Bragg and Brillouin diffraction

  1. I'm supposed to give a seminar tomorrow for my 300 level experimental physics paper. The experiments we do our reports on are pre-determined and I pulled the short straw with the most bloody complicated on here, the acousto-optic modulator. I'm trying to get my head around it and i've spend the good part of the day and not gotten very far. I understand that you require constructive interference to produce the maxima and due to the varying refractive indices of the material we will get a weird equation. But the equation we are given is

    sin (θ) = λ/2nd

    Where θ the bragg angle, d is the wavelength of the sound wave in the acousto optic modulator. So it seems odd to me firstly how are the multiple maxima produced when the light in angle must equal the light out angle shouldn't that just produce on maxima. Secondly how can we just chuck n in for the quoted refractive index of the crystal when this varies at every point. How the hell is this equation derived specifically for brillouin scattering. There are no good sources of this on the internet.

    Please help, thanks.
  2. jcsd
  3. DrDu

    DrDu 4,640
    Science Advisor

    You are doing kind of a perturbation expansion here. The small changes of n lead to the Bragg scattering, but the condition for the maxima can be determined using the average value of n.
  4. Oh okay because the crystal came with a designated n and the Bragg equation seemed useless then because this would vary. Unlike normal Bragg diffraction though we change the frequency of the sound wave causes a change in the angle does this sound right?
  5. Sorry worded poorly. A Bragg angle is a defined angle for constructive interference of the incoming and outgoing wave. But in this case I'm guessing we can change the Bragg angle by varying the frequency of the sound wave is this correct?
  6. DrDu

    DrDu 4,640
    Science Advisor

    Yes, because changing the frequency of the sound wave you change also the wavelength of the sound which modulates the diffractive index and acts as a lattice for Bragg scattering.
  7. Sweet but for brag diffraction the theta in must equal the theta out. Seeing as you can keep the theta in constant and change the theta out I'm assuming it's fundamentally different?
  8. DrDu

    DrDu 4,640
    Science Advisor

    Ok, it isn't a Bragg scattering, rather refraction at a grating.
  9. Well that's what I was taught. It's what my textbook introduction to optical electronics and this website says, so now I'm just doubley confused.
  10. DrDu

    DrDu 4,640
    Science Advisor

  11. Never mind had a chat to a solid state lecturer. Well Bragg's law tells you the angle at which theta is the incident and reflected angle producing constructive interference. Changing the lattice structure will require you to change the angle to get a new Bragg angle. This is different from my setup because with my set up we had a constant incident angle and altered the frequency of the waves to alter the outgoing angle. Which obviously wasn't equal to the incident angle like Bragg's law stare, yet when we used the equation we had to define theta as the outgoing angle only.
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