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?