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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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1 Answers1

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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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