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My understanding is that all mediums have a complex iindex of refraction where the real component is the "standard" index of refraction, and the imaginary component is the extinction coefficient indicating how well light is absorbed by the medium.

When light is incident on an interface between two mediums with different refractive indices, some of the light is specularly reflected, and some is refracted into the surface. For materials with near zero extinction coefficients, this refracted light is able to continue for awhile and thus the material is said to he transparent. For materials with structural irregularities (such as grain boundaries in its crystal structure) you get secondary reflections/refraction causing some of the light to be reflect back out the surface of the material diffusely. This interpretation seems to be supported by much of what I've read, including this diagram found in the "Mechanism" section of the page on Diffuse Reflections:

enter image description here

Metals have a very large absorption coefficient (I believe because of how conductive they are) and so visible light is barely able to penetrate metals, therefore they have essentially no subsurface scattering at visible wavelengths and so appear to have only specular reflections.

But my question is, this naive understanding seems to imply a perfect crystal with a non-zero extinction coefficient, should appear perfectly black. Because no irregularities would exist to reflect light back out of the surface. Instead, light would enter the material and simply be absorbed.

Would a perfect crystal not exhibit any diffuse reflections, and thus appear black? Or can light be scattered by the atoms themselves at random? If the latter is true, why are crystal irregularities and internal boundaries so often used to depict secondary reflections/refraction in the process of subsurface scattering?

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  • $\begingroup$ Do you assume the crystal is very large (or strongly absorbing at all visible wavelengths)? If not, a significant fraction of the light passes through the crystal and exits through its opposite face. $\endgroup$
    – A. P.
    Commented Feb 26 at 8:17
  • $\begingroup$ A perfect crystal is an infinite crystal in all directions. But, if there's a boundary, light will be scattered on this boundary, so you'll at least get specular reflection, which will possibly be colored due to uneven reflectance spectrum. $\endgroup$
    – Ruslan
    Commented Feb 26 at 9:22

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One important element about crystalline structures is what they represent:

Conduction/Valence band

On a bigger picture, the crystal structure tells you, among other things, what the different bands in a material look like, which tells you how possible it is to transfer electrons from the valence band (outer electron shell) to the conduction band (where the electrons can move inside of the material). Important to note is that band structure derives from the crystalline structure and not the other way around.

This is a more general picture, but if you want to understand how electrons actually interact with external fields, you have to look into the actual structure of the crystals, which roughly tell you what are the zones in a material where translational symmetry occurs: This means that if you move by a specific factor ($\pi/a$ or the first Brillouin zone) the solution of the Schrödinger equation is periodic and therefore can be simply moved by that factor.

Uni Kiel semiconductors theory

A statement such as "perfect crystal" in the context of crystalline structure is not that useful in the context of light-matter interaction, since even in the case where there were no defects in the crystalline structure, the energy diagram of each crystal will give you a hint on which electromagnetic radiation will be absorbed in the material. If the energy of the impinging electromagnetic radiation is not within the allowed energies (remember that energy bands are quantized) radiation will be reflected (or scattered).

Would a perfect crystal not exhibit any diffuse reflections, and thus appear black?

Not necessarily black, just opaque.

Or can light be scattered by the atoms themselves at random?

I would not say at random, rather it depends on which bands allow for transmission/absorption, and the rest of the light will be reflected or scattered.

If the latter is true, why are crystal irregularities and internal boundaries so often used to depict secondary reflections/refraction in the process of subsurface scattering?

I think that is typically because the irregularities are located within the first Brillouin zone and therefore within the coherence length of impinging external radiation, and therefore they do affect how light behaves and change the energy band diagrams. Also important to note is that visible light is far too large to be comparable to the much finer crystal structure of a material.


I went through the Wikipedia article on diffuse subsurface light transport and I think that the issue here is that the explanation from Wikipedia is incomplete since the reference it takes is computational light transport and macroscopic coefficients (such as refractive index, absorption, etc.). The diagram you shared is not necessarily a complete picture that includes the crystal structure and I think there is some distance between the crystal structure and what the article on diffuse reflection means to explain. In my opinion when "subsurface transport" is not really about the crystal structure, rather it is about the macroscopic, visible light-measurable properties of surfaces, and the subsurface transport is more a quantitative measure of the behavior of materials with visible light, which as I mentioned earlier, is much larger than the crystal lattice.

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  • $\begingroup$ Are there any textbooks you might recommend for better understanding light and matter interactions? $\endgroup$
    – Chris Gnam
    Commented Feb 26 at 13:30
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    $\begingroup$ Specifically for semiconductors I highly recommend Yu/Cardona "Fundamentals of semiconductors", it is my go-to for semiconductors, and some other explanations about group theory which is not as dense as other material. For more in-depth I would go for the Rodney Loudon - The Quantum Theory of Light -- good introduction to get started with the notation and so on. $\endgroup$
    – ondas
    Commented Feb 27 at 10:38

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