The velocity of light in air is 3 × 108 m s-1 and that in water is 2.2 × 108 m s-1. The polarising angle of incidence is

Question

The velocity of light in air is 3 × 108 m s-1 and that in water is 2.2 × 108 m s-1. The polarising angle of incidence is (A) 45° (B) 50° (C) 53.74° (D) 63°

Tardigrade Admin · Accepted Answer

The refractive index of water μ = ( textspeed of light in air/ textspeed of light in water ) =(3 × 108/2.2 × 108) = 1.36 From Brewsters law, tan ip = μ = 1.36 ∴ ip = tan-1 (1.36) = 53. 74°

Q. The velocity of light in air is $3 \times 1 0^{8} m s^{- 1}$ and that in water is $2.2 \times 1 0^{8} m s^{- 1}$ . The polarising angle of incidence is

$4 5^{\circ}$

$5 0^{\circ}$

$53.7 4^{\circ}$

$6 3^{\circ}$

The refractive index of water
$μ = \frac{speed of light in air}{speed of light in water}$
$= \frac{3 \times 1 0 ^{8}}{2.2 \times 1 0 ^{8}} = 1.36$
From Brewsters law,
$t an i_{p} = μ = 1.36$
$∴ i_{p} = t a n^{- 1} (1.36)$
$= 53.7 4^{\circ}$

Q. The velocity of light in air is 3×108ms−1 and that in water is 2.2×108ms−1. The polarising angle of incidence is

45∘

50∘

53.74∘

63∘