sol1_ex2 - Massachusetts Institute of Technology Physics...

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± Massachusetts Institute of Technology Physics 8.03 Exam 2 Solutions Tuesday, November 23, 2004 Solution for Problem 1 (a) Boundary condition is E y =0 at x ,L E ( x, z, t )= E 0 sin( k x x ) cos( ωt + k z z (1) y for the n th mode, k x = L , where n =1 , 2 , 3 ... B (b) Faraday’s law ± E = ± ∇× ± ∂t ± ∂E y ˆ ± E = y z ˆ x ∂x ∂z z + k z E 0 sin( k x x )sin( + k z z = k x E 0 cos( k x x ) cos( + k z z x (2) B ( x, t k z E 0 sin( k x x ) cos( + k z z z ± x k x E 0 cos( k x x + k z z (3) ω ω again for the n th mode, k x = L , where n , 2 , 3 ( (c) ω 2 = c 2 ( k 2 + k z 2 ) ω = c L ) 2 + k 2 x z ± ² 2 ³ 1 / 2 ± ω 2 ´ ² 2 ³ 1 / 2 1 ω 2 ´ nπc (4) k z = = c 2 L c L Phase velocity ω kc ωc v p z = = = ± ´ ² 2 ³ 1 / 2 (5) k z k z ω 2 nπc L Group velocity ± ² 2 ³ 1 / 2 k z c c 2 c ω 2 ´ nπc (6) v gr z = = = k z = dk z k ω ω L πc (d) n ,k z = L (e) Now ω = k z c v p z = c and v z = c . There is No dispersion! There is No cut-off frequency!
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This note was uploaded on 04/22/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.

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sol1_ex2 - Massachusetts Institute of Technology Physics...

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