Clarification about Bragg's law explanation

Question

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?

Answer 1

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 imagine the rays being not quite parallel and thus meeting on a detector a long way away. I would say that 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.

Answer 2

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.

Answer 3

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?