= Calculated transmission Laue pattern
{title2=$\lambda=2am/(h^2+k^2+m^2)$}
Choose a <primitive cubic lattice> with $a=0.400\,\mathrm{nm}$, one identical point scatterer per <primitive unit cell>, incident direction $\mathbf s_{\rm in}=(0,0,1)$, and the continuous <wavelength> band $0.080\leq\lambda\leq0.160\,\mathrm{nm}$. 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> give
$$
\lambda=\frac{2am}{S},\qquad
\mathbf s_{\rm out}=\left(\frac{2mh}{S},\frac{2mk}{S},1-\frac{2m^2}{S}\right).
$$
A transmission <photographic plate> at $z=L>0$ intercepts only $s_{{\rm out},z}>0$. Its coordinates relative to the direct beam are
$$
X=L\frac{s_{{\rm out},x}}{s_{{\rm out},z}}=\frac{2Lmh}{h^2+k^2-m^2},\qquad
Y=\frac{2Lmk}{h^2+k^2-m^2}.
$$
For $(h,k,-m)=(2,1,-1)$ and $L=50\,\mathrm{mm}$, $S=6$, $\lambda=0.13333\ldots\,\mathrm{nm}$ and $\mathbf s_{\rm out}=(2/3,1/3,2/3)$, 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>.
\Image[/laue-polychromatic-spot-pattern.png]
{title=Calculated white-beam <Laue method> spots and their response to crystal tilt}
{description=Original calculation for the stated <primitive cubic lattice> and a square detector with $|X|,|Y|\leq75\,\mathrm{mm}$. Left: the cubic $[001]$ direction is along the incident beam. Right: rotate the same <crystal structure> by $7^\circ$ about the laboratory $y$ axis, keeping source and detector fixed. Color gives the selected <wavelength>; equal marker sizes convey positions rather than <wave intensities>. Harmonic reflections can overlap. Neither panel copies, simulates the chemistry of, or fits an experimental plate.}
{source=laue-polychromatic-spot-pattern.py}
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 $\lambda=-2av_z/|v|^2$ and $\mathbf s_{\rm out}=(0,0,1)+(\lambda/a)v$. All indices with $|(h,k,l)|\leq2a/\lambda_{\min}=10$ suffice: <elastic scattering> bounds $|\mathbf G|\leq2|\mathbf k_{\rm in}|$. 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>.
Back to article page
