Physics - Wave Optics Question with Solution | TestHub

According to Brewster's law, the tangent of the polarizing angle (Brewster's angle) is equal to the refractive index of the medium.

$\tan {\theta }_B=\mu$

Given $\mu =\frac{3}{2}=1.5$

$\tan {\theta }_B=1.5$

We know that $\tan 45^{\circ }=1$ and $\tan 90^{\circ }\to \infty$.

Since $1<1.5<\infty$, the angle ${\theta }_B$ must lie between $45^{\circ }$ and $90^{\circ }$.

$45^{\circ }<{\theta }_B<90^{\circ }$

Answer: $45^{\circ }<{\theta }_B<90^{\circ }$