JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Geometrical Optics

Since there is no reflection from the surface, the incident angle equals Brewster's angle: $\tan {\theta }_B={\mu }_B=\sqrt{3}$, giving $i={\theta }_B=60°$. At Brewster's angle, the reflected light would be polarized perpendicular to the plane of incidence, but since there's no reflection, the transmitted light is polarized in the plane of incidence. Option (A) is correct.

Applying Snell's law at entry: $\sin 60°=\sqrt{3}\sin r_1$, giving $r_1=30°$.

Using the deviation formula: $\delta =(i+e)-A$, we have $60°=60°+e-A$, so $e=A$.

At exit: $\sqrt{3}\sin r_2=\sin e=\sin A$.

Since $r_1+r_2=A$ and $r_1=30°$, we have $r_2=A-30°$.

Substituting: $\sqrt{3}\sin (A-30°)=\sin A$.

Expanding and solving gives $A=60°$, so $e=60°$. Options (B) is incorrect (should be 60°, not 45°), and (D) is correct.

For red light at minimum deviation: ${\mu }_R=\frac{\sin \frac{A+{\delta }_{\min }}{2}}{\sin \frac{A}{2}}=\frac{\sin 45°}{\sin 30°}=\frac{1/\sqrt{2}}{1/2}=\sqrt{2}$. Option (C) is correct.