Bragg "diffraction" or "reflection" As you read about X-ray (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?
Hi afrano! 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 ) The reciprocal lattice is just a mathematical transformation. I admit I'm not familiar with it … but it doesn't bend vectors in the middle, does it? And anyway, the reflection is reality … did you mean them the other way round?
Re: Bragg "diffraction" or "reflection" 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 X-Ray's wave vector, but this has nothing to do in real lattice. Any suggestions about what I said?
Hi afrano's friend! I checked in wikpedia to make sure … … Bragg diffraction is reflection, and a crystal is a real lattice. What do you mean by saying that the reflection has nothing to do with a real lattice?
Re: Bragg "diffraction" or "reflection" Many years ago, I measured the precise energy of some x-rays near 50 KeV. For this I used a bent crystal Bragg spectrometer with a 6-mm thick quartz (310 plane) crystal, and the x-rays were transmitted through (diffracted through) the crystal. In this energy region, the x-rays could also be reflected off atomic planes (no transmission).
Re: Bragg "diffraction" or "reflection" 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) ?
Re: Bragg "diffraction" or "reflection" In the mid 1960's I built a Bragg diffraction spectrometer. In this case, the x-rays were between 50 and 100 KeV, and the x-rays passed through the crystal. I believe Bragg reflection is used for lower energy x-rays, and when the x-rays exit from the same side of the crystal they entered.
Re: Bragg "diffraction" or "reflection" 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 periodic-index 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 k-vector. Saying that light will only constructively interfere at certain angles is the same as saying that only specific k-vectors can be added to the light that will result in constructive interference. The set of added k-vectors 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.
Re: Bragg "diffraction" or "reflection" 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).
Re: Bragg "diffraction" or "reflection" 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 in-phase 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 electron-photon interaction and the coherently scattered photons can constructively interact.
Re: Bragg "diffraction" or "reflection" See my posts # 6 and #8. We are discussing n*lambda = 2 d sin(theta), either Bragg diffraction or Bragg reflection. It terms of Thomson or Compton scattering, I had a very strong source (nuclear transition, not atomic transition) about 50 KeV energy, and the spectrometer intrinsic resolution (FWHM linewidth) was about 50 eV. I put 1 inch of aluminum in front of the source (to introduce Compton scattering), and could not see any change in the linewidth, or a shift of 1 eV or more in the energy.
Re: Bragg "diffraction" or "reflection" Hi all! I'm reading a paper by Allman et al. in which is described a neutron interferometer (like a Mach-Zender 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
Re: Bragg "diffraction" or "reflection" If you have a very low probability of reflection, then most of the neutrons pass through the interferometer without interaction. If the neutron has a probability P_{1} <<1 of reflection, the the reflected neutrons will exit the front face of the interferometer with about 100% efficiency, because P_{1}^{2} (the probability of reflecting again before exiting) is essentially zero. Now consider the probability of reflecting is P_{1} = 0.25. Now the probability of that neutron reflecting again before exiting is about P_{2}= 0.5 P_{1} = 0.125. So now the total reflected is P_{1}(1-P_{2}) = P_{1} (1 - 0.5 P_{1}) = P_{1} - 0.5 P_{1}^{2} = 0.25 -0.032, and the total transmitted is 1 - P_{1}+ 0.5 P_{1}^{2} = 1 - 0.25 +0.032.