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The Brewster's angle i_b for an interface should be

Asked in NEET Sep. 2020 · Brewster's law and polarisation by reflection

(1) 30°<i_b<45°
(2) 45°<i_b<90°
(3) i_b=90°
(4) 0°<i_b<30°

Answer: (2) 45°<i_b<90°

Step-by-step solution

Given: an interface between air and a denser medium of refractive index μ.

Idea: Brewster's law fixes the polarising angle through tan i_b=μ.

For any ordinary medium denser than air, μ>1.

tan i_b>1 means i_b>45°, since tan 45°=1.

The angle of incidence can never reach 90° for a real refraction.

So 45°<i_b<90°.

Why the other options are wrong

(1) tan i_b=μ and μ>1, so i_b must exceed 45°, where the tangent is exactly 1.
(3) At 90° the light grazes the surface and tan i_b would be infinite, needing an infinite refractive index.
(4) Even further below 45°, which would need μ<1 for the denser medium.

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