Question

If ${45}^o$ and $\theta$ is the polarising angle for two optical media whose critical angles are $C_1$ and $C_2$ then the correct relation is

• A. $\sin \theta =\frac{\sin C_2}{\sin C_1}$
• B. $\theta =\frac{\sin C_2}{\sin C_1}$
• C. $\tan \theta =\frac{\sin C_1}{\sin C_2}$
• D. $\sin \theta =\frac{\sin C_1}{\sin C_2}$

Answer 1

Answer: (C)

$\tan \theta =\frac{\sin C_1}{\sin C_2}$

As tani=1/sinC

Hence $\frac{tan\theta }{tan\left({45}^o\right)}=\frac{sin{C}_1}{sin{C}_2}$