= Solution
At an air–glass interface, resolve the incident <unpolarized light> into components perpendicular ($s$) and parallel ($p$) to the plane of incidence. The <Fresnel equations> give the parallel reflection coefficient
$$
r_p=\frac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t},\qquad n_1\sin\theta_i=n_2\sin\theta_t.
$$
At <Brewster's angle>, $r_p=0$. Together with <Snell's law>, this gives $\theta_i+\theta_t=\pi/2$ and $\tan\theta_i=n_2/n_1$. Thus, taking the incident medium to be air,
$$
\boxed{\theta_i=\arctan(1.5)=56.31^\circ\text{ from the surface normal}.}
$$
The remaining reflected component has <linear polarization> \b[perpendicular to the plane of incidence]. This assumes a transparent, nonmagnetic glass with negligible absorption.
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