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UltravioletPhotography

12,000 K HID spectrum


enricosavazzi

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enricosavazzi

This is the emission spectrum of an automotive HID nominally rated at 12,000 K. Visually, the light is bluish, but not extremely so. There is a moderately large relative amount of UVA, very roughly comparable to the UVA/VIS ratio of certain fluorescent tubes for reptile illumination. The HID spectrum, however, differs from fluorescent tube emission in its much more continuous (despite the numerous peaks) UVA and VIS emission. Most fluorescent tubes (except the "black light" and reptile types) emit little UVA except the 365 nm Hg line. Even the black light and reptile tubes emit UVA mainly in this line (plus a little emitted as UVB-UVC-stimulated fluorescence of the phosphor coating). Even in the HID, however, there is a peak (wider than a typical emission line) around 365 nm, which might be an Hg emission. It is also clear from the diagram that most of the energy is concentrated in the blue portion of VIS.

 

post-60-0-91774200-1493480142.jpg

 

I did not yet attempt to remove the protective glass sheath around the HID bulb, which might cut some of the UV. The HID bulb itself is rated at 40 W and has a diameter of about 5 mm. It becomes orange-hot (literally) within a few minutes of operation. Operation without the protective sheath therefore requires some other protection, especially in case the bulb should explode.

 

My UVA+UVB meter reads just a little UV emission, only 27 µW cm-2 at 50 cm from the bulb. It should also be considered that the HID bulb emits in a spherical pattern, so with a reflector and collimator the intensity should substantially increase.

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Andy Perrin
It's a little puzzling that they call that 12000K. The peak is around 475nm (by eye), and Wien's law says that would make the temp T=(2897.8 K-µm)/(0.475µm) = 6100K. If you do it the other way, 12000K corresponds to a peak of 241nm. Something doesn't add up here!
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To be fair, Wien's Law only applies to purely incandescent emission curves, which this clearly is not. Perhaps the color temperature quoted is the perceived color tint of the visible portion only.
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enricosavazzi

To be fair, Wien's Law only applies to purely incandescent emission curves, which this clearly is not. Perhaps the color temperature quoted is the perceived color tint of the visible portion only.

That is of course the main problem in trying to stick a "K temperature" label on a non-black-body-radiator source. It just does not add up if one looks at the details of the peaks and valleys of the actual spectrum. There are algorithms for assigning a K temperature label to an RGB color, which is a simpler task, but they use lookup tables rather than computing from a formula.

 

Beside that, commercial HID ratings also show a measure of "fantasy". I have on order a set rated by the maker at 30,000 K, for instance. The first problem is that no known materials can be solid or liquid at 30,000 K (or 12,000 K), so we can only theoretically calculate what a black-body radiator would emit at that temperature. And, even before testing, I am sure that will not be the spectrum of the actual HID.

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enricosavazzi

One possibility of putting a K label on a relatively continuous spectrum is computing the CCT (Correlated Color Temperature). I tried doing this with the spectrum I published earlier in this thread and the NLPIP_LightSourceColor_Script.m Octave/Matlab script (in the public domain), but the script refuses to produce a CCT value because the input data is outside the allowed range of u, v coordinates (I assume this means outside the portion of the CIE chromaticity space where it would make sense to talk about color temperature, as shown in the following image from Wikipedia - never mind that it is labeled in mireds instead of K, conversion between the two is easy and uncontroversial):

 

post-60-0-30726700-1493729450.jpg

 

This probably means that it would not make sense to assign a CCT to this spectrum because it is just too different from black-body radiation.

 

As a sanity test, since I am almost completely unfamiliar with Matlab, I computed the CCT of one of my Bowens studio strobes from one of the spectra I recorded recently. It gives a 9202.7 K value, which may be reasonable since the strobe has an uncoated tube and therefore a higher color temperature than an ordinary coated flash tube.

 

The question then remains how the manufacturer specifies the HID color temperature. It might involve extending the isotherms (i.e. the line segments perpendicular to the Planck locus along which our vision perceives approximately the same color temperature) in the CIE chromaticity space past their allowed lengths (beyond which our vision no longer perceives the same color temperature), which means that the resulting color temperature value is likely meaningless. It might also mean taking an RGB measurement instead of a spectrum and compute the CCT from there (which I did not try), but also in this case a good portion of the possible RGB values falls well ouside the isotherms and therefore no CCT can be computed for these RGB values.

 

Edit - just for curiosity, mired=0 (the point at the end of the Planck locus) corresponds to infinite K. The scale in K along the Planck locus is highly non-linear, so the K temperature goes up very quickly approaching the infinity point. The jump between 12,000 K and 30,000 K therefore gives only a small change in perceived color temperature. It is therefore not unreasonable to have a 30,000 K color temperature light source. It should look only slightly more bluish than 12,000 K, or perhaps just the same. Whether this also means that the 30,000 K HID emits significantly more UV than the 12,000 remains to be seen, but I don't think it will be almost three times more as the simple comparison between the K temperatures might suggest :D

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