# Why is sunscreen more important in the middle of the day?

I suppose that if the Earth had no atmosphere and no magnetosphere, then when standing up on the surface of the Earth -- say, on the equator -- my body would catch more radiation from the sun in the morning (after sunrise) than in the middle of the day, due to the shape of a human body. Yet, on our actual Earth with atmosphere and magnetosphere, we are advised that sunscreen is more important in the middle of the day. What exactly causes this effect and what is an effective way to think about the relative amounts of radiation? Is some of this advice in part due to the fact that we're more likely to be in the shade of buildings or trees in the morning? The contrast with there being more radiation near the poles also seems interesting.

The answer is that the characteristic optical depth for damaging UV radiation is approximately supplied by the thickness of the Earth's atmosphere, so that only a fraction $\sim e^{-1}$ penetrates to the ground when the Sun is overhead. In contrast, visible radiation passes through the thickness of the atmosphere only losing perhaps 10% to 30% of its power (depending on atmospheric conditions) in the red to blue part of the spectrum respectively.

Thus when the Sun is overhead a significant amount of UV radiation penetrates to the ground. When the Sun is nearer the horizon, then that radiation has to travel through a greater thickness (the thickness goes like $\sim \sec \theta$, where $\theta$ is the angle from the zenith$^{1}$) of the Earth's atmosphere and so the UV transmission decreases to $\sim e^{-\sec \theta}$ and so the need for sunscreen diminishes.

$^{1}$ Note this formula assumes a homogeneous slab as a model for the absorbing atmosphere. A more complicated model is required for an inhomogeneous absorbing layer and $\theta$ close to 90 degrees (see the discussion of "air mass").

EDIT: This is quite old work, but the tables and formulae back up everything that I say above. Elterman (1964) give tables containing the extinction coefficients due to absorption, aerosols and Rayleigh scattering. All of these effects are important at the UVA-UVB (280-400nm) range intended to be blocked by sunscreen. This means that the entire thickness of the atmosphere overhead is important.

For example at 300nm the optical depth of an atmosphere (up to 50km) due to Rayleigh scattering (mainly in the bottom 10 km of the atmosphere) is 1.2, which allows a fraction $e^{-1.2}$ to reach the ground. However this is bolstered by aerosol scattering and ozone absorption (mainly at 20-40 km up) such that the total optical depth is about 5. Therefore very little of this radiation reaches the ground and that is mainly due to ozone absorption.

However, at 340 nm the total optical depth of an atmosphere to 50km is $\sim 1$ (as I claimed above) and is almost equally due to Rayleigh scattering and ozone absorption and occurs mostly in the bottom 20 km of the atmosphere.

Equation (12) in this paper, to calculate the absorption on a slanted path through the atmosphere, is exactly the $e^{-\sec \theta}$ dependence I quote above.

• Isn't the absorption mostly due to the ozone layer rather than the main body of the atmosphere? If it were due to the main body of the atmosphere, then the exponential attenuation rate would itself decay exponentially with altitude.
– user4552
Sep 3, 2018 at 16:56
• @BenCrowell The UV is attenuated by some plane parallel layer somewhere in the atmosphere. I don't really care where that is for the purposes of this argument (there is certainly plenty of absorption low down because the effects of UV exposure are definitely greater when you climb a mountain). If the layer is above you then the thickness you are looking through is $t\sec \theta$, where $t$ is the thickness of the layer overhead, and the absorption goes as $e^{-t\sec \theta}$. This is not an exact calculation. Of course. Hence the $\sim$. Sep 3, 2018 at 17:03
• @BenCrowell Probably Rayleigh scattering is more effective lower down? However, your comment (and also that the Earth is spherical) is also why the $\sec \theta$ approximation doesn't work well when say $\theta>80^{\circ}$ (see en.wikipedia.org/wiki/Air_mass_(astronomy)) Sep 3, 2018 at 17:15
• Thanks for these answers and comments. Is there any place to find data to back this up? Sep 4, 2018 at 13:01
• @present see edit Sep 4, 2018 at 13:33

Without giving it too much thought i'd argue that the fact that the sun is higher on the sky is the culprit. That means that the radiation that hits us has traveled a shorter distance through the atmosphere and thus have been attenuated less if the attenuation is simplified and modeles as a $\alpha = l\cdot z$ where $\alpha$ would be total attenuation, $l$ would be distance through the atmosphere and $z$ would be something $\frac{attenuation}{meter}$.