[argyllcms] Re: Absolute light meter calibration?

  • From: Ben Goren <ben@xxxxxxxxxxxxxxxx>
  • To: argyllcms@xxxxxxxxxxxxx
  • Date: Thu, 13 Aug 2015 09:43:46 -0700

On Aug 13, 2015, at 9:15 AM, Marwan Daar <marwan.daar@xxxxxxxxx> wrote:

A true blackbody will emit 100% of input power as radiant flux. Therefore, if
you know the input power, you know the total radiant flux.

Thermodynamics says that _everything_ going into the bulb has to come out.

Tungsten is not a true black body. It has an emissivity of about 0.35, and is
therefore a gray body. Therefore, you must take this into account when
calculating total radiant flux.

Unless I'm missing something, emissivity would have to be related to spectrum
and thus influence the spectral power distribution of the radiant flux. Again,
it's all gotta come out in the wash, so something that isn't very emissive at
visible wavelengths is going to have to be even more emissive at other
wavelengths.

That's why I'm thinking that lamp black might be a good choice. (Hold the
filament over a sooty flame until the filament is evenly and thickly covered in
soot. Would require a good vacuum else the soot would combine with air at high
temperatures.) At the least, lamp black is within shouting distance of a black
body in the visible spectrum, and I _think_ it's equally black across the whole
spectrum -- but that's something I'd need to confirm.

But even if lamp black isn't (nearly) ideally emissive at other wavelengths,
that shouldn't (I hope) matter in the visible region. It'd mean, just picking
stuff out of my netherbits, that the bulb is darker than you'd expect in the UV
but equally brighter in the IR, with the visible still having the expected
brightness, or some other combination. The situation to avoid would be the
visible having other than (close to) ideal emissivity.

...but physics is an hobby for me...if I already knew all this stuff, I
wouldn't have asked the original question....

b&

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