(Response in line)
On 10/14/2014 12:51 PM, Jones Beene wrote:
*From:*David L. Babcock
>I seem to remember that the 4th power thing is due to (largely due
to?) the strongly rising "center" of the frequency as temperature
increases. Thus, the radiated power through a narrow window (visible
band is only 1 octave) is probably only proportional to the /first/
power, at least when that window is well below the peak power frequency.
What frequency are you assuming is peak power? I would have suspected
that the higher the photon frequency, the more power, such that
visible should be peak.
I get a glimmer that I'm off, here. I visualized the peak as well above
the visible, in the ultra-violet. Unlikely, as nobody even hints that
the perceived color was blue-white. So maybe I'm right, but not right
about this particular case.
ØIf true, any error contributed by that window rapidly becomes
unimportant as temperature increases.
Yes, if true - but this depends on your assumption of peak power. Can
you elaborate on why you think it would be longer wavelength rather
than shorter?
Probably a typo here, as I was thinking the peak to be much /shorter/
wavelength.
I may not be all that wrong. The cutoff is so many nm, the window
doesn't tint ordinary light (I've seen some alumina), and the experts
agree that the effect is small at the lower power used for cal. Or at
least small enough to be easily calculated-out.
Thanks
Speaking of calculated-out, wouldn't the tester be using established
equations (or tables from same) for alumina, which then would certainly
account for any likely range of T? Then the short-coming of calibrating
at the wrong temperature is reduced -magically!- to a second order
effect. Perhaps 2% or 10%. Not enough to disapear the over-unity, just
a small embarrassment.
Thank you for the polite exchange
Ol' Bab
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