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My question is physics related but the problem itself is just math. I have an expression for a refractive index depending on the wave length $\lambda$:

$$ n(\lambda)=\sqrt{1.28604141+\frac{1.07044083\lambda^2}{\lambda^2-1.00585997\text{×}10^{-2}}+\frac{1.10202242\lambda^2}{\lambda^2-100}} $$

I know that the wave length $\lambda$ and the angular frequency are related via

$$ \omega(\lambda)=\frac{137}{18.9 \, n(\lambda) \, \lambda} $$

Now I want to exchange $\lambda$ with $\omega$ in the expression for the refractive index $n$. My approach was to put $n(\lambda)$ into the expression for $\omega$, calculate $\lambda(\omega)$ and put this into $n(\lambda)$ receiving $n(\omega)$.

Unfortunately the expression for $n(\lambda)$ is quite nasty. I don't know how to invert $\omega(\lambda)$. I tried it analytically by hand and also with software (i.e. maxima). No success...

I need the inverse only on the interval $[0.2, 2]$. There the function $n(\lambda)$ is one-to-one.

What do you think? Is there a possibility to get a (or one of the many) solution?

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1 Answer 1

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I think I can turn it into a cubic in $\lambda^2$:

$$A = 137/18.9, B=1.28604141, C=1.07044083\\D=0.0100585997, E=1.10202242, F=100\\ \left(\frac{A}{\omega}\right)^2=n^2\lambda^2\\ =\lambda^2\left(B+\frac{C\lambda^2}{\lambda^2-D}+\frac{E\lambda^2}{\lambda^2-F}\right)\\ \left(\frac{A}{\omega}\right)^2(\lambda^2-D)(\lambda^2-F)=\lambda^2(B(\lambda^2-D)(\lambda^2-F)+C\lambda^2(\lambda^2-F)+E\lambda^2(\lambda^2-D)) $$ Is it $\omega$ in the interval $[0.2,2]$, or is $\lambda$ in that range? $$(\lambda^2)^3(B+C+E)-(\lambda^2)^2\left(BD+BF+CF+ED+(A/\omega)^2\right)+\lambda^2[BDF+(A/\omega)^2(D+F)]-DF(A/\omega)^2=0$$ There is an exact solution of this cubic that will give you $\lambda^2$, but it involves the cube-root of complex numbers. (look up Tartaglia's formula) A fair approximation is $$(A/\omega)^2=2.3\lambda^2+0.08-(\lambda^2-2.5)^2/100$$

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  • $\begingroup$ It's $\lambda$ in that range. $\endgroup$
    – WolfgangM
    Commented Oct 12, 2015 at 9:33
  • $\begingroup$ Thank you for the explanation! $\endgroup$
    – WolfgangM
    Commented Oct 12, 2015 at 13:09

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