Complex refractive index. > A linearly polarized wave with the vacuum wavelength $\lambda_0$ falls perpendicular on a material with the complex refractive index $\bar{n}=1.5+i\cdot 0.15$ with that wavelength. > > After what distance does the intensity fall to $1/e$th of the original value? I'm going through all the exercises I can find in my textbook as a preparation for my upcoming exam and I'm stuck on this one for now. I tried looking up all the possible chapters connected to this problem but I can't seem to come up with an approach on solving this. I hope someone could me out here. Or am I missing something and there is actually a general way of solving this type of problem?

Assuming no dispersion, the electric field amplitude of the wave in the initial vacuum is

$$E(z) = E_0 e^{i 2 \pi (z - c t)/\lambda_0} $$

so that the original value of the intensity is $|E_0|^2$.

The (intensity) transmission coefficient into the material is

$$T = 1-\left |\frac{\hat{n}-1}{\hat{n}+1} \right |^2$$

The intensity of the field at a distance $z$ into the material is given by

$$T |E_0|^2 e^{-2 \operatorname{Im}{\hat{n}} 2 \pi z/\lambda_0} $$

Thus, the distance $z_0$ at which the intensity falls to $1/e$ of the original value is

$$2 \operatorname{Im}{\hat{n}} 2 \pi z_0/\lambda_0 = \log{T}$$