According to wikipedia: "UV radiation constitutes about 10% of the total light output of the Sun, and is thus present in sunlight" It also states that "at ground level sunlight is 44% visible light, 3% ultraviolet". Another article states that at ground level "3 to 5 percent is ultraviolet (below 400 nm)". Further, the energy emitted from the sun is about 1,400 watt per square meter (on the ground level).

Based on this data I understand how to calculated the luminous efficacy of visible light. The sun has a efficacy of 93 lumens per watt of radiant flux, which works out to about 98,000 lux of visible light per square meter.

However how do you calculate the irradiance (or radiant flux?) for the non visible spectrum (such as violet light)? In other words, while I know the number of watts per meter how do I calculate the actual amount of light (in the UVA spectrum) that reaches the ground by the sun per square meter? How much of this energy is actually converted into UVA light? Or how efficiently does the sun convert energy in the non visible spectrum?


closed as unclear what you're asking by Emilio Pisanty, Jon Custer, heather, ZeroTheHero, Daniel Griscom Sep 25 '17 at 1:06

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ The atmosphere is strongly absorbent in the UV range. As such, comparing the Sun's radiation output with that measured on the ground will lead to the kinds of contradictions you note. $\endgroup$ – Emilio Pisanty Sep 19 '17 at 16:34
  • $\begingroup$ This question is extremely unclear. What is it, exactly that you want to know? per square meter where - on the ground? How would energy get "converted" into UVA light? What does "efficiently" mean as regards the Sun, and why do you think that's related to the light that makes to the Earth's surface? I do hope you realize that the concept of luminous efficacy is completely meaningless in the UV range. $\endgroup$ – Emilio Pisanty Sep 19 '17 at 16:38
  • $\begingroup$ @EmilioPisanty yes on the ground.. Basically, I can easily calculate the amount of light that reaches the earth in the visible spectrum (works out to about 98,000 lux) based on the fact that the sun converts every watt into about 93 lumens of visible light and there's approx. 1,000 watt of energy that reaches the earth & gets converted into visible light. But I want to know the irradiance (or radiant flux) for light in the UVA spectrum. How much irradiance (or radiant flux) of UVA reaches earth on a bright sunny day for instance? $\endgroup$ – Joe Steven Sep 19 '17 at 16:43
  • 1
    $\begingroup$ The sun "converts" heat into a broad spectrum; it makes no sense to talk of the "efficiency" with which it does that - I suppose the Planck law tells you exactly what the intensity is at each wavelength: is that what you are after? Because of the strong absorption by the atmosphere, the angle that the sun makes (day of the year, time of day, latitude) and composition of the atmosphere (clouds, particles, etc) all strongly affect the answer "how much UV is seen on earth". It remains unclear to me what you really want to know. $\endgroup$ – Floris Sep 19 '17 at 16:57
  • 2
    $\begingroup$ Statements of the form "approx. 1,000 watt of energy that reaches the earth & gets converted into visible light" are deeply wrong. Energy doesn't reach Earth and then gets "converted" into visible light. Energy reaches Earth in the form of visible light. $\endgroup$ – Emilio Pisanty Sep 19 '17 at 16:59

You state that

I want to know the irradiance (or radiant flux) for light in the UVA spectrum. How much irradiance (or radiant flux) of UVA reaches earth on a bright sunny day for instance?

where the irradiance is to be taken at the surface.

This can be found from the ASTM G173-03 Solar Spectral Irradiance Reference Spectra, which is derived from the SMARTS model, and which is the base data for the stock Wikipedia solar irradiance spectrum,

A quick-and-dirty numerical integration of that data gives $1348\:\mathrm{W/m^2}$ over the entire spectrum for the extraterrestrial irradiance (i.e. at the top of the atmosphere), $1000\:\mathrm{W/m^2}$ for the whole-spectrum radiation over the entire sky, i.e.

"Global Tilt" = spectral radiation from solar disk plus sky diffuse and diffuse reflected from ground on south facing surface tilted 37 deg from horizontal

and $45\:\mathrm{W/m^2}$ for that same radiation integrated over the UVA range between $315\:\mathrm{nm}$ and $400\:\mathrm{nm}$.

It's worth stressing that these are highly variable figures, and they will change with latitude, time of day, direct and indirect cloud cover, amount of haze, and a slew of other factors. However, the ballpark estimate is roughly reliable.

  • $\begingroup$ So this is approximately equal to the UV produced by one of these amazon.com/… (roughly 50 watt) correct? Basically how many of these would I need to buy (if I placed it 3 ft away from me) to mimic the natural UVA radiation present on a bright sunny day.. $\endgroup$ – Joe Steven Sep 19 '17 at 17:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.