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I was wondering about the common saying that on a sunny, cloud-free day, depending on location, you can get a sunburn even when spending the whole day in the shadows.

Already our blue sky provides ample evidence that there is some photon intensity being scattered out of the direct solar irradiation path into the whole celestial half-sphere very efficiently. One can formalize this, by defining the ratio of energy in a certain waveband B coming from the sky minus the sun divided by the energy coming from the sun as $$\rm \eta_B \equiv \frac{E_{B,full sky}-E_{B,sun}}{E_{B,sun}}$$

Also, as is known, Rayleigh-scattering efficiency goes with $\rm 1/\lambda^4$. Therefore I would expect it to be in the ultraviolet $\rm \eta_{\rm UV} > \eta_{\rm Blue}$.

But my question would be: What is $\rm \eta_{UV}$ exactly? And how does it depend on lat/alt/time of the day...? I bet there are measurements as well as radiative transfer computations on this, but I have zero familiarity with the Earth science literature on this, so I'm asking here.

Related in astronomy: Recent question

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