Example 1 solution

# Example 1 solution - 1 = 1.69 0 x 1 = 0 2 = 1.44 0 2 = 0 z...

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x z i k G i H G i E G 80 o in θ = 10 1.69 εε = 20 1.44 = μμ = = (a) 11 1.69 1.3 r n ε == = and 22 1.44 1.2 r n = . Critical angle is 1 2 1 sin c n n ⎛⎞ = ⎜⎟ ⎝⎠ 67.38 o c = . Since angle of incidence is larger than the critical angle, there is total internal reflection. (b) The incident wave is TE polarized since its electric field vector is perpendicular or transverse to the plane if incidence. (c) 0 xi zi jk x jk z iy Ea E e e −− = G G , 0 xr zr jk x jk z ry E aE e e G G , 00 xt zt xt jk x jk z jk x z ty y E aTEe e e α G G G Under total internal reflection i e φ Γ= and 1 T = +Γ . () 1 1 2 sin sin x ix rx t i n i n n kkkk π λ === = 1 zi xi kk k =− , 1 zr xi k , 2 2 1 zt xi xi k k kj = −= , since 2 xi > under total internal reflection. So 2 2 2 12 21 2 2 sin sin xi in in nn n n ππ αθ λλ = = Using given numbers we find
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## This note was uploaded on 01/28/2012 for the course ECE 135 taught by Professor Dagli during the Spring '08 term at UCSB.

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