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Optics: Dispersion

  1. Mar 29, 2010 #1
    Hi

    I am reading about dispersion and index of refraction, and I have encountered the term "anomalous dispersion". This is where dn/dω<0, and where the absorption is strongest because of damping. Now, I can't seem to connect those two dots (and my book does not explain it nor Wiki): Why does dn/dω<0 imply strong absorption and vice versa?
     
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  3. Mar 29, 2010 #2

    Andy Resnick

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    Briefly- it's the Kramers-Kronig relations. Anamolous dispersion (and high dispersion generally) occurs in the same spectral region as absorption peaks.
     
  4. Mar 29, 2010 #3
    Thanks.

    I have another question related to optics, so I will ask it here rather than creating a new thread. When we solve Maxwells equations inside a conductor, we eventually get the Helmholtz equation, where the wavevector k is complex.

    In order to find the index of refraction, my teacher told me that n2=k, where n is the complex index of refraction and k is the complex wavevector. Did he make an error? I really have to idea where that formula comes from.
     
  5. Mar 29, 2010 #4
    Sounds like a mistake to me--the quickest way to see this is to note that n is unitless while k has units of inverse length. If you're treating n as the complex index of refraction, and k as the complex wavevector, then the formula k = n*omega/c still holds.
     
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