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Q.
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
Wave Optics
Solution:
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) $
$= 53. 74^{\circ}$