Question

The velocity of light in air is $3\times {10}^8$ m/s and in a medium is $\sqrt{3}\times {10}^8$ m/s. The Brewster’s angle of the medium is

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

$\mu =\frac{C_a}{C_m}=tan{i}_p$
$\frac{3\times {10}^8}{\sqrt{3}\times {10}^8}=tan{i}_p$
$tan{i}_p=\sqrt{3}$