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Bragg's Law/Bragg reflection

  1. Feb 9, 2015 #1
    I don't think I understand Bragg's law, but I'd quite like to! There's loads of stuff online, but most of it is too complicated for me as a first year undergrad. My understanding was that x-rays get diffracted, but then my textbook says something about electric dipole moments (never heard of them) and makes it sound like the wave is actually incident on the atoms. Apparently it's not the same as a diffraction grating at all.
    There wasn't really a question there... I'm just quite confused! I'd appreciate any help :)
     
  2. jcsd
  3. Feb 10, 2015 #2
    Your understanding that x-rays get diffracted is correct. yes, It is similar to diffraction grating. All that business about electric dipole moments is for later years, trying to explain the underlying physics.

    Bragg's Law -
    Bragg found that when x-rays are scattered from a crystal lattice, the scattering gives rise to peaks of scattered intensity observed in certain directions which meet the following conditions:
    1. The angle of incidence of the x-rays = angle of reflection of the scattered x-rays.
    2. When the path length difference between rays scattered by successive planes of the lattice is equal to an integral number of wavelength of the incident x-rays, the rays interefere constructively to produce peaks of intensity.
    The condition for maximum intensity is known as Bragg's law which allows us to calculate the atomic or molecular structure of the atoms or molecules constituting the crystal lattice. Bragg’s Law: n λ = 2d sin θ

    Hope the above makes sense to you?

    upload_2015-2-10_15-2-59.png
     
  4. Feb 11, 2015 #3
    n*(lambda)=2dsin(theeta) which is braggs eqn
     
  5. Feb 13, 2015 #4
    Thank you, that's a really good summary, definitely makes it clearer! I'll ignore dipoles for now then :)
     
  6. Feb 13, 2015 #5
    That is reality for whatisreality! I am glad you liked my explanation.
     
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