Question

The velocity of light in air is $3 \times 10^8 \,m s^{-1}$ and that in water is $2.2 \times 10^8\, m s^{-1}$. The polarising angle of incidence is

• A. $45^{\circ}$
• B. $50^{\circ}$
• C. $53.74^{\circ}$
• D. $63^{\circ}$

Answer 1

Answer: (C)

$53.74^{\circ}$

Answer 2

The refractive index of water $\mu = \frac{\text{speed of light in air}}{\text{speed of light in water }}$ $=\frac{3 \times 10^8}{2.2 \times 10^8} = 1.36$ From Brewsters law, $tan \, i_p = \mu = 1.36$ $\therefore i_p = tan^{-1} (1.36)$ $= 53. 74^{\circ}$