## 29Interference

### 29–1Electromagnetic waves

In this chapter we shall discuss the subject of the preceding chapter more 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 the field in mathematical detail, not just qualitatively.

We have already physically analyzed the meaning of
formula (28.6) quite 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 very
small amplitude, the field at some angle $\theta$ 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
\begin{equation}
\label{Eq:I:29:1}
E(t)=\frac{-qa(t-r/c)\sin\theta}{4\pi\epsO c^2r},
\end{equation}
where $a(t - r/c)$ 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, $\theta = 90^\circ$, and plot the field graphically. What we had been thinking of before is that we stand in one position and ask how the field there changes with time. But instead of that, we are now going to see what the field looks like at different positions in space at a given instant. So what we 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, a little bit later, we measure the field at a different place. Then we may assert that the 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” plot of the acceleration as a function of time; the distance is related to time by a constant scale factor $c$, which we often take as unity. This is easily seen by considering the mathematical behavior of $a(t - r/c)$. Evidently, if we add a little time $\Delta t$, we get the same value for $a(t - r/c)$ as we would have if we had subtracted a little distance: $\Delta r = -c\,\Delta t$.

Stated another way: if we add a little time $\Delta t$, we can restore
$a(t - r/c)$ to its former value by *adding* a little
distance $\Delta r = c\,\Delta t$. That is, as time goes on *the 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 constant $x_0$, the magnitude of the oscillation, times $\cos\omega t$. Then the acceleration is \begin{equation} \label{Eq:I:29:2} a=-\omega^2x_0\cos\omega t=a_0\cos\omega t, \end{equation} where $a_0$ is the maximum acceleration, $-\omega^2x_0$. Putting this formula into (29.1), we find \begin{equation} \label{Eq:I:29:3} E=-q\sin\theta\, \frac{a_0\cos\omega(t-r/c)}{4\pi\epsO rc^2}. \end{equation} Now, ignoring the angle $\theta$ and the constant factors, let us see what that looks like as a function of position or as a function of time.

### 29–2Energy 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.

This 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 $r_1$ (Fig. 29–4), and we do the same at
another distance $r_2$, 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 wave
within a given conical angle is the same, no matter how far away we
are! In particular, the total energy that we could take out of the
whole wave by putting absorbing oscillators all around is a certain
fixed amount. So the fact that the amplitude of $E$ varies as $1/r$ is
the same as saying that there is an energy flux which 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 away without diminution. So if we are far enough
away that our basic approximation 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
Chapter 32.

Let us now consider more carefully how the wave (29.3) varies as a function of 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.

### 29–3Sinusoidal waves

First let us fix the position $r$, and watch the field as a function
of time. It is oscillatory at the angular frequency $\omega$. The angular
frequency $\omega$ 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 $2\pi/\omega$, because $\omega$ 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) is 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
$\omega$, we can define a quantity called the *wave
number*, symbolized as $k$. This is defined as
*the rate of change of phase with distance* (radians per meter).
That is, as we move in space at a fixed time, the phase changes.

There is another quantity that corresponds to the period, and we might call it the period in space, but it is usually called the wavelength, symbolized $\lambda$. The wavelength is the distance occupied by one complete cycle. It is easy to see, then, that the wavelength is $2\pi/k$, because $k$ times the wavelength would be the number of radians that the whole thing changes, being the product of the rate of change of the radians per meter, times the number of meters, and we must make a $2\pi$ change for one cycle. So $k\lambda = 2\pi$ is exactly analogous to $\omega t_0 = 2\pi$.

Now in our particular wave there is a definite relationship between the frequency and the wavelength, but the above definitions of $k$ and $\omega$ 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 $\phi = \omega(t - r/c)$ the phase, and differentiate (partially) with respect to distance $r$, the rate of change, $\ddpl{\phi}{r}$, is \begin{equation} \label{Eq:I:29:4} \biggl|\ddp{\phi}{r}\biggr|= k = \frac{\omega}{c}. \end{equation} There are many ways to represent the same thing, such as

\begin{align} \label{Eq:I:29:5} \lambda &= ct_0 &\\[1.5ex] \label{Eq:I:29:6} \omega &= ck & \end{align} \begin{align} \label{Eq:I:29:7} \lambda\nu &= c &\\[1.5ex] \label{Eq:I:29:8} \omega\lambda &= 2\pi c & \end{align}

Why is the wavelength equal to $c$ times the period? That’s very easy, of course, because if we sit still and wait for one period to elapse, the waves, travelling at the speed $c$, will move a distance $ct_0$, and will of course have moved over just one wavelength.

In a physical situation other than that of light, $k$ is not necessarily related to $\omega$ 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 frequency $\omega$ will be written in general as $\cos\,(\omega t - kx)$.

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 $r^2$, and others which vary even faster. It would be worthwhile to know in what circumstances the $1/r$ 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 $1/r$ term. How far is “far enough”? The answer is, qualitatively, that the other terms are of order $\lambda/r$ smaller than the $1/r$ term. Thus, so long as we are beyond a few wavelengths, (29.1) is an excellent approximation to the field. Sometimes the region beyond a few wavelengths is called the “wave zone.”

### 29–4Two dipole radiators

Next let us discuss the mathematics involved in combining the effects of two oscillators to find the net field at a given point. This is very easy in the few cases that we considered in the previous chapter. We shall first describe the effects qualitatively, and then more quantitatively. Let us take the simple case, where the oscillators are situated with their centers in the same horizontal plane as the detector, 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 proportional to the square of the field, averaged in time. So
the thing to look at, when 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 Chapter 31.) 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 Fig. 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 N or 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 particular
intermediate angle (in fact, at $30^\circ$) the intensity is $2$, and
it falls off, $4$, $2$, $0$, 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 $\alpha$ 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 $30^\circ$ is still $2$, as
we can prove later.

Now we come to an interesting case which shows up a possibly useful
feature. Let us remark that one of the reasons that phase relations of
oscillators are interesting 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 in
Fig. 29–5(a) and we broadcast with our two antennas in
phase, because Hawaii 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 reverse the phase of one of our
antennas, and we can broadcast to the north. 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 changing 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 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 nature the result is
always going to be symmetrical. So let us consider a case that comes out
unsymmetrical, to show the possible variety.

If the antennas are separated by one-quarter wavelength, and if the N
one 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 $90^\circ$ later in
*time*, but it is already $90^\circ$ behind in its built-in
phase, therefore it arrives, altogether, $180^\circ$ 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 $90^\circ$ in time,
because it is a quarter wavelength closer. But its phase is set so
that it is oscillating $90^\circ$ *behind* in time, which just
compensates the delay difference, and therefore the two signals appear
*together* in phase, making the field strength twice as large,
and the energy four times as great.

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 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 Hawaii again, where we are sending the beam east and west but it is spread over quite an angle, because even at $30^\circ$ 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 is ten 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 fraction of a wavelength. Here the picture is quite different.

If 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 differ by $180^\circ$ and the intensity is
zero. To be precise, if we draw a line from each oscillator to a
distant point and the difference $\Delta$ in the two distances
is $\lambda/2$, half an oscillation, then they will be out of phase. So
this first null occurs when that happens. (The figure 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. Unfortunately for practical
purposes, if we were thinking of making a radio broadcasting array and
we doubled the distance $\Delta$, then 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
$2\tfrac{1}{2}\lambda$ 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 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 $10\lambda$ apart, but between them, say every $2\lambda$, 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 six 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 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 $\Delta$ 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 $\tfrac{1}{2}$ a wavelength out of phase with $1$ and $6$, and
although $1$ 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 $2\lambda$, it is possible to
find an angle where the distance $\delta$ 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 $360^\circ$, 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 spacing is bigger than one wavelength, is a
very interesting and useful phenomenon in other applications—not
radio broadcasting, but in *diffraction gratings*.

### 29–5The 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 find the effect of two sources at some particular angle in the most general case, where the two oscillators have some intrinsic relative phase $\alpha$ from one another and the strengths $A_1$ and $A_2$ 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 difference; it 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 sum $R$ of two waves: $R = A_1 \cos\,(\omega t + \phi_1) + A_2 \cos\,(\omega t + \phi_2)$. How do we do it?

It is really very easy, and we presume that we already know how to do
it. However, we shall outline the procedure in some 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
$A_1$ and $A_2$ are equal, let us say they are both equal to $A$. In
those circumstances, for example (we could call this the trigonometric
method of solving the problem), we have
\begin{equation}
\label{Eq:I:29:9}
R = A[\cos\,(\omega t+\phi_1)+\cos\,(\omega t + \phi_2)].
\end{equation}
Once, in our trigonometry class, we may have learned the rule that
\begin{equation}
\label{Eq:I:29:10}
\cos A+\cos B=2\cos\tfrac{1}{2}(A+B)\cos\tfrac{1}{2}(A-B).
\end{equation}
If we know that, then we can immediately write $R$ as
\begin{equation}
\label{Eq:I:29:11}
R=2A\cos\tfrac{1}{2}(\phi_1-\phi_2)\cos\,(\omega t+\tfrac{1}{2}\phi_1+
\tfrac{1}{2}\phi_2).
\end{equation}
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 a phase
difference $\phi_R$, called the resultant phase. In view of this, our particular
case has the following result: that the resultant amplitude is
\begin{equation}
\label{Eq:I:29:12}
A_R=2A\cos\tfrac{1}{2}(\phi_1-\phi_2),
\end{equation}
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 $\omega t$ can be considered as
the horizontal projection of a *rotating vector*. Suppose there
were a vector $\FLPA_1$ of length $A_1$ rotating with time, so that
its angle with the horizontal axis is $\omega t + \phi_1$. (We shall
leave out the $\omega t$ 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 $\omega$ (Fig. 29–9). The projection of $\FLPA_1$ along
the horizontal axis is precisely $A_1\cos\,(\omega t + \phi_1)$. Now
at $t= 0$ the second wave could be represented by another vector,
$\FLPA_2$, of length $A_2$ and at an angle $\phi_2$, and also
rotating. They are both rotating with the same angular
velocity $\omega$, and therefore the *relative* positions of the two are
fixed. The system goes around like a rigid body. The horizontal
projection of $\FLPA_2$ is $A_2\cos\,(\omega t + \phi_2)$. 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 $\FLPA_R$, the $x$-component 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 $A_1=$ $A_2 =$ $A$. In this case, we see from
Fig. 29–9 that $\FLPA_R$ lies midway between $\FLPA_1$
and $\FLPA_2$ and makes an angle $\tfrac{1}{2}(\phi_2 - \phi_1)$ with
each. Therefore we see that $A_R = 2A\cos\tfrac{1}{2}(\phi_2 -
\phi_1)$, as before. Also, as we see from the triangle, the phase
of $\FLPA_R$, as it goes around, is the average angle of $\FLPA_1$
and $\FLPA_2$ 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
problem.

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 particular case the waves could be written in
this way: $A_1e^{i(\omega t + \phi_1)}$ [the real part of this is
$A_1\cos\,(\omega t + \phi_1)$] and $A_2e^{i(\omega t + \phi_2)}$. Now
we can add the two:
\begin{equation}
\label{Eq:I:29:13}
R=A_1e^{i(\omega t + \phi_1)}+A_2e^{i(\omega t + \phi_2)}=
(A_1e^{i\phi_1}+A_2e^{i\phi_2})e^{i\omega t}
\end{equation}
\begin{align}
R&=A_1e^{i(\omega t + \phi_1)}+A_2e^{i(\omega t + \phi_2)}\notag\\[1ex]
&=(A_1e^{i\phi_1}+A_2e^{i\phi_2})e^{i\omega t}
\label{Eq:I:29:13}
\end{align}
or
\begin{equation}
\label{Eq:I:29:14}
\hat{R}=A_1e^{i\phi_1}+A_2e^{i\phi_2}=A_Re^{i\phi_R}.
\end{equation}
This solves the problem that we wanted to solve, because it represents
the result as a complex number of magnitude $A_R$ and phase $\phi_R$.

To see how this method works, let us find the amplitude $A_R$ which is the “length” of $\hat{R}$. To get the “length” of a complex quantity, we always multiply the quantity by its complex conjugate, which gives the length squared. The complex conjugate is the same expression, but with the sign of the $i$’s reversed. Thus we have \begin{equation} \label{Eq:I:29:15} A_R^2=(A_1e^{i\phi_1}+A_2e^{i\phi_2})(A_1e^{-i\phi_1}+A_2e^{-i\phi_2}). \end{equation} In multiplying this out, we get $A_1^2 + A_2^2$ (here the $e$’s cancel), and for the cross terms we have \begin{equation*} A_1A_2(e^{i(\phi_1-\phi_2)}+e^{i(\phi_2-\phi_1)}). \end{equation*} Now \begin{equation*} e^{i\theta}+e^{-i\theta}= \cos\theta+i\sin\theta+\cos\theta-i\sin\theta. \end{equation*} That is to say, $e^{i\theta} + e^{-i\theta} = 2\cos\theta$. Our final result is therefore \begin{equation} \label{Eq:I:29:16} A_R^2=A_1^2+A_2^2+2A_1A_2\cos\,(\phi_2-\phi_1). \end{equation}

As we see, this agrees with the length of $\FLPA_R$ in Fig. 29–9, using the rules of trigonometry.

Thus the sum of the two effects has the intensity $A_1^2$ we would get
with one of them alone, plus the intensity $A_2^2$ we would get with
the other one alone, plus a correction. This correction we call the
*interference effect*. It is really only the difference between
what we get simply by adding the intensities, and what actually
happens. We call it interference whether it is positive or
negative. (Interference in 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* interference.

Now let us see 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, $\phi_2 - \phi_1$, 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 distance $d$ and have an intrinsic relative
phase $\alpha$. (When one is at phase zero, the phase of the other
is $\alpha$.) Then we ask what the intensity will be in some azimuth
direction $\theta$ from the E–W line. [Note that this is *not*
the same $\theta$ as appears in (29.1). We are torn between
using an unconventional symbol like $\cancel{\text{U}}\!\!,$ or the conventional
symbol $\theta$ (Fig. 29–10).] The phase relationship is
found by noting that the difference in distance from $P$ to the two
oscillators is $d\sin\theta$, so that the phase difference
contribution from this is the number of wavelengths in $d\sin\theta$,
multiplied by $2\pi$. (Those who are more sophisticated might want to
multiply the wave number $k$, which is the rate of change of phase
with distance, by $d\sin\theta$; it is exactly the same.) The phase
difference due to the distance difference is thus $2\pi
d\sin\theta/\lambda$, but, due to the timing of the oscillators, there
is an additional phase $\alpha$. So the phase difference at arrival
would be
\begin{equation}
\label{Eq:I:29:17}
\phi_2-\phi_1=\alpha+2\pi d\sin\theta/\lambda.
\end{equation}
This takes care of all the cases. Thus all we have to do is substitute
this expression into (29.16) for the case $A_1 = A_2$, and
we can calculate all the various results for two 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 $30^\circ$ in Fig. 29–5 is the following: the two oscillators are $\tfrac{1}{2}\lambda$ apart, so at $30^\circ$, $d\sin\theta=\lambda/4$. Thus $\phi_2 - \phi_1 =$ $2\pi\lambda/4\lambda =$ $\pi/2$, and so the interference term is zero. (We are adding two vectors at $90^\circ$.) The result is the hypotenuse of a $45^\circ$ right-angle triangle, which is $\sqrt{2}$ 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.