# If parallel rays never meet why does Braggs law work?

1. Apr 9, 2015

### davidbenari

Ive seen similar questions elsewhere and people say that you can consider a ray that comes not from the atom directly beneath, but an atom beneath and to the left. That ray will coincide with some ray. I don't like that interpretation because it seems contrived.

Also that interpretation does not seem to work for the case of low-energy electron diffraction on a crystal where only the first layer of atoms is touched. In which case the equation $dsin\phi=n\lambda$ and not Bragg's law works.

So my question is why does Braggs law work and the e- diffraction formula work if parallel rays never meet?

I'm sorry if my question is too dumb. Here's what I mean by the way (with the first paragraph):

- x -
x - - Consider those two atoms with rays coming out towards the upperright corner of your screen. THe idea is that those rays do in fact coincide.

2. Apr 9, 2015

### sophiecentaur

The Bragg law refers to what happens for a plane wavefront when it comes across a crystal lattice (and not multiple rays, iirc). A plane wave front is a useful fiction and it assumes that the 'ray' it represents is wide enough and coherent enough and can be treated as coming from infinity (i.e. the source is many wavelengths away). This sort of assumption is made for many diffraction / interference calculations.
The second part of the question, about what happens when the wave does not penetrate far, seems to be a separate issue and I can't see how the parallel rays come into it. Perhaps there are situations where the whole lattice is not involved and the Bragg law becomes modified.

3. Apr 9, 2015

### davidbenari

I'm sorry but I think thats incorrect. See this image http://en.wikipedia.org/wiki/Bragg's_law#/media/File:Braggs_Law.svg In the case that you are right, I still don't see how Braggs law will give constructive interference. Sure the paths yield a difference $n\lambda$ but were not talking about the same point on my detector, so how was there a bigger electric field at that point?

Maybe I think I have the wrong interpretation of what a "ray" means, but still I know about interference and my question still confuses me.

Parallel rays come into it, see this image

http://www.chemistry.mcmaster.ca/esam/Chapter_1/fig1-3.jpg

Here the reflected rays haven't touched other planes other than the superficial ones.

4. Apr 9, 2015

### sophiecentaur

The two diagrams you are using as evidence for your perception are really not suitable. The squiggly lines are just plain misleading and the 'electron beams' are meaningless because, unless the electrons were 'coherent' (!!!!???!!!) you would get no interference effects.

If you look at any (I mean pretty well any) text on the way two slits and diffraction gratings work, you may see lines from sources to screen but these are not 'beams' . They are lines to show the paths through the system and show the path differences between them - which gives vector addition to produce maxes and mins. Step back from the specific examples you are using and go back to the basics. That will help you understand what's going on.

5. Apr 9, 2015

### davidbenari

Oh okay... Whats happening then? whats the correct way to view interference from crystals? How is the path difference relevant if I'm seeing two "rays" that do not converge? They're not "rays", but what are they then?

Thanks.

6. Apr 9, 2015

### nasu

You can take them to converge in some point situated about half a meter from the sample.
Calculate by how much will the angle between the rays will change, to convince yourself that the parallel rays are a very good approximation. It is just easier to calculate that way and considering them to converge complicates the calculations but give the same result.

Anyway Bragg formula results from the more rigorous treatment due to Laue. It's just an easy way to derive the formula which we now it is right from Laue's theory.

7. Apr 9, 2015

### sophiecentaur

It is not a good idea to use the word "ray" in this context. It may have been used in a badly composed School lesson but using the word 'ray' implies an ideal beam of light with a width of many wavelengths or diffraction will not produce a ray, travelling (mainly) in one direction. The lines that have been mis-termed 'rays' are merely lines which are normal to the wave front. (Not the same thing at all) Ray tracing, as used in lens calculations is not at the atomic level .
If you want to show interference occurring, it is necessary that there is coherence across the wave front and that will not be what you get with a set of 'rays'. If there's any doubt about the point I'm making then look at this wiki link, about simple two slit interference. There is no mention of the word Ray. Instead, the explanation is given in terms of wave fronts.

What happens when there are multiple layers of scattering points is more complicated but it can't suddenly be allowed to involve the use of 'rays'. It is based entirely on the differences between path lengths through the crystal lattice. But, first things first. Get Young's Slits sorted out in your mind before moving on to Bragg.

Last edited: Apr 9, 2015
8. Apr 9, 2015

### davidbenari

I'll read the article, however (as you predicted) I don't really se the point your making. I see what you say about rays, so I'll talk about "trajectories" or "lines".

Suppose I'm working out Young's double slits for light:

I choose a point on the wall where the light is producing the pattern and trace "lines" all the way back to my original slits.

These lines are the trajectories of two sine waves and if the difference in length of these two "lines" is $n\lambda$ I know both sine waves arrive in phase. Since the electric field obeys superposition I know I'll have a higher amplitude at that point.

Now I bring that analysis back to Bragg's law. And everything breaks down! The two "lines" are not intersecting at a point on my detector.

Wheres the main flaw in my perspective of interference?

Thanks.

9. Apr 10, 2015

### sophiecentaur

No it doesn't. The assumption of parallel paths merely introduces a tiny error when going from infinite throw to a throw of a few tens of cm. The production of maxima in certain directions only occurs for certain angles of incidence when you have multiple layers too. The diagrams showing how Braggs law is derived are pretty simple. (That one with squiggles is hardly the best one to choose). The far field / near field approximation has very little effect except when you go very close.