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

2024
Views:

195

Pages:

6
Author:

Jabin F.

Ideal Conductor (σ = ∞): Reflection Coefficients: Eor/Eoi = Eot/Eoi = -1, R = R' = 1 Transmission Coefficients: Tr = Ti = 0 Non-Ideal Conductor (σ ≠ ∞): Reflection Coefficients: R ≠ 1, R' ≠ 1 Transmission Coefficients: Tr ≠ 0, Ti ≠ 0 Example: Silver for visible light: R = 93%, T = 7% Diagram: Incoming wave (Eoi) Reflected wave (Eor) Transmitted wave (Eot) Normal component (N) Reflection coefficient (R) Transmission coefficient (T) Left-circular polarization (LCP) Right-circular polarization (RCP) Page 82 Equations: E₁ = E₀n₁cos(ωt - β₁) E₂ = E₀n₂sin(ωt - β₂) Phase Difference: Δφ = β₁ - β₂ (not defined in the text) Diagram: A circle with arrows pointing outward in all directions (representing the electric field E) A horizontal line with arrows pointing to the right (representing the direction of propagation) A vertical line with arrows pointing upward (representing the z-direction) Page 85 Relationship: ω ↑ → R ↑ Question: Which region of the electromagnetic spectrum has the highest reflectivity? Options: (a) Infrared (b) Visible (c) Ultraviolet (d) Entire Statement: For an electromagnetic wave, if the material is opaque, then it's a good conductor. If the material is transparent, then it's a poor conductor. Solution: LCP → RCP λ = 600 nm t = λ / 2(me - mo) t = 600 * 10^-3 / 2(1.551 - 1.542) μm t = 33.33 μm Bottom Section Problem: A laser has an output power (OLP) of 150 mW, a beam diameter of 10 mm, and a wavelength of 630 nm. What is the value of the electric field amplitude in units of V/m? Solution: Intensity (I₀) = (1/2) * c * ε₀ * E₀² = Power / Area E₀ = √(2 * Power / c * ε₀ * Area) E₀ = √(150 * 10^-3 * 2 * 8.854 * 10^-12 / 3 * 10^8 * π * (5 * 10^-3)^2) E₀ = [result of calculation] V/m Equation: E₀ = √(100 * 4 * 9 * 10^9 * 10^-5) Calculation: E₀ = 10 * 2 * 3 * 10² = 6000 V/m Answer: Ans