How much of Ultraviolet is cut out when Sun moves to the Horizon?

EDIT: Original question was whether or not we need to wear sunglasses when Sun is at the horizon. But it was confusing to many users whether it is on-topic or not. So the details are adjusted.

From the title this may not look like a Physics question. But hear me out.

As I read in Phil Plait's Bad Astronomy and many other books. Rayleigh Scattering is the effect where our atmosphere bounces off most of the light with lower wavelength (blue, ultraviolet) thus the Sky looks blue. And when the Sun is in the horizon, the light from it goes through extra thick layer of atmosphere (as opposed to when it is at the zenith); when that happens, we see Sun in red color because most of the blue (and ultraviolet) is scattered elsewhere.

So that begs the question, is Ultraviolet during late afternoon hours or early morning hours still prevalent?

That will also answer whether humans need Sunglasses during that.

• It's still less a physics question and more a question about the human eye. – ACuriousMind Mar 23 '16 at 16:10
• @ACuriousMind How about, How much of Ultraviolet light is cut out when Sun is in horizon compared to when its at the zenith? – fahadash Mar 23 '16 at 16:11
• This seems the sort of thing some determined Googling would pin down. Have you attempted any research of the area before posting? – John Rennie Mar 23 '16 at 16:27
• Wikipedia has a good plot of the spectrum at various times of day, and references. – tfb Mar 23 '16 at 17:06
• Please define "prevalent". Include a threshold value in some objective unit such as mW/square meter, below which UV is not a problem. Until you do, the question is unanswerable. – WhatRoughBeast Mar 23 '16 at 17:39

This is straightforward to answer in principle, but not so easy to calculate in practice. The problem is that atmospheric extinction is a very strong function of wavelength and also quite dependent on how much dust and aerosols are in the atmosphere. This turn depends on where you are and how high up you are. The "extinction curve" (a plot of extinction per airmass versus wavelength) for the Roque de los Muchachos observatory at 2400m on an island in the atlantic will be considerably lower than one at sea level in a city - by factors of 2-3 I would think.

The plot below from Schuster & Parrao (2001), shows modelled contributions to the average atmospheric extinction at an observatory in southern California. The units on the y-axis are magnitudes per airmass $A_{\lambda}$. i.e. By how many factors of 2.5 the flux is reduced by when a star is at zenith.

At lower altitude we can approximate the number of airmasses as $\sec (90^{\circ}-\alpha)$, where $\alpha$ is the altitude in degrees.

Thus the factor by which any signal is attenuated is $10^{-A_{\lambda} \sec (90^{\circ}-\alpha)/2.5}$. e.g. a source 90, 30, 10 degrees above the horizon will have its light at 320nm ($A_{\lambda}\simeq 0.6$_ attenuated by factors of 0.57, 0.33, 0.04 (only 57, 33, 4% gets through). I chose this wavelength because (i) it is about as short a wavelength as the plot goes (it does keep increasing steeply at shorter wavelengths), but (ii) it is also the dividing line between "UVA" and shorter wavelength "UVB" radiation, which are regarded as damaging and highly damaging to the eyes respectively.

At very small altitudes ($\alpha < 5^{\circ}$) the simple formula for airmass is insufficient because of refraction and because the vertical structure of the atmosphere starts to matter. You can find various approximations on the wikipedia page on airmass, but at a degree or two above the horizon you are talking about airmasses of 20-30 and therefore attenuation factors of 5 orders of magnitude or more (at 320 nm).

In ideal conditions the UV transmission is largely controlled by Rayleigh scattering. However, even at good sites UV extinction can be heavily influenced by global dust contributed from major volcanoes - which can easily increase extinction by factors of a few at UV wavelengths. 