An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then-

To find the correct properties of the reflected and transmitted light when unpolarized light is incident at Brewster's angle, we can use the wave theory of light and polarization principles.

First, let's understand Brewster's law. According to Brewster's law, when unpolarized light is incident on the boundary of a medium of refractive index $\mu$ at a specific angle called Brewster's angle (${\theta }_p$ or $i_p$), the reflected light is completely plane-polarized.

The relation between the refractive index of the medium $\mu$ and Brewster's angle $i_p$ is given by:
$\tan \left(i_p\right)=\mu$

Given:
Refractive index of the medium, $\mu =1.73$ (which is close to $\sqrt{3}\approx 1.732$).

Substituting this value into Brewster's formula:
$\tan \left(i_p\right)=1.73\approx \sqrt{3}$

Therefore, the angle of incidence (Brewster's angle) is:
$i_p={\tan }^{-1}\left(\sqrt{3}\right)=60°$

According to the laws of reflection, the angle of reflection ($r$) is equal to the angle of incidence ($i$):
Angle of reflection = $i_p\approx 60°$

Furthermore, by definition of Brewster's angle, the reflected light is completely polarized perpendicular to the plane of incidence, whereas the refracted (transmitted) light is only partially polarized.

Thus, the reflected light is completely polarized and the angle of reflection is close to 60°.