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Answer :
a) The Brewster’s angle for both interfaces is 57.2° and 46.3° respectively. b) amber oil interface will serve the critical angle. c) The transit time is calculated to be 2.46 × 10⁻⁹ s.
Brewster’s angle is also referred to as the polarization angle. It is the angle at which a non-polarised EM wave (with equal parts vertical and horizontal polarization)
a) For air-amber pair,
μ = nₐ/n
μ = 1.55
brewster angle
θair amber = tan⁻¹(1.55)
= 57.2°
ii) For amber oil pair
μ = nₐ/n₀ = 1.55/ 1.48
= 1.047
Brewster angle θ oil amber = tan⁻¹ (1.047)
= 46.3°
b) The interface amber oil will serve for critical angle and
θc = sin⁻¹ = 1.48/1.55 = 72.7°
c) As θ₁ = 48°, na = sinθ₁ /sin θ₂
θ₂ = sin⁻¹(sinθ₁/na)
= sin⁻¹ ( sin 48/1.55)
= 28.65°
Now sinθ₂/sinθ₃ = 1.48/1.55
sinθ₃ = 1.48/1.55 × sin(28.65)
θ₃ = 30
The time taken to reach p to q
= 1/c [n₁/sinθ + t × nₐ/ sin θ₂ +n₂× n₀/sin θ3
= 2.46 × 10⁻⁹ s.
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