Question

The relation between the polarising angle and the critical angle is

• A. $i_p={\tan }^{-1}\left(\text{cosec}{\theta }_c\right)$
• B. $i_p={\tan }^{-1}\left(\sin {\theta }_c\right)$
• C. $i_p={\cot }^{-1}\left(\sin {\theta }_c\right)$
• D. $i_p={\cot }^{-1}\left(\cos ec{\theta }_c\right)$

Answer 1

The correct option is A $i_p={\tan }^{-1}\left(\text{cosec}{\theta }_c\right)$

If the polarizing angle is $i_p$and ${\theta }_c$ is the critical angle, the relation between the two is given by

$\tan \left(i_p\right)=\frac{1}{\sin \left({\theta }_c\right)}⟹i_p={\tan }^{-1}\left(\arcsin \right({\theta }_c\left)\right)$