University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

423

Pages:

6
Author:

Jabin F.

Content: Equation: 2ut * cos(θt) = (2m + 1) * λ/2 Condition for destructive interference: 2ut * cos(θt) = (2m + 1) * λ/2 Problem: A thin film of water (n = 1.33) floats on the surface of a beaker of silicone oil (ns = 1.40). The arrangement is illuminated by 600 nm light incident normally from the top. A large region of the film appears bright red. What is the minimum possible thickness of the film (in nm)? Diagram: Air Water (n = 1.33) Silicone oil (ns = 1.40) Solution: Since the light is incident normally, θi = 0°. Using Snell's Law: n1 * sin(θ1) = n2 * sin(θ2) 1 * sin(0°) = 1.33 * sin(θt) sin(θt) = 0 θt = 0° For bright interference (constructive interference): 2ut * cos(θt) = m * λ Substituting the values: 2 * 1.33 * t * cos(0°) = m * 600 nm Simplifying: t = (m * 600 nm) / (2 * 1.33) For the minimum thickness, m = 1. Therefore, the minimum thickness is t = (1 * 600 nm) / (2 * 1.33) ≈ 225.56 nm Top Section 2mt + λ/2 + λ/2 = mλ 2mt = (m - 1) * λ For m = 2: 2t = λ / (2 * 1.33) t = 225.56 nm Bottom Section For normal incidence: 2nt + λ/2 = mλ → For bright fringes: 2nt = (m - 1) * λ (i) Reflection Coefficient (R): R = ((n₁ - n₂) / (n₁ + n₂))^2 R = ((1 - 1.5) / (1 + 1.5))^2 = 4% % of energy reflected: 4% % of energy transmitted: 96% (iii) Brewster's Angle (θB): θB = tan⁻¹(n₂) θB = tan⁻¹(1.5) θB ≈ 53°