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au10_312ps6_sol

# au10_312ps6_sol - ECE312 Autumn 2010 Problem Set 6...

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ECE312 Autumn 2010: Problem Set 6 Solutions Problem 1 : 3 GHz plane wave ˜ E i = (3ˆ x - y - z ) e jA (4 x +3 z ) (a) By examining e jA (4 x +3 z ) = e j 5 A ( 4 x 5 + 3 z 5 ) = e jk 1 ( x sin θ i + z cos θ i ) We find sin θ i = 4 / 5, cos θ i = 3 / 5, so that θ i = 53 . 13 (b) From part (a), 5 A = k 1 = ω μǫ , therefore A = 2 π (3 × 10 9 ) / (5(3 × 10 8 )) = 4 π rad/s . (c) In our coordinate system, ˆ y components of the electric field are perpen- dicular polarized because they are perpendicular to the plane formed by the incident vector ˆ k = ˆ x sin θ i + ˆ z cos θ i and the surface normal ˆ z . Thus ˜ E i = - ye j 4 π (4 x +3 z ) (d) Parallel polarized part of ˜ E i lies in the xz -plane, ˜ E i bardbl = (3ˆ x - z ) e j 4 π (4 x +3 z ) (e) The magnetic field can be found by ( η 377Ω) ˜ H i = 1 η ˆ k × ˜ E i = 1 η parenleftbigg x 5 + z 5 parenrightbigg × (3ˆ x - y - z ) e j 4 π (4 x +3 z ) = x + 25ˆ y - z 5 η e j 4 π (4 x +3 z ) Problem 2 : Fiber optic ǫ = 2 . 2 ǫ 0 , f = 420 THz. (a) The critical angle θ c = sin 1 radicalbig ǫ 2 1 = sin 1 (1 / 2 . 2) = 42 . 39 .

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