Past exam of the optics and electrodynamics course of the University of Cambridge / 2017 / 2 / a / 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 satisfiesThis 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.