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can send the power all in one direction. But still it is distributed over a great
range of angles. Can we arrange it so that it is focused still more sharply in
a particular direction? Let us consider the case of Hawali 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 the power. Can
we do better than that? Let us take a situation in which the separation 1s ben
wavelengths (Fig. 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 small raction of a wavelength. Here the picture is quite diÑerent.
To distant point
Fig. 29-7. The intensity pattern for two dipoles separated by 10À.
TÍ the oscillators are ten wavelengths apart (we take the in-phase case to make
it 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 difÑfer by 180” and the
intensity is zero. To be precise, iŸ we draw a line from each oscillator to a distant
point and the diference A in the two distances is À/2, half an oscillation, then
they will be out of phase. So this firsb nuÌl occurs when that happens. (The
fgure is not drawn to scale; it is only a rough sketch.) This means that we do
indeed have a very sharp beam in the direction we want, because If we just move
over a little bit we lose all our intensity. Ủnfortunately for practical purposes,
1ƒ we were thinking of making a radio broadcasting array and we doubled the
distance A, then we would be a whole cycle out of phase, which is the same as
being exactly #n phase againl Thus we get many successive maxima and minima,
just as we found with the 23A spacing in Chapter 28.
Now how can we arrange to get rid of all these extra maxima, or “lobes,” as
they are called? We could get rid of the unwanted lobes in a rather interesting
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6+ _A = 5ố
10A 2 0°
30°
Fig. 29-8. A six-dipole antenna array and part of its intensity pattern.
way. Suppose that we were to place another set of antennas between the bwo
that we already have. 'That is, the outside ones are still 10À apart, but between
them, say every 2À, 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 fñeld would be six times and the intensity thirty-six times as great (the
square of the feld). We get 36 units of intensity in that direction. Now if we
look at neighboring points, we fnd a zero as before, roughly, but If we go farther,
to where we used to get a big “bump,” we get 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 in phase and are cooperating in trying to get some strength in that direction.
But numbers 3 and 4 are roughly 3 a wavelength out oŸ phase with 1 and 6, and
althoupgh 1 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 particular example, something else will happen: namely, since the
distance between successive dipoles is 2À, it is possible to find an angle where
the distance ổ between successiue đipoles 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 3607, 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. IÝ we put in more antennas,
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closer than one wavelength apart, then this cannot happen. But the fact that
this can happen at certain angles, if the spacing is bigger than one wavelength,
is a very interesting and useful phenomenon in other applications——not radio
broadcasting, but in đjffraction gratings.
29-5 The mathematics of interference
Now we have finished our analysis of the phenomena of dipole radiators
qualitatively, and we must learn how to analyze them quantitatively. To ñnd the
efect of two sources at some particular angle in the most general case, where
the two oscillators have some intrinsic relative phase œ from one another and
the strengths 4q and 4s are not equal, we fnd that we have to add two cosines
having the same frequenecy, but with diferent phases. Ït is very easy to find this
phase diference; it is made up of a delay due to the diference in distance, and
the intrinsic, built-in phase of the oscillation. Mathematically, we have to fnd
the sum ?# of two waves: Ï = Ái cos (2£ + ôi) + Áa cos (0É + j2). How do we do
Tt is really very easy, and we presume that we already know how t$o do it.
However, we shall outline the procedure in some detail. Eirst, 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 4 and 4a are cqual, let us say
they are both equal to A. In those cireumstances, for example (we could call this
the trigonometric method of solving the problem), we have
Tỳ = Alcos (‡ + ở1) + cos (ø‡ + óa)]. (29.9)
Once, in our trigonometry class, we may have learned the rule that
cos A + cos 8 = 2cos 5(A + B) cos 3(A — 8). (29.10)
Tf we know that, then we can Immediately write l as
l= 2Acos 3(di — 9a) cos (0É + Sới + 392). (29.11)
So we find that we have an oscillatory wave with a new phase and a new amplitude.
In general, the result œ2 be an oscillatory wave with a new amplitude Áp, which
we may call the resultant amplitude, oscillating at the same frequency but with
--- Trang 508 ---
a phase diference óp, called the resultant phase. In view of this, our particular
case has the following result: that the resultant amplitude 1s
An = 2Acos š (di — óa), (29.12)
and the resultant phase is the average of the two phases, and we have completely
solved our problem.
⁄⁄2——T
œ $n ° *
Fig. 29-9. A geometrical method for combining two cosine waves.
The entire diagram ¡is thought of as rotating counterclockwise with
angular frequency 0.
Now suppose that we cannot remember that the sum of Ewo cosines is twice
the cosine of half the sum times the cosine of half the diference. 'Phen we may
use another method of analysis which is more geometrical. Any cosine function