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but if there is a large number of them, it seems that the word difraction is more
often used. 5o, we shall not worry about whether it is interference or difraction,
but continue directly from where we left off in the middle of the subject in the
last chapter.
Thus we shall now discuss the situation where there are + equally spaced
oscillators, all of equal amplitude but diferent from one another in phase, either
because they are driven diferently in phase, or because we are looking at them at
an angle such that there is a difference in time delay. Eor one reason or another,
we have to add something like this:
T = Alcos œ£ + cos (uÉ + ở) + cos (É + 29) + - - - + cos (2£ + (n — 1))], (50.1)
where ở is the phase diference between one oscillator and the next one, as seen
in a particular direction. Specifically, ¿ = œ + 2rdsinØ/A. Ñow we must add all
the terms together. We shall do this geometrically. The frst one is of length A,
and ít has zero phase. “The next is also of length 4 and it has a phase equal to ó.
The next one is again of length A and it has a phase equal to 2ø, and so on. So
we are evidently going around an equiangular polygon with ø sides (Eig. 30-1).
Now the vertices, of course, all lie on a circle, and we can fñnd the net amplitude
mmost easily if we fnd the radius of that cirele. Suppose that @ is the center of
--- Trang 513 ---
A6
Q °
O Ai S my x
Fig. 30-1. The resultant amplitude of n = 6 equally spaced sources
with net successive phase differences ý.
the circle. Thhen we know that the angle Q6 is just a phase angle . (Thịs is
because the radius Q9 bears the same geometrical relation to 4a as QO bears
to Ai, so they form an angle ó between them.) Therefore the radius r must
be such that A = 2rsin 2/2, which fixes r. But the large angle Ó@Q7' is equal
to mó, and we thus fnd that Ág = 2rsinno2/2. Combining these bwo results to
eliminate r, we get
sin n@/2
An=A————. 30.2
" sin @/2 (30.2)
The resultant intensity is thus
sinˆ „j/2
T=lo——=_. 30.3
" sin? ø/2 (80.3)
Now let us analyze this expression and study some of its consequences. Ïn
the first place, we can check it for ø = 1. It checks: σ = Tạ. Next, we check it
for ø= =2: writing sin @ = 2sin @/2cos @/2, we find that An = 2A cos @/2, which
agrees with (29.12).
Now the idea that led us to consider the addition of several sources was that
we might get a much stronger intensity in one direction than in another; that the
nearby maxima which would have been present if there were only ©wo sources will
have gone down in strength. In order to see this efect, we plot the curve that
comes rom (30.3), taking œ to be enormously large and plotting the region near
=0. In the first place, iŸ ở is exactly 0, we have 0/0, but iŸ ở is inũnitesimal,
the ratio of the two sines squared is simply n2, since the sine and the angle are
--- Trang 514 ---
approximately equal. 'Phus the intensity of the maximum of the curve is equal
to n2 times the intensity of one oseillator. That is easy to see, because if they
are all in phase, then the little vectors have no relative angle and all œ of them
add up so the amplitude is ø times, and the intensity n2 times, stronger.
As the phase ó increases, the ratio of the Ewo sines begins to fall of, and the
first time it reaches zero is when #d/2 = z, because sin 7 = 0. In other words,
@ = 2#/n corresponds to the first minimum in the curve (Fig. 30-2). In terms
of what is happening with the arrows in Fig. 30-1, the first minimum occurs
when all the arrows come back to the starting point; that means that the total
accumulated angle in all the arrows, the total phase diference between the first
and last oscillator, must be 2z to complete the circle.
1.0
H Ñ =
; \ rN TT _
Z———-`.ø⁄“.——`-ò-s.⁄ ~ ——=-
0 1 2 3 4 nộ/2m 5
Fig. 30-2. The Intensity as a function of phase angle for a large
number of oscillators of equal strength.
Now we go to the next maximum, and we want to see that it is really much
smaller than the first one, as we had hoped. We shall not go precisely to the
maximum position, because both the numerator and the denominator of (30.3)
are variant, but sin 2/2 varies quite slowly compared with sinnø/2 when øw is
large, so when sinnd/2 = I we are very close to the maximum. “The next
maximum of sin2eở/2 comes at œÓ/2 = 37/2, or ó = 3Z/n. This corresponds
to the arrows having traversed the circle one and a half times. On putting
ó = 3z/n into the Íormula to fnd the size of the maximum, we fñnd that
sin” 3x/2 = 1 in the numerator (because that is why we picked this angle), and
in the denominator we have sin? 3z/2n. Now if ø is sufficiently large, then this
angle is very small and the sine is equal to the angle; so for all practical purposes,
we can put sỉin 3/2n = 3z/2n. Thus we find that the intensity at this maximum
--- Trang 515 ---
is l = Ia(dn2/9z?). But øØỞlạ was the maximum intensity, and so we have
4/92 tỉmes the maximum intensity, which is about 0.045, less than 5 percent, of
the maximum intensity! Of course there are decreasing intensities farther out.
So we have a very sharp central maximum with very weak subsidiary maxima on
the sides.
Tt is possible to prove that the area of the whole curve, including all the little
bumps, is equal to 2wïÏo, or bwice the area of the dotted rectangle in Eig. 30-2.
ð= A/n= dsin60 \
L_——+zz.Ì '
T2 3 s n
Fig. 30-3. A linear array of n equal oscillators, driven with phases œ;
Now let us consider further how we may apply Eq. (30.3) in diferent cireum-
stances, and try to understand what ¡is happening. Let us consider our sources
to be all on a line, as drawn in Fig. 30-3. There are ø of them, all spaced by a
distance đ, and we shall suppose that the intrinsic relative phase, one to the next,
is œ. Then if we are observing in a given direction Ø from the normal, there is an
additional phase 2xđsin Ø/À because of the time delay between each successive
two, which we talked about before. Thus
= œ+ 2rdsin Ø/À
? : / (30.4)