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Unformatted text preview: 7.3 Problem 7.3 We have two semiin nite slabs of dielectric material with = and equal indices of refraction n > 1, with an air gap ( n = 1) of thickness d between them. Let the surfaces be in the x; y plane, with the gap being z 2 [0 ; d ] and the incident wave coming from z < 0. In the rst material we have an incident wave ~ E ( ~x ) = ~ E e i ~ k ¡ ~x i!t ~ B ( ~x ) = ~ k ¢ ~ E ( ~x ) =! but we also have a re ected wave ~ E R ( ~x ) = ~ E R e i ~ k R ¡ ~x i!t ~ B R ( ~x ) = ~ k R ¢ ~ E R ( ~x ) =! In the air gap, we may have two waves with oscillatory behavior or exponential behavior in z . We may write either case as ~ E ( ~x ) = ~ E g 1 e i ~ k g 1 ¡ ~x i!t + ~ E g 2 e i ~ k g 2 ¡ ~x i!t ~ B ( ~x ) = ~ k g 1 ¢ ~ E g 1 ( ~x ) =! + ~ k g 2 ¢ ~ E g 2 ( ~x ) =! but remembering that the z components of the wavenumbers may be imaginary. The second slab has only an outgoing wave ~ E ( ~x ) = ~ E 2 e i ~ k 2 ¡ ~x i!t ~ B ( ~x ) = ~ k 2 ¢ ~ E 2 ( ~x ) =! The squares of the wavenumbers are determined by the indices of refraction and ! : k = k R = k 2 = n!=c k 2 g 1 = k 2 g 2 = ! 2 =c 2 but we must keep in mind that ~ k g 1 z may be imaginary, in which case k 2 g 1 = k 2 g 1 x + k 2 g 1 y j k g 1 z j 2 . As for the single interface discussion, we may chose x so that the incident wave is in the x; z plane. As the wave equations and the boundary conditions are invariant under translations in the x and y directions, we can Fourier transform in those directions and see that the equations involve only the same values for the k x 's and for the k y 's, so k x = k R x = k g 1 x = k g 2 x = k 2 x k y = k R y = k g 1 y = k g 2 y = k 2 y = 0 1 From the equality of the k x 's and the relations among the k 2 '2 we have k z = k R z = k 2 z = k cos i , and k g 1 z = k g 2 z = q k 2 g 1 k 2 x = k p n 2 = sin 2 i = k cos r=n , with the angle of re ection for ~ k R and the angle of ~ k 2 equal to the angle of incidence i , and the angle of re ection, r given by Snell's law n sin i = sin r . Note k gi x will be imaginary if n sin i > 1, and r will then be complex. Finally, we can divide the problem into a part ( E ? ) for which ~ E is perpendicular to the plane of incidence ( ~ E k ¦ ^ e y ) and part (E ) in which it lies in the plane of incidence ( E y = 0), in which case ~ B k ¦ ^ e y in the rst material. As the problem is invariant under re ection in the y = 0 plane, in the ( E ? ) case all elds are reversed, so all of the ~ E 's are in the ¦ y direction, and all the ~ B s are in the x; z plane. In the ( E k ) case re ection in the y = 0 plane changes none of the incident elds, and therefore none of the others, so all the ~ E y 's vanish, and all the ~ B 's are perpendicular to the plane of incidence. The boundary conditions are continuity of D z , B z , E x , E y , H x , and H y at each of the two boundaries....
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This note was uploaded on 12/08/2011 for the course PHYSICS 505 taught by Professor Liu during the Winter '08 term at University of Michigan.
 Winter '08
 LIU

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