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Choose an original transparent-band model, defined only for $400\le \lambda \le 700\mathrm{nm}$:This chosen refractive index is not a fit to water, glass or the historical liquids. It decreases from $1.3425$ to approximately $1.300408$, so the threshold $1/n$ increases across the band. For $\beta =0.75$, solving $n({\lambda }_{\ast })=1/\beta$ givesOnly $400\le \lambda <{\lambda }_{\ast }$ emits. Faster particles with $\beta =0.78$ and $0.90$ emit throughout this chosen band. The Cherenkov cone angle and the Frank-Tamm formula are both evaluated with the same $n(\lambda )$.

For a $1\mathrm{cm}$ track the chosen model predicts about $0.589$, $27.40$ and $142.85$ emitted photons for $\beta =0.75$, $0.78$ and $0.90$, respectively. An expected count below one is meaningful over repeated equivalent tracks; it is not a fractional photon observed on one track. The near-threshold example illustrates why extending a constant-index approximation across an entire optical band can give the wrong emitting interval.

Finite bandwidth is essential. The $1/{\lambda }^2$ factor cannot be extrapolated down to arbitrarily small wavelength: the chosen material model stops at its stated band. Real optical dispersion changes the threshold and direction, and appreciable absorption requires a complex material response beyond the simple transparent-medium expression. A detected count must also include transport, collection geometry and the detector's wavelength-dependent response. Even a correct produced spectrum therefore does not, on its own, predict perceived colour or observed glow intensity.

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.

Take $N$ equally spaced scattering sites at $x_j=jd$ on a line, a normally incident plane wave, equal site scattering amplitudes, and observation far enough away that outgoing directions are effectively parallel. Retain only one scattering event per amplitude, as in the Born approximation. For angle $\varphi$ from the outward normal, write the adjacent-site quantum phase difference as $\delta =kd\sin \varphi$, with $k=2\pi /\lambda$; reversing its overall sign leaves the intensity unchanged. Summing a geometric series givesAt $\delta =2\pi m$ use the continuous limit $I/I(0)=1$, rather than treating $0/0$ as a missing peak. The integer $m$ is the diffraction order: $d\sin \varphi =m\lambda$. This is a surface-row spacing, not the plane spacing in Bragg's law. The plotted intensity is normalized to its central value; it is not an integrated detection probability or an absolute collector electric current.

At $100\mathrm{V}$, $\lambda \simeq 0.12264\mathrm{nm}$ and the positive first-order angle is $\arcsin (\lambda /d)\simeq 29.4^{\circ }$. At $400\mathrm{V}$, the wavelength halves and this angle becomes about $14.2^{\circ }$. Increasing $N$ narrows each peak through more selective interference, while increasing momentum brings a fixed-order peak nearer the normal. Unequal site amplitudes, multiple scattering, a three-dimensional crystal lattice structure factor, and finite detector acceptance can change this simple picture; it is an explanatory model with hypothetical inputs.

For the ideal uniform wave equation on a string, the wave speed is $c=\sqrt{\tau /\mu }$, where $\tau$ is constant tension and $\mu$ is linear mass density. Assume small transverse slopes, negligible bending stiffness and no damping. Fix both ends at zero displacement, giving the homogeneous Dirichlet boundary conditions $y(0,t)=y(L,t)=0$. Our standing wave already satisfies the first. For a nonzero amplitude, the second requires $\sin (kL)=0$, henceThe value $n=0$ gives zero displacement in this sine family, not a vibrating normal mode. Each positive $n$ fits $n$ half-wavelengths into the string: it has $n$ antinodes, $n-1$ interior nodes of a standing wave, and two endpoint nodes of a standing wave. For $L=1\mathrm{m}$ and $c=20\mathrm{m}{\mathrm{s}}^{-1}$, $f_n=10n\mathrm{Hz}$; the third normal mode has $f_3=30\mathrm{Hz}$ and wavelength $2/3\mathrm{m}$.

The equally spaced eigenfrequencies follow from this string's uniform wave speed and fixed-end boundary conditions. They are not a universal spectrum for every vibrating object. Changing an endpoint constraint changes the allowed mode shapes; changing tension changes wave speed. At fixed $L$ and $\mu$, quadrupling $\tau$ doubles every $f_n$. The more general construction of arbitrary initial motion belongs to normal mode and wave equation on a string.

Stipulate a model receiver held at $T=5.4\mathrm{K}$ with a reference at $T_b=5.0\mathrm{K}$. Choose $G=5\mathrm{mW}{\mathrm{K}}^{-1}$, so outward thermal conduction is $G(T-T_b)=2\mathrm{mW}$. Ignore other heat transfer, take $P_{\mathrm{par}}=0$ in both states, and stipulate perfect optical/electrical thermal equivalence. These values specify an original example; they are not measurements or claimed performance of a NIST instrument.

With the beam blocked, supply $P_{h,d}=2\mathrm{mW}$. With light admitted, the required heater power is $P_{h,l}=1.005\mathrm{mW}$. For stipulated radiometric absorptance $a=0.995$ at the chosen single wavelength,The illuminated receiver still gets $1.005+0.995=2\mathrm{mW}$ of heating. The remaining $0.005\mathrm{mW}$ of incident radiant flux escapes without heating it. Calling the measured heater reduction “incident power” would underestimate the stipulated incident radiant flux by $0.5\%$.

Two variations expose different pitfalls. First, if admitting the beam also adds $0.010\mathrm{mW}$ of stray heating while the same $0.995\mathrm{mW}$ is absorbed from the intended beam, the heater falls to $0.995\mathrm{mW}$. The apparent reduction is now $1.005\mathrm{mW}$; subtracting $\Delta P_{\mathrm{par}}=0.010\mathrm{mW}$ recovers the beam's absorbed power. Second, keep the original heater readings but stipulate a light-state receiver offset $\delta T=+1\mathrm{mK}$ at fixed reference temperature. The link carries an extra $G\delta T=0.005\mathrm{mW}$; the inferred absorbed power becomes $1.000\mathrm{mW}$, and incident radiant flux becomes $1.000/0.995\simeq 1.005025\mathrm{mW}$. This second variation deliberately violates exact fixed-temperature substitution.

The heater can supply only nonnegative power in this model. Consequently an absorbed beam above the available dark heater power cannot be compensated at the chosen setpoint merely by reducing the heater. The operating point, link, or admitted radiant flux must change; zero heater demand is a range boundary, not evidence of greater radiant flux sensitivity.

Use the schematic below to count the water traversals explicitly. Each stationary tube has laboratory length $\ell$. In the displayed setting, upper-tube water flows right and lower-tube water flows left, each at speed $u>0$. Beam A goes right through the upper tube, exchanges tubes at the far optics, and returns left through the lower tube: it follows the flow twice. Beam B goes right through the lower tube and returns left through the upper tube: it opposes the flow twice. Each complete beam path therefore contains total water length $L=2\ell$, not $\ell$ or $4\ell$. The connecting air paths and fixed optics are assumed to contribute no flow-dependent phase difference.

Let $v_{+}$ and $v_{-}$ be positive phase velocity magnitudes along and against the water flow:For a laboratory source of frequency $\nu =c/{\lambda }_0$, where ${\lambda }_0$ is its vacuum wavelength, define the signed interference fringe offset $N$ by the propagation phase difference $\Delta \varphi =2\pi \nu \Delta t=2\pi N$. This is an equivalent phase delay between steady beams; no stopwatch resolves their transit-time difference. Subtract the zero-flow baseline and expand to first order:Thus one shifted fringe corresponds to one cycle of phase difference. If the stationary pattern has spatial interference fringe spacing $d$, a simple readout model is $I(x)=I_0[1+\mathcal{V}\cos (2\pi x/d+2\pi N)]$, where $\mathcal{V}$ is the interference visibility. The maxima shift by $-dN$ with this choice of screen coordinate. A sign assigned to “rightward” motion depends on the image orientation; the magnitude in fringe units does not.

Reversing both flows gives $N(-u)=-N(u)$. The displacement between equal and opposite settings therefore has magnitudeThe factor $2$ from comparing opposite flows is additional to the factor $2$ already included in $L=2\ell$. Repeating a round trip in just one flowing tube would instead put a beam with the flow once and against it once; that geometry does not produce the two same-sign contributions drawn here.

Choose $\ell =1.50\mathrm{m}$, $n=4/3$, ${\lambda }_0=550\mathrm{nm}$ and $u=6.00\mathrm{m}{\mathrm{s}}^{-1}$, with $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. These are illustrative choices, not measured values or a reconstruction of Hippolyte Fizeau's data. Each beam traverses $L=3.00\mathrm{m}$ of water. The nondispersive prediction isThe first number is the offset from rest to one flow setting; the second compares equal opposite settings. The very short equivalent delay explains why interference is useful: a measurable fraction of a wavelength replaces an unattainable direct timing measurement.

The Galilean transformation would give phase velocity magnitudes $c/n\pm u$, hence $N_{\mathrm{full}}\simeq 4\ell u{n}^2/(c{\lambda }_0)$. At the chosen values this is $0.38815$ fringe from rest, larger than the special relativity prediction by $n^2/(n^2-1)=16/7$. A model with no flow-induced change of phase velocity instead gives $N=0$. The three alternatives are experimentally distinguishable even when $u/c$ is tiny.

Hippolyte Fizeau's positive displacement favoured Augustin Jean Fresnel's partial-drag law over zero or full drag. He explicitly distinguished support for that law from establishing its underlying luminiferous aether hypothesis.^[ref] The Fizeau experiment therefore supplies a nonzero relative-motion measurement, not a null test proving an luminiferous aether's absence. The derivation above exhibits the same first-order law within special relativity, showing why agreement with the coefficient alone cannot select an physical interpretation.

Two limitations matter for precision. The velocity profile sampled by the beam need not equal the cross-sectional average inferred from volumetric flow rate. Also, the Doppler effect gives opposite rest-frame frequencies for the counterpropagating beams. With optical dispersion, the coefficient $n^2-1$ in the interference fringe formula becomeswhere the refractive index and its derivative are evaluated for stationary water at the laboratory angular frequency $\omega =2\pi c/{\lambda }_0$. A wavelength-independent $n$ therefore defines an approximation, not a precision fit to real water.^[ref] Flow reversal removes a fixed phase difference, but a mechanical or temperature change that reverses with the flow can still imitate the signal and must be controlled.

Fluorescence is photon emission following photon absorption by matter: an excited state releases some of its energy as light. A fluorophore can therefore mark where something is, while its fluorescence emission spectrum and fluorescence lifetime report different aspects of its environment. Fluorescein is one example; FRET exploits a competing route for excitation energy rather than requiring a second emitted photon.

The usual molecular sequence is excitation, relaxation, then emission, as in the Jablonski diagram. The emitted photon commonly has lower energy and longer wavelength, giving a Stokes shift; the exceptions are discussed below.^[ref]

A fluorescence emission filter transmits the chosen emitted wavelength band while rejecting excitation leakage. Finite rejection, overlap of the fluorescence emission spectrum with excitation, and the photodetector's wavelength-dependent response can all affect the count. A darker filtered image alone does not establish a changed fluorescence lifetime or increased photobleaching.

A fluorescence emission spectrum resolves emitted photons by wavelength at fixed excitation settings. Subtract a matched sample-container and solvent background, and correct the wavelength-dependent throughput and photodetector sensitivity when estimating the emitted distribution. A camera image through one fluorescence emission filter instead measures a weighted band integral, not a whole fluorescence emission spectrum. A fluorescence lifetime resolves delay after excitation; its horizontal axis is time, not wavelength.

A fluorescence excitation filter selects the illumination wavelength band reaching the specimen. Its finite transmission changes the absorbed photon rate and thus image brightness. Specify that band when interpreting a fluorescence excitation spectrum or comparing images.

A fluorescence excitation spectrum scans excitation wavelength while detecting fluorescence in a fixed emission band. A fluorescence emission spectrum does the converse. The inspected spectrofluorometer example uses an excitation monochromator scan with the emission monochromator fixed, and reverses their roles for emission.^[ref]

An operational model explains why this need not reproduce absorption. For one dilute, unsaturated emitting species, write $a(\lambda )$ for absorbed fraction, $R_{\mathrm{in}}(\lambda )$ for incident photon rate, $\Phi (\lambda )$ for fluorescence quantum yield, and ${\eta }_{\mathrm{band}}(\lambda )$ for the fraction of emitted photons counted in the selected band. The background-subtracted count rate isOnly after correcting incident photon rate, and with constant yield and emission-band collection, is this proportional to absorbed fraction. A nonfluorescent absorber contributes to absorption but not this readout. Thick samples, multiple emitting species, and reabsorption of fluorescence complicate the model.