White laser?

Not true; human-perceived color is modeled as a 3-d space, with red-green-blue axes, or luma-hue-chroma, or for inks, cyan-magenta-yellow. A mixture of cyan and magenta inks can create any color on the line between the cyan 'pole' and the magenta 'pole' defined by those two ink colors. Between is VERY meaningful as it applies to colors.

On Sun, 9 Aug 2015 23:05:23 -0700 (PDT), whit3rd Gave us:

Like the brown color between your ears?

Oh wait... That's an opaque! :-)

You're picking nits the same way Jasen was. The laser article wasn't talking about the human perceptual system (which is indeed 3D, and therefore "in between" has a *3-dimensional* meaning). It said that the laser can produce red, green, blue (not the reference to single colors!) and "any color in between", implying any single color (following on the three pure colors listed). My questions were trying to find out whether they really meant that - and Steve has confirmed that they don't.

Then Jasen piped up with his typical argumentativeness and said color not wavelength" which raises the possibility of meaning either a continuous spectrum across the visible range, or the human 3D perceptual system. Though it' not what they meant, the definition of "in between" of the latter has some validity (3D containment on some color space, as you point out) but there's no way to even define "in-between" for comparing spectra. Which is what *I* said.

The upshot is that the original article was being deliberately vague and/or misleading. "color" is not a well-defined term; I've defined three different meanings for it, and they could have intended any of the three.

Now can we drop this unnecessary quibbling? We all know that wavelengths are continuous, that spectra are complex, and color perception is also complex. None of us has learned anything new, so there's little value in discussing it here.

Clifford Heath.

Fun thanks! 160 GHz is not from our sun though... I've heard the peak for our ~6000 K sun is not in the green (as it is if plotted vs wavelength) but somewhat redder. Lunch time math task, take derivative and find the peak for

6000K (or 5800).

George H.

Depends if you plot it in per-wavelength or per-frequency units.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

Yeah, that's what I wanted to check. So lunch math fun, I took the derivative of the (linked to) BB formula. Set equal to zero and found,

e^x = 3/(3-x), where x is h*f/(k*T) (f= freq.) (say is there a way to solve that exactly? I just plugged in numbers till I got close.. a little less than 2.822)

From which I found a peak frequency of 3.411E14 Hz (3.411x10^14) or about 880 nm.. the near IR but certainly not the peak of the visible spectrum.

George H.

The most generally useful BB formula, I think, is the spectral radiance leaving a cavity in per-hertz units.

2h n**2 nu**3 cos theta L_nu = -------------------------------- c**2 (exp(h nu/kT) - 1)

whose derivative is

L_nu dot = nu*L_nu*(3 - h nu/kT exp(h nu/kT)/(exp(h nu/kT)-1))

whose zero is at h nu/kT ~= 2.822 as you say.

However, in per-wavelength units, there's another factor of nu**2 in the formula (from the derivative of a quotient), so the thing that has to be zero is

(5-nu/kT exp()/(exp()-1)

and the constant comes out to h nu/kT = 4.965 by my reckoning, which puts the per-wavelength peak of a 5500K black body at 569 THz (527 nm).

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

per log units it's the same either way.

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  \_(?)_