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I know the fact that a material appeared as black when all wavelength photons were absorbed. I also know that opaque object is the result of reflection and scattering of light wave. Is it right? However, how can I explain about opaque black object? If all waves are absorbed, is there still available light at the surface of the material to reflect or to scatter? If there still light to reflect and to scatter, what is ratio of light in absorption and reflection/scattering? Is there really black object exist? Obviously, I have very limited understanding of physics.

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  • $\begingroup$ Why can't you think it as an object which transforms all light into heat? $\endgroup$ – jinawee Jun 4 '14 at 19:27
  • $\begingroup$ I think it will ease your mind to think of a black object as a very dark shade of grey. $\endgroup$ – Davidmh Jun 4 '14 at 21:45
  • $\begingroup$ @Davidmh I prefer "A Whiter Shade of Pale" :-) $\endgroup$ – Carl Witthoft Jun 5 '14 at 13:33
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What you 'know' is misleading. The object will appear black to our eyes if it absorbs all photons in the visible portion of the spectrum, and further is not emitting significant numbers of photons (due to the Planck/blackbody radation laws, it's always emitting a few visible photons).

An object is opaque because of scattering, or because of absorption. All "opaque" means is that light doesn't travel through the object. Thus, an object which appears to our eyes to be black just isn't allowing any photons in the visible range to reach our eyes. In fact, if (thought example), a perfectly reflecting mirror is placed in a pitch-black room and a bright light is aimed at the mirror at an angle such that the reflected beam doesn't reach us, we'll still see the mirror as being black.

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  • $\begingroup$ But the mirror will start emitting light according to the Planck spectrum after a while, because it is impossible to reflect perfectly. $\endgroup$ – auxsvr Jun 5 '14 at 12:19
  • $\begingroup$ @auxsvr Yes, as I said in the first paragraph-- there will be some, but very few, visible photons emitted. $\endgroup$ – Carl Witthoft Jun 5 '14 at 13:33
  • $\begingroup$ I thought that it is black if material absorb all photons and no remaining at surface because there is no photon. OK, I think I'm not sure about "reflection", but if there is no photon/light, what is "scattering" from surface? I'm confused. $\endgroup$ – user47546 Jun 5 '14 at 18:36
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A surprisingly efficient absorber is a stack of razor blades. Nowadays the paint scrapers and utility knife refills at the hardware store tend to come individually wrapped in cardboard, but the refills for X-Acto knives come in packages of five or ten blades stuck together in a package with a little light machine oil to keep them from sliding apart.

three x-acto blades

Here's a photo of three blades, stuck together and held upright using a binder clip. You can see from the broad side on the left that the finish on the blades is quite smooth, almost mirrorlike. But where the three razor edges run parallel to each other the photo is quite dark and featureless. With a wider stack it really looks like remarkably dark black velvet. (I advise you to resist the urge to check with your finger whether it feels soft; it doesn't.)

How do we get from the excellent reflectivity of a single blade to the near-total absorption of the stack? It's due to the small angle between adjacent surfaces: in order for a light ray to turn around, it must make many, many reflections; a small loss at each reflection translates to a very large overall absorption.

light path entering razor blade stack

I have heard that stacks of razor blades were frequently used as beam stops for high-power lasers in decades past, although I think there are better options available today.

I am not completely sure that I understand your question, but this is the essence of what makes opaque, black objects. Opaque objects are optically thick — for instance, a kilometer-long column of water is opaque, and there is no sunlight in the deep ocean. In physics, "black" is an idea that's a little more complicated; I would say, perhaps, that an object is "black" if the spectrum of light that you get from it has little or no correlation with the spectrum of light that you use to illuminate it.

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  • $\begingroup$ I think you are explaining this. If there is very deep hole made with white material or mirror, the deepest part seems black to our eyes even though around the opening is shinny bright because light does not reach their. Am I right? I may be not enough explaining my question because I'm still confused. $\endgroup$ – user47546 Jun 5 '14 at 18:56
  • $\begingroup$ Continue from above…. I don't get this "an object is "black" if the spectrum of light that you get from it has little or no correlation with the spectrum of light that you use to illuminate it." For me, it's not easy to understand. $\endgroup$ – user47546 Jun 5 '14 at 18:59
  • $\begingroup$ It's "black" if all the light that you shine on it vanishes; however, even a perfectly black object emits light with a spectrum that depends on its temperature. For instance, the sun's atmosphere is a blackbody — any light that you put into it scatters many times and gets absorbed. But the sun's atmosphere is hot enough that it emits visible light. $\endgroup$ – rob Jun 5 '14 at 19:14

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