Complex refractive index and absorption I have a quite plain and simple question: Why is the imaginary part of the refractive index negative? I read it is to allow a positive sign in case of absorption. But how/why? I don't see why it should be positive when absorbing? What does the sign change?
 A: The refractive index of a material tells you how to get the wavevector $k$ from the angular frequency $\omega$, via the dispersion relation
$$
k=n\frac{\omega}{c},
$$
which directly determines the relationship between the spatial and temporal dependence of a plane wave
$$
f(x,t)=e^{i(kx-\omega t)}.
$$
If the refractive index is complex, then the wavevector will be complex, which means that what used to be an oscillatory imaginary exponential over position now contains, additionally, a decaying real exponential:
\begin{align}
f(x,t)
&=\exp(i((k_\mathrm{re}+ik_\mathrm{im})x-\omega t))
\\&=e^{-k_\mathrm{im}x}\exp(i(k_\mathrm{re}x-\omega t))
.
\end{align}
Here the plane wave $f(x,t)=e^{i(kx-\omega t)}$ represents a wave that is travelling to the right, with power flowing from negative $x$ to positive $x$ (and some power getting absorbed), so we expect there to be less signal the more positive that $x$ gets, and this requires the exponential dependence $e^{-k_\mathrm{im}x}$ to be decaying, i.e. it requires $k_\mathrm{im}=\frac\omega c\mathrm{Im}(n)$ to be positive (or at least nonnegative). 
