Clouds can be a variation of grays, from white to some dark gray, almost black. They are clearly grayscale, so it means they reflect all colours uniformly, but some clouds are whiter than others, so the reflectance varies. What makes a cloud to be darker than an another? Typically rain clouds are darker, is the tone a measure of how much "water" they are carrying?
[Physics] The colour of clouds
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Related Solutions
Wrong:
"since infrared waves have a shorter wavelength"
Infrared has longer wavelength than visible and visible longer wavelength than ultraviolet .
White is a term for visible light mixed wavelengths. In the plot you can see that almost half of the sun's radiated energy arrives as visible light. The white buildings reflect this visible light which otherwise,impinging on the surfaces would be absorbed and turned into infrared by the interactions, adding to the arriving infrared.
What is absorbed and what is reflected depends on the chemical bonds of the surfaces, whether the incoming radiation can excite molecular states of the materials. Infrared is in frequencies/wavelengths of the black body radiation of bodies in the temperature ranges comfortable for the human body, so they easily raise the vibrational and rotational levels of solids and liquids and the kinetic energy of gasses.
37C curve seen here practically all in infrared, and lower temperatures more so.
Thus white paint will not reflect infrared as efficiently as visible, a large part of infrared will be absorbed as also some part of visible will scatter at the surface and degrade to infrared. Infrared can be reflected by metal mirrors, from the collective fields in metals . If you put aluminum foil in front of a heater you are sheltered from most of the heat which is reflected, but some of it is absorbed as can be seen by touching the foil.
If I expose an object to EM radiation only from the infrared spectrum, will it only reflect back infrared?
Yes, but most of it will be absorbed ( except by mirror metal surfaces) because the materials have the receptors for these wavelengths. This is due to the fact that larger wavelengths have photons with less energy which cannot excite higher energy levels.The energy of the photons goes as h*c/lamda where h is plancks constant, lamda the wavelength and c the velocity of light.
Is this true for other types of EM radiation?
No.Visible and ultraviolet by scatterings degrade their energy down to infrared frequencies, depending on the material.
Is it possible to make an object that looks white and absorbs a lot of infrared radiation?
Usually most of the infrared will be absorbed except by mirror metal surfaces.
If an object reflects most of the EM radiation that it receives of a particular wavelength λ, will it also reflect most of the radiation it receives of wavelengths less than λ (and absorb most of the radiation of wavelengths larger than λ)? Is this why objects that reflect most visible light (and hence look white) also reflect most infrared radiation (since infrared waves have a shorter wavelength)?
There is no such rule. It depends on the material and its chemical bonds.
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
You are just seeing differently illuminated clouds, as should be clear from this picture of an aircraft flying over a thunderstorm:
Clouds are made of water droplets or ice crystal, which are transparent. When light is incident upon a cloud, it is scattered at the same wavelength (a process known as Mie scattering). This means that it remains of the same color.
Look at this picture of clouds at sunset: they are red because they are scattering red sunlight. The upper part appears more gray-ish because it receives less light. It is clear from the illumination of the clouds that light is coming from below (the upper part is darker) because the Sun's position is low.