## 10 - Lecture notes for GEOL3010

Nesse p. 163

## Bragg's Law

If we consider an incident beam approaching scattering centers
at some angle (q),
it can be shown that the only place the scattered beam will be
in phase is at the same or "reflected" angle that leaves
the scattering points.

Geometrically, the conditions of constructive interference
are met only when DC = CE therefore,

q = q' and AC = DCsin
q.

Under these conditions there is zero path length difference
between rays 1 and 2.

Unlike light, which can be reflected at all angles, X-rays
are "reflected" only at specific angles.

The wave fronts that pass through a crystal must have path-length-differences
exactly 1,2,3...n integers away or they will destructively interfere.

Compare Rays 1 and 3.

Note path length difference into the plane of atoms is the
distance FB + BG.

This additional distance must be equal to some integer distance
(i.e., FB + BG = nl), but it does not.

Note that the distance AC is the interplanar d-spacing.

Under "reflecting" conditions then q = q' and sin q = AC / DC
and sin q'
= AE / CE

Let: d = AC and nl= DC + CE

then: 2AC sin q
= DC + CE

By substitution, we get Bragg’s law.

nl
= 2dsin q

For more on X-ray diffraction click
here.