wave phase becomes $-\varphi +\arg \gamma$, so the local wavevector changes from $\nabla \varphi$ to $-\nabla \varphi$. For example, $U=A{e}^{i(k_{x}x+k_{z}z)}$ with $k_z>0$ becomes $\gamma A{e}^{-i(k_{x}x+k_{z}z)}$: both components of the wavevector reverse. In a transparent stationary medium with isotropy and positive refractive index, light rays follow the wavevector, giving counterpropagation. The same wavefronts now have opposite propagation normals. This is why phase conjugation can return a distorted beam toward its origin: it reverses the detailed spatial wave phase, rather than requiring that wave phase to be flat.

The attractive application is correction of optical aberrations accumulated on an outward trip. The return uses the distortion itself as the prescription for correction, as the original thin phase-plate cancellation by phase conjugation example makes explicit. An experimentally accessible implementation measures the spatial wave phase and uses a spatial light modulator to generate its negative; digital optical phase conjugation describes an inspected 2010 demonstration of recovery through multiple scattering.

Phase matching makes wave amplitudes emitted at different depths add coherently. Consider collinear plane waves in a uniform transparent material of length $L$, with fixed polarization of waves and nonzero selected nonlinear optical susceptibility. Assume negligible pump depletion, negligible absorption, an initially absent harmonic, and a wide beam for which diffraction is negligible. The pump's wave amplitude is then constant throughout the interaction.

The pump electric field has spatial wave phase $k_{\omega }z$; its quadratic source has spatial wave phase $2k_{\omega }z$. The freely propagating harmonic has wave number $k_{2\omega }$. Define the second-harmonic phase mismatch byFor the chosen complex-field convention, the outgoing harmonic wave amplitude, up to an overall constant and propagation phase, is proportional toChoosing the conjugate complex-field convention reverses the exponent's sign without changing the wave intensity. Taking the squared complex modulus givesHere $C$ is fixed for the chosen modes and material; its dimensions accommodate the stated wave intensity convention. The quadratic pump dependence follows because the source wave amplitude is proportional to $E_0^2$, whereas $I_{\omega }$ is proportional to $E_0^2$. The sinc function comes from summing contributions along the whole material, just as a finite aperture produces diffraction through addition of complex amplitudes.

At $\Delta k=0$, the integral is $L$ and the harmonic wave intensity grows as $L^2$. At $\Delta kL=2\pi$, all those contributions cancel. This cancellation needs no absorption: their wave phases wind through a full turn. At high conversion, the constant-pump assumption fails, so the $L^2$ law cannot continue indefinitely. Tight focusing, finite optical pulses, absorption, and transverse separation of modes also require a more complete model.

Optical dispersion generally makes $n_{2\omega }\ne n_{\omega }$. In a suitable crystal, birefringence allows modes with different polarization of waves to experience different refractive indices, making their selected values equal. The appropriate crystal direction and allowed nonlinear optical susceptibility component must both work.^[ref]

For a worked calculation, choose $L_0=1\mathrm{mm}$ and $\Delta k=\pi \mathrm{m}{\mathrm{m}}^{-1}$. At $L=L_0$, the harmonic wave intensity is $4/{\pi }^2\simeq 0.4053$ of its perfectly matched value. At $L=2L_0$ it vanishes in this model. More material can therefore reduce the output. If instead $\Delta k=0$, doubling $L$ multiplies output by four. The two curves retain that common normalization, so their different peak heights are meaningful.

For the overlapping monochromatic wave components, let $d$ be the thickness and ${\lambda }_0$ the vacuum wavelength. Define $\Delta n=n_s-n_f>0$ using the slow and fast refractive indices. With complex amplitude convention $e^{i(kz-\omega t)}$, the accumulated slow-minus-fast phase difference isThis is the optical path difference $\Delta nd$ expressed as a wave phase. It is a signed accumulation convention; with the opposite time convention the sign reverses. The outgoing polarization of an electromagnetic wave depends on $\delta$ modulo $2\pi$. Here the positive slow-minus-fast $\Delta n$ differs from the signed principal-index $n_e-n_o$ used in the parent birefringence entry when $n_e<n_o$.

For an original worked example, stipulate $\Delta n=0.0100$ at ${\lambda }_0=600\mathrm{nm}$ and a uniform thickness $d=15.0\mu \mathrm{m}$. These are design assumptions, not measured material data. The optical path difference is $150\mathrm{nm}$, soIt is a quarter-wave plate. At the same wavelength and refractive indices, doubling $d$ to $30.0\mu \mathrm{m}$ gives $\delta =\pi$, a half-wave plate. A $1.0\mu \mathrm{m}$ thickness increase changes $\delta$ by $\pi /30$, or $6^{\circ }$; retaining the same birefringence at $500\mathrm{nm}$ would instead give $\delta =3\pi /5$ for the $15.0\mu \mathrm{m}$ plate. Real refractive indices also vary with wavelength, so this last calculation explicitly neglects optical dispersion.

For these equal transmitted wave amplitudes, orthogonal projection onto a polarizer crossed with the initial $45^{\circ }$ linear polarization gives a simple independent readout. After removing a common wave phase, the component complex amplitudes are proportional to $(1,e^{i\delta })/\sqrt{2}$; orthogonal projection onto $(1,-1)/\sqrt{2}$ gives $(1-e^{i\delta })/2$. Squaring its modulus yields $I_{\perp }/I_{\mathrm{in}}={\sin }^2(\delta /2)$: zero at $d=0$, one half at $15\mu \mathrm{m}$, and one at $30\mu \mathrm{m}$. This assumes ideal polarizers, negligible specular reflection and wave absorption, overlapping components and stable relative wave phase. It measures a periodic function of phase retardation, so one brightness measurement cannot determine the thickness uniquely.

The phase velocity is the motion of a constant-wave phase surface. For a one-dimensional plane wave $u=a\cos (kx-\omega t+{\varphi }_0)$ with real $k\ne 0$ and real $\omega$, fixing $kx-\omega t+{\varphi }_0$ givesEach wave crest of this monochromatic travelling wave advances at that signed phase velocity. Its magnitude is the phase speed $|c_p|=\lambda /T$, where $\lambda =2\pi /|k|$ is the wavelength and $T=2\pi /|\omega |$ is the temporal period when $\omega \ne 0$. Choosing $\omega >0$ makes positive $k$ mean rightward phase velocity and negative $k$ mean leftward phase velocity. Reversing both signs leaves the real plane wave and its phase velocity unchanged. At $k=0$ there are no spatial crests to track, so $\omega /k$ is undefined; a finite limit as $k\to 0$ is a limiting phase speed, not the speed of a spatially uniform oscillation.

In several dimensions, a plane wave has wave phase $\mathbf{k}\cdot \mathbf{x}-\omega t+{\varphi }_0$. Constant-wave phase planes obey $\mathbf{k}\cdot \mathbf{v}=\omega$. Their normal phase velocity isTangential motion within a plane does not move that wavefront, which is why the normal direction fixes the vector convention. In an anisotropic medium the group velocity need not point along this wavevector. Likewise the horizontal phase velocity of a tilted wavefront is its crossing speed along a horizontal line, rather than a component of the normal phase velocity.

For the nondispersive wave branch $\omega =ck$, a wave crest and a wave envelope both move at $c$. For deep-water surface gravity waves, $\omega =\sqrt{gk}$ gives $c_p=\sqrt{g/k}$ and group velocity $c_g=c_p/2$. With the illustrative choices $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\lambda =1\mathrm{m}$, the model predicts $c_p\simeq 1.25\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.625\mathrm{m}{\mathrm{s}}^{-1}$. These are calculated values for negligible surface tension and sufficiently large $kh$, not measurements. A wave crest moves forward through the wave packet, so timing a crest does not time the packet's journey.

There is a direct laboratory meaning to this distinction. MIT's 2005 surface-wave lab uses a long water tank, a controllable hydraulic paddle, and an absorbing beach. Resistance probes convert submerged length into voltage; calibration converts their readouts into free surface elevation versus time. Two probes a known distance apart, backed by a ruler and stopwatch, let students time a wave crest and separately time a feature of the wave packet. The lab records depth and varies frequency and wave amplitude.^[ref]

For a clean measurement of phase velocity, fit the periodic portion of both records or follow one identifiable wave crest. A phase difference is only known modulo $2\pi$: probe spacing below half the shortest relevant wavelength, or continuity while sweeping frequency, avoids assigning the wrong whole-cycle delay. Then compare the timed phase velocity with $\lambda /T$. Tracking pepper grains instead measures fluid-particle motion, which has a different meaning from either phase velocity or group velocity.

The Klein-Gordon packet velocities example makes the distinction especially sharp: its phase velocity exceeds $c$, yet its narrow wave envelope moves below $c$. A carrier wave labels successive oscillations; its speed alone does not establish information velocity. The front velocity depends on the full causal model, including angular frequencies far outside the carrier's narrow range.

A point spread function is an imaging system's response to an optical point source. Here it means the distribution of image light intensity: an ideal point object becomes a finite spot, often with faint rings, because of diffraction. An unresolved star and a sufficiently small bead emitting fluorescence are practical approximations to an optical point source. Their image tells us how nearby object details will overlap, and reveals blurring from optical aberrations, defocus, and, for an optical telescope, astronomical seeing.

A useful convention separates shape from throughput. At a fixed image plane, let $p(\mathbf{x})\ge 0$ be the point spread function with ${\int }_{{\mathbb{R}}^2}p(\mathbf{x})d^2\mathbf{x}=1$. Its units are inverse area. A point delivering detected radiant flux $F$ at image position ${\mathbf{x}}_0$ produces light intensity $Fp(\mathbf{x}-{\mathbf{x}}_0)$; this point spread function normalization makes the integral over the image equal to $F$. A profile divided by its own peak is convenient for comparing widths, but is a different point spread function normalization and does not by itself compare total radiant flux.

In incoherent imaging, different object points have negligible mutual spatial optical coherence over the measurement, so their image light intensities add. If the response also obeys translation invariance over the chosen field of view, the expected continuous image is the convolutionHere $O$ is the object brightness distribution expressed in image-scaled coordinates, with magnification already included, and $\tau$ is constant transmission. For $O(\mathbf{u})=F_0\delta (\mathbf{u}-{\mathbf{u}}_0)$, where $\delta$ is the two-dimensional Dirac delta function, this gives $I=\tau F_{0}p(\mathbf{x}-{\mathbf{u}}_0)$. A changing response across the field of view instead requires a position-dependent integral kernel $p(\mathbf{x};\mathbf{u})$; one measured point spread function is then only a local calibration.

In coherent imaging, the superposition principle applies to the complex wave amplitudes. With amplitude spread function $h$, the corresponding ideal model isThus $|h\ast E{|}^2$ generally differs from $|h{|}^2\ast |E{|}^2$: coherent contributions produce wave interference terms that depend on their relative wave phases. Their time average vanishes for mutually incoherent emitters, which explains the intensity convolution. Partial optical coherence requires the correlations between object points as well.^[ref]

The figure is an original diffraction model, not measured data. All panels have the same total radiant flux: each member of a pair contributes half the single-source radiant flux. Dashed curves show the separate light intensity contributions; the solid curve is their sum. The point spread functions have tails, so even sources satisfying the Rayleigh criterion overlap. As separation decreases, the image changes continuously; there is no abrupt disappearance of all information at one separation.

A motivating measured example is Measurement of point-spread function (PSF) for confocal fluorescence microscopy by InCheon Song and others (2005). A $543.8\mathrm{nm}$ laser, delivered through an optical fiber, excited fluorescent beads on a slide through a $100\times$, numerical aperture $1.3$ oil-immersion objective. A moving stage scanned the bead; emitted light returned through the objective, an emission filter and a pinhole to a photodiode. The readout was fluorescence light intensity against scanned position. Under the same setup, nominal bead diameters $0.17$, $0.5$, and $1\mu \mathrm{m}$ produced progressively broader measured spots. The size comparison controls for the finite object: the measured response is the convolution of bead emission with the effective point spread function, rather than a pure point response. Inverting this convolution model gave a response compared with the theoretical one. The smaller bead gave a weaker signal relative to noise; uncertainty in actual bead diameter also affected the restored width. These are limits of the calibration. In confocal microscopy, excitation and pinhole detection jointly determine the effective point spread function; this experiment does not measure the ideal widefield Airy disk plotted above.^[ref]

For a new calibration, useful additional controls are to measure detector background, avoid saturating the photodetector, repeat the scan to check stability, and compare isolated beads at several positions in the field of view. Record wavelength, numerical aperture, pinhole setting, sampling interval and optical focus. These are practical recommendations, not extra measurements claimed for the published experiment.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

For an isolated weakly damped resonator mode, $Q=\omega E/P_{\mathrm{loss}}$. If $E(t)=E(0)e^{-t/{\tau }_E}$, differentiating gives $P_{\mathrm{loss}}=E/{\tau }_E$, so $Q=\omega {\tau }_E$. Since energy is quadratic in wave amplitude, the wave amplitude exponential relaxation time is ${\tau }_A=2{\tau }_E$. A decay by an energy factor $D$ in a measured interval $t_d$ gives $Q=\omega t_d/\ln D$. For a conventional reverberation time, $Q=2\pi f{T}_{60}/[6\ln (10)]$ under this exponential single-rate approximation.

A quarter-wave plate gives a quarter-cycle phase retardation. Equal transmitted wave amplitudes initially in linear polarization at $45^{\circ }$ become circular polarization; unequal wave amplitudes generally give elliptical polarization. Reversing the sign reverses the traversal of the same circle.

The simplest off-resonant explanation is a vibrating polarizability. Let $Q(t)=Q_0\cos (2\pi {\nu }_{v}t)$ be a small normal mode coordinate of molecular vibration, and use one scalar component of polarizability, $\alpha (Q)\simeq {\alpha }_0+{\alpha }_0^{\prime }Q$. For incident electric field $E(t)=E_0\cos (2\pi {\nu }_{0}t)$, the induced electric dipole moment isThe product-to-sum formula produces the two displaced frequencies. A nonzero derivative ${\alpha }_0^{\prime }$ gives vibrational Raman scattering in this linear model. Real polarizability is generally a tensor, and the selected components depend on incident and observed wave polarization. This condition differs from vibrational infrared activity, which requires a changing molecular electric dipole moment without an applied optical electric field. A vibration can therefore appear in one measurement and be absent in the other.

This classical modulation argument explains line locations but supplies equal sideband wave amplitudes; it does not explain their thermal asymmetry or the nonzero Stokes Raman scattering from a vibrational ground state. Those require the quantum harmonic oscillator population and transition strengths. The far-off-resonance drawing uses dashed virtual intermediate levels: they represent terms in a scattering probability amplitude, not a long-lived occupied electronic energy eigenstate. Near an electronic resonance, a more detailed treatment of the intermediate response is needed.

The Rayleigh criterion calls two equal, mutually incoherent optical point sources just resolved when one circular-pupil Airy disk's centre coincides with the other's first zero. It is a specified discrimination convention, distinct from full width at half maximum and from precision in estimating one source's position.

For this circular optical pupil, the Rayleigh criterion uses angular separation $s_{\mathrm{R}}={\theta }_1\simeq 1.22\lambda /D$, or $s_{\mathrm{R}}=r_1\simeq 1.22\lambda f/D$ in the paraxial approximation to the focal plane. The point spread functions still overlap.^[ref]

For the Airy disk design example with $D=5.00\mathrm{mm}$, $f=100\mathrm{mm}$ and $\lambda =550\mathrm{nm}$, the angular convention gives $s_{\mathrm{R}}=1.34164\times 10^{-4}\mathrm{rad}$, or about $27.67$ arcseconds, and the image-plane separation is $13.4164\mu \mathrm{m}$. Two equally bright, mutually incoherent optical point sources at separations $0.7$, $1$ and $1.5$ times this scale would have image separations $9.391$, $13.416$ and $20.125\mu \mathrm{m}$ respectively. These calculated cases are compared in incoherent two-source Airy profile.

The Rayleigh criterion is useful because it ties an easily recognized two-peak convention to a predictable optical scale. It does not assert that all information vanishes below that separation. Inferring two unknown positions also depends on photon shot noise, background, unequal brightness and uncertainty in the point spread function; locating a single isolated source is a different task. A theoretical analysis shows that alternative optical measurements can estimate the separation of two incoherent optical point sources below this scale.^[ref]

The mutual optical coherence condition is essential. In incoherent imaging, time-averaged light intensities add. In coherent imaging, wave amplitudes add before squaring, leaving a wave interference term controlled by relative wave phase. Two holes illuminated by the same coherent laser therefore do not implement the equal-incoherent-source convention merely because their individual patterns resemble Airy disks. The coherent single-aperture laboratory test establishes the response of each aperture; it does not establish the pair's mutual incoherence.