(Sometime before I mentioned a personal estimate of requiring 700 Kelvin to see with dark-adapted eyes something visibly incandescing.)
I have heard of people seeing single photons. But if you dark-adapt yourself and see individual flashes, I suspect some of those will come from thermal agitation as well as from photons. But I'll make a guesstimate of how much light it takes to see...
I have heard, I think I remember, that the human retina utilizes (at favorable wavelengths) 20% of photons hitting it. In extreme darkness, I am guesstimating persistence ov vision to be effectively .2 second (varies slightly inversely with lighting level). And I feel like arbitrarily saying that to see things you want to see 5 times a second a photon from every square 1/2 degree of what you are looking at. (I picked that because the angular diameter of the sun and moon seen from Earth are slightly over half a degree).
A very highly dilated human pupil is, as seen when magnified by the cornea, about 10 millimeters in diameter. (I know that 21 CFR 1040.1, regulating laser exposure, says to use 7 mm when pupil diameter affects exposure. I am guessing that people who write regulations are assuming you won't be working with lasers in such extreme darkness as to have your pupils to be as big as possible.)
A surface 1 meter away would have a piece 8.73 mm wide appear half a degree wide. And 1 photon out of every 40,000 emitted would pass through an effectively 10 mm diameter pupil from that distance. 200,000 photons per second from a square 8.73 mm wide is about 26.2 million photons per second emitted per square meter.
The wavelength that scotopic vision is most sensitive to is 508 nm, and one photon of that wavelength has an energy of about 2.44 electron volt. Divide that by Faraday's number (6.3 E18) to get joules - that is
3.87E-19. Multiply by the above 26.2 million photons per second per square meter, and I get a rough "calculation" of 10.1 picowatts per square meter.Human vision is about 30% as sensitive to "white light in general" (counting only visible wavelengths) as to the most vuisible wavelengths - so now my rough "first calculation" becomes 33.5 picowatts per square centimeter.
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I try again: I have heard of Venus being able to illuminate ground brightly enough to cast a shadow. Venus has a mgnitude mostly around -4 and the Sun has one around -27 (I forget exactly). 5 magnitudes (the astronomic measurement of brightness of a celestial object) is a factor of
100. That makes Venus having a brightness around that of the Sun times 6.3E-9. The Sun typically delivers about 1 kilowatt per square meter to a perpendicular surface at Earth's surface, with about half of that being visible. 500 times 6.3E-9 is 3.15E-6. Since Venus is almost always within 30 degrees of the horizon when it's really dark, I would say more like 1.5 microwatts of visible light from Venus per square meter of ground on a good evening or pre-dawn.==================
Another attempt: Illumination to match the "surface brightness" of a hypothetical 6th magnitude hypothetical celestial object that is square and half a degree wide. Astronomy books like to say 6th magnitude is the dimmest brightness at which celestial objects are visible to the naked eye, but I have heard reports of 8th magnitude in favorable circumstances.
With my above calculation of Venus delivering 3.15E-6 watt of visible light per square meter of a perpendicular surface, a 6th magnitude object delivers 1/10,000 of that or somewhere around 3.15E-10 watt per square meter. An 8.73 mm wide square (half a degree wide at 1 meter away) doing this would radiate that times pi, divided by .00873 squared, or *roughly* 13 microwatts per square meter. 8th magnitude would make that roughly 2 microwatts per square meter.
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Temperature that a blackbody needs have brightness equivalent to that of radiating 1 microwatt per square meter in the 400-700 nm (visible spectrum) range: 1 microwatt per square meter of "white light" (having an efficacy of 243 lumens per watt for equal-power-per-unit-wavelength band from 400-700 nm) illuminates to an extent of 2.43E-4 lux, and such illuminated surface has a surface brightness (in candela per square centimeter) of that divided by
10,000 and by pi - which is 7.7E-9. A blackbody does this at about 687 Kelvin. 702 Kelvin gets double the brightness of 687 Kelvin (same surface brightness as 2 microwatts of "white light" per square meter).If that 33.5 picowatt per square centimeter figure I got at first has any truth to it, then surface brightness would be 2.7E-13 candela per square centimeter. A blackbody achieves this at about 521 Kelvin.
These temperatures are for calculating for photopic vision. Since scotopic vision requires shorter wavelengths than photopic vision does, the temperature required to make a blackbody visibly glow would be a little higher than these temperature figures.
- Don Klipstein ( snipped-for-privacy@misty.com)