I have been trying to demonstrate that for a refractive index $n=n_R+in_I$ we have $$n_I=-\frac{Ne^2\gamma\omega}{2m\epsilon_0[(\omega_0^2-\omega^2)^2+\gamma^2\omega^2]},$$ $$n_R=1+\frac{Ne^2(\omega_0^2-\omega^2)}{2m\epsilon_0[(\omega_0^2-\omega^2)^2+\gamma^2\omega^2]}.$$ I start by using the fact that the susceptibility is given by $$\chi=\frac{Ne^2}{m\epsilon_0(\omega_0^2-\omega^2+i\gamma\omega)}.$$ Considering the polarisation of the dielectric as $\vec{P}=\epsilon_0\chi\vec{E}$ and $n^2=1+\chi$ $\implies n\approx1+\frac{\chi}{2}$, we get $$n=1+\frac{Ne^2}{2m\epsilon_0(\omega_0^2-\omega^2+i\gamma\omega)}.$$ Can anyone see where to go from here?


1 Answer 1


You are almost done, you just have to separate the real and the imaginary part of $n$. In general,

$$ \frac{a + ib}{c + id} = \frac{ac + bd}{c^2+d^2} + i \frac{bc - ad}{c^2+d^2} \;. $$

In your case, $b=0$. If you use this formula, you will get the correct results almost immediately.


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