# Why can you have shiny black objects?

Knowing black is supposed to be the "color" (I don't want to get into the color/hue/shade debate, please) that absorbs light. how does one manage to have shiny black surfaces? I know about "gloss black" versus "matte black" finishes, but shouldn't the light passing through the gloss (if they didn't pass through the gloss, you wouldn't see the black, right?) be absorbed by the underlying black object? Then there are black gemstones like jet and opal.

How do black objects shine?

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I guess a black object isn't actually black just a very very dark grey. As complete black is a complete lack of light being emitted or reflected. – Jonathan. Nov 10 '10 at 23:54
Every smooth surface is glossy. Rough dark surfaces are 'darker' because of multiple reflections. – Piotr Migdal Nov 11 '10 at 0:27
@Jonathan, an object can still be "black" (absorb completely in the visible spectrum) and reflect light. – ptomato Nov 11 '10 at 7:16
@ptomato, good point ;) – Jonathan. Nov 11 '10 at 7:46

At any surface (at least one which has a different index of refraction from air) some light is reflected, depending on the angle at which the light hits the surface and the polarization of the light; the Fresnel equations will tell you what fraction of your light is reflected and what fraction is transmitted. When you see a black object "shine", you are seeing the reflected light. But since the object is black, all the "transmitted" light is simply absorbed.

The difference between a matte black and a gloss black finish is one of index of refraction, I guess, and possibly of rough/smoothness.

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A perfectly black body is not something you see on your daily life, so when you see a "black object" it's actually a almost black object (black enough, for our perception) but it is reflecting some light, wich we are able to perceive as specular.

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Ptomato's answer is right. I just wanted to add that many glossy black objects in daily life reflect light because they have a thin transparent layer on top. If its index of refraction is high and its surface is smooth, you have a glossy surface. Underneath that, you have your black (and possibly rough) material. This is the case in many cheap plastic objects.

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"Underneath rough" is is a funny concept. – Georg Nov 19 '11 at 17:42
@Georg It is very well possible to have a rough surface and cover it with a thin layer of, say, polymer. This polymer can have a smooth upper surface. I don't see what's funny or why my answer is not relevant. – Arnoques Nov 19 '11 at 18:56
When the former surface is covered, it is no longer surface. Therefore that ex-ruoghness is no longer relevant. I can "imagine" a lot of rough buried surfaces in the depth of any material. BTW, did You notice tha age of this thread? – Georg Nov 19 '11 at 19:31
Well, maybe the word "surface" was not very clear. Is "interface" better? You have a boundary between two regions of space that have very different optical properties. In the case I was talking about, this means that light coming from this interface will be able reach your eyes and you can see a black object underneath the glossy smooth surface. – Arnoques Nov 19 '11 at 19:36
""of space that have very different optical properties."" That is not really possible in plastics and laquers, with the exeption of the interface between some pigments and the binder. But this is present always in some laquer with such pigments. That is not underneath rough, that is eg a film of laquer containing eg titanium dioxide crystals. – Georg Nov 19 '11 at 19:56

Reflection of radiowaves is the result of electrical currents or dielectric charge/discharge induced on the surface of conductors, dielectrics and everything in between. For visible light the model is still valid, but may be more affected by atomic/molecular properties of surface.

So black glossy dielectric partially reflects (horizontally) polarized white light and absorbs the rest of energy. The proportion depends on angle. Black matte dielectric or white matte conductor (say PLatinum with porous surface) has more complex path with more than 1 changes of light path, so more energy is absorbed. The purely metal (white) conductor with very complex surface can look like completely matte black.

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As ptomato points out, even a "very black" object reflects some light. If the surface is rough (at a small scale), the light is reflected off in all directions, and the object appears, matte. If the surface is smooth, then all of the reflected light will bounce in the same way, so that you can see the reflection of the light source; this reflection will be easier to see against the black surface than it would be on a similar white (and shiny) object.

If my model is correct, then shiny black should look shiny under direct light, but shiny black and matte black should look very similar under indirect light (in monotonous surroundings).

Also, matte black objects are shiny when wet, so if my model is correct then that shine must be from the liquid surface. So I predict that the color of the shine will be the color of the liquid. For example, if a white light shines on a matte maroon object, if you wet it with water the reflections will be white, but if you polish it the reflections will be maroon.

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Nope, even if you polish it, the shine is always the color of the light source. – ptomato Sep 15 '11 at 17:40