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This solves the problem that we wanted to solve, because it represents the result as a complex number of magnitude AR and phase R To see how this method works, let us find the amplitude AR which is "length"of R. To get the"length"of a complex quantity, we always multiply e quant by its complex conjugate, which complex conjugate is the same expression, but with the sign of the is reversed A2=(A1e+A2e2)(A1e-的+A2e-2) In multiplying this out, we get Ai+ A2(here the e's cancel), and for the cross terms we have A1A2(e11-2)+e2-) el+e-°=cos8+isinθ+cosθ-isin6 That is to say, e+e-ie= 2 cos 8. Our final result is therefore AR= Ai+A2+2A1 A2 cos(o2-P1) As we see, this agrees with the length of AR in Fig. 29-9, using the rules of trigonometry. Thus the sum of the two effects has the intensity ai we would get with one of them alone, plus the intensity As we would get with the other one alone, plus a correction. This correction we call the interference efect. It is really only the differ- ence between what we get simply by adding the intensities, and what actually happens. We call it interference whether it is positive or negative.(Interference ordinary language usually suggests opposition or hindrance, but in physics we often do not use language the way it was originally designed! If the interference term is positive, we call that case constructive interference, horrible though it may sound to anybody other than a physicist! The opposite case is called destructive ence Now let how to apply our general formula(29. 16) for the case of two oscillators to the special situations which we have discussed qualitatively. To apply this general formula, it is only necessary to find what phase difference, PI-2, exists between the signals arriving at a given point. (It depends only on the phase difference, of course, and not on the phase itself )So let us consider the case where the two oscillators, of equal amplitude, are separated by some dis ance d and have an intrinsic relative phase a. (When one is at phase zero, the phase of the other is a Then we ask what the intensity will be in some azimuth direction 8 from the E-w line. [Note that this is not the same 8 as appears in (29. 1). We are torn between using an unconventional symbol like B, or the con- ventional symbol A(Fig. 29-10). The phase relationship is found by noting that e difference in distance from P to the two oscillators is d sin 0, so that the phase difference contribution from this is the number of wavelengths in d sin 8, multiplied by 2T(Those who are more sophisticated might want to multiply the wave number k, which is the rate of change of phase with distance, by d sin 8; it is exactly the ame. The phase difference due to the distance difference is thus 2md sin 0 /, but Fig. 29-10. Two oscillators of equa due to the timing of the oscillators, there is an additional phase a. So the phase amplitude, with a phase difference a difference at arrival would be 中2-中1=a+2 rd sin6/A (29.17) This takes care of all the cases. Thus all we have to do is substitute this expression to(29, 16) for the case A1=A2, and we can calculate all the various results for wo antennas of equal intensity. Now let us see what happens in our various cases. The reason we know, for example, that the intensity is 2 at 300 in Fig. 29-5 is the following: the two oscilla tors areλ apart,soat309,dsin0=M/4.Thusφ2-中1=2丌入/4入=丌/2,and so the interference term is zero. (We are adding two vectors at 90 The result is the hypotenuse of a 45 right-angle triangle, which is v2 times the unit amplitude squaring it, we get twice the intensity of one oscillator alone. all the other cases can be worked out in this same way
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