Why does the Complex Index of Refraction take part in the Reflection?

I'm aware that in Optics, the complex index-of-refraction $$\eta = n+ik$$ is used, which famously leads to the reflection property at an incident angle, i.e. Fresnel's law:

$$R=\frac{(n-1)^2+k^2}{(n+1)^2+k^2}$$

However, $$k$$ is the absorption (edit: extinction or attenuation) coefficient, indicating a loss of energy when the wave propagates through the medium. Fresnel's law refers to the reflection, so it is unclear to me why the absorption should have any effect on the reflected ray.

1. So, why is it that $$k$$ plays a role in the reflection? (An argument using Maxwell's Equations would be perfect.)
2. Has Fresnel's law (in particular, w.r.t. measuring the reflected ray), ever been validated by an experiment?
• If you follow and believe the derivation from Maxwell's equations for a medium with purely real index, then you are almost done. The derivation is perfectly valid for complex index. Replace $n$ with $n+ik$ and calculate the modulus. Note that your equation for $R$ is only good when the incident angle is zero. – garyp Jun 20 at 23:58
• $k$ plays a role because it changes the phase of the polarization of the medium. The fields just inside the surface are not the same as they are for a pure real $n$, so by the boundary conditions, the fields just outside the surface must also be different. Fresnel's laws are tested billions of times every day in all sorts of optical equipment. If you have an anti-reflection coating on your eyeglasses, you are testing them yourself. – garyp Jun 21 at 0:00

$$k$$ is the extinction coefficient. For metals it is what drive reflection. The cause of the reflection is that the wave cannot propagate in the metal - and therefor is extinguished. Your expression for R is derived from Maxwell's equation by Fresnel in the 19th century. His solution works incredibly well.