# 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
• @user114806 I've got the exact same question. Btw the link you've added in your comment, it doesnt give the Bragg's law as$2dsin(\theta)=n\lambda$ – Kashmiri Nov 24 '20 at 13:05

Because of the beam divergence in reality the beam is not 1D like what you are seeing in the image. Having enough distance from the surface, their beam profiles will have overlap and we would see the interference. The amount of overlap and details depend on the beam profile of our source or sources. This divergence is not just for electromagnetic waves, matter-waves like electrons also have the beam divergence. The periodic nature of the scattering medium can give rise to different reflection peaks and changing the interference pattern as well.

• Can you please elaborate Which surface are you talking about here? – Kashmiri Nov 24 '20 at 14:18

You can imagine that the two rays converge at the detector. So they are not parallel but rather the sides of a triangle with base AB and the vertex at the detector. Now realize that AB is of the order of angstroms and the sides (distance to detector) are of the order of tens of centimeters. Now in this triangle draw CC'at a distance of a few angstroms from the base. Remember that the vertex is about 10^9 angstroms from the base). How would the segments AC' and BC will look like? Technically not exactly parallel to each other. But would you be able to "see" that they are not parallel? How far would be the angle between them from zero degrees?

A similar "effect" happens with the rays of light from the Sun or a distant star. They are considered parallel but they are not exactly so. How divergent are they?

It is a very common mistake to confuse the rays on a ray diagram with pencil-beams of light. They are not little pencil-beams; rather, they indicate the direction of travel of plane wavefronts.

When we treat interference effects such as from a grating or by reflection from a crystal, we are considering a given plane wave incident on the crystal from a given direction. Each wavefront is a plane orthogonal to the ray and extended 'sideways' from the ray, in principle out to infinite distance. So this means that two parallel rays separated from one another represent two plane waves that fully overlap.

The ray diagram is useful because the rays in fact track two important pieces of information:

1. the direction of the ray indicates the direction in which a plane wave is travelling (already mentioned)
2. the path along the ray indicates one path along which that plane wave accumulates phase as it goes.

The second piece of information is what we need in order to calculate the relative phase between the two overlapping plane waves at the 'output' side of the interferometer. Each wave picks up the total phase according to the path traveled by some given point on the wavefront. A wavefront is, by definition, a locus of places which all have the same phase.

To summarise: whenever you see a ray diagram used for interference calculations, draw some wavefronts onto the diagram if they have not been indicated. Either draw them neatly onto your book/lecture note, or just do it in your mind's eye, and keep in mind that these wavefronts are filling the space of the diagram.

In order to detect the interference effect, it is common practice either to place an output lens or mirror so as to focus the plane waves onto a detector, or else (e.g. for X rays) put the detector far enough away so that approximately plane waves (of finite spatial width) arrive at different places on the detector. This further information is not needed in order to calculate the interference effect, but sometimes people like to image the rays being not quite parallel and thus meeting on a detector a long way away. I would say that picture does not help with an interference calculation for things like a grating or a crystal which have a wide aperture. It is clearer, I think, to let the output rays be exactly parallel and understand that they are telling you about overlapping plane waves.