(YES) Here,the incident ray is in the denser medium (medium-$2$) and the refracted ray is in the rarer medium (medium-$1$). Hence,according to Brewster's law,
$\tan \theta_{P} = \frac{n_{1}}{n_{2}} \quad \ldots (1)$
(Where $\theta_{P} =$ polarisation angle or Brewster angle)
Here,if the critical angle of the denser medium with respect to the rarer medium is $C$,then according to Snell's law,$n_{2} \sin C = n_{1} \sin 90^{\circ}$
$\therefore \sin C = \frac{n_{1}}{n_{2}} \quad \ldots (2)$
From equations $(1)$ and $(2)$,$\tan \theta_{P} = \sin C$
$\therefore \frac{\sin \theta_{P}}{\cos \theta_{P}} = \sin C$
$\therefore \sin \theta_{P} = (\cos \theta_{P}) \sin C$
But here $0 < \cos \theta_{P} < 1$
$\Rightarrow \sin \theta_{P} < \sin C$
$\therefore \theta_{P} < C$
For a ray of light incident on the surface of the rarer medium at the polarisation angle,satisfying the above condition,the reflected light will be completely plane polarised.