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For identical mirrors with power reflectivity $\mathcal{R}$, transmission $\mathcal{T}$ and absorption $1-\mathcal{R}-\mathcal{T}$, summing round trips at normal incidence givesHere $\delta$ includes propagation and both internal reflection phases. Optical absorption reduces the resonant peak below unity, while a nonzero reflection phase shifts the resonance condition $\delta =2\pi m$. Frequency-dependent mirror phase and refractive index can also change the free spectral range.

The acoustic impedance of a fluid is $Z=\rho c$. For a progressive plane acoustic wave, pressure amplitude $P$ and velocity amplitude $U$ in its propagation direction satisfy $P=ZU$. At an oblique planar interface, zero reflection requires a matching condition involving the normal impedances $Z/\cos \theta$ or, equivalently for displacement amplitudes, $Z\cos \theta$.

Amplitude transmittance is the local ratio of transmitted to incident complex amplitude in a thin optical-element model. It includes wave phase as well as attenuation. For identical incident and transmitted media at normal incidence, the corresponding intensity transmittance is $|t{|}^2$; $t$ itself is not the transmitted power fraction. A pure phase screen has $|t|=1$.

For a film of refractive index $n$ and thickness $d$ in air, its round-trip propagation phase is $\delta =(4\pi d/{\lambda }_0)\sqrt{n^2-{\sin }^2{\theta }_i}$. Two neighbouring same-type interference fringes differ by $2\pi$. At small angular spacing in radians,The relation assumes slowly varying interface reflection coefficients and no rapidly varying reflection phase. At normal incidence the first derivative vanishes, so a local linear angular-spacing formula must be replaced by a quadratic expansion.

A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

A formal grating order must have a real outgoing angle. For normal incidence, $|m|\lambda \le d$ is necessary. Oblique incidence can access orders outside that normal-incidence range; reflection Littrow geometry has $m\lambda =2d\sin \theta$. The ideal resolving power of a diffraction grating applies only to an order that exists and has sufficient signal.

For a reciprocal ideal quarter-wave plate followed by a mirror at normal incidence, use the same fixed transverse laboratory axes for the forward and returning Jones vectors. If the axes coincide with the plate's fast and slow axes, each pass has Jones matrix $J=\mathrm{diag}(1,i)$ up to a common phase factor. A polarization-independent mirror has scalar reflection coefficient $r_m$, so the returned Jones vector is $r_m{J}^2\mathbf{e}=r_m\mathrm{diag}(1,-1)\mathbf{e}$. The plate acts twice as a half-wave plate, rather than undoing its phase retardation.

For input linear polarization at angle $\alpha$ to the fast axis, the returned line is at $-\alpha$ modulo $\pi$. At $\alpha =45^{\circ }$ it is perpendicular to the original line. An axis-aligned input instead returns along the same line. General elliptical polarization requires its full input Jones vector; no unique returned state follows from the word “polarized” alone. The propagation direction reverses, so comparing descriptions of circular polarization must also specify the viewing convention.

For a lossless nonmagnetic uniaxial optical medium, the electric displacement field of a plane electromagnetic wave is perpendicular to its wavevector. The magnetic field is perpendicular to the plane containing the electric field and wavevector, while the Poynting vector is perpendicular to the electric field in that same plane. Thus the angle between $\mathbf{E}$ and $\mathbf{D}$ equals the walk-off angle between energy flow and phase propagation. At normal incidence on a parallel slab of thickness $L$, a lateral separation $\Delta$ of the ordinary and extraordinary exiting rays gives ${\alpha }_e=\arctan (|\Delta |/L)$, while ${\alpha }_o=0$. The output rays can be parallel even though their internal energy paths differ.

For an ideal air-gap Fabry–Pérot etalon at normal incidence, the free spectral range is ${\nu }_{\mathrm{FSR}}=c/(2d)$ and the peak full width at half maximum is $\delta \nu ={\nu }_{\mathrm{FSR}}/\mathcal{F}$. To separate a pair of lines spaced by $\Delta \nu$ by at least $M$ peak widths, require $d\mathcal{F}\ge Mc/(2\Delta \nu )$. Putting both lines inside one unambiguous free spectral range additionally requires $d<c/(2\Delta \nu )$, so $\mathcal{F}>M$. In terms of their vacuum wavelengths, $c/\Delta \nu ={\lambda }_1{\lambda }_2/|{\lambda }_2-{\lambda }_1|$.

The strict order condition needs a margin if tails from adjacent periodic peaks are to be negligible. If every neighbouring peak in a scanned two-line periodic spectrum must be separated by $M$ widths, the stronger condition is $\min (\Delta \nu ,{\nu }_{\mathrm{FSR}}-\Delta \nu )\ge M{\nu }_{\mathrm{FSR}}/\mathcal{F}$. Ordinary resolving power alone does not prevent ambiguous repeated orders. A fixed etalon's transmission as a function of wavelength is a periodic-in-frequency filter; it is not automatically a plot of the incident source's spectral lines.

At normal incidence, with no optical dispersion and zero round-trip reflection wave phase modulo $2\pi$, the resonance condition in Fabry–Pérot transmission from multiple reflections isHere $\nu$ is frequency and $c$ is the speed of light in vacuum. A change of $\nu$ by one free spectral range advances the round-trip wave phase by $2\pi$. At fixed internal angle of incidence $\theta$, the same nondispersive model gives $c/(2nd\cos \theta )$. Holding the external angle of incidence fixed in a dispersive medium need not hold the internal $\theta$ fixed, because of Snell's law.

For stipulated $n=1$ and $d=10\mathrm{mm}$, the free spectral range is $14.9896229\mathrm{GHz}$ using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. A frequency difference of $0.10\Delta {\nu }_{\mathrm{FSR}}$ is therefore $1.49896229\mathrm{GHz}$. Transmission repeats every free spectral range: one isolated scan cannot distinguish that separation from $1.10\Delta {\nu }_{\mathrm{FSR}}$. Independent knowledge of the optical spectrum, an additional filter, or a changed mirror separation is needed to assign orders.

Equal spacing in frequency does not imply equal spacing in wavelength. At normal incidence the exact adjacent resonance wavelengths obeyFor example, choosing $2nd=10{\lambda }_{10}$ makes the exact gap to order 11 equal to ${\lambda }_{10}/11$, while the high-order estimate gives ${\lambda }_{10}/10$. The approximation is already ten percent larger at order 10. With optical dispersion or frequency-dependent reflection wave phase, obtain local spacings from the total phase difference instead: $\Delta \nu \simeq 2\pi /(d\delta /d\nu )$ when its derivative varies little over one order.

Consider two identical, lossless parallel mirrors with intensity reflectivity $0\le R<1$, separation $d$, and a uniform transparent medium of refractive index $n$. Fix one polarization of an electromagnetic wave and an internal angle of incidence $\theta$ measured from the mirror normal vector. Illuminate with a monochromatic wave having planar wavefronts; the emerging contributions must overlap coherently in the detected direction. Define $T=I_{\mathrm{transmitted}}/I_{\mathrm{incident}}$ using the same incident and exit media. Each mirror transmits a fraction $1-R$ of incident power. Its reflected wave amplitude has magnitude $\sqrt{R}$, so two internal specular reflections multiply the circulating complex amplitude by magnitude $R$, not $R^2$.

Let $\lambda$ be the vacuum wavelength and choose the reflection wave phase convention in which the product of the two internal amplitude reflection coefficients is real and positive. The phase difference between successive transmitted contributions, evaluated on a common outgoing wavefront, isWith other mirror coatings or phase conventions, use the total $\delta =4\pi nd\cos \theta /\lambda +{\varphi }_{\mathrm{rt}}$, where ${\varphi }_{\mathrm{rt}}$ is the sum of the two internal reflection wave phases. The stated resonance positions presume ${\varphi }_{\mathrm{rt}}=0$ modulo $2\pi$; a constant nonzero ${\varphi }_{\mathrm{rt}}$ shifts them, while a frequency-dependent one can also change their spacing.

Why does the cosine appear rather than its reciprocal? Inside the gap, the extra two legs have optical path length $2nd/\cos \theta$. But the later emerging contribution starts $2d\tan \theta$ farther along the mirror. Comparing both on one outgoing wavefront subtracts $2nd\tan \theta \sin \theta$ from that path difference, leaving $2nd\cos \theta$. Equivalently, a plane wave's mirror-normal wavevector component is $k_z=(2\pi n/\lambda )\cos \theta$, so the round-trip wave phase is $2k_{z}d$. The diagram uses the same medium outside and inside to show the common-wavefront subtraction directly; at a refracting exit interface, tangential wavevector matching in Snell's law supplies the same result.

Absorb the common first-pass wave phase into the incident complex amplitude $E_{\mathrm{in}}$. Successive transmitted contributions are $(1-R)E_{\mathrm{in}}$, $(1-R)E_{\mathrm{in}}R{e}^{i\delta }$, and so on. The superposition principle and the convergent geometric series giveThe last step expands the squared complex modulus as $1+R^2-2R\cos \delta$. This transmission profile is often called the Airy transmission function; it is distinct from the Airy disk, a diffraction image of a circular aperture. The University of Washington's lab handout gives the same transmission profile and distinguishes mirror reflectivity from other sources of peak broadening.^[ref]

For an original example, set $R=3/4$. At $\delta =0$, the amplitudes add to $(1/4)/(1-3/4)=1$, so $T=1$. At $\delta =\pi$, they alternate in sign and add to $(1/4)/(1+3/4)=1/7$, so $T=1/49$. At $\delta =\pi /2$, the amplitude is $4/25+3i/25$ and $T=1/25$. Adding the individual intensities instead would give $(1-R{)}^2{\sum }_j{R}^{2j}=(1-R)/(1+R)=1/7$, independent of wave phase; that discards the wave interference responsible for the peaks.

In general the maximum is $T_{\max }=1$ at $\delta =2\pi m$, with integer $m$, and the minimum is $T_{\min }=[(1-R)/(1+R){]}^2$ halfway between peaks. Perfect resonant transmission requires identical lossless mirrors, coherent illumination, and sufficient time to establish the steady field. Absorption, unequal mirror transmission, finite beams that fail to overlap, and fluctuations of $d$ can reduce it. The exactly reflecting case $R=1$ is excluded: no incident power enters through a perfectly reflecting mirror, and the resonant limit $R\to 1^{-}$ is singular.

Faraday rotation exhibits optical nonreciprocity: with the bias magnetic field held fixed, a retracing beam gains another rotation in the same laboratory sense. Natural optical activity in a reciprocal medium instead unwinds on retracing the path.^[ref]

To reconcile this with the signed $B_{\parallel }$ formula, retain the laboratory $x,y,z$ axes for comparing physical lines. For the backward ray choose a right-handed ray basis $x^{\prime }=x$, $y^{\prime }=-y$, $z^{\prime }=-z$. Its magnetic flux density component is $B_{\parallel }^{\prime }=-B_z$, so its local rotation increment is $-\theta$. But positive $y^{\prime }$ is negative laboratory $y$: the backward increment is therefore $+\theta$ in the laboratory orientation plot. Reversing the ray reverses both the longitudinal component and the basis used for its signed angle. Reversing only one creates the false prediction of cancellation.

In fixed laboratory transverse components, define the rotation matrixFor an ideal mirror at normal incidence that contributes the same complex amplitude factor $r$ to both transverse components, the round-trip map is $rR(\theta )R(\theta )=rR(2\theta )$. The common mirror factor does not change the oscillation line. A reciprocal natural rotator instead gives $rR(-\alpha )R(\alpha )=rI$. This comparison assumes a retraced straight path and no additional polarization-changing elements.

For example, a chosen single-pass $30^{\circ }$ Faraday rotation produces a returned line at $60^{\circ }$ to the original laboratory $x$ axis. A chosen reciprocal $30^{\circ }$ natural rotator returns it to $0^{\circ }$. The same addition makes an ideal optical isolator possible: polarizers at $0^{\circ }$ and $45^{\circ }$ surround a $45^{\circ }$ Faraday rotator. Forward light follows $0^{\circ }\to 45^{\circ }$ and passes. Returning light selected at $45^{\circ }$ reaches $90^{\circ }$ at the entrance polarizer and is rejected by Malus's law. Real transmission loss and incomplete extinction limit the isolation.

To isolate the readout mechanism, replace the processed photographic emulsion by $N$ equally spaced, identical weak scattering planes, with separation $d$. Take the readout refractive index $n$ to be real and independent of wavelength, and let $\theta$ be the internal angle from the plane normal. Adjacent specularly backscattered contributions differ in wave phase by $2k_{z}d=4\pi nd\cos \theta /{\lambda }_0$. Their geometric series givesHere $a$ is a small plane reflection wave amplitude, assumed constant in this illustrative band; $F_N=1$ at $\delta =0$ by continuity. Since incident and reflected waves occupy the same external medium, $|r_N{|}^2$ is the model reflectance when transmission factors are absorbed into $a$. Require $N|a|\ll 1$ for the weak-scattering sum to remain self-consistent. Normalizing to $F_N$ discards absolute brightness; it does not predict a highly reflecting historical plate.

The first-order peak obeysThis is Bragg's law with an angle measured from the normal, rather than the glancing Bragg angle. For external viewing angle $\alpha$ in air approximated by refractive index one, Snell's law gives $n\sin \theta =\sin \alpha$, henceThe expression describes specular readout: illumination must supply the matching incident direction. Tilting a plate under fixed narrow illumination can make the signal disappear as well as change its selected colour.

For the chosen $n=1.5$ and $d=200\mathrm{nm}$, calculated peaks are $600\mathrm{nm}$ at $\alpha =0$, $565.69\mathrm{nm}$ at $30^{\circ }$, and $529.15\mathrm{nm}$ at $45^{\circ }$. Greater obliquity selects a shorter wavelength. Uniform shrinkage to $d^{\prime }=sd$, with unchanged $n$, multiplies every peak wavelength by $s$: a chosen $s=0.95$ moves the normal peak to $570\mathrm{nm}$. More generally, ${\lambda }_B^{\prime }/{\lambda }_B=(n^{\prime }/n)s$ at normal incidence. A change of thickness alone is insufficient if processing also changes the refractive index.

Finite $N$ gives a finite spectral response. The nearest zeros satisfy $\delta =\pm 2\pi /N$, giving ${\lambda }_{\pm }={\lambda }_B/(1\pm 1/N)$. At the normal $600\mathrm{nm}$ peak, $N=30$ gives neighbouring zeros at approximately $580.65$ and $620.69\mathrm{nm}$; $N=10$ gives $545.45$ and $666.67\mathrm{nm}$. Thus greater sampled depth narrows the central band in this model. The small-detuning distance from peak to a nearest zero is approximately ${\lambda }_B/N$; it is not the full width at half maximum.

Real processed photographic emulsions have a continuous, irregular depth response rather than identical planes. Absorption, wavelength-dependent refractive index, unequal scattering strengths, a nonideal exposure reflector, changes during photographic processing and multiple scattering of light can alter both peak positions and shapes. A transfer calculation with measured material parameters would be needed for quantitative reflectance. The present model explains the phase-selection mechanism and the tradeoff between depth and spectral width; it cannot establish faithful reproduction of an arbitrary incident optical spectrum. Its useful lesson is that a photograph can record colour through sub-wavelength structure, while its eventual appearance still depends on illumination, viewing geometry and processing.

For $N$ equally spaced, coherently illuminated narrow slits at normal incidence, the phasor sum is proportional to $\sin (N\delta /2)/\sin (\delta /2)$, where $\delta =2\pi d\sin \theta /\lambda$. The first zero around the central maximum has $d\sin \theta =\lambda /N$. Principal maxima are removable zero-over-zero limits, not intensity zeros.

The free spectral range is the wavelength or frequency spacing between adjacent orders or resonances of an optical instrument. It identifies when different orders can overlap. Always specify the coordinate: a uniform spacing in frequency need not be uniform in wavelength. For an ideal Fabry–Pérot interferometer at normal incidence, with mirror separation $d$ and constant refractive index $n$, Fabry–Pérot resonance spacing gives $\Delta {\nu }_{\mathrm{FSR}}=c/(2nd)$, where $c$ is the speed of light in vacuum. The free spectral range of an echelle grating follows a different order condition.

In a monochromatic two-path interferometer, an ideal phase plate of thickness $t$ and refractive index $n$ adds phase $2\pi (n-1)t/\lambda$ at normal incidence. The interference pattern is unchanged when $(n-1)t=m\lambda$ for integer $m$, provided transmitted amplitudes and alignment are unchanged. Calling the smallest index requires specifying the admissible range; $n>1$ excludes the optically null $n=1$ solution.

A spectrometer that resolves spectral lines using the angular dispersion of a diffraction grating. A narrow slit at a collimator focal plane supplies nearly parallel illumination; a telescope or focusing lens forms separate images for different outgoing angles. At normal incidence, the grating equation is $d\sin \theta =m\lambda$. The slit width, illuminated groove count, aberrations and detector sampling jointly determine whether a calculated angular separation is actually resolved.

At a stationary, uniform planar boundary, incident and transmitted plane waves have the same variation of wave phase along the interface and in time. The figure marks places where equal-wave phase wavefronts meet the boundary. Although their perpendicular spacing differs, their intersections must advance along the boundary together. This is the local mechanism behind Snell's law.^[ref]

For a short derivation, let $x$ run along the interface and $z$ increase into the transmitted medium. Write each wave phase as ${\Phi }_j=k_{j,x}x+k_{j,z}z-{\omega }_{j}t+{\varphi }_j$. The electromagnetic boundary conditions must hold at every $x,t$ on $z=0$; their matching requires ${\omega }_i={\omega }_t$ and $k_{i,x}=k_{t,x}$. Constant offsets ${\varphi }_j$ and wave amplitudes are determined separately. Since $k_{j,x}=k_j\sin {\theta }_j$, this givesFor nonzero tangential wavenumber, the spacing of equal-wave phase intersections along the interface is ${\lambda }_i/\sin {\theta }_i={\lambda }_t/\sin {\theta }_t$. It is not the perpendicular wavelength. At normal incidence, the tangential wavenumber vanishes and the interface is uniform in wave phase.^[ref]

At normal incidence, let $z$ measure distance from the reflector into a uniform transparent layer with refractive index $n$. For a monochromatic wave of vacuum wavelength ${\lambda }_0$, the layer wavenumber is $k=2\pi n/{\lambda }_0$. Represent the incident and returning electric fields by complex wave amplitudes with time factor $e^{-i\omega t}$ suppressed:The time-averaged squared electric field, normalized to the incident contribution, isThis electric exposure pattern has maxima separated byThe wave phase of the reflector shifts the maxima's positions but leaves their spacing unchanged in this uniform-layer model. The electric pattern, rather than the zero net energy flux of an ideal standing wave, is the relevant exposure proxy. An imperfect mirror reduces its contrast: $(I_{\max }-I_{\min })/(I_{\max }+I_{\min })=2|r|/(1+|r{|}^2)$, where $I$ denotes this exposure proxy.

Choose ${\lambda }_0=600\mathrm{nm}$ and $n=1.5$. Then the medium wavelength is $400\mathrm{nm}$ and $d=200\mathrm{nm}$. For an ideal phase-reversing reflector, $r=-1$ and the normalized pattern is $4{\sin }^2(kz)$: its first two maxima are at $z=100$ and $300\mathrm{nm}$. These are chosen calculations, not measured properties of a historical photographic emulsion or the optical response of Mercury.

For multiple mutually incoherent wavelengths, time averaging adds exposure contributions rather than their instantaneous electric fields. In a linearized recording model,Here $S$ is incident spectral exposure density, $q$ a wavelength-dependent recording sensitivity and $D$ a model exposure response. An optical spectrum therefore writes overlapping depth oscillations, not necessarily visibly separated sheets. This is a useful connection to the Fourier transform: approximately, depth is paired with optical spatial frequency $2n/{\lambda }_0$. Finite recording depth, varying sensitivity and nonlinear photographic processing prevent treating this expression as an exact inversion rule for an actual plate.

Use a slit light source, a rectangular transparent block on plain paper, pencil, ruler and protractor. Trace the block outline, keep it fixed, and mark widely separated points on incident and emergent light rays. Remove the block and join the entry and exit points to reconstruct its internal light ray. Draw the surface normal vector at entry; measure $i$ outside and $r$ inside from that normal line. Choose a path exiting the opposite parallel face rather than a side edge. Repeat for a range of $i$ with the same block and source colour; record $i,r,\sin i,\sin r$ and repeat-reading spread. Plot $\sin i$ vertically against $\sin r$ horizontally: Snell's law predicts slope $n_{\mathrm{block}}/n_{\mathrm{air}}$, approximately the block's refractive index. The emergent light ray should be parallel to the incident one. Beam width, pencil marks, protractor resolution, block movement and colour spread limit precision; a narrow-colour source reduces spread from optical dispersion. These are measured angular readings, not a direct timing of phase velocity.^[ref]

A normal incidence trial checks the drawn normal and the absence of bending, but the ratio $\sin i/\sin r$ is then $0/0$. Small nonzero angles also make that ratio sensitive to reading errors. Compare several trials through the slope rather than relying on one nearly axial pair. This rectangular-block measurement complements the existing total internal reflection experiment, which probes a different boundary condition.