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The Bragg angle is the glancing angle from the incident beam to the selected lattice planes, with $0<\theta \le 90^{\circ }$. If $\alpha$ is instead the angle from the beam to the plane normal, $\alpha =90^{\circ }-\theta$ and Bragg's law becomes $n\lambda =2d\cos \alpha$. Defining the angle explicitly prevents a sine-versus-cosine error as well as a factor-of-two error in the detector deflection.

Choose $\lambda =0.154\mathrm{nm}$ and $d=0.200\mathrm{nm}$ as an illustrative model, not historical measurements. For the first diffraction order,The second diffraction order has $\sin \theta =0.770$ and $2\theta \simeq 100.707778^{\circ }$; the third would require $\sin \theta =1.155$ and is impossible. These are different diffraction orders of the same interplanar spacing at the same wavelength.

Conversely, a first-order reflection at a measured deflection $2\theta$ gives the interplanar spacingUsing $2\theta \simeq 45.287481^{\circ }$ recovers $d\simeq 0.200\mathrm{nm}$. This is the useful reversal: a known wavelength turns an angular measurement into an atomic length. An unassigned diffraction order or an unknown wavelength prevents that single measurement from determining $d$ uniquely.

Compare two rays scattered into parallel outgoing directions from neighboring lattice planes. In the diagram, $A$ lies on the upper plane and $B$ on the lower plane, directly beneath $A$ with $AB=d$. Draw through $A$ the incoming and outgoing equal-wave phase fronts, perpendicular to their respective rays. They meet the lower ray at $C$ and $D$. Each right triangle gives $CB=BD=d\sin \theta$, so the lower ray travels the extra distanceThe resulting phase difference is $\Delta \phi =2\pi \Delta \ell /\lambda$. Constructive interference requires $\Delta \phi =2\pi n$, which proves Bragg's law in this ray construction. Comparing equal-wave phase fronts, rather than arbitrary ray endpoints, is what makes the distance comparison meaningful.

The same geometry connects to the reciprocal lattice. With wavenumber $k=2\pi /\lambda$, the elastic Bragg scattering condition gives a transfer magnitude $|q|=2k\sin \theta$. A reciprocal vector normal to these lattice planes has $|q|=2\pi n/d$. Equating them again yields Bragg's law. The ray and reciprocal lattice descriptions therefore express the same wave phase matching.

Bragg's law selects directions of constructive interference in coherent elastic scattering from equally spaced lattice planes:Here $d>0$ is the perpendicular interplanar spacing, $\lambda >0$ the incident wavelength, $n$ the positive integer diffraction order, and $\theta$ the Bragg angle, measured between the incoming beam and the lattice planes. The scattering deflection, between incoming and outgoing propagation directions, is $2\theta$. Thus a plotted detector angle $2\theta$ must be halved before insertion into Bragg's law. The $\theta$ in the parent elastic Bragg scattering condition denotes the full deflection; the present $\theta$ denotes half of it. ^[ref]

The lattice planes organize wave interference throughout a crystal structure; they are not literal metal mirrors or separate reflecting surfaces. The picture assumes elastic scattering, hence an unchanged wavelength, and sufficiently stable relative wave phases across the contributing lattice planes. X-ray wavelengths comparable with atomic interplanar spacings make these directions experimentally accessible. The sine cannot exceed one: an order exists geometrically only if $n\lambda /(2d)\le 1$. For $\lambda >2d$, even the first diffraction order is inaccessible. A geometrically possible direction need not have nonzero wave intensity; the crystal lattice structure factor and crystal basis structure factor also matter. ^[ref]

Differentiate Bragg's law at fixed correctly assigned diffraction order:Divide the second equality by the first. For small signed changes,The Bragg angle error $\delta \theta$ must be in radians. An error quoted for detector deflection $2\theta$ must first be halved. The expression is a first-order Taylor expansion; an unknown diffraction order is a discrete assignment error, not an infinitesimal perturbation.

For error bounds, add magnitudes. For independent zero-mean random errors with standard deviations ${\sigma }_d,{\sigma }_{\theta }$, variance addition instead givesCorrelated errors require a covariance term. A shared calibration bias does not become independent merely because many line positions are measured.

With a chosen angular error $10$ arcseconds, $\delta \theta =4.84814\times 10^{-5}\mathrm{rad}$, the angular contribution is $8.3972\times 10^{-5}$ at $30^{\circ }$ and $1.7646\times 10^{-5}$ at $70^{\circ }$. Larger Bragg angles reduce this contribution because $\cot \theta$ decreases. For a nearby spectral doublet, the inverse relation is $\Delta \theta \simeq \tan \theta \Delta \lambda /\lambda$: a fixed fractional separation spreads farther apart angularly at large Bragg angle.

This does not permit unlimited improvement. The geometry requires $n\lambda \le 2d$; a desired wavelength might not have an accessible large-angle reflection with the chosen crystal. Higher diffraction orders can reach larger Bragg angles but their strengths depend on the crystal lattice structure factor. Finite crystal response, slit widths and source size remain relevant. An error in $d$, including uncorrected thermal expansion, remains even when the angular term is small. Very near $90^{\circ }$, the leading angular derivative vanishes and the second-order Taylor expansion matters.

A crystal X-ray spectrometer selects a narrow wavelength interval by diffraction at known interplanar spacing $d$ and Bragg angle $\theta$. For fixed diffraction order $n$, a known $d$ lets an angular measurement determine $\lambda =2d\sin \theta /n$. This reverses the crystallography problem of finding $d$ from a known wavelength.

Siegbahn's laboratory developed X-ray tubes and evacuated spectrographs for longer wavelengths. Rotating the crystal selected reflections; photographic plates recorded positions. Opposite-side exposures measured angles on both sides of the direct beam. Later tube instruments mounted a beam-defining slit opposite the plate holder. A separate double instrument compared two reflected beams with two ionization chambers. These were different instruments, not one universal assembly.^[ref]

The Nobel biography describes constant-temperature spectrographs and a precision angular scale read by diametrically opposed microscopes.^[ref] Temperature control matters because thermal expansion changes $d$; an accurately read angle cannot compensate for an unknown interplanar spacing.

The schematic separates preparation, selection and readout. An X-ray tube target supplies Characteristic X-rays with a possible bremsstrahlung background. A collimator limits incident directions; evacuation reduces air X-ray attenuation. Rotating the crystal brings different wavelengths into reflection. A photographic plate stores line positions over an exposure, whereas an ionization chamber can measure collected electric charge at successive angles. Neither readout alone establishes absolute wavelength: an angular zero, angular-scale calibration, diffraction order, $d$ and any required refraction correction enter the inference. Narrower slits reduce accepted angular spread but also reduce signal; a weak line can become harder to locate despite a narrower instrumental response.

A control reported in Siegbahn's lecture photographed a line system with copper-sulphate and calcium-sulphate crystals. Irregular relative shifts around copper's characteristic frequency supported anomalous dispersion; the arrangement excluded differing absorption across that frequency as the explanation.^[ref] A finer angular scale alone cannot remove this physical refraction correction.

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 spectral doublet comprises two neighbouring spectral lines. Resolving it requires information about the line response, relative strengths and noise, not only precision in reading an angle. Choose two equal angular responses shaped as Gaussian functions with common standard deviation $\sigma$ and separation $\Delta$. If $x$ is angular displacement from their midpoint, their summed signal isThese are assumed response shapes, not a measured Siegbahn frequency spectrum. By symmetry $f^{\prime }(0)=0$, andThe midpoint changes from a maximum to a minimum at $\Delta =2\sigma$. For this particular equal-width, equal-height model, two distinct maxima occur when $\Delta >2\sigma$. Each isolated Gaussian function has full width at half maximum $2\sqrt{2\ln 2}\sigma \simeq 2.355\sigma$; the two-maxima threshold is therefore about $0.849$ times that width.

In this chosen model, the two Bragg angles are $30.000000^{\circ }$ and $30.033085^{\circ }$. Their separation is $0.827$ times $\sigma$ for $\sigma =0.040^{\circ }$, giving one maximum, and $4.136$ times $\sigma$ for $\sigma =0.008^{\circ }$, giving two. Sharpening the response can reveal the doublet without changing either emitted wavelength.

The Rayleigh criterion uses a first-zero convention for specified diffraction patterns. A Gaussian function has no finite zero, so its two-maxima threshold is not the Rayleigh criterion. Neither convention is a universal inference limit: known response shapes and adequate signal-to-noise ratio can support estimating a separation before two visible maxima appear. Unequal strengths, background, finite sampling and uncertainty in the response can make that inference much harder. Measurement resolution, the precision of an isolated line's centre and absolute wavelength accuracy should therefore be reported separately.

Choose an original simulated model: a primitive cubic lattice with one identical point scatterer per primitive unit cell, edge $a=0.400\mathrm{nm}$, and incident X-ray wavelength $\lambda =0.200\mathrm{nm}$. These are stipulated lengths, not measurements of sodium chloride. Suppose orientation information has already assigned the two observed reflections the Miller indices $(100)$ and $(110)$. The detector deflection is $2\theta$, where $\theta$ is the Bragg angle.

Use the existing interplanar spacing formula and Bragg's law, with higher diffraction orders absorbed into the reflection indices:The simulated $(100)$ reflection has $\sin \theta =1/4$, hence $2\theta =28.9550^{\circ }$. The $(110)$ reflection has $\sin \theta =\sqrt{2}/4$, hence $2\theta =41.4096^{\circ }$. Their squared sines are $0.0625$ and $0.125$, in ratio $1:2$. Substituting the unrounded angles independently givesAgreement tests a proposed indexing and scale together; disagreement would prompt checks of the indexing, wavelength, angular zero, or the assumed unit cell geometry.

Two peak positions alone do not establish the indexing. Assigning the same two angles to $(110)$ and $(200)$ would give $N=2,4$ and $a=0.400\sqrt{2}\simeq 0.565685\mathrm{nm}$ instead. More reflections, orientation information and systematic absences help discriminate such assignments. An unobserved reflection can be inaccessible, weak or extinguished by the atomic basis; absence is not automatically evidence for a larger interplanar spacing.

Absolute scale also requires measurement calibration of the wavelength. To first order, differentiation of the inferred length giveswith angular changes in radians. A stipulated detector-zero error $\delta (2\theta )=+0.100^{\circ }$, with correct wavelength, biases the first inferred edge downward by approximately $0.338\%$ and the second by $0.231\%$. A common angular offset therefore need not cancel when two inferred lengths are compared. These are model sensitivity estimates, not historical instrument uncertainties.

Consider a chosen primitive cubic lattice $a{\mathbb{Z}}^3$ with $a=0.400\mathrm{nm}$ and one identical point scatterer per primitive unit cell. Retain the incident wavelength $\lambda =0.154\mathrm{nm}$. This is an ideal point model. Its reciprocal lattice vectors are $q=(2\pi /a)(h,k,l)$ for integer triples other than $(0,0,0)$. Write $S=h^2+k^2+l^2$. The associated interplanar spacing and first-order Bragg angle satisfyThe reflection labels $(hkl)$ use the Miller index convention, keeping unreduced triples to encode diffraction order. With that convention, set $n=1$ in Bragg's law for each label. The $(200)$ reflection is equivalently the second diffraction order of $(100)$: $d_{200}=a/2$ does not imply extra scatterers halfway between the occupied $(100)$ planes. Counting both descriptions as distinct reflections would count the same direction twice.

For $(100)$, $(110)$, $(111)$ and $(200)$, respectively, $S=1,2,3,4$ and the calculated interplanar spacings are $0.400000$, $0.282843$, $0.230940$ and $0.200000\mathrm{nm}$. Their calculated deflections $2\theta$ are $22.197435^{\circ }$, $31.594744^{\circ }$, $38.952948^{\circ }$ and $45.287481^{\circ }$. Equal steps in $S$ give equal steps in ${\sin }^2\theta$, not equal steps in $2\theta$. This regularity makes an angular pattern a test of a proposed crystal structure.

A real crystal structure has a Bravais lattice together with an atomic basis of a crystal. The crystal lattice structure factor selects the periodic directions; the crystal basis structure factor combines the scattering amplitudes of atoms within a unit cell and can cancel a reflection. Consequently Bragg's law alone predicts neither allowed reflections nor their wave intensities. ^[ref] Systematic absences further restrict observed families; for example, face-centered cubic lattice centering permits only indices all even or all odd in conventional cubic indexing. The four strips above apply to the stated primitive cubic lattice, not to sodium chloride or a face-centered cubic lattice. ^[ref]