According to Brewster's law, the refractive index of a medium is equal to the tangent of the polarizing angle (Brewster's angle), i.e., $\mu = \tan i_b$. For any transparent optical medium placed in air or vacuum, the absolute refractive index is always greater than 1 ($\mu > 1$). Therefore, $\tan i_b > 1$. Since $\tan 45^\circ = 1$ and the tangent function increases with the angle in the first quadrant, it implies that $i_b > 45^\circ$. Also, the maximum possible angle of incidence at an interface is $90^\circ$, so $i_b < 90^\circ$. Thus, the Brewster's angle for an interface must lie in the range $45^\circ < i_b < 90^\circ$.