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Blacker Than Black

An anonymous reader writes "British scientists at the National Physical Laboratory in London have invented the darkest material on Earth. 'It could revolutionise optical instruments because it reflects 10 to 20 times less light than the black paint currently used to reduce unwanted reflections. The key to the nickel and phosphorous coating's blackness is that its surface is pitted with microscopic craters.' Wonder how effective it would be as a solar heating surface ?"

8 of 432 comments (clear)

  1. Absorb or absorb not, there is no reflect by nusuth · · Score: 5, Informative

    The light that is neither transmitted nor reflected is absorbed. If it is totally opaque too it has to be also a good solar heating surface. That said, one might be a very good absorber at particular wavelengths, but transparent or reflective at others. The cavities should act as a blackbody and operate at a wide range of frequencies though.

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  2. amature astronomy by Njerd · · Score: 5, Informative

    I would love to see this used to coat the vanes that support the secondary mirrors of reflector type telescopes. Diffraction spikes (the little spikes on relatively bright stars) are really the reflection of light on these little supports. If you are into photographing nebulae, having a bright star in view can be a real photo killer.

  3. Martin Black by mikeselectricstuff · · Score: 5, Informative

    The book 'The Hubble Wars' mentions a coating called 'Martin Black' developed by Lockheed-Martin for use in spy satellites - I wonder how this stuff compares. I found some info Here : The 'Martin Black' is not a paint at all, but a specially etched aluminum surface that acts like an anechoic chamber on a microscopic scale. The surface looks like an array of very steep pyramids a few wavelengths of light apart. It's extremely fragile & expensive to produce, but was never a classified process. Mostly used in aerospace optical hardware such as star trackers & imaging systems that have to work in direct sunlight. Ball Aerospace has a version of this process. It's considered to be a 'proprietary' process, ie they won't tell you how it's done for commercial reasons.

  4. A Tad more detail by dmontreuil · · Score: 5, Informative

    There's a little more detail and a few pictures at http://www.npl.co.uk/optical_radiation/superblack. html

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  5. Well... by caveat · · Score: 4, Informative

    ...this is a Good Thing for production instruments, but it won't matter much for research/labwork/prototypes; right now I'm working on laser detection of single atmospheric particles; we needed a *black* coating for the inside of the chamber, but it didn't need to be particularly robust, just dark - so we smoked it with a flame. Carbon black is the least reflective substance known, IIRC it absorbs something like 99.996% of incident radiation...anybody who's seen the inside of an old kerosene lamp chimney knows exactly what they mean in the article when they talk about the 'black velvet' appearance. We did have some problems with it 'popcorning' as we pumped the chamber down, but a staged evacuation with good degassing periods took care of that.

    Oh, this would make a great solar heating material - somebody mentioned the specific heat of the material, but as long as you have a thin layer backed by a heatsink, the specific heat doesn't matter (it's just the amount of heat a material can contain per gram; if you have just just a tiny bit of black substance, it doesn't matter how much heat it stores); it's all about the absorbtion.

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  6. "Martin Black" has been available for years by mfago · · Score: 3, Informative

    Martin Marietta (now Lockheed Martin) has produced a proprietary "super black" coating for years now. I've seen it, and it is _very_ non-reflective. The coating mentioned in the article sounds similar.

    "Martin Black" is proprietary though, so if you want a part coated you have to send it to Lockheed.

  7. Re:10 to 20 times less? by You're+All+Wrong · · Score: 4, Informative

    If old stuff reflects X, and new stuff reflects 90% less, then new stuff reflects X-X*(90/100) = 0.1*X.

    If old stuff reflects X, and new stuff reflects 100% less, then new stuff reflects zero.

    With me still?

    100% of X _is_ one times X.

    Ten times X _is_ 1000% of X.

    With me still?

    Something that reflects 10 times less than the old stuff reflects 1000% less than the old stuff, and therefore reflects -9*X.

    With me still?

    The original wording is misleading. The original complaint against it was valid. Instead they should have put something more like:

    The new material reflects 1/10th to 1/20th of the amount that the old material reflected.

    The new material is 10 to 20 times less reflective than the old material.

    YAW.

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  8. Re:What's blacker than black... by Raiford · · Score: 3, Informative
    A little physics awnsers this. No ! If the surface is optically rough as resulting from these little microcraters then internal specular reflection will only occur for wavelengths smaller than the crater size. Diffraction will occur as you get to sizes on the order of the crater size and just plane old specular reflection from a rough surface will occur for wavelengths larger than the crater size. This is a simple explanation but it captures the idea of how this kind of thing works. If the material already has a fairly high intrinsic absorptivity then multiple internal reflection will cause the effective reflectivity to be extremely low. Longer wavelengths will have a tendancy to not even see the little craters and probably give a higher reflectivity. There are things that make the problem more complex. There are most likely a distribution of crater sizes that interact differently with the incident light and the intrinsic absorptivity of the material is most likely dispersive (dependent on the wavelength of the incident radiation).

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