# Imaginary part of semiconductors index of refraction

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?

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: $$E=E_0 e^{ink_0 x} e^{-k\,k_0\,x}$$

and

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

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

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