2 interferenee 29-1 Electromagnetic waves In this chapter we shall discuss the subject of the preceding chapt 29-1 Electromagnetic waves mathematically. We have qualitatively demonstrated that there are maxima and minima in the radiation field from two sources, and our problem now is to describe 29-2 Energy of radiation the field in mathematical detail, not just qualitatively 29-3 Sinusoidal waves We have already physically analyzed the meaning of formula(28.6)quite 29-4 Two dipole radiators satisfactorily, but there are a few points to be made about it mathematically. In the first place, if a charge is accelerating up and down along a line, in a motion of 29-5 The mathematics of interference very small amplitude, the field at some angle 0 from the axis of the motion is in a direction at right angles to the line of sight and in the plane containing both the acceleration and the line of sight( Fig. 29-1). If the distance is called r, then at time t the electric field has the magnitude E(n where a(r -r/c) is the acceleration at the time(t-r/c), called the retarded acceleration Now it would be interesting to draw a picture of the field under different conditions. The thing that is interesting, of course, is the factor a(t-r/c), and to understand it we can take the simplest case, 0=900, and plot the field graphically. What we had been thinking of before is that we stand in one position and ask how Fig. 29-1. The electric field e due the field looks like at different positions in space at a given instant. So what we acceleration/ ge whose retarded the field there changes with tiRne. But instead of that, we are now going to see what to a positive che want is a"snapshot"picture which tells us what the field is in different places Of course it depends upon the acceleration of the charge. Suppose that the charge at first had some particular motion: it was initially standing still, and it suddenly accelerated in some manner, as shown in Fig. 29-2, and then stopped. Then little bit later, we measure the field at a different place. Then we may assert that he field will appear as shown in Fig. 29-3. At each point the field is determined by the acceleration of the charge at an earlier time, the amount earlier being the delay r/c. The field at farther and farther points is determined by the acceleration at earlier and earlier times. So the curve in Fig. 29-3 is really, in a sense, a"reversed lot of the acceleration as a function of time; the distance is related to time by a Fig. 29-2. The acceleration of a constant scale factor c, which we often take as unity. This is easily seen by consider is a function of time the mathematical behavior of a(t-r/c). Evidently, if we add a little time Ar, we get the same value for a(t-r/c)as we would have if we had subtracted a little distance:△r=-c△ Stated another way: if we add a little time At, we can restore a(t-r/)to its field moves as a wave outward from the source. That is the reason why we sometimes say light is propagated as waves. It is equivalent to saying that the field is delayed or to saying that the electric field is moving outward as time goes on An interesting special case is that where the charge q is moving up and down in an oscillatory manner. The case which we studied experimentally in the last chapter was one in which the displacement x at any time t was equal to a certain Fig. 29-3. The electric field as a constant xo, the magnitude of the oscillation times cos wr. Then the acceleration is function of position at a later time. (The o cOS wf= ao cos (292) I/r variation is ignored. 291
where ao is the maximum acceleration, -wxo. Putting this formula into(29. 1), E=-q sin 8 Now, ignoring the angle 0 and the constant factors let us see what that looks like as a function of position or as a function of time 29-2 Energy of radiation First of all, at any particular moment or in any particular place, the strength of the field varies inversely as the distance r, as we mentioned previously.Now we must point out that the energy content of a wave, or the energy effects that such an electric field can have, are proportional to the square of the field, because if. for instance, we have some kind of a charge or an oscillator in the electric field then if we let the field act on the oscillator, it makes it move. If this is a linear oscillator, the acceleration, velocity, and displacement produced by the electric field acting on the charge are all proportional to the field. So the kinetic energy which is developed in the charge is proportional to the square of the field. So we shall take it that the energy that a field can deliver to a system is proportional somehow to the square of the field fare. his means that the energy that the source can deliver decreases as we get farther away; in fact, it varies inversely as the square of the distance. But that has a very simple interpretation: if we wanted to pick up all the energy we could from the wave in a certain cone at a distance r1(Fig. 29-4), and we do the same at an other distance r2, we find that the amount of energy per unit area at any one place goes inversely as the square of r, but the area of the surface intercepted by the cone goes directly as the square of r. So the energy that we can take out of the gave within a given conical angle is the same, no matter how far away we are ig. 29-4. The energy flowing within In particular, the total energy that we could take out of the whole wave by putting the cone OABCD is independent of the absorbing oscillators all around is a certain fixed amount. So the fact that the distance r at which it is measured amplitude of E varies as l/r is the same as saying that there is an energy flux hich is never lost, an energy which goes on and on, spreading over a greater and greater effective area. Thus we see that after a charge has oscillated, it has lost some energy which it can never recover; the energy keeps going farther and farther ay without diminution So if far enough away that our basic tion is good enough, the charge cannot recover the energy which has been, as we say, radiated away. Of course the energy still exists somewhere, and is available to be picked up by other systems. We shall study this energy"loss"further in apter Let us now consider more carefully how the wave(29.3)varies as a function f time at a given place, and as a function of position at a given time. again we ignore the 1/r variation and the constants 29-3 Sinusoidal waves First let us fix the position r, and watch the field as a function of time. It is oscillatory at the angular frequency a. The angular frequency w can be defined as the rate of change of phase with time(radians per second ). We have already studied such a thing, so it should be quite familiar to us by now. The period is the time needed for one oscillation, one complete cycle, and we have worked that out too; it is 2T/, because w times the period is one cycle of the cosine Now we introduce a new quantity which is used a great deal in physics.This has to do with the opposite situation, in which we fix t and look at the wave as a function of distance r. Of course we notice that, as a function of r, the wave(29.3) also oscillatory. That is, aside from 1/r, which we are ignoring, we see that e oscillates as we change the position. So, in analogy with n define symbolized as k. This is defined as the rate of of phase with distance (radians per meter). That is, as we move in space at a fixed
There is another quantity that corresponds to the period, and we might cal the period in space, but it is usually called the wavelength, symbolized X.The avelength is the distance occupied by one complete cycle. It is easy to see, then that the wavelength is 2T/k, because k times the wavelength would be the number of radians that the whole thing changes, being the product of the rate of change the radians per meter, times the number of meters, and we must make a 2T change for one cycle. So kA= 2T is exactly analogous to wto= 2T Now in our particular wave there is a definite relationship between the fre- quency and the wavelength, but the above definitions of k and w are actually quite general. That is, the wavelength and the frequency may not be related in the same way in other physical circumstances. However, in our circumstance the rate of change of phase with distance is easily determined, because if we call p=w(I-r/c)the phase, and differentiate(partially)with respect to distance the rate of change, ao/ar, is There are many ways to represent the same thing, such as (29.7) ck(296) A=2mc(29.8) ly is the wavelength equal to c times the period? That's very easy, of course ecause if we sit still and wait for one period to elapse, the waves, travelling at the eed c, will move a distance cfo, and will of course have moved over just one wavelengt In a physical situation other than that of light, k is not necessarily related to w) in this simple way. If we call the distance along an axis x then the formula for a cosine wave moving in a direction x with a wave number k and an angular fre- quency w will be written in general as cos(ot -kx). Now that we have introduced the idea of wavelength, we may say something more about the circumstances in which (29. 1)is a legitimate formula. We recall that the field is made up of several pieces, one of which varies inversely as r, another part which varies inversely as r, and others which vary even faster. It would be worth while to know in what circumstances the I/r part of the field is the most important part, and the other parts are relatively small. Naturally, the answer is if we go 'far enough'away, because terms which vary inversely as the square ultimately become negligible compared with the 1/rterm. How far is"far enough""?2 The answer is, qualitatively, that the other terms are of order x/r smaller than the 4x24 he field. Sometimes the region beyond a few wavelengths is alled the“ wave zone a=0 0 =T 29-4 Two dipole radiators Next let us discuss the mathematics involved in combining the effects of two Fig. 29-5. The intensities in various oscillators to find the net field at a given point. this is very easy in the few cases directions from two dipole oscillators idered in the previous chapter. We shall first describe the effects e-half wavelength apart. Left: in qualitatively, and then more quantitatively. Let us take the simple case, where the phase (a= O). Right: one- half period oscillators are situated with their centers in the same horizontal plane as the de- out of phase(a= t). tector. and the line of vibration is vertical Figure 29-5(a)represents the top view of two such oscillators, and in this particular example they are half a wavelength apart in a N-s direction, and are oscillating together in the same phase, which we call zero phase. Now we would like to know the intensity of the radiation in various directions. By the intensity we mean the amount of energy that the field carries past us per second, which is roportional to the square of the field averaged in time. So the thing to look at hen we want to know how bright the light is, is the square of the electric field, not the electric field itself. (The electric field tells the strength of the force felt by a
stationary charge, but the amount of energy that is going past, in watts per square meter,is proportional to the square of the electric field. We shall derive the constant of proportionality in the next chapter If we look at the array from the w side both oscillators contribute equally and in phase, so the electric field is twice as strong as it would be from a single oscillator. Therefore the intensity is four times as strong as it would be if there were only one oscillator. (The numbers in Fi 29-5 represent how strong the intensity would be in this case, compared with what it would be if there were only a single oscillator of unit strength. )Now, in either the nor S direction along the line of the oscillators, since they are half a wavelength apart, the effect of one oscillator turns out to be out of phase by exactly half an oscillation from the other, and therefore the fields add to zero. At a certain par- ticular intermediate angle(in fact, at 30%)the intensity is 2, and it falls off, 4, 2, and so forth. We have to learn how to find these numbers at other angles. It is a question of adding two oscillations with different phases Let us quickly look at some other cases of interest. Suppose the oscillators are again one-half a wavelength apart, but the phase a of one is set half a period behind the other in its oscillation(Fig. 29-5b). In the w direction the intensity is now zero, because one oscillator is"pushing"when the other one is"pulling "But in the n direction the signal from the near one comes at a certain time and that of the other comes half a period later. But the latter was originally half a period behind in timing, and therefore it is now exactly in time with the first one, and so the intensity in this direction is 4 units. The intensity in the direction at 30%is still 2, as we can prove later. Let us remark that one of the reasons that phase relations of oscillators are in teresting is for beaming radio transmitters. For instance, if we build an antenna system and want to send a radio signal, say, to Hawaii, we set the antennas up as Fig 29-5(a) and we broa as in phase, because is to the west of us. Then we decide that tomorrow we are going to broadcast toward Alberta, Canada. Since that is north, not west, all we have to do is to phase of So we can build antenna systems with various arrangements. Ours is one of the simplest possible ones; we can make them much more complicated, and by chang ing the phases in the various antennas we can send the beams in various directions and send most of the power in the direction in which we wish to transmit, without ever moving the antenna! In both of the preceding cases, however, while we are A1-2 broadcasting toward alberta we are wasting a lot of power on Easter Island, and it would be interesting to ask whether it is possible to send it in only one direction At first sight we might think that with a pair of antennas of this natur is always going to be symmetrical. So let us consider a case that comes out un- symmetrical, to show the I variety Fig. 29-6. A pair of dipole antennas If the antennas are separated by one- quarter wavelength, and if the n one giving maximum power in one direction is one-fourth period behind the S one in time, then what happens(Fig. 29-6)? In the w direction we get 2, as we will see later. In the s direction we get zero, because the signal from S comes at a certain time; that from N comes 90% later in time, but it is already 90 behind in its built- in phase, therefore it arrives, altogether, 180 out of phase, and there is no effect. On the other hand. in the n direction the n signal arrives earlier than the S signal by 90 in time, because it is a quarter wavelength closer. But its phase is set sa that it is oscillating 90 behind in time which just compensates the delay difference, and therefore the two signals appear together in phase, ma field strength twice as larg nergy Thus, by using some cleverness in spacing and phasing our antennas, we can send the power all in one direction. But still it is distributed over a great angles. Can we arrange it so that it is focused still more sharply in a particular direction? Let us consider the case of Hawaii again, where we are sending the beam east and west but it is spread over quite an angle, because even at 30 we are still getting half the intensity-we are wasting power. Can we do better th that? Let us take a situation in which the separation is ten wavelengths (Fi
29-7),which is more nearly comparable to the situation in which we experimented in the previous chapter, with separations of several wavelengths rather than a all fraction of a wavelength. Here the If the oscillators are ten wavelengths apart(we take the in-phase case to make Iox easy), we see that in the E-w direction, they are in phase, and we get a strong intensity, four times what we would get if one of them were there alone. On the other hand, at a very small angle away, the arrival times differ by 180 and the intensity is zero. To be precise, if we draw a line from each oscillator to a distant point and the difference 4 in the two distances is A/2, half an oscillation, then they not drawn to scale: it is only a rough sketch. This means that we do indeed have two dipoles separated by 10.n for will be out of phase. So this first null occurs when that happens. (The figure is s very sharp beam in the direction we want, because if we just move over a little we lose all our intensity. Unfortunately for practical purposes, if we wer thinking of making a radio broadcasting array and we doubled the distance A, th we would be a whole cycle out of phase, which is the same as being exactly in phase again Thus we get many successive maxima and minima, just as we found with the 22A spacing in Chapter 28 Now how can we arrange to get rid of all these extra maxima, or"lobe ey are called? We could get rid of the unwanted lobes in a rather interesting way Suppose that we were to place another set of antennas between the two that we already have. That is, the outside ones are still 10A apart, but between them, say lox every 2A, we have put another antenna, and we drive them all in phase. There are now six antennas, and if we looked at the intensity in the E-w direction, it would of course, be much higher with six antennas than with one. The field would be ix times and the intensity thirty-six times as great(the square of the field ).We get 36 units of intensity in that direction. Now if we look at neighboring points we find a zero as before, roughly, but if we go farther, to where we used to get big"bump ct a much smaller"bump" now. Let us try to see why The reason is that although we might expect to get a big bump when the distance A is exactly equal to the wavelength, it is true that dipoles 1 and 6 are then Fig. 29-8. A six-dipole antenna ar- in phase and are cooperating in trying to get some strength in that direction. But ay and part of its intensity pattern. numbers 3 and 4 are roughly 2 a wavelength out of phase with I and 6, and although I and 6 push together, 3 and 4 push together too, but in opposite phase. Therefore there is very little intensity in this direction-but there is something; it does not balance exactly. This kind of thing keeps on happening; we get very little bumps, and we have the strong beam in the direction where we want it. But in this particu lar example, something else will happen: namely, since the distance between suc- cessive dipoles is 2x, it is possible to find an angle where the distance 8 between successive dipoles is exactly one wavelength, so that the effects from all of them are in phase again. Each one is delayed relative to the next one by 360, so they all come back in phase, and we have another strong beam in that direction! It is easy to avoid this in practice because it is possible to put the dipoles closer than one wavelength apart. If we put in more antennas, closer than one wavelength apart, then this cannot happen. But the fact that this can happen at certain angles, if the ger than one wavelength, is a very interesting and useful ph in other applications--not radio broadcasting, but in diffraction gratings 29-5 The mathematics of interference Now we have finished our of the phenomena of dipole radiators qualitatively, and we must learn how to analyze them quantitatively. To find the effect of two sources at some particular angle in the most general case, where the two oscillators have some intrinsic relative phase a from one another and the strengths AI and A2 are not equal, we find that we have to add two cosines having the same frequency, but with different phases. It is very easy to find this phase is made up of a delay due to the difference in distance and the intrinsic, built-in phase of the oscillation. Mathematically, we have to find the R of two waves:R=A1cos(ω!+φ1)+ A. cos(ut+φ2).Ho do it?
It is really However, we shall outline the procedure in detail. First, we can, if we are clever with mathematics and know enough about cosines and sines, simply work it out. The easiest such case is the one where A I and A2 are equal, let us say they are both equal to A. In those circumstances, for example (we could call this the ric method of solving the R=A[cos(at+φ1)+cos(at+φ) Once, in our trigonometry class, we may have learned the rule that cos A cos B= 2 cos(A B)cos I(A-B) (29.10) If we know that, then we can immediately write R as R=2Acos(中1一中)cos(at+妙1+她2) (29.11) So we find that we have an oscillatory wave with a new phase and a new amplitude In general, the result will be an oscillatory wave with a new amplitude A R, which we may call the resultant amplitude, oscillating at the same frequency but with phase difference R, called the resultant phase. In view of this, our particular case has the following result: that the resultant amplitude is AR=2Acos如(中1-中2) (29.12) and the resultant phase is the average of the two phases, and we have completely solved our problem Now suppose that we cannot remember that the sum of two cosines is twice the cosine of half the sum times the cosine of half the difference. Then we may use another method of analysis which is more geometrical. Any cosine function of at can be considered as the horizontal projection of a rotating vector. Suppose there were a vector A 1 of length A 1 rotating with time, so that its angle with the horizontal axis is wt pI. ( We shall leave out the at in a minute and see that it makes no difference. Suppose that we take a snapshot at the time t=0, although in fact, the picture is rotating with angular velocity w( Fig. 29-9). The projection Fig. 29-9. a geometrical method of A 1 along the horizontal axis is precisely A1 cos(ot p1). Now at t=0 the combining two cosine waves. The entire second wave could be represented by another vector, A2, of length A2 and at an diagram is thought of as rotating counter- angle p2, and also rotating. They are both rotating with the same angular velocity clockwise with angular frequency w w, and therefore the relative positions of the two are fixed. The system goes around like a rigid body. The horizontal projection of a2 is A2 cos(t + 2). But we know from the theory of vectors that if we add the two vectors in the ordinary way, by the parallelogram rule, and draw the resultant vector AR, the x-componen of the resultant is the sum of the x-components of the other two vectors. That solves our problem. It is easy to check that this gives the correct result for the special case we treated above, where A,= A2= A. In this case we see 29-9 that AR lies midway between A1 and A 2 and makes an angle 2(o with each. Therefore we see that AR 2A cos 2( 2-1), as before. Al an of A1 and A2 when the two amplitudes are equal. Clearly, we can also solve for the case where the amplitudes are not equal, just as easily. We can call that the geometrical way of solving the proble There is still another way of solving the problem, and that is the analytical way. That is, instead of having actually to draw a picture like Fig. 29-9, we can write something down which says the same thing as the picture: instead of drawing the vectors, we write a complex number to represent each of the vectors. The real parts of the complex numbers are the actual physical quantities. So in our par ticular case the waves could be written in this way: Ale at+op[the real part of this is A, cos(ot u] and A2e of+ea. Now we can add the two R
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
