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-

The correct option is C 53.74^∘ The refractive index of water, μ= Speed of light in air #160;Speed of light in air= 3×10^82.2× 10^8=1.36 From Brews ...

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Standard XII

Physics

Brewster's Angle

The velocity ...

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:

45^{\circ}

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50^{\circ}

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{53.74}^{\circ}

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63^{\circ}

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Solution

The correct option is C ${53.74}^{\circ}$
The refractive index of water, $μ = \frac{Speed of light in air}{Speed of light in air} = \frac{3 \times 10^{8}}{2.2 \times 10^{8}} = 1.36$

From Brewster's Law:-
$t a n i_{p} = μ = 1.36$
$∴ i_{p} = t a n^{- 1} (1.36) = {53.74}^{o}$

So correct option is $C$

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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:

The correct option is C 53.74∘The refractive index of water, μ=Speed of light in air Speed of light in air=3×1082.2×108=1.36From Brewster's Law:-tan ip=μ=1.36∴ip=tan−1(1.36)=53.74oSo correct option is C

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: