 Calculated transmission Laue pattern

Choose a primitive cubic lattice with $a=0.400\mathrm{nm}$, one identical point scatterer per primitive unit cell, incident direction ${\mathbf{s}}_{\mathrm{in}}=(0,0,1)$, and the continuous wavelength band $0.080\le \lambda \le 0.160\mathrm{nm}$. This is a hypothetical model, not the zinc blende specimen of the Laue X-ray diffraction experiment of 1912. Write a candidate reciprocal lattice vector as $\mathbf{G}=(2\pi /a)(h,k,-m)$ with $m>0$, and set $S=h^2+k^2+m^2$. The Laue equations and elastic scattering giveA transmission photographic plate at $z=L>0$ intercepts only $s_{\mathrm{out},z}>0$. Its coordinates relative to the direct beam areFor $(h,k,-m)=(2,1,-1)$ and $L=50\mathrm{mm}$, $S=6$, $\lambda =0.13333\dots \mathrm{nm}$ and ${\mathbf{s}}_{\mathrm{out}}=(2/3,1/3,2/3)$, so $(X,Y)=(50,25)\mathrm{mm}$. Its length is one, and each component difference satisfies the corresponding Laue equations. This verifies a direction and a selected wavelength, without predicting an exposure or measured wave intensity.

The first panel has fourfold symmetry because its chosen Bravais lattice, beam axis and square detector share that symmetry. For the tilted panel, apply a rotation matrix $R$ to each index vector, set $v=R(h,k,l)$, and use $\lambda =-2a{v}_z/|v{|}^2$ and ${\mathbf{s}}_{\mathrm{out}}=(0,0,1)+(\lambda /a)v$. All indices with $|(h,k,l)|\le 2a/{\lambda }_{\min }=10$ suffice: elastic scattering bounds $|\mathbf{G}|\le 2|{\mathbf{k}}_{\mathrm{in}}|$. Thus the plotted spectral and angular selection is complete within this ideal point model and detector window. Multiples of an index vector select inversely scaled wavelengths and can land on the same spot; a white-beam spot need not identify a unique diffraction order.