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The differential cross-section of Rayleigh scattering is inversely proportional to the fourth power of wavelength of the radiation incident on the scatterers. More precisely, $$\frac{d\sigma}{d\Omega}=\frac{8\pi}{3}\bigg(\frac{e^2}{4\pi\epsilon_0mc^2}\bigg)^2\frac{(1+\cos^2\theta)}{\lambda^{4}}$$ where $\theta$ is scattering angle.

This means that the small wavelength components of light are strongly scattered than the large wavelength components. Thus the white light coming from the sun will suffer more scattering by the atmospheric scatterers in the blue end of the spectrum than the red end. Therefore, when we look at the sky, it appears blue because the scattered light is rich in blue. Here, the conventional explanation stops but it seems to me to be incomplete. Here is why:

According to the explanation above, the blue light is scattered away from the observer. But then blue light is not reaching the eyes of the observer. Then why should we see the sky as blue?

Solidification
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  • An important note: If the Rayleigh formula were accurate (which it is), then the color of visible light most scattered wouldn't be blue, but rather violet, which is of a shorter wavelength. We should by all means seeing a purplish sky instead, but it just so happens that our eyes are more sensitive to blue light than violet so the blue color predominates in our vision. – Yejus Oct 01 '20 at 05:14
  • Stephen Wolfram gave a pretty good discussion of this topic recently on his physics Q&A youtube livestream (starting around 23:04): https://youtu.be/whYI1_ezWqw?t=1384 . I hope this helps. – ad2004 Oct 01 '20 at 05:28
  • @Yejus It's unrelated to the question, but that explanation at scientificamerican.com is wrong. There's no reason to expect the spectrum of the sky, which has substantial energy over the whole visible range, to be perceived in the same way as narrowband high-frequency light. They're different stimuli that lead to different cone outputs. The auditory system does identify fundamental frequencies, but vision only gets the outputs of three types of cones. It can't and doesn't identify a peak in the spectrum; it has no concept of frequency at all. – benrg Oct 01 '20 at 06:13

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If you look directly at the sun, blue light is preferentially scattered away. So the sun appears redder than it would.

But some sunlight is not on a direct path to you. Some of it will be scattered from anywhere in the sky toward you. You will see blue light appearing to originate from a direction where there is no direct light source. The sky will appear blue.

mmesser314
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  • If the line of vision is away from the Sun (say, opposite to the Sun), since the scatterers do not scatter red light at large angles as strongly as blue, light from the red side of the spectrum can barely reach us. Did I get it right? – Solidification Oct 01 '20 at 05:54