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I'm trying to make sense of how the fresnel equations apply to metals. Here are a few of the things I believe I understand:

  • All reflections of light occur specularly. When an object appears to reflect diffusely, this is because at a microscopic level, the surface is rough and so the light is being specularly reflected in many misaligned directions.

  • For dielectrics, the refracted light that is transmitted into the material can continue propagating. If the atoms of the medium do not resonate with the frequency of light, the EM waves can propagate through and the medium is transparent to that wavelength.

  • For dielectrics, if the medium has an irregular crystal structure with internal surfaces (such as grain boundaries), you'd get secondary fresnel reflections/refractions resulting in subsurface scattering. If the scattering causes the light to be reflected back out of the material very close to the surface (that is to say, it doesn't penetrate deep into the material), then this appears like a diffuse reflection.

  • For metals (conductors), there are so many free electrons that can oscillate from incident EM waves, they any light transmitted into a metallic medium is immediately absorbed. This is why metals only exhibit specular reflection with no diffuse component (there is no change for even shallow subsurface scattering. The surface of the metal, can of course, still be rough)

This seems to make sense, however I have also read other sources stating that metals have no refracted beam at all. Is this meant as short-hand for saying that the refracted light is immediately absorbed by the metal? Or does it actually mean that the fresnel equations do not apply?

Further, why is it that metals tint the colors of their reflections while dielectrics do not? Is it that a given metal (such as gold) has a structure which will absorb specific wavelengths and not reflect them at all? (This also feels like a deviation of fresnel's equations, as I did not think them to be wavelength dependent)

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This seems to make sense, however I have also read other sources stating that metals have no refracted beam at all. Is this meant as short-hand for saying that the refracted light is immediately absorbed by the metal? Or does it actually mean that the fresnel equations do not apply?

This isn't true, the Fresnel equations do apply to metals. In particular, light can pass through metal films that are sufficiently thin. It's hard to comment further on the claims you speak of without context.

Further, why is it that metals tint the colors of their reflections while dielectrics do not? Is it that a given metal (such as gold) has a structure which will absorb specific wavelengths and not reflect them at all? (This also feels like a deviation of fresnel's equations, as I did not think them to be wavelength dependent)

Note that the Fresnel equations include the refractive index of the medium (or equivalently, the wave impedance in the medium). The normal reflectance from a smooth metal surface can be written as $$ R_\text{normal} = \left|\frac{\bar n - 1}{\bar n + 1}\right|^2 = \frac{(n-1)^2+\kappa^2}{(n+1)^2+\kappa^2}$$ where $\bar n$ is the complex refractive index, and $n$ and $\kappa$ are its real and imaginary components. These parameters are frequency dependent for metals. Their variation in real metals can be quite complicated, but a commonly observed feature is that below a certain plasma frequency, the conductivity of the metal is high, leading to $n < 1$ and large $\kappa$, and hence high reflectance according to the equation above.

Above the plasma frequency, $n$ approaches $1$ and $\kappa$ is small (sufficiently thin metal films are therefore approximately transparent), leading to a small reflectance. A consequence of all this is that gold has high reflectance at low frequencies (reddish colors) and lower reflectance at high frequencies (bluish colors), resulting in its characteristic color.

Dielectrics can also have a similar tint in color if they have a resonance frequency in or close to the visible spectrum, but it may not be as easy to discern especially if they are not opaque.

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