# How is the imaginary part of angular frequency omega related to the imaginary part of the refractive index?

I'm trying to find the attenuation constant (leak rate) $\alpha$ from the imaginary part of the refractive index of a lossy material. I have the eigen frequency $\omega$ for my structure which has an imaginary part. If I understand correctly, this imaginary part expresses that the medium is lossy.

How can I use this this $\omega$ to find the (attenuation coefficient) $\alpha$ ?

Losses are usually taken into account by considering the complex relative permittivity $\epsilon_r$ where the real part determines the polarization of the medium, while the imaginary part is related to to the losses in the material. This description is equivalent to the classical optical description in term of complex refractive index. As in non magnetic materials the refractive index is the square root of the dielectric constant, the complex refractive index can be written as \begin{equation} N=n-jk=\sqrt {\epsilon^{'}_r-j\epsilon^{''}_r} \end{equation} where n is the usual real refractive index, while the quantity k is the extinction coefficient. It is possible to show that the extinction coefficient and the optical absorption are dependent on each other, and \begin{equation} \alpha_o=\frac{4 \pi k}{\lambda_c} \end{equation} where $\lambda_c$ is the free-space wavelength. Please note that the optical absorption $\alpha_o$ is different from the electrical attenuation constant $\alpha_e$ since $\alpha_o=2 \alpha_e$. Both real and imaginary part of the refractive index depend on the angular frequency. The absorption coefficient is therefore described as the reciprocal of the depth of penetration of radiation into a bulk solid, i.e., it is equal to the depth at which the energy of the radiation has decreased by the factor of $e^{-\alpha x}$. If you know the propagation constant as a function of $\omega$ you can compute its imaginary part and then determine $\alpha_o$ as a function of the angular frequency.