Bragg's law and the distribution of electromagnetic energy.

by siddharth5129
Tags: bragg, distribution, electromagnetic, energy
siddharth5129 is offline
Dec24-13, 11:23 PM
P: 58
While I get the coherent and incoherent scattering process that leads to the bragg diffraction condition, I don't really understand the physical mechanism behind the transmission and reflection. Now, as I understand it, the bragg diffraction condition is satisfied only for one or two particular X-ray wavelengths, and this translates to it's reflections from successive crystal planes coherently interfering to producing an intense reflected beam. 1) Is it true that you would get a similar diffraction pattern for every X-ray wavelength by varying the angle of diffraction for each wavelength. 2) For a fixed angle, you should get coherent interference for only one particular wavelength. But this wouldn't mean a sharp peak in a graph of intensity vs. wavelength(at that fixed angle), would it? And how is it that this simple fact forces the rest of the wavelengths to get transmitted through the crystal. The way I see it, this should simply lead to a redistribution of the energy with peak intensities for different wavelengths occurring at different angles.The amount of reflected energy should be independent of whether or not bragg's condition is satisfied, should it not? Or does Bragg's condition somehow force energy to be reflected at certain wavelengths and transmitted at the rest?
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edguy99 is offline
Dec25-13, 01:31 AM
PF Gold
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P: 290
1/ Sure, within a reasonable range. Your wavelength must be appropriate for the atomic spacing you are measuring.
2/ Yes, you get a peak in the intensity vs. wavelength for a fixed angle when the wavelength matches the spacing adjusted for the angle chosen. The rest of the wavelengths are not necessarily transmitted through the crystal, its just that they are not reflected with a peak at that angle but shoot off in a more random pattern.

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