Spyder III Pro Arctic - Questions Answered

"It doesn't. If someone with a LPM measures the power with and without the lens, isn't the difference of the two measurements the reflected power projected back from the lens?"

Did you read the word "absorbed" in my post?

Markus Unread said:

Did you read the word "absorbed" in my post?

Click to expand...

I did but I never really thought about glass absorbing photons. So I did some searching and found some interesting information.

Apparently is happens at the sub-atomic level with the electrons where it is usually a combination of absorption, transmission, and reflection of photons.

But I did find that a lot of glass blocks out ultraviolet (UV) light and that electrons in the glass absorb the energy of the photons in the UV range while ignoring the weaker energy of photons in the visible light spectrum.

So does that mean hardly, if any, of the photons of the 445nm wavelength would be absorbed?

If the electrons absorb the energy of any portion of the visible spectrum, the light that transmits through will appeared colored according to the portion of the spectrum absorbed.

Well this was a good day, I learned something new!! :beer:

I've seen non-reflective loss in optics by re-radiance (you can see it a lot more often with a 405nm beam because it doesn't saturate your vision like a blue or green). In a thicker lens, it shows as a faint line passing through, often a different color. Then there's edge loss by scattering, where the sides of the optics light up.

So it's not really absorption but re-radiation as heat. Of course, the same could be said for a black aluminum heat sink. Slam 455nm in, get heat out, and most folks would call it absorption.


Updated- I found an old 0.25" square silicon photodetector, dialed my 445nm down to 300ma and measured 130/11 uA... Meh. I really need to find someone who has an Arctic front window and REAL LPM. I don't trust this crufty test setup.