Chapter 9Reflection, Refractionand PolarizationChapter 9IntroductionWhen you solved Problem 5.2 using the standing-waveapproach, you found a rather curious behavior as the wavepropagates and meets the boundary. A new wave emergesfrom the boundary, traveling in the opposite direction andinverted, as seen in Fig. 111111111110000000000111111110000000011111111000000001111111111000000000011111111000000001111111100000000 After reflection t = 0 Before reflectionFigure 1 The shape of a traveling pulse before and afterreflection.The wave emerging from the boundary is called a reflectedwave. The existence of a reflected wave and its invertednature is ÒautomaticÓ when you solve the problem using thestanding-wave approach of Problem 5.2, because the solu-tions you found there are valid for all times t > 0. On theother hand, we saw in Chapter 6 that the solution to thisproblem for t > 0 can be written asζ ( x , t ) = 1 2 a ( x − ct) + 1 2 a ( x + ct) (1)In the figure above, a(x) is a square shape. From the compu-phy 241 MenŽndez 168Chapter 9tational point of view, this solution is much easier toimplement than the standing-wave approach we used inProblem 5.2. However, while the standing wave solution toProblem 5.2 is valid for all times, it is apparent that Eq. 1 isnot valid when the pulse reaches the boundary. In fact, it isobvious that Eq. 1 doesnÕt know anything about the bound-ary: according to it, the pulses would keep traveling, undis-turbed, to the right and to the left.If we want to use the traveling wave solutions for times afterthe pulse bounces off the boundary, it is clear that we mustmodify Eq. 1. In the particular case of Fig. 1, it is easy to seewhat happens: If a section of the wave a(x-ct) hits the rightboundary at time t1, it is transformed into the wave Ra(x+c(t-t1)), where R = -1. Similarly, if a section of the wavea(x+ct) hits the left boundary at time t1, it is transformedinto the wave R a(x-c(t-t1)), where R = -1. The constant R iscalled the coefficient of reflection.In the next few sections, we will develop a theory of thecoefficient of reflection, allowing for cases in which R is notnecessarily of magnitude 1. This occurs when the boundaryis not a rigid wall. In fact, we will derive the most generalexpression for the coefficient of reflection by considering thecase of two cords joint at one interface. In this case, a wavevibrating in a cord will propagate into the other cord, sothat we will have to define a coefficient of transmission T.Reflection andtransmissionThe wave equation at a discontinuityLet us consider two different cords (for example, cords withmass per unit length ρ1 and ρ2) with a common end at x =0.Each segment satisfies a wave equation of the formM 2 ξ M t 2 = v1 M 2 ξ M x 2 M 2 ξ M t 2 = v2 M 2 ξ M x 2 (2)phy 241 MenŽndez 169Chapter 9 x = 0 (1) (2)Figure 2 Cord with a discontinuity at x = 0.where v1 is the wave speed in segment (1) and v2 the wavespeed in segment (2). Exactly at the boundary x = 0 the waveequation is not satisfied because there is a discontinuity inthe density. All quantities entering a differential equationmust be continuous and differentiable. On the other hand,the displacement ξ, the cord velocity ¶ξ/¶t, and the trans-verse force T¶ξ/¶x (see Chapter 6) are obviously continuousacross the interface. Hence no matter what the solutions tothe two wave equations are, we must havelimε 6 0 ξ ( − ε , t ) = limε 6 0 ξ ( + ε , t ) limε 6 0 M ξ M t ( − ε , t ) = limε 6 0 M ξ M t ( + ε , t ) limε 6 0 TM ξ M x ( − ε , t ) = limε 6 0 TM ξ M x ( + ε , t ) (3)Let us consider the case of a wave traveling from the left tothe right. Suppose that we write the solutions asξ ( 1 ) = A i sin ω ( 1 ) ( t − x v 1 ) , ξ ( 2 ) = A t sin ω ( 2 ) ( t − x v 1 ) (4)where the subscripts ÒiÓ and ÒtÓ mean ÒincidentÓ andphy 241 MenŽndez 170Chapter 9Òtransmitted,Ó respectively. This equations contain four freeparameters: Ai, At, ω(1), and ω(2). However, we should havefive free parameters, since we must satisfy three boundaryconditions plus we are free to select any frequency and anyamplitude for the incident wave. Let us see where theproblem arises. When we apply the first two boundaryconditions at x = 0, we obtainA i sin ω ( 1 ) t = A t sin ω ( 2 ) t ω ( 1 ) A i cos ω ( 1 ) t = ω ( 2 ) A t cos ω ( 2 ) t (5)which can be satisfied for Ai = At, and ω(1) = ω(2). It is notsurprising that the transmitted frequency turns out to beequal to the incident frequency. We can consider theincident wave as an external force acting on medium (2). Aswe learned in Chapter 2, if a system is acted upon by a forceof angular frequency ω, the system responds by oscillating atthe same frequency. This is a very general result for systemsthat satisfy linear equations of motion.When we subject our proposed solutions to the thirdboundary condition, we find at x = 0T A i v 1 cos ω ( 1 ) t = TA t v 2 cos ω ( 2 ) t . (6)This cannot be satisfied for Ai = At unless v1 = v2, in whichcase there is no discontinuity. Clearly, if there is a disconti-nuity our solutions in Eq. (4) cannot be valid. Something ismissing. The missing term is precisely the reflected wave. Inorder to satisfy all boundary conditions we must thereforepropose solutions of the formξ ( 1 ) = A i sin ω ( 1 ) ( t − x v 1 ) + A r sin ω ( 1 ) ( t + x v 1 ) ξ ( 2 ) = A t sin ω ( 2 ) ( t − x v 1 ) (7)where the subscript ÒrÓ in the second term of ξ(1) meansÒreflected.Ó Using these solutions we find that all boundaryconditions can be satisfied providedphy 241 MenŽndez 171Chapter 9ω ( 1 ) = ω ( 2 ) A t = 2 v 2 v 1 + v 2 Ai A r = v 2 − v 1 v 1 + v 2 Ai (8)We can thus define the reflection and transmission coeffi-cients asA r A i = R = v 2 − v 1 v 1 + v 2 A t A i = T = 2 v 2 v 1 + v 2 (9)Let us now consider some limit cases. If the two velocitiesare equal, i.e., there is not discontinuity, we obtain T = 1and R = 0. Hence the incident waves travels through x = 0unperturbed. Suppose now that the cord in medium (2) isinfinitely heavy. Then we approach the boundary conditionof problem 5.2, where the cord (1) is attached to a …