# Clarification about Bragg's law explanation

The Wikipedia has this illustration of Bragg's law and then says

The two separate waves will arrive at a point with the same phase, and hence undergo constructive interference, if and only if this path difference is equal to any integer value of the wavelength, i.e. $(AB+BC) - (AC') = n\lambda$

What I don't understand is the "will arrive at a point with the same phase" part. Aren't the points $C$ and $C'$ separated in space, by a distance of roughly the same order as $d$? To constructively interfere, these two rays ($AC'$ and $BC$) must continue on to some detector, and somehow meet at the same point in space. How does that happen?

• The diagram is for illustrative purposes of the interference condition only. The actual diffraction requires many layers of the crystal to produce a sharp angle dependence. Beyond that the finite sample size and the finite aperture size of the x-ray source have to be compensated for. See e.g. web.stanford.edu/group/glam/xlab/MatSci162_172/LectureNotes/… for actual instrument geometries. – CuriousOne Oct 6 '14 at 6:51
• Try drawing two incoming rays such that line $AC'$ from one ray is coincident with line $BC$ from the other, and keep in mind all rays are from an incident parallel wavefront. – Carl Witthoft Oct 6 '14 at 13:17
• @CarlWitthoft : Do you mean something like this? In that image, the coincident rays (which will interfere at a single point at the detector) come from atoms one down and to the left of each other. But I still have the question - why are virtually all other illustrations of Bragg's law like the one in my post, showing only the wavefront (CC') of the scattered wave. To be detected at the detector, wouldn't this (plane wave) wavefront need to be focused by, say, some lens, to a single point? – user114806 Oct 7 '14 at 0:27