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(C) $1$. Since there is no reflection,the light must be incident at the Brewster's angle $	heta_B$. Thus,$i = 	heta_B$,where $	an 	heta_B = \mu_B

Vedclass · Accepted Answer

$A, B, D$

Vedclass · Answer

(C) $1$. Since there is no reflection,the light must be incident at the Brewster's angle $	heta_B$. Thus,$i = 	heta_B$,where $	an 	heta_B = \mu_B = \sqrt{3}$. This gives $i = 60^{\circ}$.$2$. The light is polarized in the plane of incidence (parallel to the plane of incidence) for Brewster's angle reflection condition to be satisfied. Thus,statement $(A)$ is correct.$3$. Using Snell's law at the first surface: $1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r_1 \implies \sin r_1 = 1/2 \implies r_1 = 30^{\circ}$.$4$. Given $\delta = 60^{\circ}$ and $\delta = i + e - A$,we have $60^{\circ} = 60^{\circ} + e - A \implies e = A$.$5$. At the second surface,$\sqrt{3} \sin r_2 = 1 \sin e = \sin A$. Since $r_1 + r_2 = A$,$r_2 = A - 30^{\circ}$.$6$. Substituting: $\sqrt{3} \sin(A - 30^{\circ}) = \sin A$. Expanding: $\sqrt{3}(\sin A \cos 30^{\circ} - \cos A \sin 30^{\circ}) = \sin A \implies \sqrt{3}(\sin A \cdot \frac{\sqrt{3}}{2} - \cos A \cdot \frac{1}{2}) = \sin A \implies \frac{3}{2} \sin A - \frac{\sqrt{3}}{2} \cos A = \sin A \implies \frac{1}{2} \sin A = \frac{\sqrt{3}}{2} \cos A \implies 	an A = \sqrt{3} \implies A = 60^{\circ}$. Thus,statement $(B)$ is correct.$7$. Since $e = A = 60^{\circ}$,statement $(D)$ is correct.$8$. For red light,$\mu_R = \frac{\sin((A + \delta_{	ext{min}})/2)}{\sin(A/2)} = \frac{\sin((60^{\circ} + 30^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$. Thus,statement $(C)$ is correct.

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