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A developed grain is a visible deposit of silver produced during photographic processing. The chain of developed grains in a nuclear emulsion records interactions along a charged particle's path. Its width is a property of the recording and viewing process, not a measurement of the particle's diameter.

An emulsion range measurement follows a stopping charged particle through the coating to estimate its complete particle range. A projected length on one image can underestimate the distance travelled. In an original geometry example, a straight segment with projected length $L_{xy}=240\mu \mathrm{m}$ and corrected depth change $\Delta z=180\mu \mathrm{m}$ has lengthUsing the projection alone would underestimate this length by $20\%$. A bent track requires a sum over reconstructed segments. These are stipulated coordinates, not measurements of a historical track.

Suppose photographic processing contracts the coating only in depth, with shrinkage ratio $s=t_{\mathrm{developed}}/t_{\mathrm{exposed}}$. Then an observed depth difference $\Delta z_{\mathrm{developed}}$ corresponds to $\Delta z=\Delta z_{\mathrm{developed}}/s$. For $s=0.60$, an observed $108\mu \mathrm{m}$ becomes the $180\mu \mathrm{m}$ used above. This correction assumes uniform contraction and unchanged lateral dimensions; it cannot repair a track that leaves the coating before stopping.

A simple continuous-loss model connects particle range to kinetic energy $T_0$ through stopping power $S(T)=-\mathrm{d}T/\mathrm{d}x>0$:This describes the mean trend for a specified particle and material. Fluctuating energy loss produces a spread of stopping distances, called range straggling. An exit point supplies only a lower bound on particle range. Consequently, a narrow distribution from fully contained tracks and one assembled by treating exits as stops answer different questions.

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.

A latent image is the pattern of exposure-induced changes in a photographic emulsion before photographic processing makes it visible. A missing visible mark before development therefore says nothing about whether a charged particle passed through the coating.

In his 14 December 1908 Nobel lecture, Gabriel Lippmann described a photographic plate bearing a transparent, even, extremely fine sensitive layer. A holder put Mercury against that layer as a mirror during exposure. Incident and reflected light formed a standing wave within the photographic emulsion; after photographic processing and drying, white light revealed colour by selective reflection. The reflector served during recording; the developed structure supplied the later colour selection.^[ref]

The lecture's control was to project a spectrum photograph, wet it, and project it again. Its colours disappeared as the gelatin swelled, then returned during drying, red before the shorter wavelengths. This tested a spacing-dependent explanation. Lippmann also reported roughly one-minute sunlight exposures and identified portrait exposure time as a limitation.^[ref]

A reversible colour change supports a structural explanation, but does not independently measure every fringe spacing or the refractive index. For a conceptual comparison, record a narrow band and compare reflected optical spectra at fixed illumination and viewing angles; an unexposed processed plate tests background. A second comparison changes only viewing geometry. A third compares the dry state before and after a reversible spacing change. These are proposed discriminating controls, not additional experiments attributed to Lippmann. Background surface reflection must be separated from the selective signal. A photograph's appearance also depends on the illuminating optical spectrum; viewing under an arbitrary lamp is not a calibrated spectral measurement.

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.

In photography, photographic processing uses chemical reactions to develop and stabilize an exposed photographic emulsion.