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Beam divergence derivation

  1. Oct 23, 2014 #1
    For a Gaussian beam, which has 86% of its power within its beam diameter (spot size 2w0), I've read that beam (angular) divergence is given by

    2θ = 4λ/(π[2w0])

    Where does this come from? I hate memorizing equations. It makes me feel stupid.
     
  2. jcsd
  3. Oct 24, 2014 #2
    θ is the half-angle divergence for z→∞ so that :

    θ=limz→∞ {ω(z)/z} = ω0 / zR = λ/(π.ω0)

    with zR = π.ω02 / λ : the Rayleigh range.
     
  4. Oct 25, 2014 #3
    Why is it θ=limz→∞ {ω(z)/z} = ω0 / zR = λ/(π.ω0)?

    Why not θ=limz→∞ atan{ω(z)/z} first of all? And second, as z, the distance from the source, goes to infinity, I would expect the beam divergence to be infinitely wide... so... 180 degrees maybe?
     
  5. Oct 26, 2014 #4
  6. Oct 26, 2014 #5
    There is no atan because Gaussian beams are implicitly in the paraxial approximation, i.e. small values of theta.

    The beam divergence is an angle. As z --> infinity, the beam size becomes infinite, but the divergence converges to the value quoted above.

    For small values of z, the beam waist plays a role in the beam size. For large z this contribution is negligible. In the limit z --> infinity you simply forget about the contribution from the waist. The beam size is then given by the divergence times z.
     
  7. Oct 27, 2014 #6
    I understand the approximation that tan(θ) is about θ, but not why Gaussian beams are seen as some kind of paraxial approximation. Gaussian beams are basically intensity versus distance plots that are distributed like a Gaussian. But, I guess that doesn't say anything about which direction anything is in? I'm confused. Could you explain further please? :)

    I see, the beam divergence is an angle, so no matter how much you scale it, it still remains that fixed angle. I think I get the intuition based on the last thing you said. The angle is "set" by a triangle you could draw with legs w and z. So there's a constant "slope" to each of those legs. I'm not quite sure why it bows out like that though. Why do the (real) wavefronts do that?

    Thanks for the input. Hope this also helps people in the future.
     
  8. Oct 28, 2014 #7
    In principle every beam can have a Gaussian profile. The typical Gaussian beam math you find in textbooks and on the web, however, is in the paraxial approximation.
     
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