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Let the source remain at $(0,0,-p)$ and let the observation point be $(X,Y,q)$. The paraxial approximation to the source-to-aperture-to-observer optical path length isCompleting the squares givesThe point $(x_0,y_0)$ is where the straight source-observer line intersects the aperture. Define the four dimensionless edge positionsOn the Cornu spiral $F(u)=C(u)+iS(u)$, draw the directed chord from $F(u_{x,-})$ to $F(u_{x,+})$, and a second directed chord from $F(u_{y,-})$ to $F(u_{y,+})$. Denote their complex differences by $D_x$ and $D_y$. The complex amplitude is proportional to their product: multiply their lengths and add their arguments, including the common propagation phase factor.

More explicitly, normalize by the unobscured field ${\psi }_0(X,Y)$ at this same observation point. The constant wave phase from completing the square is then already included in ${\psi }_0$, andEach directed chord is a one-dimensional Fresnel diffraction factor. One must multiply the two factors, rather than add the chords or use just one chord length for the square aperture. Setting $X=Y=0$ recovers $D_x=D_y=2F(w)$ and the on-axis answer.

Reflection and central spot. At a boundary from lower to higher refractive index, the near-normal electric-field reflection coefficient is negative, $r=(n_1-n_2)/(n_1+n_2)$. This gives a $\pi$ wave phase reversal on reflection from the glass plate through the air film. Reflection at the lens's glass-to-air underside has no such reversal. These are consequences of the Fresnel equations for the reflected electric field.

At ideal contact the film thickness is zero, so the two reflected contributions have zero geometrical optical path difference but a relative $\pi$ phase reversal. They produce destructive interference, and the central spot is dark. This conclusion concerns the film reflection; other lens surfaces can add a weak background.

Rings and their radii. At radial distance $r$, let the air thickness be $t$. The two relevant rays reflect at the upper and lower film interfaces. The lower reflection accumulates an extra round-trip optical path length $2t$, plus the relative $\pi$ reflection shift. As $t$ increases away from contact, alternating constructive and destructive thin-film interference forms Newton's rings. At near-normal incidence, a bright ring satisfiesThe extended source can illuminate these localized film fringes: each incident contribution interferes with its own two reflected paths; unrelated source points need not be mutually coherent.

For the spherical lens surface, geometry gives $r^2+(R-t{)}^2=R^2$, so $r^2=2Rt-t^2$. Since $t\ll R$, $t\simeq r^2/(2R)$. Combining the equations yieldsThe $m=0$ bright ring is the first bright ring around the dark contact point.

Curvature from ring diameters. Writing $D_m=2r_m$, subtract the squared diameters to eliminate the unknown initial order:Using SI units,

Liquid-filled film. With film refractive index $n=1.4$, still smaller than both glass indices, there remains exactly one relative reflection phase reversal. The round-trip optical path length is now $2nt$, soHere $\lambda =500\mathrm{nm}$ is the illuminating vacuum wavelength, approximately its air value, rather than its shorter value inside the liquid. The ring labelled $m=2$ therefore has diameterFilling the gap contracts each ring radius by $1/\sqrt{n}$ at fixed order and curvature.

A specimen whose leading optical effect is a spatially varying phase, rather than absorption. A thin specimen with refractive index $n_s$ in a medium of refractive index $n_m$ has, in a straight-path approximation,Here $\lambda$ is the vacuum wavelength; the integral is an optical path length difference. For the chosen values $\lambda =550\mathrm{nm}$, thickness $1\mu \mathrm{m}$ and $n_s-n_m=0.005$, the phase is $\pi /55\simeq 0.0571$ radians. This calculation does not identify thickness separately from refractive index contrast.^[ref]

A photon gas is a collection of photons described statistically as radiation. Here the equilibrium model is homogeneous, isotropic electromagnetic radiation in vacuum at temperature $T>0$: each photon has energy $E=c|\mathbf{p}|=h\nu$, with momentum $\mathbf{p}$ and frequency $\nu$, and there are two independent transverse states of electromagnetic polarization. An absorbing cavity with walls at a common temperature provides a way to establish this thermal equilibrium. The bulk formulas assume cavity dimensions much larger than the typical wavelength $hc/(k_{B}T)$, so the electromagnetic mode density in a box can replace individual cavity modes by a continuum. ^[ref]

The crucial constraint is whether photon emission and photon absorption, or other number-changing processes, can equilibrate the photon population. When they can, photon number is not an independently conserved quantity and the equilibrium photon chemical potential is ${\mu }_{\gamma }=0$. For each polarization and frequency, the mean occupation number then follows the Bose-Einstein distribution:If number-changing processes are ineffective but number-conserving interactions redistribute energy, a constrained Bose-Einstein distribution can instead have $f=[\exp ((h\nu -{\mu }_{\gamma })/(k_{B}T))-1{]}^{-1}$ with chemical potential ${\mu }_{\gamma }\le 0$ fixed by the number constraint. Thus thermal equilibrium with respect to energy exchange alone does not establish ${\mu }_{\gamma }=0$. The zero value can also persist in an already prepared Planck photon distribution during suitable subsequent evolution; continual photon emission is not required to preserve it. ^[ref]

Combining $f$ with the two-polarization density of states, $g_{\nu }/V=8\pi {\nu }^2/c^3$, gives the Planck photon distribution and the spectral photon energy density:Here $B_{\nu }$ is the spectral radiance in Planck's law; $u_{\nu }$ is stored energy density per unit frequency, not outgoing power per area. Their different units are connected by the full-sphere solid angle and the speed of light $c$. The Boltzmann constant is $k_B$, and $h$ is the Planck constant. ^[ref]

The equilibrium equation of state is particularly simple:The radiation constant $a_R$ is the same constant called $a_{\mathrm{rad}}$ in black-body radiation; $\sigma$ is the Stefan-Boltzmann constant. An outline of the energy density calculation substitutes $x=h\nu /(k_{B}T)$ and uses ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation pressure follows because the angular average of the squared directional component of photon momentum is $1/3$ for isotropy; the thermodynamic derivation of blackbody radiation pressure supplies an independent route. These formulas describe the radiation contribution, not the cavity walls. The isotropic massless relation $p=u/3$ survives a different isotropic spectrum, whereas the coefficient in $u=a_R{T}^4$ requires the zero-chemical potential Planck photon distribution.

For a computed example with stipulated $T=2.725\mathrm{K}$, the photon energy density is $u\simeq 4.17174\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$ and the radiation pressure is $p\simeq 1.39058\times 10^{-14}\mathrm{Pa}$. The photon number density in the International System of Units isHere $\zeta$ is the Riemann zeta function. The mean photon energy differs from the energy at the peak of a spectral curve: counting photons and weighting them by energy emphasize different frequencies. The calculations use the exact International System of Units values of the Planck constant, Boltzmann constant and speed of light. ^[ref]

A motivating observation is the cosmic microwave background spectrum measured by the Far Infrared Absolute Spectrophotometer (FIRAS) aboard COBE. Its apparatus used a polarizing Michelson interferometer, with one input viewing a roughly seven-degree sky patch and the other an internal blackbody. Scanning mirrors changed the optical path length difference; heat-sensitive detectors recorded the varying power, and Fourier transform spectroscopy recovered the spectral difference. Operation outside the terrestrial atmosphere removed atmospheric emission and absorption, while shielding and cooling the optics in a helium cryostat suppressed the instrument's own radiation. ^[ref]

Absolute calibration came from moving an external blackbody into the sky aperture, filling the same beam used for the sky. This was an iron-loaded epoxy absorber shaped as a re-entrant cone. Repeated reflections made it a close approximation to a blackbody, while thermometers supplied its temperature. Mather and colleagues explicitly evaluated reflections, temperature gradients, radiation leaking through the material and leakage around its edge. These are limits on how faithfully the calibrator realizes thermal equilibrium radiation, rather than details that can be eliminated merely by calling it black. ^[ref]

During sky observations the internal blackbody temperature was adjusted to nearly null the spectral difference, reducing gain and drift errors. Fixsen and colleagues' 1996 analysis used calibrated spectra over wavenumbers $2$ to $21\mathrm{c}{\mathrm{m}}^{-1}$, about $60$ to $630\mathrm{GHz}$. They separated the sky's spatially uniform component, its dipole and Galactic emission, and varied sky cuts and foreground templates as controls. The resulting cosmic microwave background spectrum agreed with a single Planck function, with reported RMS deviations below 50 parts per million of its peak brightness. This is a residual statistic over the analyzed spectrum, not a pointwise relative error bound at every frequency. ^[ref]

The observation motivates using a photon gas as a physical model rather than merely a formal counting exercise. Its interpretation has a clear boundary: agreement with Planck's law constrains departures from an equilibrium spectral shape within the calibrated band and foreground assumptions; it does not directly measure radiation pressure or entropy density of a photon gas. Nor does it establish that today's cosmic microwave background is continually re-equilibrating with matter. A previously thermal spectrum can retain its Planck function shape as expansion of the universe lowers every photon frequency in the same proportion. The usefulness of the model is that one measured spectral temperature, together with its equilibrium assumptions, predicts energy density, radiation pressure, photon number density and entropy density of a photon gas consistently.

The refractive index compares the phase velocity of light in a medium with the vacuum speed of light $c$. For a transparent, uniform medium with isotropy and a positive-index propagation branch,Here $\omega >0$ is the angular frequency, $k>0$ is the wavenumber, and $v_{\mathrm{ph}}=\omega /k$ is the speed of a surface of constant wave phase. The refractive index is a dimensionless quantity; it is a property of a specified material response at a specified frequency, rather than a universal number attached to the word “glass”. More general absorbing or negative-index media require the conventions discussed in refractive index and electromagnetic response. ^[ref]

For an original model example, choose $n=3/2$ at vacuum wavelength ${\lambda }_0=600\mathrm{nm}$. The phase velocity is $v_{\mathrm{ph}}=2c/3\simeq 1.999\times 10^{8}m/s$, the frequency is $f=c/{\lambda }_0\simeq 4.997\times 10^{14}\mathrm{Hz}$, and the medium wavelength is $400\mathrm{nm}$. These are stipulated model values, not measured properties of glass or water. A value at this single frequency does not determine a pulse's group velocity.

The usefulness of refractive index is that one measured response connects several different observations: Snell's law predicts refraction, optical path length predicts accumulated wave phase, and changes with frequency describe optical dispersion. Even at normal incidence, where a light ray does not bend, a changed refractive index changes wavelength and wave phase. The following comparison isolates that effect; the oblique geometry is developed in wavefront.

For a small sinusoidal optical path length error $h_0\cos (2\pi x/\Lambda )$, expand its phase factor to first order. The two exponential harmonics each have field amplitude $i\pi h_0/\lambda$ relative to the unperturbed field, so each speckle has contrast $C=(\pi h_0/\lambda {)}^2$. Its angular displacement is $\lambda /\Lambda$. Here $h_0$ is optical path error, not physical mirror displacement; a reflecting mirror introduces twice its surface displacement in the near-normal limit.

Model a thin transparent plate by a lossless reciprocal phase screen with transmission $t(x)=e^{i\psi (x)}$ at the chosen angular frequency. Both passages multiply the local field by the same $t(x)$: reversing travel does not by itself change the sign of an accumulated optical path length. Assume the plate is stationary, the two passages use corresponding transverse positions, and the plate is adjacent to the return surface or perfectly reimaged onto it. Ignore propagation between these reference surfaces, wave absorption, clipping and intensity-dependent changes. Take the incident boundary envelope to be $U_0(x)=A(x)e^{i{\varphi }_0(x)}$. Outward transmission gives $t{U}_0$. Ideal phase conjugation and the second transmission then giveThis is a complete short proof of cancellation in the specified phase screen model. For an ideal ordinary mirror with constant reflection factor $r$, the corresponding envelope is $rt(x{)}^2{U}_0(x)$: the extra wave phase is $2\psi (x)$, rather than zero. A constant reflection wave phase changes neither conclusion.

For an original toy optical aberration, let $\xi =x/a$ on $-1\le \xi \le 1$, take ${\varphi }_0=0$, and chooseThe even term adds symmetric curvature and the odd term makes the two sides unequal. The outward wave phase is $\psi$; immediately after ideal phase conjugation it is $-\psi$; after the second passage it is zero. At $\xi =-1,0,1$, the outward values are $\pi /4,0,3\pi /4$, while an ordinary mirror's completed return has values $\pi /2,0,3\pi /2$. All wave phases here are unwrapped relative to the central ray and are otherwise equivalent modulo $2\pi$. At a stipulated vacuum wavelength $\lambda =532\mathrm{nm}$, these outward errors correspond to optical path length errors $h=\lambda \psi /(2\pi )$ of $66.5,0,199.5\mathrm{nm}$. These are chosen values; they are not observations or physical plate thicknesses.

The wave phase plots make a subtle distinction visible: the field is still curved immediately after phase conjugation, and becomes flat only after crossing the same plate again. More generally, conjugation restores $U_0^{\ast }$, not a universally flat wavefront. If the plate instead has transmission $t=\alpha (x)e^{i\psi (x)}$ with real $0\le \alpha \le 1$, the return is $\gamma \alpha (x{)}^2{U}_0^{\ast }$. Its plate-induced wave phase cancels but its lost wave amplitude does not recover. If the return plate has changed to wave phase ${\psi }_{\mathrm{r}}$, the residual factor is $e^{i({\psi }_{\mathrm{r}}-\psi )}$. Those two variations distinguish correction of a stationary optical aberration from universal reversal of loss or material evolution.

A wavefront connects points with the same wave phase at one specified time. It is a surface, rather than the trajectory of a particular piece of light. For a scalar monochromatic wave written as $u(\mathbf{r},t)=A(\mathbf{r})\cos \Phi (\mathbf{r},t)$, a wavefront is a connected level set of the phase function $\Phi$; different surfaces can share the same wave phase modulo $2\pi$. A plane wave has planar wavefronts. An outward wave from an ideal point source in a uniform medium has spheres as wavefronts, while an optical lens can turn nearly planar wavefronts into converging ones. The wavefront describes wave phase, not wave amplitude: drawing it does not say how bright the beam is. ^[ref]

The relation to light rays becomes precise in a stationary medium with isotropy, negligible loss and a real positive scalar refractive index $n(\mathbf{r},\omega )$. In the geometrical optics limit, spatial changes in the refractive index and wave amplitude are slow on the local wavelength scale. For a single smooth propagation branch, writeHere $\omega >0$ is the angular frequency, $c$ is the vacuum speed of light, $S$ has units of length, and the last relation is the optical eikonal equation. The local wavevector is $\mathbf{k}=\nabla \Phi =k_0\nabla S$. Its direction is normal to a smooth wavefront, since a gradient is normal to its level sets. A forward light ray has unit tangent vector $d\mathbf{r}/ds=\nabla S/n$, so it follows this normal under the stated assumptions. Along that light ray, $dS/ds=\nabla S\cdot (d\mathbf{r}/ds)=n$: the change in $S$ is the optical path length. This is the short geometric argument linking wavefronts, light rays and optical path length. ^[ref]

The diagram shows incident and transmitted branches separately, with the reflected branch omitted. Its stipulated refractive indices are $n_1=1$ and $n_2=3/2$, and its incident angle is $30^{\circ }$ to the interface normal. At a stationary interface, the frequency and tangential wavevector agree on the two sides: Snell's law gives $\sin {\theta }_2=1/3$, hence ${\theta }_2\approx 19.4712^{\circ }$. The wavefronts turn while their normal spacing decreases from ${\lambda }_0$ to $2{\lambda }_0/3$, where ${\lambda }_0$ is the vacuum wavelength. More explicitly, with $x$ normal to the interface and $y$ tangential, take $S_1=\sqrt{3}x/2+y/2$ for $x<0$ and $S_2=\sqrt{2}x+y/2$ for $x>0$. These functions agree at $x=0$; the drawn wavefronts are $S_j=m{\lambda }_0$ at $t=0$ with ${\varphi }_0=0$. This is an exact piecewise plane wave phase construction for the stipulated transparent media, not measured material data. ^[ref]

The normal-to-wavefront rule requires its assumptions. In a medium with anisotropy, the wavevector remains normal to a smooth wavefront, but the group velocity and energy flux need not follow it. ^[ref] Near a wave caustic, where multiple light rays meet, one smooth phase function may no longer describe the field; diffraction and the full wave amplitude matter. A wavefront is also not generally a surface of equal pulse arrival time. In a medium with optical dispersion, a monochromatic wave phase moves with the phase velocity, while a sufficiently narrow wave packet follows the group velocity. Even identical optical path lengths at one frequency can give different group delays. The usual equal-travel-time picture needs a common launch reference and a nondispersive ray model; an arbitrary broadband disturbance has no single monochromatic wavefront.^[ref]

A wavefront error is the difference in optical path length from a reference wavefront, often an ideal plane or sphere. Its root mean square should specify the illuminated pupil and which modes, such as constant phase or image displacement, have been removed.