homework_9_solution

homework_9_solution - ° e x i ° E ° e y e − i ω t ik...

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E 0 E ° 0 E °° 0 ° E E 0 E ° 0 = E °° 0 ° B E 0 + E °° 0 = n · E ° 0 E 0 = n +1 n 1 E °° 0 R = 1 2 E °° 2 0 1 2 E 2 0 = | 1 n 1+ n | 2 ∇× ( ∇× ° E )+ 1 c 2 2 t 2 ° D =0 ° k × ( ° k × ° E )+ ω 2 c 2 ° D =0
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° k = k ° n ° n ( ° n · ° E ) ° E + v 2 c 2 ° D =0 ± ij [( n i n j δ ij )+ v 2 ( δ ij /v 2 j )] E j =0 det(( n i n j δ ij )+ v 2 ( δ ij /v 2 j )) = 0 = v 2 v 2 1 v 2 2 v 2 3 [ n 2 1 ( v 2 v 2 2 )( v 2 v 2 3 )+ n 2 2 ( v 2 v 2 1 )( v 2 v 2 3 )+ n 2 3 ( v 2 v 2 1 )( v 2 v 2 2 )] ∇× ( ∇× ° E )+ ω 2 c 2 ° D = ∇× ( ∇× ° E )+ ω 2 μ 0 ( ± 0 ° E + γ ∇× ° E )=0 | ° k | 2 ° E + ω 2 c 2 ( ° E + γ ± 0 ° k × ° E )=0 ( k 2 ± ω 2 c 2 )( E x ± iE y )=0 k ± k 2 ± ω 2 c 2 = ± ω 2 c 2 γ ± 0 k ± k ± = ω c (1 ± γω 2 ± 0 c ) n ± =1 ± γω 2 ± 0 c
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° E ( z,t )= ° E R ( z,t )+ ° E L ( z,t )=( E
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Unformatted text preview: ° e x + i ° E ° e y ) e − i ω t + ik − z +( E ° e x + i ° E ° e y ) e − i ω t + ik + z E y E x = tan[ ( k + − k − ) 2 L ] φ = ( k + − k − ) 2 L = γ L ± ω 2 c 2...
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This note was uploaded on 12/29/2011 for the course PHYSICS 606 taught by Professor Bedaque during the Spring '11 term at Maryland.

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homework_9_solution - ° e x i ° E ° e y e − i ω t ik...

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