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Conductors are opaque because, when hit by a Maxwellian wave, the free charges on their surface create another wave which destructively interferes with the former in the region of space beyond said surface. This doesn't happen in dielectrics; why then are some of them, say, wood or rubber, opaque still?

I would very much appreciate, if possible, an answer that uses classical electromagnetism instead of QM, photons, absorption spectra, etc.

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    $\begingroup$ Bandgaps, as such, only exist in perfect crystals. An amorphous, granular, or otherwise macroporous or matrix material has no reciprocal lattice -- boundary or other effects will dominate the optical properties. Pure dielectric crystals will generally be transparent to photon energies lower than their bandgap. $\endgroup$
    – J...
    May 25, 2018 at 17:14
  • $\begingroup$ Reference : Assumptions and limits of band structure theory $\endgroup$
    – J...
    May 25, 2018 at 17:18
  • $\begingroup$ Are you asking why materials which are dielectric at near-DC frequencies are not transparent at visible frequencies or are you asking whether materials are simultaneously dielectric and transparent at the same frequencies? $\endgroup$ May 25, 2018 at 21:34
  • $\begingroup$ @Eric Towers I'm not sure what DC stands for, but I would say the latter. $\endgroup$
    – user115153
    May 27, 2018 at 21:53
  • $\begingroup$ DC = "direct current", meaning at usual frequencies for conductivity measurements. $\endgroup$ May 28, 2018 at 1:18

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Just because the material doesn't conduct currents on a macroscopic scale, does not mean it doesn't contain any movable charges at all. In fact, as the very name “dielectric” suggests, such a material contains charges which can be to some degree separated – electrons move a bit to one side or the other, never actually bidding their parent atom farewell but still generating a significant field.

When light hits such a material, the electrons are inclined to vibrate in sync with the incomming EM wave. This creates a secondary electromagnetic field of the same frequency, which adds to the original field, much like the larger-scale currents below the surface of a conductor do. Unlike with a conductor, the electrons aren't movable freely so this field can't completely cancel the incoming wave, but it also doesn't pass through completely unimpeded:

  • Some of the energy that the secondary wave gives back is directed not in the original propagation direction but backwards: a bit of the light is reflected. (This also happens for conductors, almost perfectly, which is why metals are shiny.)
  • The secondary wave isn't completely in sync with the incoming one, but generally has a bit of a phase lag (like when you shake a pendulum). As a result, the wave looks “delayed”, as if it has travelled a longer way through the material than it actually has – from an outside point of view that means the wavelength is shortened despite constant frequency. But because the wavefront still has to be in sync everywhere, you'll get refraction.

Both effects are very notable for clear glass: some of the light is reflected instead of passing through, and the light that is passing through is refracted. Mind, for a glass pane the refraction doesn't do much, it just changes the direction a bit when the light enters and then back again to original when it leaves. But for materials that aren't so homogeneous, like styrofoam, you'll get not just one reflection and refraction but instead have lots and lots of microscopic surfaces, all aligned in different directions. As a result, even if each microsurface lets most of the light past, it doesn't actually manage to get very far into the material but is “reflected” in a completely scattered manner. And that's basically what's happening for all materials which look white. Most of them are still somewhat transparent, but if the components are sufficiently fine and refractive, it won't get further than a few micrometres.

For materials that look coloured or even black, something else happens in addition: the vibrating electrons actually take away some of the energy of the incoming wave and don't re-transmit it at all with that frequency, but instead “convert” to other forms of energy – usually, either lower-frequency light (fluorescence) or charge-neutral lattice vibrations (which manifest as heat). Explaining how this conversion works without quantum mechanics is problematic, but you can basically picture it as a damped vibration – there is “friction” in the oscillation. Often, this only happens to a significant level in a particular frequency band, due to molecular resonances; that's then how colour comes about, because different light-frequencies are damped to different fractions.

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  • $\begingroup$ „... material contains charges which can be to some degree separated – electrons move a bit to one side or the other, never actually bidding their parent atom farewell but still generating a significant field.” Isn’t this explanation applicable to sharp edges and the phenomenon of light refraction behind such edges? $\endgroup$ May 25, 2018 at 20:37
  • $\begingroup$ I enjoy this answer because it doesn't just answer the question about transparency, but also explains things that appear to be colored or black or white. $\endgroup$ May 25, 2018 at 23:17
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You cannot totally avoid quantum mechanics, but it may suffice to say that reflections by free electrons are not the only way to prevent transmission. Any situation where light can promote an electron from a low energy state to a higher one will cause absorption, regardless of DC conductivity. Or even with little absorption, massive scattering of light by many small particles or many disordered interfaces can also prevent transmission. A couple of quick examples:

  1. Semiconductor with band gap smaller than the light energy (e.g. silicon, graphite).

  2. Materials made of many small scatterers (e.g. wood, paint, skin).

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    $\begingroup$ Paint is a good example. Titanium Dioxide is a transparent crystal. When crushed to a fine powder, it makes white dye for paint, with excellent "hiding" properties. The paint base is transparent, by itself also. It works because the refractive index of TiO2 is so high relative to the paint base that the powder still scatters light. $\endgroup$
    – Kaz
    May 25, 2018 at 15:27
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    $\begingroup$ Indeed. Snow is white, but it is the same stuff as ice and as water: transparent. Wood is cellulosa (cellophane) with many cell walls that scatter the light. $\endgroup$
    – user137289
    May 25, 2018 at 23:32
  • $\begingroup$ And this incidentally also gives the OP the answer to the question "why is paper white?" as a bonus! $\endgroup$
    – Vendetta
    Jun 5, 2018 at 16:38

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