I understand that the index of refraction is complex and can be expressed as such: $ \widetilde{\eta} = \eta + i \kappa $. However I’ve been searching for a bit and I am unable to find the derivation of why the imaginary part of the refractive index in semiconductors is as follows $$k = \frac{\lambda \alpha}{4 \pi}$$ Can someone demonstrate?


1 Answer 1


If $\alpha$ is the attenuation coefficient, such that $|E|^2 \propto e^{-\alpha x}$ it is, by pure identification, the definition of $\alpha$.

Let's write:

$$ E=E_0 \exp\big(i (n+ik)k_0 x\big) $$ where $k_0=\frac{2\pi}{\lambda}$ is the vacuum wave number. You get then: $$ E=E_0 e^{ink_0 x} e^{-k\,k_0\,x}$$


$$|E|^2=|E_0|^2 e^{-2k\,k_0\,x}$$

Hence $\alpha=2k\,k_0= \frac{4\pi\, k}{\lambda}$.

  • $\begingroup$ Thank you very much!! $\endgroup$ Apr 1, 2020 at 15:57
  • 1
    $\begingroup$ If you agree my answer, as your comment suggest, could you please consider to mark it as "accepted " ? $\endgroup$
    – Jhor
    Apr 1, 2020 at 17:45

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