[argyllcms] Re: Absolute light meter calibration?
- From: Richard Kirk <richard@xxxxxxxxxxxxxxxx>
- To: argyllcms@xxxxxxxxxxxxx
- Date: Fri, 14 Aug 2015 08:37:39 +0100
On 13 Aug 2015, at 17:43, Ben Goren <ben@xxxxxxxxxxxxxxxx> wrote:
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.
I am on holiday, so this is going to be a bit sketchy. Anyhow, if I was at work
I doubt if my lab books would be much help. But, I have remembered a bit more,
so here is a quick overview on emissivity...
Suppose you have a large furnace at a constant temperature. It will be a cavity
filled with photons, with each waveband in dynamic equilibrium. The emissivity
is a measure of the probability of a photon being absorbed at a surface
(varying with material, temperature, angle of surface, and wavelength). If you
have equilibrium, then the emissivity will also be probability of a photon
being emitted at the same surface, or things wouldn't balance.
Suppose, now, you make a hole in the side of this furnace. It is a small hole
so you don't disturb the equilibrium significantly. You will see the same
spectrum wherever you look: the surfaces with low emissivity will be putting
out less light, but will be reflecting more, so everything looks the same.
Standard light sources looked like this up until about the 1970's - you would
have a big, hot chamber and a tiny window to let a small amount of the standard
light out. This is why you can't get them on e-bay.
A lightbulb is the reverse of this situation: you have a very exposed hot
element that is far from equilibrium. However, you then have something called
the 'grey-body' approximation, that the emission spectrum is often much the
same. This fits our experience - things always seem to glow the same colours.
At this point, I have departed from the 'absolute light meter calibration'
because I am depending on other experiments to prove that the 'grey-body'
approximation holds for tungsten filaments, but it does. For example, have a
look at...
http://pyrometry.com/farassociates_tungstenfilaments.pdf
So, we can reasonably assume that the energy put into a tungsten filament comes
out as a blackbody-like curve. However, can we suppose the approximation holds
when integrated over the long infra-red tail of the emission curve? Well, we
can correct for this by taking measurements at two different filament
temperatures. Have a look at the top log-log plot in...
http://how-it-looks.blogspot.com/2010/01/infrared-radiation-black-bodies-and.html
We can see each temperature has a blackbody emittance with a smooth curve on
the right, and a quantum cut-off on the left. If we had two measurements at
3000K and 3200K they would probably be almost the same for 1 micron and larger
wavelengths if you scaled up the 300K curve by a ratio that we can calculate
from the blackbody laws. So, you could subtract the scaled 3000K blackbody
spectrum from the 3200K blackbody spectrum and get a difference that peaked in
the visible wavelengths and was almost zero for wavelengths above 1 micron. Or,
if you really want to be classy, you could move the bulb closer to the detector
to scale the spectrum, so you are using the same part of your spectrometer A/D.
The 'grey-body' emissions will be the same curves, but multiplied by the
emissivity, which will be a function of wavelength and temperature. However, if
the emissivity is a weak function of temperature (and it seems to be) then the
differences will cancel for grey bodies.
The 'grey-body' approximation has a few hidden bodges. We cannot easily measure
the emissivity of a surface at a high temperature at a particular wavelength.
We cannot usually measure the temperature directly - we deduce that from the
spectrum, which is the very thing we are trying to investigate. Most real
materials have an emissivity which varies slowly for the visible wavelengths,
which adds a gentle tilt to the visible part of the spectrum, which looks just
like a small shift in temperature (which we haven't independently measured), so
we don't really know. So, it is not an 'Absolute' calibration with a capital
'A' - I think you would have to build a synchrotron for that - but it might be
a recipe for making a standard emitter.
While I have a letter open, here's a possible alternative...
I have been wondering whether we can do some calorimetry with LEDs or laser
pointers. You mount the LED on a heat sink with a resistor and a temperature
sensor. You run the LED until the heat sink reaches a steady state and record
the temperature difference with the surroundings. You turn off the LED, and
vary the current through the resistor until you get the same temperature. You
have the electrical power into the LED from the LED volts*amps. You have the
heat power from the LED from the matching resistor volts*amps. The difference
should be the power of light emitted. Unfortunately you cannot get 555nm LEDs
to match the definition of a lumen.
Actually, come to think of it, it doesn't really matter what my old lab-books
say. The important thing is to see whether other people can try the same thing
and get the same results.
Cheers.
Richard Kirk
---
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Tel: +44 (0)20 7292 0400 Fax: +44 (0)20 7292 0401
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