= Solution
Sunlight is approximately unpolarized before its interaction with the environment. For <Rayleigh scattering> by small isotropic scatterers, the induced <electric dipoles> radiate only their components transverse to the viewing direction. For unpolarized incident <light> scattered through angle $\chi$, the perpendicular and parallel <intensities> are proportional to 1 and $\cos^2\chi$. The degree of <linear polarization> is therefore
$$
\boxed{\mathcal P=\frac{\sin^2\chi}{1+\cos^2\chi}.}
$$
At $90^\circ$ single scattering is fully linearly polarized perpendicular to the <scattering plane>. This explains the polarized sky away from the <Sun>; multiple scattering and aerosols reduce the ideal contrast.
<Reflection> at a <dielectric> surface treats the two <linear polarizations> differently. At Brewster incidence, $\tan\theta_B=n_2/n_1$, the reflected p component vanishes for ideal transparent nonmagnetic media. The remaining reflected glare is s polarized, perpendicular to the incidence plane, approximately horizontal for a horizontal water surface. Away from this angle the <reflection> is generally partially polarized. The transmitted beam also has unequal s/p transmission.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2013/paper-1-oe2013-environmental-polarization.png]
{title=Original diagrams of polarization perpendicular to a single-scattering plane and of s-polarized reflection near Brewster incidence at a dielectric surface}
Refraction and internal <reflection> in water droplets likewise have polarization-dependent Fresnel coefficients, producing strong <linear polarization> in rainbow <light>. These are generic scattering and boundary-condition mechanisms; no biological detection mechanism is required.
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