Page 448 - Engineering Electromagnetics, 8th Edition
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430                ENGINEERING ELECTROMAGNETICS

                                                                      3
                                     polarization, or is said to be s-polarized. E is also parallel to the interface, and so
                                     the case is also called transverse electric, or TE polarization. We will find that the
                                     reflection and transmission coefficients will differ for the two polarization types, but
                                     that reflection and transmission angles will not depend on polarization. We only need
                                     to consider s- and p-polarizations because any other field direction can be constructed
                                     as some combination of s and p waves.
                                        Our desired knowledge of reflection and transmission coefficients, as well as how
                                     the angles relate, can be found through the field boundary conditions at the interface.
                                     Specifically, we require that the transverse components of E and H be continuous
                                     across the interface. These were the conditions we used to find   and τ for normal
                                     incidence (θ 1 = 0), which is in fact a special case of our current problem. We will
                                     consider the case of p-polarization (Figure 12.7a) first. To begin, we write down the
                                     incident, reflected, and transmitted fields in phasor form, using the notation developed
                                     in Section 12.4:
                                                                      + − jk · r
                                                               E = E e     + 1                       (51)
                                                                 +
                                                                 s1
                                                                      10
                                                                      − − jk · r
                                                               E = E e     − 1                       (52)
                                                                 −
                                                                 s1
                                                                      10
                                                               E s2 = E 20 e − jk 2 · r              (53)
                                     where
                                                           k = k 1 (cos θ 1 a x + sin θ 1 a z )      (54)
                                                            +
                                                            1
                                                          k = k 1 (− cos θ a x + sin θ a z )         (55)
                                                           −


                                                                       1
                                                                                1
                                                           1
                                                           k 2 = k 2 (cos θ 2 a x + sin θ 2 a z )    (56)
                                     and where
                                                                                                     (57)
                                                                r = x a x + z a z
                                                                      √                        √
                                     The wavevector magnitudes are k 1 = ω   r1 /c = n 1 ω/c and k 2 = ω   r2 /c =
                                     n 2 ω/c.
                                        Now, to evaluate the boundary condition that requires continuous tangential elec-
                                     tric field, we need to find the components of the electric fields (z components) that
                                     are parallel to the interface. Projecting all E fields in the z direction, and using (51)
                                     through (57), we find
                                                               +
                                                  E zs1  = E +  e − jk · r  = E cos θ 1 e − jk 1 (x cos θ 1 +z sin θ 1 )  (58)
                                                    +
                                                                      +
                                                               1
                                                          z10
                                                                      10
                                                                              jk 1 (x cos θ −z sin θ )
                                                                −
                                                   E −  = E  −  e − jk · r  = E cos θ e  1    1      (59)
                                                                       −
                                                                1
                                                                            1
                                                    zs1
                                                                      10
                                                          z10
                                                  E zs2 = E z20 e − jk 2 · r  = E 20 cos θ 2 e − jk 2 (x cos θ 2 +z sin θ 2 )  (60)
                                     3  The s designation is an abbreviation for the German senkrecht, meaning perpendicular. The p in
                                     p-polarized is an abbreviation for the German word for parallel, which is parallel.
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