= Bragg path difference
{c}
{title2=$\Delta\ell=2d\sin\theta$}
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 distance
$$
\Delta\ell=CB+BD=2d\sin\theta.
$$
The resulting <phase difference> is $\Delta\varphi=2\pi\Delta\ell/\lambda$. <Constructive interference> requires $\Delta\varphi=2\pi n$, 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.
\Image[/bragg-rays-and-extra-path-geometry.png]
{title=Bragg rays, extra path, and the distinction between the glancing angle and scattering deflection}
{description=Original schematic of <coherent elastic scattering>. The orange segments give the extra path $CB+BD$; dashed segments compare equal-<wave phase> fronts. The separate angle sketch shows why the outgoing beam is deflected by $2\theta$ when the <Bragg angle> is $\theta$. The illustrated angle is schematic.}
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.
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