An unpolarized beam of light travelling in a transparent dense...

For light travelling from a dense medium of refractive index $n_1$ to a rarer medium of refractive index $n_2$, Brewster's angle $\theta_B$ is given by:

$\tan \theta_B = \dfrac{n_2}{n_1}$

Given $n_1 = \sqrt{3}$ and $n_2 = 1$ (air), we have:

$\tan \theta_B = \dfrac{1}{\sqrt{3}} \Rightarrow \theta_B = 30^{\circ}$

At Brewster's angle, the reflected light is completely polarized perpendicular to the plane of incidence.

The angle of incidence is $i = \theta_B = 30^{\circ}$. By the law of reflection, the angle of reflection is also $30^{\circ}$.

To find the angle of refraction $r$, we use Snell's law:

$n_1 \sin i = n_2 \sin r$

$\sqrt{3} \sin 30^{\circ} = 1 \times \sin r$

$\sqrt{3} \left(\dfrac{1}{2}\right) = \sin r \Rightarrow \sin r = \dfrac{\sqrt{3}}{2}$

$\Rightarrow r = 60^{\circ}$

Alternatively, at Brewster's angle, the reflected and refracted rays are mutually perpendicular, so $i + r = 90^{\circ}$.

$30^{\circ} + r = 90^{\circ} \Rightarrow r = 60^{\circ}$

Thus, the reflected light is completely polarized and the angle of refraction is $60^{\circ}$.

Answer: The reflected light is completely polarized and the angle of refraction is $60^{\circ}$.