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Unformatted text preview: ECE 222B, Winter 2010 Homework Set 4 Solutions 1. We wish to determine what fraction of light in glass (for simplicity, let the permittivity of the glass be ε g = 2 ε ), incident from the left upon an air gap with width ‘ is transmitted across the gap into an identical piece of glass. Assume the incident and reflected waves have frequency ω and magnetic fields that are ˆ y-polarized (TM polarization), and that the incidence angle θ i is the critical angle (angle at which total internal reflection first occurs for a single boundary). You may assume that μ = μ in both the glass and gap. (a) What is the input impedance in the gap region? For this part, first write down Z l for arbitrary θ i , then take the limit as θ i approaches the critical angle. Solution: The easiest way to solve this problem is to use the trans- mission line equivalent shown in the figure. The equivalent line wave numbers and impedances are β g = ω √ ε g μ cos θ i and Z g = η g cos θ i for the incident and transmitted regions, and β = ω √ ε μ cos θ and Z = η cos θ in the gap region. We have used θ i = θ t (a consequence of Snell’s Law), and defined η g = p μ /ε g and η = p μ /ε . Finally, θ (angle of propagation in the gap region) is related to θ i by Snell’s Law; namely √ ε g sin θ i = √ ε sin θ The input impedance evaluated at z =- ‘ for the equivalent trans- mission line problem is Z in (- ‘ ) = Z • Z g + jZ tan( β ‘ ) Z + jZ g tan( β ‘ ) ‚ = η cos θ • η g cos θ i...
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This note was uploaded on 10/09/2010 for the course ECE 230 at UCSD.