Question

At the polarising angle $\left({\theta }_B\right)$, angle of refraction is given by :

• A. ${90}^{\circ }$
• B. ${90}^o+{\theta }_B$
• C. ${90}^o-{\theta }_B$
• D. $\frac{{90}^o}{{\theta }_B}$

Answer 1

The correct option is C ${90}^o-{\theta }_B$

At polarizing angle ${\theta }_B$
$\frac{sin{\theta }_B}{cos{\theta }_B}=\frac{sin{\theta }_B}{sin\gamma }$
we get $sin\gamma =cos{\theta }_B$
$sin\gamma =sin(90-{\theta }_B)$
Thus $\gamma =90-{\theta }_B$