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Bragg diffraction or reflectionby afrano
Tags: reciprocal lattice 
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#1
Mar1809, 03:38 AM

P: 10

As you read about Xray (or neutron) diffraction in solids, you come across authors who use the term "Bragg diffraction" and others that use the term "Bragg reflection". Some authors even use both terms in the same edition of the book (Kittel, 5th Ed.,, I think..). My question is:
Isn't there a difference b/w the two terms? Or better yet, why are they used equivalently? Some discussion with colleagues suggested that reflection occurs in the reciprocal lattice, and diffraction occurs in the real lattice. Is this the case? 


#2
Mar1809, 05:21 AM

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They're the same … Bragg diffraction is by reflection … exactly the same way that you can get reflection off a diffraction grating (or an ordinary CD!). With an ordinary diffraction grating, it would be silly to have a different name for what is essentially the same effect from both reflection and transmission … so they're both called diffraction. Also, "diffraction" comes from the Latin frango frangere fregi fractum, meaning to break, and in this case refers to the breaking of light into fractions, or different colours. (Not sure how refraction fits into that ) And anyway, the reflection is reality … did you mean them the other way round? 


#3
Mar1809, 11:57 AM

P: 397

I'm afrano's friend, and would like to join the discussion, we've been discussing this long time.
I believe that it's reflection over the Miller indices, i.e. over the reciprocal lattice, where miller indices are obtained through the reciprocal lattice, and actually when we use Laue's method we use the dot product between miller's indices wave vector and XRay's wave vector, but this has nothing to do in real lattice. Any suggestions about what I said? 


#4
Mar1909, 06:56 AM

P: 397

Bragg diffraction or reflection
Hey guys! is the question that complicated?



#5
Mar1909, 09:12 AM

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I checked in wikpedia to make sure … What do you mean by saying that the reflection has nothing to do with a real lattice? 


#6
Mar1909, 03:12 PM

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#7
May1209, 07:45 PM

P: 16

Its clear from above replies how diffraction is reflection but noone has touched the relationship of the reciprocal lattice and the real lattice particularly. I'm currently still trying to get my head around it but the way i understand it id that when you examine the diffraction pattern from a crystal, the spacing you get from Bragg's law gives you the spacing of the planes in the reciprocal lattice (by the diffraction condition k.(.5G)=(.5G)^2 where G is the reciprocal lattice vector). So to obtain the actual crystal plane spacings you need to transform back to real lattice (Fourier transform i think).
Is this correct ( or helpful :p) ? 


#8
May1209, 09:14 PM

P: 4,663

In the mid 1960's I built a Bragg diffraction spectrometer. In this case, the xrays were between 50 and 100 KeV, and the xrays passed through the crystal. I believe Bragg reflection is used for lower energy xrays, and when the xrays exit from the same side of the crystal they entered.



#9
May1609, 02:12 AM

P: 2

I would probably use "diffraction" to refer to the type of experiment you are talking about and "reflection" to refer to a Bragg mirror in, e.g., a DFB or DBR laser (that would be a periodicindex structure that causes waves of a certain wavelength to reflect through destructive interference of the forward wave). Nonetheless, Bragg diffraction/reflection is as you say: light coming into a crystal will reflect off of planes of atoms. When it does, only light reflecting at certain angles will constructively interfere with itself, and this depends on the atom spacing in that plane. Now, reflected light changes direction and so has a change in kvector. Saying that light will only constructively interfere at certain angles is the same as saying that only specific kvectors can be added to the light that will result in constructive interference. The set of added kvectors that gives you constructive interference is the reciprocal lattice. Thus, when you look at a diffraction pattern on the wall, the lattice (or any image, really) you see is the reciprocal of the real lattice from which the light is diffracting. To get the real lattice, you need to take the Fourier transform of the pattern. Incidentally, this is how a hologram works.



#10
May2909, 11:29 AM

P: 1

str8 is right, holograms operate through Bragg diffraction/reflection (volume holograms specifically). When talking in technical circles, it is always discussed as "Bragg diffraction," which is the most correct. When talking in practical terms about the actual creation of holograms, often the term "Bragg reflection" will be used for discussing reflection holograms (where the incident light and diffracted light enter and leave through the same side of the hologram plate, thus "reflecting"), while "Bragg diffraction" would be used more often for transmission holograms (where the incident light and diffracted light enter and exit through opposite sides of the hologram plate, thus "transmitting" through the hologram).



#11
Jun1009, 09:56 AM

P: 33

The phrase "reflection" or "diffraction" is oversimplified.
This incoming photon interacts with the electrons in the crystal and scatters in a coherent manner (Thompson scattering). Incoherent scattering (Compton scattering) will not result in diffraction. The scatter is a very small amplitude event and happens in all directions. In select directions (due to the regular lattice of atoms) the coherently scattered photons are inphase and constructively interact. In these directions you observe a large intensity of scattered photons. in the other directions there are still photons being scattered, but they are much lower intensity. You can call it whatever you want, but fundamentally this is an electronphoton interaction and the coherently scattered photons can constructively interact. 


#12
Jun1009, 01:04 PM

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#13
Jun1909, 11:29 AM

P: 17

Hi all!
I'm reading a paper by Allman et al. in which is described a neutron interferometer (like a MachZender one). The neutron beam (monochromatic E=14meV ) is divided in two coherent beams (transmitted and reflected) using Bragg diffraction. From this discussion I understand that the probability of transmission and reflection depends on the energy of the particles... I've always thought that bragg diffraction was "in reflection" but here I have transmission, too. Maybe I'm in a bit of confusion about that: why is there transmission?? Can you help me or show me a reference to understand it better? Thanks 


#14
Jun1909, 12:34 PM

P: 4,663




#15
Jul1009, 02:39 PM

P: 17

I think its both, reflection and defraction



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