= Solution
The <electric field> of the reflected beam is perpendicular to the vertical <plane of incidence>, hence parallel to the water surface. At <Brewster's angle> the reflected <p-polarization> <amplitude> vanishes; only the <s-polarization> component remains. For incidence from air, approximated by <refractive index> one,
$$
\tan\theta_B=n_w=1.33,\qquad \theta_B=53.1^\circ.
$$
Incidence is measured from the surface normal, whereas solar elevation is measured from the horizontal. The <Brewster solar elevation> is therefore
$$
\boxed{h=90^\circ-\theta_B=\tan^{-1}(1/1.33)=36.9^\circ.}
$$
This assumes an ideally flat, transparent, nonmagnetic interface; ripples, absorption and diffuse light would spoil complete <polarization>.
Back to article page
