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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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  • $\begingroup$ 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. $\endgroup$ – Jonathan. Nov 10 '10 at 23:54
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    $\begingroup$ Every smooth surface is glossy. Rough dark surfaces are 'darker' because of multiple reflections. $\endgroup$ – Piotr Migdal Nov 11 '10 at 0:27
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    $\begingroup$ @Jonathan, an object can still be "black" (absorb completely in the visible spectrum) and reflect light. $\endgroup$ – ptomato Nov 11 '10 at 7:16
  • $\begingroup$ @ptomato, good point ;) $\endgroup$ – Jonathan. Nov 11 '10 at 7:46
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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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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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  • $\begingroup$ "Underneath rough" is is a funny concept. $\endgroup$ – Georg Nov 19 '11 at 17:42
  • $\begingroup$ @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. $\endgroup$ – Arnoques Nov 19 '11 at 18:56
  • $\begingroup$ 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? $\endgroup$ – Georg Nov 19 '11 at 19:31
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    $\begingroup$ 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. $\endgroup$ – Arnoques Nov 19 '11 at 19:36
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    $\begingroup$ ""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. $\endgroup$ – Georg Nov 19 '11 at 19:56
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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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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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The light that you see reflected off of a solid surface is a mix of specular reflection and diffuse reflection. For most surfaces, the apparent color of the object is due almost entirely to the diffuse component.

I can't explain the mechanisms behind the two types of reflection, but when you see a shiny black object, "shiny" means the surface gives a strong specular reflection, and "black" means it only gives a very weak diffuse reflection.

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As I was reading other people's answers I noticed that they were very... complicated. A simpler answer would be that if you have a shiny black object, the shine could be coming from a thin transparent coating, not the black object itself.

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Simply, shiny 'black' objects are not black. They may be close to black and will appear black compared to other colours, but, as soon as they reflect some electromagnetic radiation (even a few photons), these surfaces have colour. The difference between a matte black and a shiny black with radiation is that when the photons directly hit a polished object, they are more likely to be absorbed than if the hit is at an angle. The difference between a matte finish and a shiny finish is that, no matter the angle, the light will always hit a part directly as matte is just loads of smooth surfaces facing different directions.

You can get into refraction indices and the Fresnel Equations, but there is really no point as it is just a mathematical way of saying the same thing: matte black is just a polished surface that is always at the correct angle to absorb maximally and, when it does reflect, it can reflect into itself as the photons are obviously small enough to collide with multiple 'bumps'.

Emission of matt surfaces is purely that if you consider a bumpy surface, the surface area is higher (like the villi in the gut), meaning that a larger amount of radiation can be emitted.

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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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    $\begingroup$ Nope, even if you polish it, the shine is always the color of the light source. $\endgroup$ – ptomato Sep 15 '11 at 17:40
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Black is all coulers, not a shade, when learning about the spretrem of light from the sun, And haw objects only can reflect there own couler of its surface but with a different couler that is not of there the object can not reflect other coulers there for no light bounces of and we do not see a red object but a black object that doesn't repct any light at all. But we all know black object reflect of light for why we can see the couler that it is, witch cpuld mean black absorbs all coulers from the sun rather than one or two but all of them, when looking at black objects we can still see that light is reflected and we can see the object and know the colour , what we do know is from what we witness, and the only way for black to be able to be seen and have a reflective surface is that all Spectrum of light is Absorbed then revealing the object to the naked eye and light can bounce of it becouse it is all colours from Spectrum of light,rather being just a black blob or object that dose not reflect light. And when a object becomes black becouse it can only reflect its own Colour there for it has its coulée but can not be seen by any other Spectrum of light so it is not a black object like a b object that is black and Reflects light all the other objects.
So yes my Theory is that black is every colour .

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    $\begingroup$ Hi and welcome to Physics SE! I don't see how this answer addresses the question. Also, the text is hard to read and full of misspellings. You can check How do I write a good answer? for general advice on providing better answers. $\endgroup$ – stafusa Sep 22 '17 at 18:05
  • $\begingroup$ @stafusa - Spinal Tap's “Smell the Glove” album cover was ultimate black. As guitarist Nigel Tufnel pointed out: "how much more black could this be?" and the answer is "None. None more black." $\endgroup$ – Peter4075 Apr 16 '18 at 5:54

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