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Intensity of stokes and anti-stokes lines?

  1. Mar 16, 2006 #1
    According to Raman effect, the intensity is directly proportional to the 4th power of the wavelength. Then how come stokes lines, which have higher wavelengths than anti-stokes lines, are more intense than the anti-stokes lines?
  2. jcsd
  3. Mar 16, 2006 #2
    uh? I think you answered your own question. If I is propotional to lambda to power 4 and the stokes line has higher wavelengths of course they're more intense then.
  4. Mar 16, 2006 #3
    If I remember well, the anti-stokes line is near absorption bands. Its gain is too low there, so it is less intense than stokes line.
  5. Mar 18, 2006 #4
    Sorry intensity is inversely proportional to the 4th power of wavelength.:biggrin:
  6. Mar 19, 2006 #5

    Claude Bile

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    Isn't the lambda^-4 dependance characteristic of Rayleigh scattering?

  7. Mar 20, 2006 #6
    To have a Raman anti-stoke diffusion, we need to have a transition of an atom initially in a excited vibrationnal level to the ground level (if we forget the intermediate virtual state). The stoke diffusion, instead, relies on a transition from the ground level to an excited vibrationnal level.

    At thermal equilibrium, the populations of the different levels follows the Boltzmann distribution. So the population of the ground level is higher that the excited level. So the stokes transition is much more probable that the anti-stokes transition.

    Does it answer to the question ?

  8. Mar 20, 2006 #7


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    To add a touch to Claude's and Barth's posts : vibrational eigenmodes (or phonons in solids) are bosons. Their distribution among different levels is given by the Bose occupation number. At low temperatures, the different levels have significantly different populations.
    Last edited: Mar 20, 2006
  9. Mar 25, 2006 #8
    yes it does answer my question. Thanx
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