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MHT CET 2020 (19 Oct, Shift 2) · Polarisation

Polarisation question from MHT CET 2020 (19 Oct, Shift 2)

Q48MHT CET 2020 · 19 OCT · SHIFT 2Medium

Polarisation

The Brewster's angle for the glass-air interface is (54.74)∘(54.74)^\circ. If a ray of light passing from air to glass strikes at an angle of incidence 45∘45^\circ, then the angle of refraction is

[tan⁡(54.74)∘=2,sin⁡45=12]\left[\tan(54.74)^\circ = \sqrt{2}, \sin 45 = \frac{1}{\sqrt{2}}\right]

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Answer: (c)

  1. Brewster's law: μ=tan⁡54.74∘=2\mu = \tan 54.74^\circ = \sqrt{2}
  2. Snell's law: sin⁡r=sin⁡45∘2=1/22=0.5\sin r = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = 0.5
  3. So the answer is (c) sin⁡−1(0.5)\sin^{-1}(0.5).
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