BRAGG SCATTERING


We can think of the crystal as being composed of thousands of mirrors that reflect the X-rays. These "mirrors" are called Bragg planes.
When light is reflected from a mirror, the angle of incidence (the angle at which it strikes the plane of the mirror) is equal to the angle of reflection. The same is true of Bragg planes, and the reason is that when the angle of incidence is equal to the angle of reflection, light rays hitting the plane (mirror) in phase exit in phase, regardless of where they hit the mirror.
Notice, by the way, that the incoming and outgoing rays differ in direction by a total angle of 2θ.
If rays reflected from a plane have identical pathlengths, then rays reflected from different planes must have different pathlengths. We can easily work out how far apart the planes must be for the difference in pathlength to be equal to the wavelength of the incoming radiation, so that the scattered rays from the two planes would again be in phase. It turns out that the difference in pathlength depends on the angle of incidence (and reflection). The image below shows how to work out the relationship, which is called Bragg's law.

The difference in pathlength between the rays reflected from the two planes is twice the distance l. Simple geometry tells us that the upper angle in the little triangle must be q, because the sum of the angles inside a triangle is 180 degrees, and the other two angles are 90 degrees and 90-θ. Then simple trigonometry tells us that the distance l is equal to d sinθ.For the two rays to be diffracted in phase, twice l must be equal to the wavelength, so we have the relationship:
λ = 2 d sinθ

In Bragg's law, as the angle increases, d must become smaller for the pathlength to remain equal to one wavelength. We can show this by various common rearrangements of the equation:
sinθ/λ = 1/(2 d)

d = λ/ (2 sinθ)



We can obtain same phase of different waves by calculating the optical path deference between two waves:
Δloptic=2π/λ*2dsinθ
.
If we want that two waves will have the same phase and will get maximum intensity we need that the condition that the relative phase will be a multiple of 2π : 2dsinθ=λm.
If we want that the intensity of the diffracting waves will be 0 we need to demand Δφ=π so the optical path deference between two waves will satisfy the condition : 2dsinθ=λ(m+1/2).
Finally ,in the general case we have the condition that the optical path deference between two waves will be arbitrary angel : 2dsinθ=λ(m+Δφ/(2π))
the Bragg scattering from two Bragg planes can be visualize in the function: scanning_angels_func(h,f,d_teta,final_teta) ,which can be downloaded in the download section.
Here we can see an example of waves diffracting in phase for cubic latice with a=3 and wave with λ=3 :
An example of waves diffracting in of phase

An example of waves diffracting out of phase

And an example of waves diffracting in phase Δφ :

And a graph of I vs. θ for d=3 and wave with λ=3

notice that for angels 0o and 90o are not physical so the only positive refraction is for 30o.
negative diffraction is for 14.48o and 48.59o.

This is another examples with d=3 and wave with λ=2

Positive refraction is for 19.47o and 41.81o.
Negative diffraction is for 9.59o , 30o and 56.44o.

More graphs like that can be plotted using I_teta_graph_func.for more details see downloads section






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