Are Fresnel equations valid in conductors?

Question

The derivation of Fresnel equations assume no surface charge density or current density is present at the interface, hence making the normal and tangential components of D and H continuous, respectively. I understand this is the case for dielectrics, but what about conductors? After deriving Fresnel coefficients some books (ex: Introduction to Optics by Pedrotti) present the complex permittivity of conductors and then plug this expression in Fresnel coefficients. I would like to understand why this is correct due to the fact that Fresnel coefficients where derived only for the case where sigma = 0 and j = 0 at the interface, in the first place.

Answer 1

I’m not sure which derivation you’re looking at, but (a) the tangential component of $D$ is virtually always continuous (absent a static charge density which simply contributes a static field), and (b) including a current density poses no problem. Take a look at Ampere’s Law (time-harmonic approximation): $$ \nabla\times H = J + i \omega D $$ It’s clear that the free current and displacement current terms both contribute to the curl of H, they just have a relative phase shift: $$ \nabla\times H = (\sigma + i \omega \epsilon)E $$ So we can bundle $\sigma$ (the conductivity, not the surface charge density) into $\epsilon$, giving it a complex value, but nothing else about Ampere’s Law changes. Therefore, the same derivation for the Fresnel coefficients applies in the case of a complex $\epsilon$.

Just in case you don’t believe your book, trust that the validity of the Fresnel coefficients for complex-valued $\epsilon$ has been empirically verified countless times over decades. I have used it myself to calculate the transmission spectra of thin metal films, which I used to successfully model my data.