final_exam_222A_ 2000

final_exam_222A_ 2000 - ECE 222A Final Exam Dec 8, 2000...

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ECE 222A Final Exam Dec 8, 2000 Name Instructions. This a is three hour, closed book, examination. Calculators are permitted, but not needed. 1. (10 points) An exotic material has the properties μ = , σ c = 0, and ² = 0. What boundary conditions does such a surface impose on ~ E , ~ H , ~ B , and ~ D in an adjacent free space region at nonzero frequencies? Hint: Do not allow fields with infinite magnitudes. Also, do not allow currents or charges other than those owing to conduction effects. Solution : First, we know that since μ = , it follows that ~ H = 0 inside the material. We can’t say anything about ~ B because μ ~ H = ∞ · 0 is indeterminant. We also have ~ D = 0 inside the material, a consequence of ² = 0 . The electric field inside the material is unrestricted. Finally, there are no currents (or conduction electrons) since σ c = 0 . The four boundary conditions (valid for all materials) are ˆ n · h ~ D 2 - ~ D 1 i = 0 ˆ n · h ~ B 2 - ~ B 1 i = 0 ˆ n × h ~ E 2 - ~ E 1 i = 0 ˆ n × h ~ H 2 - ~ H 1 i = 0 where region “1” refers to the exotic material and region“2” is the vacuum. If we make use of the specific material properties of region “2,” we can refine the boundary conditions to ˆ n · ~ D 2 = ˆ n · ~ E 2 = 0 ˆ n · h ~ B 2 - ~ B 1 i = 0 ˆ n × h ~ E 2 - ~ E 1 i = 0 ˆ n × ~ H 2 = ˆ n × ~ B 2 = 0 2. (15 points) A beam of light is propagating inside glass with an index of refraction n i = p ² g 0 = 2, where ² g and μ g are the permittivity and permeability of the glass, respectively, and where we assume that μ g = μ 0 . The beam is incident upon a planar interface with a vacuum ( n 0 = 1) at an angle θ that exceeds the critical angle of total reflection θ c (see Figure 1
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2). The resulting evanescent wave in the vacuum decays as exp[ - z/ Δ]. What is the smallest possible value for the penetration depth Δ? Your answer should express Δ as a function of the wavenumber β g p ω 2 ² g μ 0 Solution : For the incident wave in the glass, we know that β 2 g = ω 2 ² g μ 0 = β 2 x + β 2 z where β x = β g sin θ i , and β z = β g cos θ i , where the direction normal to the surface is z. We assume that the incidence angle exceeds the critical angle, so it follows that in the vacuum we have β 2 0 = β 2 x - Δ - 2 where we have used Snell’s Law in equating
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final_exam_222A_ 2000 - ECE 222A Final Exam Dec 8, 2000...

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