Complex Refractive Indices, Absorption, and Transparency

Question

A complex refractive index is defined as $n = n_0 + \kappa$ where $n_0$ is the "standard" refractive index, and $\kappa$ is the optical extinction coefficient. The optical extinction coefficient "indicates the amount of attenuation when the electromagnetic wave propagates through the material". The larger the value, the more light is absorbed by the material.

The fresnel equations tell us how much light is reflected, vs how much is transmitted into a material. Because metals have a high $\kappa$ value, this means that they absorb light really well and so light cannot propagate through the metal. Therefore, metals cannot be transparent and any EM waves that enter them are absorbed (I understand they can be transparent to higher frequencies, as the refractive indices are dependent on frequency, but for right now I'm focusing just on ~visible light).

Naively, this would lead me to believe that if a material doesn't have a high $\kappa$ value, then it must allow light to propagate through it reasonably well. But this sounds like it means the material should be transparent, which obviously isn't always the case. Many compounds do not have a complex refractive index (or at least, have a very tiny imaginary component), and yet are completely opaque.

From some reading, it seems like what's happening is that the light is scattered in the material somehow. But shouldn't the light still eventually escape? And why would the material cause light to scatter to begin with? I've seen it suggested that imperfections in the material, such as the grain boundary between two grains of the material, form an interface which cause fresnel reflection/refraction to occur again. This seems like it could be reasonable but it feels incomplete, as it'd suggest a perfect crystal would be transparent then. And it still wouldn't explain how the light ends up getting absorbed by the material anyways.

Answer 1

A wave propagating in a uniform medum in the $+z$ direction can be expressed as (the real part of) $$E = E_0e^{i(\omega t - nk_0 z)}$$ where $\omega$ is the angular frequency, $k_0 = 2\pi/\lambda$ is the wavenumber in vacuum and $\lambda$ is the wavelength in vacuum. With $n = n_0 - i\kappa$ (the sign of the imaginary component depends on convention), this becomes $$E = E_0e^{i\omega t}e^{-in_0 k_0 z}e^{-\kappa k_0 z}.$$ This represents a wave with wavelength $\lambda/n_0$ that decays in amplitude along $z$. Its intensity (given by the real part of the Poynting vector) is proportional to the modulus squared of the wave amplitude, that is, $$S=S_0e^{-2\kappa k_0z}=S_0e^{-\alpha z}=S_0e^{-z/\delta}\propto|E|^2$$ where $\alpha = 4\pi\kappa/\lambda$ is known as the absorption coefficient, and $\delta=1/\alpha=\lambda/(4\pi\kappa)$ is the penetration depth. $\delta$ is the characteristic decay length of intensity: its significance is that moving a distance $\delta$ in the $+z$ direction causes intensity to drop by $1 - 1/e=63\%$. If the thickness of the medium is more than a few times $\delta$, it will not permit waves to pass through, so it will be opaque.

For the example you've provided ($\text{Fe}_2\text{O}_3$ with $\kappa = 0.0288$ at $\lambda = 633\text{ nm}$), we have $ \delta = 1.75\ \mu\text{m}$, so a film that is thicker than a few μm will be opaque at this wavelength. Note that $\delta$ is smaller for shorter (bluish) wavelengths, where $\text{Fe}_2\text{O}_3$ is consequently more opaque. In summary, it is not enough to look at how large or small $\kappa$ is relative to unity to tell whether a film will be opaque. Instead, we must compare the penetration depth to the film thickness. $\kappa^{-1}$ can loosely be interpreted as the number of wavelengths of thickness it takes for the film to be completely opaque.

All this discussion was for a uniform medium. In the case of a non-uniform medium, such as a material that has impurities or defects, is porous or is in powder form, light is also attenuated by scattering, which can also make a material opaque. For $\text{Fe}_2\text{O}_3$ powder for instance, incident light is scattered every which way in the powder, where much of it is forced to travel long distances and get absorbed. The part that makes it out consists predominantly of wavelengths with longer penetration depth, so $\text{Fe}_2\text{O}_3$ powder appears reddish in color.

It's not true that "only metals are ever reported with complex valued refraction indices". Take a look at RefractiveIndex.INFO, which lists the extinction coefficients of many materials, metallic or otherwise. The extinction coefficient can be quite small, but not always. All dielectrics have characteristic resonance frequencies near which they absorb, and thus have non-zero extinction coefficients.

Finally, sometimes the optical response is described/reported in terms of complex permittivity $\epsilon = \epsilon' - i\epsilon''$ or susceptibility $\chi = \chi' - i\chi''$ instead of refractive index (once again, the signs of the imaginary components depend on convention). These are directly related via $n = \sqrt\epsilon = \sqrt{1 + \chi}$.