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we must have

                                                           θ i = θ r .
                        This is known as Snell’s law of reflection. We can similarlydefine the transmission angle
                        to be the angle between the direction of propagation of the transmitted wavefronts and
                        the interface normal. Noting that k · ˆ z = cos θ t and k · ˆ x = sin θ t , we have from (4.281)
                                                                      ˆ
                                                      ˆ
                                                                       2
                                                       2
                        and (4.282)
                                                               t
                                                              τ cos γ t
                                                                            ,                 (4.292)
                                               cos θ t =
                                                             2
                                                         2
                                                                    t 2
                                                                         2
                                                        β sin θ i + (τ ) cos γ  t
                                                         1
                                                              β 1 sin θ i
                                                                            ,                 (4.293)
                                               sin θ t =
                                                             2
                                                         2
                                                                    t 2
                                                                         2
                                                        β sin θ i + (τ ) cos γ  t
                                                         1
                        and thus

                                                    θ t = tan −1  β 1 sin θ i  .              (4.294)
                                                                t
                                                              τ cos γ t
                          Depending on the properties of the media, at a certain incidence angle θ c , called the
                                                                                           ˆ
                        critical angle, the angle of transmission becomes π/2. Under this condition k has only
                                                                                            2
                        an x-component. Thus, surfaces of constant phase propagate parallel to the interface.
                        Later we shall see that for low-loss (or lossless) media, this implies that no time-average
                        power is carried bya monochromatic transmitted wave into the second medium.
                          We also see that although the transmitted field maybe a nonuniform plane wave, its
                        mathematical form is that of the incident plane wave. This allows us to easilygeneralize
                        the single-interface reflection problem to one involving manylayers.
                        Uniform plane-wave reflection for lossless media.  We can specialize the preceding
                                                                                        c
                        results to the case for which both regions are lossless with ˜µ = µ and ˜  =   real and
                        frequency-independent. By (4.224) we have
                                                              √
                                                         β = ω µ ,
                        while (4.225) gives
                                                           α = 0.
                        We can easilyshow that the transmitted wave must be uniform unless the incidence angle
                        exceeds the critical angle. By(4.279) we have
                                                      2
                                                           2
                                                              2
                                                 A = β − β sin θ i ,  B = 0,                  (4.295)
                                                          1
                                                      2
                        while (4.280) gives


                                                       2 1/4    2    2  2
                                                 τ = A     =   |β − β sin θ i |
                                                                2    1
                        and
                                                            1
                                                         t      −1
                                                       γ =    tan (0).
                                                            2
                                                                                          t
                                                         t
                        We have several possible choices for γ . To choose properlywe note that γ represents
                                                                √
                                                             t
                        the negative of the phase of the quantity k =  A.If A > 0 the phase of the square root
                                                            z
                                                                              t
                        is 0.If A < 0 the phase of the square root is −π/2 and thus γ =+π/2. Here we choose
                                        t
                        the plus sign on γ to ensure that the transmitted field decays as z increases. We note
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