Question: If I understand correctly, an object's color is determined by the optical frequency which it does not absorb. So an apple appears red because it only reflects red. My question is, if this is true, why do we get mirror effects from dark surfaces? For example, one can often see multicolored reflections from a dark vehicle, or a dark computer screen. If an object is black, it should absorb all visible frequencies, so how come we can see colors (sometimes even white light) reflecting from a dark object as if it were a mirror?

My attempt: optical frequencies are not my specialty, but from thinking about it, the only thing I can conclude is that these objects are not truly "black". Rather, they are some combination of colors that seems almost black until white light shines on it. If this is true though, it also raises the question of a truly black object being possible to fabricate.

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    $\begingroup$ A truly black surface wouldn't look like a surface, at all. In order for us to tell that there is a surface, it has to reflect enough light so we can make use of our stereoscopic depth perception. cs.virginia.edu/~cab6fh/renderer/images/ctemp5k.png shows the rendering of a dark sphere. A completely black sphere would look like a perfect circle but have the same shadow! I think that would look extremely disorienting to us. The shiny appearance of bodies is due to specular reflection: en.wikipedia.org/wiki/Specular_reflection. $\endgroup$
    – CuriousOne
    Commented Jun 14, 2016 at 22:45

2 Answers 2


If an object is black, it should absorb all visible frequencies

This is too simplistic. Objects appear black if they absorb "most" of the light that hits it. We can't make materials that absorb 100%. What happens to that small fraction that is not absorbed is critical to how you perceive it.

Most surfaces we encounter scatter reflected light well. Light that is not absorbed leaves in directions that are not strongly correlated with the incidence angle. That's why you don't see your reflection in a wall.

But if the surface is smooth, then the portion of the light reflected from there appears like a mirror.

A dark rock might absorb 90% of the light incident, and reflect only 10%. If it's rough, you would probably call it "black". Some light is reflecting, but because it is diffuse, you don't notice it. Now cut and polish one surface of it smooth and the same amount of light is reflecting, but because it can image the light that is falling on it, you notice it.

  • $\begingroup$ It's not simplistic, it's the truth: black is (by definition) the complete absorption of light. Maybe what is simplistic is our classification of anything "dark enough" as black. But I see your angle. Good point about the surface geometry! I'm quite familiar with smooth/rough scatter, I just didn't think to bring that into this problem. +1 for the insight, thanks! $\endgroup$
    – M Barbosa
    Commented Jun 14, 2016 at 23:38
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    $\begingroup$ @MBarbosa If the definition of black is the complete absorption of light, then absolutely nothing in the universe is black. That's not a very useful definition. $\endgroup$
    – Mark H
    Commented Aug 23, 2016 at 20:27

Part of the light reflected from an object comes from the surface itself and is due to the abrupt change in refractive index at the surface. The refractive index of a glass plate changes a bit with color of the light, but not very much. Therefore the reflection from the surface is relatively free from color. Let us call the variation of index as you move from air to glass $f(x)$, where $x$ points along the normal to the glass surface. It is interesting that if one could make the refractive index change smoothly from air to glass, then the surface reflection would be much reduced. The residual reflection is dependent on which derivative of $f(x)$, $\frac{d^nf(x)}{dx^n}$, is discontinuous.

Normally, the value of $f(x)$ itself is discontinuous and we get a strong reflection, about $4\,\%$ for ordinary glass. If $f(x)$ is continuous, but not the first derivative, the reflection is reduced, and if only the second derivative is discontinuous, the reflection is reduced even further. The so-called "moths eye" anti-reflection coating consists of a layer of microscopic cones pointing out from the surface. This makes for a gradual change of refractive index, and is a very efficient anti-reflection coating.


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