I have read this question:

High-scattering barriers are opaque for the same reasons that the undersides of clouds (which are made of transparent water droplets or ice crystals) are dark. There is some thickness of scatterer where the incident radiation is equally likely to be transmitted or backscattered. Many of these thicknesses, and the incident radiation is exponentially attenuated, even with negligible absorption.

Why is water a good neutron absorber?

Now I did not find any answer on this site about this topic specifically.

As clouds build vertically and get thicker, such as a cumulonimbus cloud, less light can pass through the cloud. This will give it a darker appearance. This is also the reason why the bottom of clouds sometimes appear darker than the top.


But this doesn’t specifically explain in detail why the bottom would be darker. One thought might be gravity, accumulating more mass (water molecules) to the bottom, but that still does not explain the darkening on the bottom completely. There is a hint of reasoning on this site though about why wet (more water molecules) objects are darker.

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  1. Why are the undersides of clouds (which are made of transparent water droplets or ice crystals) dark?
  • 1
    $\begingroup$ Because you have the bright upper side of the cloud, that's directly sunlit, as white reference and much less light reaches the underside. On an overcast day the underside of the clouds mostly looks white (except for places where the cloud cover is thicker, and thereby more light is scattered in other directions than us – as explained in one of your quotes). $\endgroup$ Feb 2, 2023 at 0:01
  • $\begingroup$ You can even see this in the upper part of the cloud in your image: where it's not directly sunlit it appears grey as well. So the primary reason is probably not some stratification, or different properties of the different layers of the cloud. $\endgroup$ Feb 2, 2023 at 0:03

1 Answer 1


Since I wrote the quoted paragraph, I guess I can quantify. Let's imagine a transparent piece of glass, where incident light is reflected with amplitude $r$, absorbed with amplitude $a$, and transmitted with amplitude $t=1-a-r$. If we put two of these pieces parallel to one another, the transmission falls to $t^2$. Actually, it's a little more than $t^2$, because some of the internally-reflected light makes it through after bouncing around in the gap between the two windows.

Now suppose that we have a thousand of these transmitting layers, each with $r=\frac{1}{100}$. The direct transmission $t^{1000}=(0.99)^{1000}$ is only about forty parts per million. I wrote a little tool to keep track of the light sloshing around in the interior layers, and found that the total reflectivity of this stack is about 91%:

Plot of reflected and transmitted intensity

Here the transmission is zero until the light has gone through a thousand transmit/reflect processes, because the light must cross all thousand surfaces to reach the other side. By that point, at the 1000th interaction, about 75% of the incident light has already leaked out the incident side as a reflection. A big pile of high-quality transmitting windows is a terrible transmitting window.

You can also see that a very small absorption coefficient, with all of the back-and-forth-ing that takes place in such a system, will have a big effect on the total transmitted power. Compare to discussions of blackbody radiation which demonstrate a startlingly dark small aperture into a big box, only to reveal that the inside of the box is lined with reflective foil. This is a demo that isn't effective on a computer video, because "startlingly dark" on a computer screen is just "off."

  • $\begingroup$ Thank you so much! $\endgroup$ Feb 19, 2023 at 6:37

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