= Solution
Molecules in clean air are much smaller than visible wavelengths. Their induced <electric dipole moments> undergo <Rayleigh scattering>. Away from molecular resonances, the induced dipole amplitude is approximately $p=\alpha E$, with only weak <wavelength> variation in the <polarizability> $\alpha$. The radiation-zone <electric field> is proportional to $\ddot p$, so the scattered <intensity> satisfies
$$
\boxed{I_{\rm sc}\propto\omega^4|\alpha|^2 I_{\rm in}\ \simeq\ \lambda^{-4}I_{\rm in}.}
$$
This is the <Rayleigh scattering wavelength dependence>. For equal incident <intensities>, $450\,\mathrm{nm}$ light scatters about $(650/450)^4=4.35$ times as strongly as $650\,\mathrm{nm}$ light. Thus light scattered out of the direct solar beam and into our line of sight is enriched in blue wavelengths.
The dipole mechanism also determines the <Rayleigh scattering angular pattern>. For <unpolarized light>, its angular factor is $1+\cos^2\vartheta$; near $90^\circ$ scattering, the scattered light has strong <linear polarization>. The sky's perceived colour depends additionally on the solar spectrum, atmospheric transmission and the eye's spectral response, so the $\lambda^{-4}$ law alone does not predict a violet sky. Over a long near-horizontal path the direct beam loses more short-wavelength light, explaining red sunsets. Large aerosol particles instead produce <Mie scattering>, which is less strongly wavelength-selective and can make haze look whitish.
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