# Reflectance and Reflectance Color of Metals

I do not understand the layman's physics behind the reflectance curve of a metal based on the complex index of refraction. Figure 24 from Background: Physics and Math of Shading by Naty Hoffman graphs the RGB reflectance of copper and diamond. For reference this is the complex refraction of copper, and this is the complex refraction of diamond. And lastly, the higher the ratio of refractive indices at a boundary, the more light is reflected and less refracted. (Note: this last is a preconception so I searched for a supporting source).

Starting conditions:

1. Hoffman, Figure 12: refracted light + reflected light = incident light
2. Hoffman, Figure 16: In metals, refracted light is immediately absorbed and lost.
3. The higher the ratio of index of refraction between materials, the more light is reflected.

Questions:

1. Why do metals have a specular reflectance color when only the extinction coefficient really varies with wavelength, but the extinction coefficient describes refracted light which is lost? In other words, what explains the difference between the complex refraction graph of copper and the reflectance graph of copper?

2. Diamond has such a high F0 specular reflection (0.17) because the real component of its index of refraction is so high. How can copper have a much higher F0 specular reflection (0.95-0.54/RGB) despite having a smaller real component.

3. Can answers to [1] and [2] be derived solely from the complex index of refraction, ignoring the underlying physics?

Notes:

1. I've found a source that states a "perfectly absorbing surface will reflect all light". But that is not how a black body radiation works, so I don't understand this statement.

Your questions on specular reflection from a smooth boundary can indeed be answered by the Fresnel equations (that is the underlying physics!). Plug in the complex refractive index for your materials and calculate with your favorite scientific computing software.

But short of actually going through with the calculations (not that they’re terribly hard, if you’re comfortable with that sort of thing), allow me some comments:

Both real and imaginary components of refractive index, $$n$$, vary with wavelength for metals. Thinking of the imaginary part as an extinction coefficient is only relevant for a wave propagating in a single medium. For reflection at a boundary, both real and imaginary parts combine in a complex way (pun intended!)—just look at the equations. Take the normal-incidence equation for field reflection at a vacuum/material interface: $$r=\frac{1-n}{1+n}.$$ This is the simplest possible case. Is it immediately obvious what $$r$$ will be when you choose some arbitrary complex number for $$n$$? My guess is no. But let’s think about your final point: Let $$n \to \infty$$. Then $$|r| \to 1$$. No surprise! But what about $$n \to -i\infty$$ (“perfect absorber”)? Same thing!

(As an aside, a black body is not a smooth interface with $$n=-i\infty$$. It’s a cavity or extremely rough surface within which light gets trapped and gradually entirely absorbed.)

Keep in mind that when you’re calculating the power reflection, it is the absolute value of the complex field reflection coefficient which is relevant. So, the imaginary part contributes just as much as the real part.

• Wow this is a really intuitive answer from the math side for [3]. For [1]/[2] the "underlying physics" explanations I've read tend like this. Incident EM wave generates an inverse wave via the metal's free electrons that cancels out the incoming wave (1). That's why metals are so reflective. At some frequencies the free electrons are excited to a new band and that's why some colors don't reflect and metals have colored reflections (2). Commented Aug 28, 2021 at 13:40
• Anyway I don't see why an inverse wave along the incident direction would also generate one in the reflected direction. Don't the excited electrons also generate a wave (they're still free electrons...), and what happens when they return to their original state? Commented Aug 28, 2021 at 14:23
• To nitpick, isn't that air/material interface, and $r \to -1$? Also I'm not clear why $R = \lvert r \rvert ^2$. Commented Aug 28, 2021 at 14:25
• Oops, air is 1.0027, vacuum is 1. Anyway this page describes how electrons can refract light. One electron re-radiates light along the incident direction. A group of electrons along a line re-radiates or refracts light at an angle. It behaves almost like diffraction. No explanation for reflection but I would assume it is even more complicated. I find it interesting that it ascribes back-EMF in an inductor to momentum. I had never really thought why the voltage would be 180° out of phase. Commented Aug 29, 2021 at 1:22
• @user19087 to answer your nitpicks, (A) vacuum is the only medium for which n is actually 1. Air’s index is close to 1, but not exactly (see, e.g., mirages). (B) yes, $r \to -1$; I was being fast and loose. Edited. But that sign is only a phase of reflection, and power reflected is the same...which brings us to (C) power is proportional to electric field times magnetic field (read about “Poynting vector”), so power reflection is field reflection squared (actually absolute value squared, which matters for complex refractive index). Commented Aug 29, 2021 at 1:57