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In the double-slit experiment, each photon is detected as a localized event carrying energy $h\nu$, while repeated events build the interference pattern of coherent probability amplitudes from the two slits. Acquiring information that distinguishes the paths suppresses the interference; splitting the classical intensity into two incoherent alternatives would not reproduce the fringes.

Put one plate before each slit at normal incidence, with the same known refractive index $n$ and illumination of vacuum wavelength $\lambda$. Relative to the plate-free reference, their thickness difference $\Delta d=d_1-d_2$ produces an optical path difference $(n-1)\Delta d$ and a phase shiftObserve the displacement of the full interference fringe pattern, or scan a controllable reference phase and estimate $\Delta \varphi$. This compares the plates directly; a detectable phase displacement establishes unequal thickness under the common-material assumption. For fringe spacing $\lambda D/b$, the displacement is $-D(n-1)\Delta d/b$, where $b$ is slit separation and $D$ the screen distance, with sign set by the slit labeling.

A differential phase measurement can resolve a small difference sensitively and avoid determining two large absolute thicknesses. It does not guarantee a universally faster or more photon-efficient test: noise, coherence, alignment, and required accuracy matter. Moreover a null phase shift at one wavelength only impliesso equality requires a prior bound, a broadband or multiple-wavelength measurement, or continuously counting interference fringes as the plates are introduced. If the refractive indices differ, this apparatus measures optical thickness difference rather than geometrical thickness difference.

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.

A Michelson interferometer turns displacement into a count of interference fringes: a moving mirror changes both the outward and return paths. This lets a wavelength act as a ruler whose divisions need no engraving. Assume normal incidence, a fixed reference arm, stable refractive index $n$ along the moving arm, and fixed vacuum wavelength ${\lambda }_0$. Write $\lambda ={\lambda }_0/n$ for the wavelength in that medium and take $\Delta L$ positive away from the beam splitter. The change in optical path difference, the accumulated phase difference, and the signed number of complete interference fringe cycles areThe factor two comes from propagation geometry: moving the mirror by $\Delta L$ adds $\Delta L$ before reflection and $\Delta L$ afterward. One bright-to-bright cycle means mirror travel $\lambda /2$; bright-to-dark means $\lambda /4$, half a cycle. All lengths must use compatible units, and $\Delta N$ is dimensionless. A constant reflection phase difference changes the initial fringe position but cancels from this change formula.

For an original hypothetical example, suppose $\lambda =600\mathrm{nm}$ in the actual surrounding medium and a continuous scan gives $40.25$ cycles, including the final fractional cycle. Then $\Delta L=(600\mathrm{nm})(40.25)/2=12075\mathrm{nm}=12.075\mu \mathrm{m}$. With the wavelength treated as exact, a fringe-reading measurement uncertainty of $0.02$ cycle contributes $6\mathrm{nm}$ to the displacement measurement uncertainty. An uncertainty of $0.01\mathrm{rad}$ in a local phase difference instead corresponds to $600(0.01)/(4\pi )\mathrm{nm}\approx 0.477\mathrm{nm}$. This illustrates measurement resolution, not a complete claim of accuracy: wavelength calibration, air changes, mirror tilt and missed cycles can dominate.

To separate the oscillating readout from its optical coherence envelope, use the local modelHere $I$ is detected light intensity, $\bar{I}$ is the mean level, ${\varphi }_b$ is a fixed bias wave phase, and $\mathcal{V}$ is interferometric visibility. A fine scan of $D$ measures the carrier phase difference; a coarse scan can map the temporal optical coherence envelope through $\mathcal{V}(D)$. The envelope maximum locates matched delays only under the source and apparatus assumptions; it is not a bright fringe identifying an arbitrary mirror position. Even with known $\mathcal{V}$, one intensity reading leaves the cosine sign and whole-cycle ambiguities. Continuous motion of known direction, or an independently obtained pair of signals in phase quadrature, is needed to recover signed travel through reversals.

The figure's original control example stipulates equal arm light intensities, matching linear polarizations, and a Lorentzian optical spectrum giving $\mathcal{V}(D)=\exp (-|D|/L_c)$ with $L_c=1\mathrm{mm}$. In vacuum, arm mismatches $|L_2-L_1|=0.60\mathrm{mm}$ and $0.10\mathrm{mm}$ give $\mathcal{V}=e^{-1.2}\approx 0.301$ and $e^{-0.2}\approx 0.819$. Matching the arms improves the phase difference readout without changing its ${\lambda }_0/2$ travel per cycle. A fine mirror scan of one cycle changes $D$ by only $600\mathrm{nm}$ for the stipulated $600\mathrm{nm}$ source, much less than $L_c$. These are calculated values, not historical data.

A proposed measurement sequence follows from this model. Overlap the returning beams, match their linear polarizations, find a high-contrast delay and use a small calibrated mirror scan to determine the sign and fringe period. Read a fixed spatial sample of the interference fringes with a photodiode; record each arm separately and the dark signal, avoid detector saturation, and retain enough time resolution to count every cycle. Approach mechanical settings consistently and repeat the return scan to expose lost motion or drift. For very small changes, bias near phase quadrature, where $|dI/d\varphi |=\bar{I}\mathcal{V}$ is largest; an intensity extremum has zero first-order sensitivity. A known drive voltage alone does not provide measurement calibration for mirror travel: compare the actual travel with the fringe count using an independently known wavelength.

Control temperature, pressure, vibration and air currents because the instrument measures optical path difference, not mirror motion alone. For a fixed reference arm and fixed ${\lambda }_0$, the first-order change of the moving arm's round-trip path is $\Delta D\approx 2n\Delta L+2L\Delta n$. Thus an unrecognized $\Delta n$ mimics $\Delta L_{\mathrm{app}}=L\Delta n/n$. A hypothetical $L=0.10\mathrm{m}$, $n\approx 1$, and $\Delta n=10^{-7}$ give $10\mathrm{nm}$ apparent travel. Holding the mirror still while varying or monitoring the air conditions tests this confounder. Separate checks of alignment, wavelength stability and thermal expansion are necessary before sub-fringe measurement resolution becomes a defensible measurement uncertainty.

A one-way arm-length difference $\delta$ gives optical path difference $2\delta$ and phase difference $2k\delta$. An ideal constructive output port has intensity $I=I_{\mathrm{in}}[1+\cos (2k\delta )]/2$. The other output is complementary.

Consider an original, explicitly normalized model of primary coma, rather than a prescription for a particular optical lens. Let $(u,v)$ be Cartesian coordinates in a circular optical pupil divided by its radius $a$, with $u^2+v^2\le 1$. Let $h=\alpha /{\alpha }_0$ be a signed small field angle divided by an arbitrary positive reference angle. With polar coordinates $u=\rho \cos \theta$, $v=\rho \sin \theta$, prescribe the wavefront errorHere $W_0$ is an arbitrary positive optical path difference scale, not a measured error. The cubic optical pupil dependence and linear field dependence are the classical Seidel form.^[ref]

In the paraxial approximation, light ray direction errors follow the transverse gradient of wavefront error. For a reference sphere a distance $L$ from the exit optical pupil in a uniform image medium of refractive index $n$, the corresponding image-error scale is $s=L{W}_0/(na)$. Choose image-error coordinate signs so the normalized displacement is $\nabla w$; reversing the convention reverses the plotted orientation. Image coordinates below are errors relative to the ideal image of this particular object point, rather than positions of different objects:The partial derivatives explain the asymmetry: opposite optical pupil points $(u,v)$ and $(-u,-v)$ land together, rather than on opposite sides of the ideal image. For $h>0$, every $X$ is nonnegative, while $Y$ can have either sign. The figures stipulate this single wavefront error, omit other optical aberrations and diffraction, and assign no instrument dimensions.

For a ring of fixed optical pupil radius $\rho$, the double-angle formulas for trigonometric functions giveThus each ring maps into a circle of radius $|h|{\rho }^2$ whose centre is displaced by $2h{\rho }^2$. Doubling $\rho$ quadruples both sizes; for $h=1$, the edge ring has centre $(2,0)$ and radius $1$, while the half-radius ring has centre $(1/2,0)$ and radius $1/4$. Superposing these displaced circles produces the comet-like envelope.

For uniform irradiance over the circular optical pupil, equal area samples give an equal-weight geometrical image. Its centroid is $(h,0)$ because $⟨u^2⟩=⟨v^2⟩=1/4$ and $⟨uv⟩=0$. Moving the coordinate origin to that centroid translates the image; it does not remove its asymmetric shape. Adding a linear term to $w$ likewise translates all light rays equally. These origin choices must be distinguished from changing the optical aberration itself.

The model's full transverse extent is $3|h|$ along $X$ and $2|h|$ along $Y$. Halving the field angle halves the blur, and reversing it reverses the pattern about the ideal image. This isolates why coma matters for an optical camera's usable field of view: collecting more light does not ensure that off-axis points remain compact. Near diffraction-limited resolution, a physical optics calculation of the point spread function is needed instead of interpreting the geometrical tip as an infinitely sharp image.^[ref]

For two equal-intensity mutually incoherent narrow spectral lines with wavenumbers $k_1,k_2$, a Michelson interferometer with optical path difference $x$ has normalized intensity $I/I_0=1+\cos (\bar{k}x)\cos (\Delta kx/2)$, where $\bar{k}=(k_1+k_2)/2$ and $\Delta k=|k_1-k_2|$. With mirror displacement $d=x/2$, the fast period is $\pi /\bar{k}\simeq \bar{\lambda }/2$; successive interference visibility revivals are separated by $\pi /\Delta k\simeq {\bar{\lambda }}^2/(2\Delta \lambda )$. The first visibility zero is halfway to the first revival. A signed envelope has twice the period of its absolute value. Unequal line strengths prevent perfect zeros.

Thin-film interference combines coherent light contributions reflected or transmitted at the two surfaces of a layer. For a film of refractive index $n$, thickness $d$ and internal angle ${\theta }_t$, the round-trip optical path difference is $2nd\cos {\theta }_t$. The relative reflection phase shift must also be included. For a nonabsorbing higher-index film surrounded by air, one of the two leading reflected contributions has a $\pi$ phase shift and the other does not. Reflection maxima therefore satisfy $2nd\cos {\theta }_t=(m+1/2){\lambda }_0$. A smooth reflection spectrum filters many visible wavelengths at once; varying thickness and angle produce spatially varying mixed colours.