A transparent liquid has Brewster angle 53.13° for light incident from air. Its density is related to its refractive index by ρ = 750μ kg·m⁻³. The liquid flows steadily through a horizontal pipe whose cross-sectional area decreases from A to A ÷ 2. Its speed in the wider section is 2 m·s⁻¹. Neglect viscosity. If the pressure drop is N kPa, find N. Use tan53.13° = 4 ÷ 3.
Physics · Wave Optics · polarisation
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Question

A transparent liquid has Brewster angle 53.13° for light incident from air. Its density is related to its refractive index by ρ = 750μ kg·m⁻³. The liquid flows steadily through a horizontal pipe whose cross-sectional area decreases from A to A ÷ 2. Its speed in the wider section is 2 m·s⁻¹. Neglect viscosity. If the pressure drop is N kPa, find N. Use tan53.13° = 4 ÷ 3.

Physics · Wave Optics · polarisation

G1 Physics · Accepted Answer

From Brewster's law, μ = tan i(B). Therefore, μ = 4 ÷ 3. The liquid density is ρ = 750μ = 750×4 ÷ 3 = 1000 kg·m⁻³. By continuity, A₁v₁ = A₂v₂. Since A₂ = A₁ ÷ 2, v₂ = 2v₁ = 4 m·s⁻¹. For horizontal flow, Bernoulli's equation gives P₁+(1 ÷ 2)ρv₁² = P₂+(1 ÷ 2)ρv₂². Hence, P₁−P₂ = (1 ÷ 2)ρ(v₂²−v₁²). Substitute: P₁−P₂ = (1 ÷ 2)(1000)(4²−2²) = 500(16−4) = 6000 Pa = 6 kPa. Therefore, N = 6.