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Derivation of complex refractive index

 
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Sep15-12, 11:49 AM   #1
 

Derivation of complex refractive index


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.
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Sep17-12, 12:58 AM   #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.
Sep17-12, 04:26 AM   #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!
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complex algebra, optics physics, refractive index
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