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Advanced Electromagnetics

Jackson 7.4 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell UMASS LOWELL C.S.BAIRD PROBLEM: A plane-polarized electromagnetic wave of frequency w in free space is incident normally on the flat surface of a non-permeable medium of conductivity o and dielectric constant e. (a) Calculate the amplitude and phase of the reflected wave relative to the incident wave for arbitrary o and e. (b) Discuss the limiting case of a very poor and a very good conductor, and show that for a good conductor the reflection coefficient (ratio of reflected to incident intensity) is approximately R~1-2"g c where is the skin depth. SOLUTION: (a) If we set up the problem in the usual way with the incident wave E traveling in the positive z direction, the transmitted wave E' traveling in the positive z direction, and the reflected wave E" traveling in the negative z direction, we get from the boundary conditions: E,'=E.+E." k'E.'=kE.-kE." We can use both of these to solve for the reflected wave in terms of the incident wave: k k-k' k+k' c The wave number in the other material obeys: k '=Ve(w) o w For a conducting material we can use e(w)=e+i - so that we have 3 toa Plugging these in above leads to: E E. Eo m3 The problem asks us to explicitly expand this in terms of amplitude and phase. The amplitude of a complex number is defined as |z|= V z * z so that we have E E VEO Eo E m3 m3 E E m3 m3 g E Eo m 3 Ec E The phase of a complex number is Arg(z)=tan-1 (b) Discuss the limiting case of a very poor and a very good conductor, and show that for a good conductor the reflection coefficient (ratio of reflected to incident intensity) is approximately R~1-2"s where is the skin depth. For a very poor conductor o << &o which means the same as o/2o << 1. This leads to: E. E. Eo E +2 VE E E. Now recognize: 2 =ly/x so that ( Jvx+iy= 2 We must be careful because the arctan leads to some ambiguity of what quadrant the phase should be in. We know from physical arguments that the phase difference should be for a perfect dielectric, so that our answer should reduce to 1 for o->0. The answer must be in the third quadrant: 9 Arg Eo For a very good conductor o >> &oW V2 m3 m 03 1+ m3 E." which leads to E. after dropping small terms m 3 E. 0