Page 446 - Engineering Electromagnetics, 8th Edition
P. 446

428                ENGINEERING ELECTROMAGNETICS

                                     on the figure. The phase velocity along x is the velocity of the intersection points
                                     between the phase fronts and the x axis. Again, from the geometry, we see that this
                                     velocity must be faster than the velocity along k and will, of course, exceed the
                                     speed of light in the medium. This does not constitute a violation of special relativity,
                                     however, since the energy in the wave flows in the direction of k and not along x or z.
                                     The wave frequency is f = ω/2π and is invariant with direction. Note, for example,
                                     that in the directions we have considered,
                                                                  ν p  ν px  ω
                                                              f =   =     =
                                                                  λ    λ x  2π

                   EXAMPLE 12.6
                                     Consider a 50-MHz uniform plane wave having electric field amplitude 10 V/m. The
                                     medium is lossless, having   r =   = 9.0 and µ r = 1.0. The wave propagates in the

                                                                r
                                     x, y plane at a 30 angle to the x axis and is linearly polarized along z. Write down
                                                   ◦
                                     the phasor expression for the electric field.
                                     Solution. The propagation constant magnitude is
                                                              √                 6
                                                      √      ω   r  2π × 50 × 10 (3)      −1
                                                 k = ω µ  =       =                 = 3.2m
                                                              c         3 × 10 8
                                     The vector k is now
                                                  k = 3.2(cos 30a x + sin 30a y ) = 2.8a x + 1.6a y m −1
                                     Then

                                                                r = x a x + y a y
                                     With the electric field directed along z, the phasor form becomes
                                                       E s = E 0 e − jk · r  a z = 10e − j(2.8x+1.6y)  a z


                                        D12.4. For Example 12.6, calculate λ x , λ y , ν px , and ν py .

                                                              8
                                                                         8
                                        Ans. 2.2 m; 3.9 m; 1.1 × 10 m/s; 2.0 × 10 m/s

                                     12.5 PLANE WAVE REFLECTION AT
                                             OBLIQUE INCIDENCE ANGLES

                                     We now consider the problem of wave reflection from plane interfaces, in which
                                     the incident wave propagates at some angle to the surface. Our objectives are (1) to
                                     determine the relation between incident, reflected, and transmitted angles, and (2) to
                                     derive reflection and transmission coefficients that are functions of the incident angle
                                     and wave polarization. We will also show that cases exist in which total reflection or
                                     total transmission may occur at the interface between two dielectrics if the angle of
                                     incidence and the polarization are appropriately chosen.
   441   442   443   444   445   446   447   448   449   450   451