When I was young I read one book in which is written that you get more tan on the morning than on evening even light angle from Sun to Earth is the same. Don't remember exact reason, I think because ultraviolet is more absorbed on the evening because air is more humid or something like this. Is it true?
[Physics] Why is getting a tan in the evening less likely than getting a tan in the morning
atmospheric scienceeveryday-lifesunvisible-light
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First of all, atmospheric attenuation in the visible region is primarily due to scattering, not molecular absorption as in the infrared and microwave regions. This is perhaps not that important to your question, but a good thing to keep in mind. The ligth doesn't disappear, it only changes direction.
If you have the density profile of the atmosphere, let's denote it $\rho(h)$, where $h$ is the height above sea level, you can calulate what you call mass-thickness, (I will call it slant mass column and denote it $C$), using this integral:
$$ C=\int^{\infty}_{0}\frac{\rho(h)dh}{\sqrt{1-\left(\frac{\cos{\theta_0}}{1+h/R}\right)^2}} $$
or this one:
$$ C=\int^{\pi/2}_{\theta_0}\rho\left(R\left(\frac{\cos{\theta_0}}{\cos{\theta}}-1\right)\right)\frac{R\cos{\theta_0}}{\cos^2{\theta}}d\theta $$
where $\theta_0$ is the elevation angle of the sun and $R$ is the earth radius. These formulas are for an obeserver located at sea level.
This assumes that the light travels straight through the atmosphere. This assumption works reasonably well for high elevation angles, but for lower angles (think sunsets) you have to take atmospheric refraction into account and perhaps also some additional scattering effects.
Your digital camera experiment might work in theory but it will perhaps not be as easy as you would like. First of all you need to account for refraction when calculating the slant mass column as I explained above. Refraction is wavelength-dependent so you will have to do it for each color. Secondly, only Rayleigh scattering will be proportinal to $C$. You will also have Mie scattering due to aerosols. This will depend on the aerosol loadings in the atmosphere which will be variable. Mie scattering will also be more important for low elevation angles since a larger part of the path through the atmosphere will be in the lower parts, which have higher aerosol loadings.
Let's try to validate and quantify the conjecture first raised by Carl Witthoft in a comment to the question, which is basically that the sky only appears less blue in the second picture because a lot more light is scattering off of the windows towards your camera.
If this is true, we ought to be able to see it. The first thing to do is convert the pictures from the relatively useless RGB colorspace to the much more useful XYZ colorspace which is built on a model of the actual receptors in the human eye. The $Y$ coordinate corresponds to the perceived luminance of the image (i.e. the average human response across the visible spectrum) and the $Z$ coordinate corresponds to our blue receptor response. The $X$ coordinate is set to be pick up the slack and doesn't necessarily have a clear physical interpretation. See here the responses across the visible spectrum: (from wikipedia):
So, that is the first thing I did. I obtained:
Above you will see the two original pictures, as well as their $Y$ and $Z$ values. Here we can clearly see that the total illumination ($Y$) in the Gray picture has gone up, and the blue content of the image ($Z$) has gone up as well.
Let's try to take a closer look. To do that I will next look at a histogram of the $Y$ and $Z$ values in the images:
Looking at this histogram of values, we can clearly see that at the middle levels (near ~ 0.5) both of the images have a blue hump. Let's assume that is the sky (we'll check in a second). But notice also that if anything that blue hump has shifted up a bit in activation. Sorta nearby the blue hump is a hump in the luminance ($Y$), which appears to move a lot. But there is a lot going on in the image, and if the conjecture is right and there is more light coming in through the windows, we would expect everything in the picture to be brighter, including the columns and wall. So, we need to try to filter the sky, so let's make a cut on the image given by those humps in the blue. I've shown my choices for the cuts as the vertical dashed lines in the image. Applying that cut to the original image we obtain:
Absolutely wonderful! We've just developed a nearly perfect sky filter. Now that we know which pixels correspond to the sky, we can look again at our histograms, but this time only for "sky" pixels.
And now it would appear as though there is no denying Carl Witthoft's explanation, the sky appears less blue, in the "Gray Sky" picture, not because any of the blue has gone away (in fact if anything there is more blue content in it) but because there is just so much more light coming from those points beyond just the blue, and so it doesn't look blue anymore. For completeness, let's look at the histograms in the RGB channels of just the sky pixels:
Here we can clearly see that it is not that the blue went away, we just have a heck of a lot more red and green coming from the windows now.
But why does it look so much less blue, when the values of the red and green channels are still smaller than the blue?
That is entirely an effect of human perception. We are a lot less sensitive to blue light than we are to green. If you take a look at the plot at the top of this answer again, remember that the $Y$ curve was chosen to be the perceptual sensitivity of human subjects across the visible spectrum. Notice how little it overlaps with blue.
In fact, a common formula people use to convert images to grey scale (that is worse than the XYZ transformation, but easy to do) is:
$$ L = 0.21 R + 0.72 G + 0.07 B $$
This demonstrates the issue with just three numbers. Roughly 72% of what we perceive as brightness comes from the green channel, 21% comes from the red, and only 7% comes from the blue. This is why, when the sun shines on those windows in your building, even though there is more blue light coming in, and the blue components still dominate the other colors, it suddenly looks very drab indeed.
All of the code used to make these figures is available as an ipython notebook here.
Best Answer
This sounds false to me if it's proposed as a general statement rather than something about weather patterns that operate on a 24-hour cycle. Disregarding weather, the intensity of sunlight depends on (1) the angle at which the solar radiation intercepts the landscape and the surface of your skin, and (2) the thickness of the atmosphere traversed by a ray of sunlight, which causes absorption. Both of these factors depend only on the height of the sun in the sky. The claim would obviously be silly, for example, if applied to the arctic -- you wouldn't be able to get a tan in October, when the sun was setting, but you would be able to get one in April, when it was rising??
And if it is a statement about weather patterns, then it's certainly not something that can hold true in all places. E.g., on a lot of the California coast a common summer weather pattern is for there to be fog in the morning, which burns off later in the day -- this would be the opposite of what was claimed.