Physics - Wave Optics Question with Solution | TestHub

Brewster's angle is defined as the angle of incidence at which the reflection of particular polarized light is zero, and it perfectly transmitted through a transparent surface. 

Brewster's angle($i_{\mathrm{b}}$) is given by, $\tan \left(i_{\mathrm{b}}\right)=\frac{{\mu }_2}{{\mu }_1}=\frac{{\mu }_2}{1}$, where ${\mu }_2$ is the refractive index of the dielectric and ${\mu }_1$ is the refractive index of air.
We know that, ${\mu }_2>1$. (refractive index of any dielectric will be greater than refractive index of air.)
$\Rightarrow \infty <\tan i_b>1$
$\Rightarrow {90}^{\mathrm{o}}>i_b>{45}^{\mathrm{o}}$