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A curved transparent surface resting on a plane traps a thin film whose thickness increases radially, producing circular interference fringes. For near-normal reflected light and one relative $\pi$ reflection wave phase change, contact is dark and bright rings obey $2nt=(m+1/2){\lambda }_0$ for $m=0,1,\dots$. Here $n$ is the film's refractive index and ${\lambda }_0$ the vacuum wavelength. For spherical curvature radius $R$, $t\simeq r^2/(2R)$ gives $r_m^2=R{\lambda }_0(m+1/2)/n$. Consequently $D_{m+s}^2-D_m^2=4sR{\lambda }_0/n$ for ring diameters $D_m=2r_m$. A contact gap, altered reflection phase, or significant oblique incidence changes the formula.

Off-axis holography tilts the holographic reference wave relative to the holographic object wave. Its interference fringes acquire a carrier spatial frequency; the two cross terms occupy different spatial frequency regions when the carrier is sufficiently large compared with the object's signal bandwidth. This permits filtering a single image term.

Gabor's lecture credits Emmett Leith and Juris Upatnieks with the later skew-reference method using helium–neon laser illumination, separating reconstructed images angularly.^[ref]

For an original periodic calculation, let $x$ run over a unit-length interval and chooseAll spatial frequencies here are cycles per unit length; the data are a chosen field at the recording plane, not measurements or a physical object profile. Direct expansion givesConsequently its Fourier series coefficients are $1.05$ at zero, $0.02$ at $\pm 8$, $0.2$ at $\pm 17$ and $0.1$ at $\pm 25$. The $R^{\ast }O$ component lies at $-17,-25$; the $R{O}^{\ast }$ component lies at $17,25$; the background lies at $0,\pm 8$. Filtering the negative-frequency cross term and multiplying by $R$ recovers $O$, including its relative wave phase.

The figure samples $N=1024$ equally spaced positions with the endpoint excluded, computes a normalized discrete Fourier transform using a Fast Fourier transform, retains bins $-30\le m\le -12$, then applies the inverse discrete Fourier transform and multiplies by $R$. The recovered real and imaginary parts coincide with the chosen $O$ to numerical precision. This exact periodic example avoids leakage by placing every frequency on an integer bin; finite nonperiodic records need allowance for broadening and an aperture.

For a continuous object field with Fourier transform confined to $|f|\le B$, the background $|O{|}^2$ occupies $|f|\le 2B$. A carrier $f_c>3B$ is a sufficient separation condition for the sideband centred at $-f_c$, and $f_c+B$ must remain below the Nyquist spatial frequency. Here $B=5$, $f_c=20$ and the Nyquist spatial frequency is $512$. This sufficient bound describes a whole bandwidth, rather than exploiting the gaps in this particular two-component field.

The same example also makes optical playback explicit. Choose $t_0=0.3$ and $\beta =0.01$. Since $|O|\le 0.3$, the bounds $0.49\le I\le 1.69$ imply $0.3049\le t\le 0.3169$, and $|\beta I|/t_0\le 0.0564$. The affine response is stipulated; its weak-modulation interpretation is approximate. In $Rt$, the desired field $0.01O$ has frequencies $3,-5$; the conjugate term $0.01R^2{O}^{\ast }$ has frequencies $37,45$; the background has frequencies $20,12,28$. Thus the same carrier that separates the recorded cross terms separates the playback components. Reillumination restores a field through diffraction; computational filtering extracts the corresponding encoded field through Fourier analysis.

An optical path length measures how much propagation wave phase a path accumulates, expressed as the distance that would accumulate the same wave phase in vacuum. A centimetre in a uniform medium with refractive index $3/2$ has optical path length $1.5\mathrm{cm}$, although its geometric length remains $1\mathrm{cm}$. This makes optical path length useful whenever two contributions must arrive with matching wave phases, from an optical lens to a Michelson interferometer. ^[ref]

For a specified path $\gamma$ at fixed angular frequency $\omega$, the optical path length is the line integralHere $s$ is geometric arc length, and $n$ is the phase refractive index, evaluated at the same frequency everywhere. The propagation interpretation assumes a stationary linear medium with isotropy, a real positive scalar refractive index and negligible loss, on a valid geometrical optics branch. The path can curve; at sharp interfaces apply the appropriate matching conditions to each segment. A prescribed curve has a mathematical line integral, but it predicts a propagation wave phase only when it represents the physical branch. ^[ref] Optical fibers can require a mode wavenumber rather than the bulk refractive index along a guessed light ray. ^[ref]

Fix the vacuum wavelength ${\lambda }_0=c/f=2\pi c/\omega$, with frequency $f=\omega /(2\pi )$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$. In a locally uniform medium the wavelength is $\lambda ={\lambda }_0/n$, while the frequency stays unchanged at a stationary interface. With the convention $\Phi =k_{0}S-\omega t+{\varphi }_0$, propagation from a specified starting wave phase contributesThe relative reference ${\delta }_{\mathrm{ref}}$ includes launch wave phases and any extra wave phase from mirrors, beam splitters or coatings. Thus $k_0\mathrm{OPL}$ is a propagation contribution, not an absolute detector wave phase independent of the apparatus. If the reference contributions stay fixed, a change in optical path difference $h$ changes the phase difference by $2\pi h/{\lambda }_0$. Using the material wavelength again in this formula would count the refractive index twice. ^[ref]

A motivating measurement turns invisible optical oscillations into a slowly changing electrical signal. MIT's teaching experiment uses a laser, a beam splitter, two returning mirrors, an amplified photodiode, an oscilloscope and a calibrated mirror actuator driven by a function generator. The returning beams overlap on the photodiode; one channel records detector voltage and another the actuator drive. The measured observable is the changing light intensity, including its maxima and minima, rather than direct timing of an optical cycle. A calibrated displacement per drive voltage and the spacing of the extrema relate geometric motion to the phase difference. The guide infers the laser wavelength from known actuator calibration; conversely, with independently known vacuum wavelength and ambient refractive index, the same relation calibrates mirror motion. These two unknowns cannot both be inferred from fringe spacing alone. ^[ref]

For a proposed change-of-medium measurement, keep the mirrors fixed and put a transparent cell of internal length $L$ in one arm. Compare its contents at refractive indices $n_a$ and $n_b$, leaving its windows in place. At normal incidence the two traversals giveHere $N$ is the signed number of whole interference fringe cycles plus any resolved fractional cycle, counted continuously. A mirror displacement $x$ instead gives $\Delta \mathrm{OPL}=2n_{\mathrm{amb}}x$; distinguish this double-pass factor from a single pass in a Mach-Zehnder interferometer. OpenStax's gas-cell worked example provides the same double-pass reasoning; it is a teaching calculation rather than a published measurement dataset.^[ref]

An exact toy calculation makes the sensitivity visible. Stipulate ${\lambda }_0=600\mathrm{nm}$, $L=10\mathrm{mm}$ and $n_b-n_a=3/100000=0.00003$. Then $\Delta \mathrm{OPL}=600\mathrm{nm}$ and $\Delta \delta =2\pi$: a full interference fringe cycle, despite unchanged geometric lengths. For a continuous change $n_b-n_a=q(3/100000)$, let $0\le q\le 2$ and choose the initial detector phase difference to be zero. Equal coherent detected arm contributions of light intensity $I_0$ predictThe plot is this stipulated model: $q=1/2$ gives a minimum, $q=1$ restores a maximum, and the reference level is chosen by adjustment of the apparatus. It contains no measured values or assigned properties of a real gas. The equivalent mirror travel in vacuum is $300\mathrm{nm}$ per cycle. A single endpoint light intensity cannot distinguish $q=0$ from $q=1$; the continuous fringe count carries the missing information.

For an interpretable cell measurement, establish the initial phase difference with a small calibrated mirror scan, then track slow changes through successive interference fringes. Measure the dark signal and each arm separately, keep photodiode gain within its linear range, align both beams on the same region and match their linear polarizations. Monitor interferometric visibility and keep arm delays within the useful optical coherence time. Record cell length, contents, pressure and temperature; return to the starting condition to check drift. Changes in window deformation, cell length, external air or mirror position also change optical path length, so a reference cell or an independently monitored mechanical path is needed to isolate the contents' refractive index. These controls are deductions from the model, not a claim that the illustrated apparatus was operated. Loss of fringe count leaves a whole-cycle ambiguity; at an intensity extremum the first derivative with respect to phase difference vanishes, so a scan around phase quadrature is more sensitive to small changes. Broadband illumination, unstable optical coherence, vibration and strong loss can invalidate a simple one-frequency inference.

Put the plate behind the slit at transverse coordinate $+d/2$, and let positive screen coordinate $x$ point toward that slit. The extra optical path length isIn the paraxial approximation, the covered path minus the uncovered path has total optical path difference $\Delta -dx/L$. Its zero determines the shifted central interference fringe:toward the slit covered by the plate. Moving toward that slit shortens its geometric path and compensates the extra delay in the plate.

More generally, the phase difference is $(2\pi /\lambda )(\Delta -dx/L)$, so every interference fringe translates by the same $x_0$ in this approximation. The spacing $\lambda L/d$ is unchanged, and the shift corresponds to $\Delta /\lambda$ fringe periods. For an ideal uniform transparent plate that changes only the wave phase, the envelope and interferometric visibility are unchanged. Reflection losses or unequal transmission through a real plate could additionally reduce the interferometric visibility; those are separate amplitude effects. No numerical wavelength is needed for the physical displacement of the zero-order fringe.

Write $R_z(\theta )=e^{-i\theta Z/2}$ and $R_x(\theta )=e^{-i\theta X/2}$, using the Pauli operators. The Hadamard gate obeys $HZH=X$, so unitary conjugation of the matrix exponential gives $H{R}_z(\beta )H=R_x(\beta )$. Therefore the symmetric z-x-z beam splitter factorization isThe rightmost operator acts first. Each factor is unitary, which also proves the given matrix is unitary. Choosinggives $U_{\mathrm{BS}}=-iH$, so it is exactly a Hadamard gate up to global phase.

For the specified identical splitters with $\alpha =0$, put $c=\cos (\beta /2)$ and $s=\sin (\beta /2)$. Here $U_{\mathrm{BS}}=R_x(\beta )$, and direct multiplication givesThus the upper-port probability iswith $P_1={\sin }^2\beta {\cos }^2(\varphi /2)$. These identical beam splitters in a two-path interferometer give a nonconstant interference fringe whenever $\sin \beta \ne 0$. With this matrix and port convention, two identical balanced splitters swap the bright and dark ports relative to the two-Hadamard model: at $\varphi =0$, $R_x(\pi /2{)}^2=-iX$. The reflection phases must remain in the calculation.

The upper-port maximum and minimum are $1$ and ${\cos }^2\beta$. Hence the peak-to-peak modulation is ${\sin }^2\beta$ and its normalized interferometric visibility isBoth measures of contrast are maximal precisely forThese are balanced beam splitters, with equal single-input transmission and reflection probabilities. When $\beta \in \pi \mathbb{Z}$, only one route contributes to the output and the fringes disappear.

Let $I_0$ be the intensity from either fibre alone at the observation point, assuming equal output intensities, matching polarizations and negligible drift. The screen samples the relative phase $\Phi (X)$ between the two outputs. For an ideal monochromatic laser, first-order optical coherence is complete andThe interference fringes are approximately equally spaced in the small-angle far field, with a slowly varying diffraction envelope. If the output separation is $a$ and the screen distance is $D$, the interference fringe spacing is approximately ${\lambda }_{0}D/a$. Equal phases at the two input ends do not ensure a central bright fringe: the extra fibre contributes carrier phase $2\pi n_{\mathrm{eff}}(l_2-l_1)/{\lambda }_0$ in addition to the free-space optical path difference.

For a finite linewidth, the time-averaged interference signal becomesHere $g^{(1)}$ may be defined for the slowly varying field envelope, with the carrier phase already included in $\Phi$. A realistic laser therefore gives reduced interferometric visibility; for $V=1/2$, its extrema are $I_0$ and $3I_0$.

For a Lorentzian line shape of frequency full width at half maximum $\Delta \nu$, temporal optical coherence givesThe relevant index is the group index. Using the intended nondispersive approximation $n_g\simeq n_{\mathrm{eff}}=2$,This delayed-fibre linewidth measurement assumes the loss of contrast comes from linewidth. Unequal output intensities, polarization mismatch or mechanical phase fluctuations also lower interferometric visibility. If dispersion matters, the stated phase index alone does not specify the necessary group index.

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

Take $f(t)=a(t)e^{-i\omega t}$ at a fixed reference point, and define the full-field temporal optical coherence function byThe angle brackets denote an ensemble expected value, or a sufficiently long time average when justified. The factor $C$ puts $\Gamma$ in light intensity units for the chosen peak complex-field convention. The normalized first-order optical coherence is $g^{(1)}(\tau )=\Gamma (\tau )/\Gamma (0)$ for a stationary source.

“Independent of absolute time” means the second moment depends on the time difference $\tau$ but not on the starting time $t$: shifting both observation times by the same amount leaves it unchanged. Together with a constant mean this is second-order or weak stationarity; it does not mean that the random electric field is constant in time. Put $I_0=\Gamma (0)=C⟨|a{|}^2⟩$. The full-field convention is important: an envelope-only correlation omits $e^{-i\omega \tau }$ and must have that carrier factor restored before deriving optical interference fringes.

For each beam in vacuum, its magnetic field is ${\mathbf{B}}_j={\hat{\mathbf{k}}}_j\times {\mathbf{E}}_j/c$. Write the two complex scalar contributions, including the beam splitter factor, as $u_1=f(t-t_1)/\sqrt{2}$ and $u_2=f(t-t_2)/\sqrt{2}$. Since their electric fields are both along $z$,The optical-cycle and ensemble-averaged Poynting vector isThe mean individual light intensities are equal, soThus energy flows along the bisector of the two beam directions, crossing a detector plane normal to $x$. Such a sensor can in principle record the variation across its $y$ coordinate. Its actual normal energy flux is $I(y)/\sqrt{2}$ in the normalization used above; this fixed factor does not change interferometric visibility.

There is a practical resolution condition in addition to the direction of energy flow. The interference fringe spacing $\lambda /\sqrt{2}$ is very small, so direct pixels would need spacing no larger than $\lambda /(2\sqrt{2})$ as required by the Nyquist–Shannon sampling theorem, or optical magnification and sufficient angular aperture to preserve both incident directions. An ordinary optical camera with much larger effective pixels averages the interference fringes. The Poynting-vector argument establishes the allowed measurement geometry, not guaranteed resolution with any optical camera.

Each mutually incoherent patch of an extended star produces a displaced double-slit interference pattern. The detector adds their light intensities. For two apertures separated by a projected interferometric baseline $D$, a source direction offset by a small angle ${\vartheta }_x$ changes their phase difference by $2\pi D{\vartheta }_x/\lambda$. The normalized sum over the source's angular brightness distribution is its first-order optical coherence, whose magnitude is the interferometric visibility for equally bright channels.

For a uniform circular stellar disk of full angular diameter $\alpha$, the normalized angular Fourier transform gives the uniform-disk stellar visibilityHere $J_1$ is the order-one Bessel function of the first kind. This follows by integrating $e^{i(2\pi D/\lambda ){\vartheta }_x}$ over the disk and dividing by its area. In particular, its power series gives $2J_1(x)/x=1-x^2/8+O(x^4)$.

For an ordinary slit separation $d=1\mathrm{mm}$, using the paper's $\alpha =22.6\times 10^{-8}\mathrm{rad}$ and observing wavelength $\lambda =570\mathrm{nm}$,Even $d=10\mathrm{mm}$ gives a visibility loss of only $1.94\times 10^{-5}$. Thus millimetre separation leaves the interference fringes almost indistinguishable from those of a point source; it gives very little sensitivity to the diameter.

The movable mirrors in the Michelson stellar interferometer collect light at a large adjustable separation $D$ and redirect it into closely spaced output beams. They increase the sampling interferometric baseline independently of the fixed output slit spacing $d$: $D$ sets sensitivity to source structure, whereas the output optics set the screen's interference fringe spacing. The first Bessel function zero is $x_1=3.83170597$, henceThe interference fringes disappear at this baseline even though each beam separately remains bright.

The value $3.08\mathrm{m}$ assumes a uniform disk. The original question supplies a diameter without a surface brightness law; real limb darkening changes the precise visibility zero. A one-dimensional uniformly bright strip of the same full angular width instead has a sinc function visibility and first zero $D=\lambda /\alpha =2.522\mathrm{m}$. That is a different source model, not the circular-disk result. The robust conclusion without a surface brightness model is that a baseline of a few metres, rather than millimetres, is required.

A mercury-vapor lamp is an extended, spontaneously emitting source, so practical interference requires spatial filtering and spectral selection. A narrow illuminated source slit gives sufficient spatial coherence; selecting one strong spectral line improves temporal coherence and avoids overlapping differently spaced interference fringes.

In the classical double-slit experiment, both apertures are illuminated by the same narrow source. The arrangement is symmetric, permits direct measurement and adjustment of slit separation, and has equal optical materials in the two paths. Its main drawbacks are low transmitted intensity through narrow apertures, the need for a well-made closely spaced slit pair and alignment, and the finite single-slit diffraction envelope. A broad source washes out interference fringes.

Lloyd's mirror uses the direct wave from a narrow source and the reflected wave, which appears to come from its virtual image behind a plane mirror. It requires only one physical source slit; the mirror can be adjusted to make a small effective source separation without manufacturing a double aperture. Only the direct-reflected overlap region shows interference fringes. Finite mirror size, alignment, surface quality, unequal intensities and the polarization-dependent reflection coefficient can reduce visibility. For the usual grazing reflection from glass, the reflected electric wave acquires approximately a $\pi$ phase shift, so the zero path-difference position is dark rather than the bright central fringe of the symmetric classical double-slit experiment. Both arrangements still require adequate spatial coherence and temporal coherence.

Illuminate a narrow entrance slit with the mercury-vapor lamp, use a collimating lens to illuminate two parallel narrow slits coherently, and insert a bandpass filter selecting the green line. Place a screen or scanning detector a measured distance $D$ beyond the double slit. The source slit and the two apertures should be parallel. The aperture widths should be appreciably smaller than their separation to provide an extended diffraction envelope; the green filter should reject the other strong mercury lines. The layout is drawn below.

For slit separation $d$ and observation angle $\theta$, the optical path difference in air is $\Delta =d\sin \theta$. At a bright interference fringe, $\Delta =m\lambda$. For a distant screen and small angles, $\sin \theta \simeq \tan \theta \simeq y/D$, givingMeasuring several interference fringe spacings and dividing by their number reduces relative position error. The derivation assumes the slits have equal phase on illumination and that the screen region lies within their common diffraction envelope.

The two wavelengths have independent interference patterns. Their interference fringe spacings areDifferent optical frequencies do not maintain a fixed mutual phase over the detector's integration time, so their intensities add. For narrow equal slits, suppressing the slowly varying diffraction envelopes, one may writeThe relative line strengths are not supplied; the figure illustrates equal peak intensities rather than inferring a quantitative brightness ratio. At $y=0$ both colours are bright, producing a blue-green mixture. Away from the centre the green and blue interference fringes progressively separate, creating alternating colour and brightness variations; places where one line is dark can still be bright in the other. Neither two-line mixing nor its central maximum is generally white.

Across $-15\mathrm{mm}\le y\le 15\mathrm{mm}$, the green maxima occur at $y=2.730m\mathrm{mm}$ for $m=-5,\dots ,5$, and the blue maxima at $y=2.185n\mathrm{mm}$ for $n=-6,\dots ,6$. Near $y=\pm 10.92\mathrm{mm}$, the fourth green and fifth blue maxima nearly coincide: $4{\beta }_g=10.920\mathrm{mm}$ and $5{\beta }_b=10.925\mathrm{mm}$. The $5\mu \mathrm{m}$ mismatch shows that these are near coincidences, not exact ones for the given wavelengths. The plotted central interval includes all these maxima.

Use a two-element radio interferometer: couple two small dishes coherently and align the projected baseline along the angular separation of the neighbouring satellites. Each dish's broad antenna response should include both satellites. Match the electronic delays so that the desired satellite arrives in phase at the combiner. For baseline length $b$ perpendicular to the central line of sight, a small angular displacement $\theta$ produces geometric path difference $b\sin \theta$ and phase differenceFor equal signal amplitudes from one source, the output wave intensity is $4I_0{\cos }^2(\varphi /2)$. Put one satellite on a maximum and its neighbour on the first null: $\varphi ({\theta }_R)=\pi$. The smallest useful projected baseline is thereforeThe angles here are measured near the normal to the baseline. At a general pointing angle ${\theta }_0$, replace $b$ in the small-separation estimate by $b\cos {\theta }_0$. Longer odd-half-fringe baselines can also null the neighbour but create more angular ambiguity. This arrangement provides discrimination along one direction; two collectors alone have repeated interference fringes and do not produce an unambiguous full image.

Single travelling wave. Differentiation of the plane wave gives ${\partial }_x\psi =ik\psi$, so its probability current isThis is probability density times the free-particle velocity $\hslash k/m$.

Counterpropagating waves. Remove the common time phase and set $z=B^{\ast }A{e}^{2ikx}$. Direct multiplication givesConsequentlyThe interference term $z-z^{\ast }$ is purely imaginary in the last real-part operation. The probability density can have interference fringes even though this net probability current is spatially constant.

Scattering amplitudes. Locate the step at $x=0$ and first take $E>V$. With $k_1=\sqrt{2mE}/\hslash$ and $k_2=\sqrt{2m(E-V)}/\hslash$, useThe common factor $e^{-iEt/\hslash }$ has been suppressed. A finite step with the same mass on both sides requires continuity of $\psi$ and ${\psi }^{\prime }$, giving $1+r_a=t_a$ and $k_1(1-r_a)=k_2{t}_a$. HenceIf $0<E<V$, choose $k_2=i\kappa$, $\kappa =\sqrt{2m(V-E)}/\hslash$, so the transmitted wave decays. The same amplitude formulae continue to hold, but the semi-infinite forbidden region carries no outgoing probability current: $R=1$ and $T=0$.

Flux balance. In the propagating case,In particular,The velocity ratio in $T$ is essential: the squared transmission amplitude alone is not the transmission probability when the two velocities differ.

Numerical transmission. The mass cancels from the ratio, and the energies can remain in the same electronvolt units:Thus about $92.8\%$ is transmitted and $7.18\%$ reflected, despite the incident energy exceeding the step.

With two coherent paths, the probability amplitude at a screen is ${\psi }_1(x)+e^{i\varphi }{\psi }_2(x)$. The Born rule givesThe final term shifts when the relative phase changes, producing interference fringes. For slit separation $d$ in the far field, the phase difference is approximately $2\pi d\sin \theta /\lambda$; constructive interference occurs at $d\sin \theta =n\lambda$. Finite slit widths add a diffraction envelope, so an ideal equally bright, infinitely narrow slit model does not predict the full observed envelope.

single-electron interference pattern buildup starts with isolated localized detector events at sufficiently low source intensity. Repeated events accumulate the interference probability distribution. Thus detection is discrete while propagation is described by coherent probability amplitudes. The effect cannot be explained by interference between different particles if the apparatus is occupied by at most one particle at a time. A motivating realization used a field-emission electron source, an electron biprism to recombine two paths, electron lenses and a position-sensitive electron counter; it recorded individual arrivals building up the fringes. This was a biprism realization of two-path interference, not a plate with two literal slits.^[ref]

The crucial comparison is between both coherent paths open and a preparation that supplies which-path information. If the paths become correlated with detector states $|D_1⟩,|D_2⟩$, the joint state is ${\psi }_1|D_1⟩+e^{i\varphi }{\psi }_2|D_2⟩$. Ignoring that detector multiplies the interference term by $⟨D_1|D_2⟩$. Orthogonal detector states remove the term completely, while identical detector states preserve it. This is a consequence of entanglement and the partial trace; a human need not read the detector.

In a real test, check the single-path distributions, detector background, source coherence and stability of the relative phase. Classical phase drift can wash out fringes without supplying which-path information. Conversely, localized detections alone do not establish coherent two-path propagation; the controlled phase dependence of their accumulated distribution provides the discriminating evidence.

Represent equal narrow slits at $x=\pm D/2$ by two contributions with equal complex amplitudes. In a direction $\theta$ the path difference is $D\sin \theta$, soTherefore double-slit interference givesThe bright interference fringes satisfy $D\sin \theta =m\lambda$, with a central maximum. In the small-angle region,Dark fringes lie halfway between bright ones. Exactly, spacing is uniform in $\sin \theta$, not in $\theta$ at large angles. The first panel of the pattern figure below shows the narrow-slit result.

A relative phase $\pi$ changes the sign of one slit field. Their sum becomesThe central interference fringe is now dark. The maxima move to $D\sin \theta =(m+1/2)\lambda$, while the minima occur at $D\sin \theta =m\lambda$. Thus the pattern shifts by half a fringe, $2.00\mathrm{mrad}$ near the axis, but its angular fringe separation remains $\boxed{4.00\mathrm{mrad}}$. This assumes the glass changes phase uniformly and introduces negligible absorption or beam deflection.

A Michelson interferometer splits a collimated beam at a partially reflecting beam splitter. Two mutually perpendicular arms reflect from mirrors and recombine at the same beam splitter; the superposed returning beams produce interference fringes at the detector. Translate one mirror along its arm while holding the other fixed.

If that mirror moves a measured distance $\Delta x$, the round-trip optical path difference changes by $2\Delta x$ in air, approximated as vacuum. Every change of one wavelength moves one complete bright-dark-bright fringe cycle past a fixed detector point. Count $N$ such cycles to obtainUse a spectral line selected with a suitable optical filter, and count many interference fringes to reduce the fractional uncertainty of the distance and count. Keep the two arms within the source's optical coherence length and maintain overlap and alignment. If the vacuum wavelength is required for an arm filled with air of known refractive index $n$, the corrected relation is ${\lambda }_0=2n\Delta x/N$ at stable $n$.

Take the quoted separation to mean adjacent slit centre spacing $d=50\mu \mathrm{m}$. For three equal very narrow slits at $-d,0,d$, the Fraunhofer diffraction complex amplitude in the transverse direction isWith $I_1$ denoting one narrow slit's contribution, the light intensity isThe quotient is interpreted by its limit where both numerator and denominator vanish. At small angles on a screen at $R=1\mathrm{m}$, $\sin \theta \simeq X/R$ and $\beta =\pi X/(10\mathrm{mm})$. Principal interference fringes occur atBetween successive principal maxima there are two zeros, at $X=(m+1/3)(10\mathrm{mm})$ and $X=(m+2/3)(10\mathrm{mm})$. Halfway between them is a smaller maximum $I=I_1$, one ninth of a principal peak. The pattern is symmetric about $X=0$. The positions quoted in millimetres use the paraxial approximation; exactly $d\sin {\theta }_m=m\lambda$ and $X_m=R\tan {\theta }_m$.