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= œ+ kdsin 8.
First, we shall take the case œ = 0. 'That ïs, all oscillators are in phase, and we
want to know what the intensity is as a function of the angle Ø. In order to ñnd
out, we merely have to put @ = kdsin Ø into formula (30.3) and see what happens.
In the first place, there is a maximum when ở = 0. 'PThat means that when all
the oscilators are in phase there is a strong intensity in the direction Ø = 0. Ôn
the other hand, an interesting question is, where is the first minimum? “Phat
occurs when @ = 27/n. In other words, when 2zdsinØ/A = 2m/n, we get the
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ñrst minimum of the curve. lf we get rid of the 27ˆs so we can look at it a little
better, it says that
ndsin 8 = À. (30.5)
Now let us understand physically why we get a minimum at that position. nd
is the total length Ù of the array. Referring to Eig. 30-3, we see that nđsin Ø =
LsinØ = A. What (30.5) says is that when A is equal to one tuauelength, we
get a minimum. Now why do we get a minimum when A = À? Because the
contributions of the various oscillators are then uniformly distributed in phase
from 0° to 360°. The arrows (Fig. 30-1) are going around a whole circle—we are
adding equal vectors in all directions, and such a sum is zero. So when we have
an angle such that A = À, we get a minimum. That is the first minimum.
There is another important feature about formula (30.3), which is that if
the angle ø is increased by any multiple of 2z, it makes no diference to the
formula. So we will get other strong maxima at ở = 27, 4m, 6z, and so forth.
Near cach of these great maxima the pattern of Fig. 30-2 is repeated. We may
ask ourselves, what is the geometrical circumstance that leads to these other
great maxima? 'Phe condition is that ô = 2m, where m is any integer. That is,
2zdsin 8/À = 2m. Dividing by 27, we see that
đsỉin 8 = mÀ. (30.6)
Thịis looks like the other formula, (30.5). No, that formula was nđsin Ø = À. The
diferenee is that here we have to look at the 7md?uidual sources, and when we say
đsin Ø = mÀ, that means that we have an angle Ø such that ổ = ?mÀ. In other
words, each source is now contributing a certain amount, and successive ones
are out of phase by a whole multiple of 360”, and therefore are contributing 7n
phase, because out of phase by 360” is the same as being in phase. So they all
contribute in phase and produce just as good a maximum as the one for rn = Ö
that we discussed before. 'Phe subsidiary bumps, the whole shape of the pattern,
1s jus$ like the one near ¿ = 0, with exactly the same minima on each side, etc.
Thus such an array will send beams in various directions—each beam having a
strong central maximum and a certain number of weak “side lobes.” 'The various
strong beams are referred to as the zero-order beam, the first-order beam, etc.,
according to the value of ?m. ?n is called the order of the beam.
W© call attention to the fact that if đ is less than À, Eq. (30.6) can have
no solution except rm = 0, so that If the spacing 1s too small there is only one
possible beam, the zero-order one centered at Ø = 0. (Of course, there is also
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a beam in the opposite direction.) In order to get subsidiary great maxima, we
mmust have the spacing đ oŸ the array greater than one wavelength.
30-2 The difraction grating
In technical work with antennas and wires it is possible to arrange that all the
phases of the little oscillators, or antennas, are equal. The question is whether
and how we can do a similar thing with light. We cannot at the present time
literally make little optical-frequency radio stations and hook them up with
Infnitesimal wires and drive them all with a given phase. But there is a very
easy way to do what amounts to the same thing.
Suppose that we had a lot of parallel wires, equally spaced at a spacing đ, and
a radiofrequency source very far away, practically at infñnity, which is generating
am electric fñeld which arrives at each one of the wires at the same phase (it is
so far away that the từìme delay is the same for all oŸ the wires). (One can work
out cases with curved arrays, but let us take a plane one.) Then the external
electric fñeld will drive the electrons up and down in each wire. 'That is, the
fñeld which is coming from the original source will shake the electrons up and
down, and in moving, these represent + genera‡ors. Thịs phenomenon is called
scattering: a light wave from some source can induce a motion of the electrons
in a piece of material, and these motions generate their own waves. Thherefore all
that is necessary is to set up a lot of wires, equally spaced, drive them with a
radiofrequency source far away, and we have the situation that we want, without
a whole lot of special wiring. Tf the incidence is normal, the phases will be equal,
and we will get exactly the cireumstance we have been discussing. Therefore, If
the wire spacing is greater than the wavelength, we will get a strong intensity of
scattering in the normal direction, and in certain other directions given by (30.6).
This can dalso be done tuíth líghH Instead oŸ wires, we use a flat piece of glass
and make notches in it such that each of the notches scatters a little diferently
than the rest of the glass. If we then shine light on the glass, each one of the
notches will represent a source, and if we space the lines very fnely, but not
closer than a wavelength (which is technically almost impossible anyway), then
we would expect a miraculous phenomenon: the light not only will pass straight
through, but there will also be a strong beam at a finite angle, depending on
the spacing of the notchesl Such objects have actually been made and are in
common use—they are called đjfƒraction gratings.
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In one of its forms, a difraction grating consists of nothing but a plane glass
sheet, transparent and colorless, with scratches on it. There are often several
hundred scratches to the millimeter, 0ery carefully arranged so as to be equally
spaced. 'Phe efect of such a grating can be seen by arranging a projectOr so as
6o throw a narrow, vertical line of light (the image of a slit) onto a screen. When
we put the grating into the beam, with its scratches vertical, we see that the
line is still there but, in addition, on each side we have ønother strong patch
of light which is colored. “This, of course, is the slit Image spread out over a
wide angular range, because the angle Ø in (30.6) depends upon À, and lights of
diferent colors, as we know, correspond to diferent frequencies, and therefore
diferent wavelengths. 'The longest visible wavelength is red, and since đsin Ø = À,
that requires a larger 0. And we do, in fact, fnd that red is at a greater angle
out from the central imagel 'There should also be a beam on the other side, and
indeed we see one on the sereen. Then, there might be another solution of (30.6)
when ?m = 2. We do see that there is something vaguely there—very weak——and
there are even other beams beyond.
We have just argued that all these beams ought to be of the same strength,
but we see that they actually are not and, in fact, not even the first ones on the
right and left are equall “The reason is that the grating has been carefully built to
do just this. How? If the grating consists of very fine notches, inÑnitesimally wide,