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Do surfaces that reflect visible light efficiently also reflect UV light? If not, are there surfaces that do?

(I have a large array of UV LEDs that I need to make larger and more diffuse, so I'm considering reflecting it off nearby surfaces and into some diffuse material.)

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Probably, but how easy or difficult it will be depends on the wavelength. The shorter the wavelength, the harder it will be. If you're talking "Vacuum ultraviolet" it will be quite difficult to make a diffusor out of anything. But because you've got LED's, I'm sure you're not talking about vacuum UV. What is the wavelength you care about? And how efficient of a reflection do you need? – Anonymous Coward May 10 '11 at 0:40
390-400nm -any reflection is a bonus, so how efficient I need it would be a function of how cheap it is. If 50% is 5 dollars worth of material and 60% is 20 dollars and 80% is 75 dollars or something, it'd probably be the 50%. – rfusca May 10 '11 at 0:45
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Then I'll just repeat what Martin Beckett said: aluminum is a decent reflector at 390-400nm as well as in the visible. The folks who professionally make mirrors out of aluminum can get you R>90% at 400 nm, and R>80% for most of the visible. If you wanted 99% reflectivity, you'd need a broadband dielectric, but from the sound of it aluminum foil would do the job for you. It'll get you R>50% (and even do your diffusing if you use the "other side") and costs less than $5. – Anonymous Coward May 10 '11 at 1:19

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Aluminium is about the best coating for mirrors below 400nm.

Most commercial small mirrors for makeup etc will be aluminium behind glass. Although the glass isn't very good in the UV it's thin enough not to have too much absorption

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Wiki's reflectivity site illustrates the albedo for Al, Au and Ag. The reflectivity of Al is ~92% over the 200-400 nm range. – Michael Luciuk May 9 '11 at 0:26
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In physics, the reflectivity of a surface is described by Fresnel Equations. One of the parameters of this equations is the refractive index which in general depends on the wavelength of light used. So surfaces that reflect visible light need not be good reflectors in another wavelength. A very good example of this is the phosphorescent coating that you find inside fluorescent lights. These coatings reflect visible light very efficiently (hence appear white to our eyes). But they absorb UV and re-emit it as visible light (this is essentially what happens when you switch the light on).

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Do the phoshorescent coating reflect visable light well? Or are they just good a re-emitting it? I assume they give off white(ish) light when excited by UV, which doesn't say much about the reflectivity. – Omega Centauri Apr 30 '11 at 3:21
-1 for not reading the question! – Georg Apr 30 '11 at 8:58
@Omega Centauri: The phosphorescent coating appears white to our eyes, so it reflects visible light rather well. But when illuminated by UV, it absorbs it and re-emits visible light, making it a poor reflector of UV. – Bernhard Heijstek Apr 30 '11 at 17:05
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Most of materials that highly reflect the visible spectrum of light are highly absorbing materials for UV- spectrum, like noble metals(Au, Ag and Cu), and also alkali metals. To have a mirror that reflect light in a long spectrum we can use multilayer mirrors that have constructed from complex of several different layers with thickness of several nanometers,(coated on a substrate) that known as VUV-UV Mirrors . You can search for “multilayer uv mirrors:pdf” in net to find something like this:

Design of multilayer extreme-ultraviolet mirrors for enhanced reflectivity

Whatever the range of spectrum required for reflection increases, the number of layers increase. enter image description here

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