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Derivation of complex refractive index

  1. Sep 15, 2012 #1
    Hi,

    Got a problem with the following derivation:

    Coming from the Helmholtz equation one gets:

    [itex]\textbf{n}^2[/itex]=[itex]\mu[/itex][itex]c^{2}[/itex]([itex]\epsilon[/itex]+i[itex]\frac{\sigma}{\omega}[/itex])

    which is of course something like:

    [itex]\textbf{n}[/itex]=n+i[itex]\kappa[/itex]

    My question is, how do you obtain the following relations?

    [itex]n^{2}[/itex]=[itex]\frac{1}{2}[/itex][itex]\mu[/itex][itex]c^{2}[/itex][itex]\epsilon[/itex]([itex]\sqrt{1+(\frac{\sigma}{\epsilon\omega})^{2}}[/itex]+1)
    [itex]\kappa^{2}[/itex]=[itex]\frac{1}{2}[/itex][itex]\mu[/itex][itex]c^{2}[/itex][itex]\epsilon[/itex]([itex]\sqrt{1+(\frac{\sigma}{\epsilon\omega})^{2}}[/itex]-1)

    Maybe it's obvious, but I am arriving at everything but this. Enlighten me guys and thanks if you do.
     
  2. jcsd
  3. Sep 17, 2012 #2
    [tex]n[/tex]=n+ik, put into [tex]n^2[/tex] and you will have two eqn ,one equating the real part and other equating the imaginary part which you can solve to get.if that is what you are asking.
     
  4. Sep 17, 2012 #3
    That's right, but the problem is you cannot separate n or κ out. For example you get something like

    [itex]n^{2}[/itex]([itex]n^{2}[/itex]-[itex]\mu[/itex][itex]c^{2}[/itex][itex]\epsilon[/itex])=([itex]\frac{μc^{2}σ}{2ω})^{2}[/itex]

    and I don't see how to get to n2.

    Oh, OK, just have to solve the quadratic equation after substitution and one gets to the results, duh! Thanks!
     
    Last edited: Sep 17, 2012
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