Practice question
Question
What is the critical angle for light passing from a medium with refractive index 1.9 to air (refractive
index 1.0)?
Explanation
**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. sin i_c = (n₂/n₁) , where n₁ = 1.9 , n₂ = 1.0 . sin i_c = (1.0/1.9) ≈ 0.526 , i_c = sin⁻¹(0.526) ≈ 31.8° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.