Diffraction 30-1 The resultant amplitude due to n equal oscillators This chapter is a direct continuation of the previous one, although the name 30-1 The resultant amplitude due to as been changed from interference to Diffraction. No one has ever been able to n equal oscillators define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between 30-2 The diffraction grating them. The best we can do, roughly speaking, is to say that when there are only a few 30-3 Resolving power of a grating ources,say two, interfering, then the result is usually called interference, but if there is a large number of them, it seems that the word diffraction is more often 30-4 The parabolic antenna used. So, we shall not worry about whether it is interference or diffraction, but 30-5 Colored films; crystals continue directly from where we left off in the middle of the subject in the last chapter 30-6 Diffraction by opaque screens Thus we shall now discuss the situation where there are n equally spaced os- 30-7 The field of a plane of cillators, all of equal amplitude but different from one another in phase, either oscillating charges because they are driven differently in phase or because we are looking at them at an angle such that there is a difference in time delay. For one reason or another we have to add something like th R=A[cost+cos(t+刂+cos(at+2φ)+…+cos(at+(n-1)) (30.1) where p is the phase difference between one oscillator and the next one, as seen in a particular direction. Specifically, a+ 2md sin 8/. Now we must add all he terms together, We shall do this geometrically. The first one is of length A and it has zero phase The next is also of length A and it has a phase equal to The next one is again of length A and it has a phase equal to 2o, and so on. So we are evidently going around an equiangular polygon with n sides(Fig. 30-1) Now the vertices, of course, all lie on a circle, and we can find the net amplitude most easily if we find the radius of that circle. Suppose that g is the center of the circle. Then we know that the angle oQS is just a phase angle p.(This is because he radius Qs bears the same geometrical relation to A 2 as g0 bears to Al,so they form an angle o between them. Therefore the radius r must be such that A= 2r sin $/2, which fixes r. But the large angle OQT is equal to no, and we thus find that AR= 2r sin nd /2. Combining these two results to eliminate r, we Fig. 30-1. The resultant amplitude A sin№/2 (30.2) of n =6 equally spaced sources with net successive phase differences The resultant intensity is thus in no /2 of the first place, we can check it for n= 1. It checks: I=Io. Next, we check it forn=2: writing sinφ=2sinφ/2cosφ/2, we find that Ar=2Acos中/2 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 two sources gol from(30.3) enormously large and plotting the
p=0. In the first place, if o is exactly 0, we have 0/0, but if o is infinitesimal, the ratio of the two sines squared is simply n, since the sine and the angle are approxi- mately equal. Thus the intensity of the maximum of the curve is equal to n2 times the intensity of one oscillator. That is easy to see, because if they are all in phase then the little vectors have no relative angle and all n of them add up so the ampli tude is n times, and the intensity n- times, stronger. first time i hase p increases, the ratio of the two sines begins to fall off, and the 9= 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 accumu lated angle in all the arrows, the total phase difference between the first and last oscillator, must be 2 to complete the circle 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 maxi mum position, because both the numerator and the denominator of (30. 3)are variant, but sin /2 varies quite slowly compared with sin no/2 when n is large so when sin np/2= 1 we are very close to the maximum. The next maximum of sin2nφ/2 comes at np/2=3丌/2,orφ=3丌/n. This corresponds to the arrows having traversed the circle one and a half times. On putting o= 3T/n into the formula to find the size of the maximum, we find that sin 23 /2=I in the nu merator(because that is why we picked this angle), and in the denominator we have sin 3T/2n. Now if n is sufficiently large, then this angle is very small and sine is equal to the angle; so for all pra put sin 3T/2n n+/2r 3T/2n. Thus we find that the intensity at this maximum is I=1.(4n2/972).But nlo was the maximum intensity, and so we have 4 9T2 times the maximum Fig. 30-2. The intensity as a function tensity, which is about 0.047, less than 5 It, of the maximum intensity! Of of phase angle for a large number of course there are decreasing intensities farther out. So we have a very sharp centra oscillators of equal strength maximum with very weak subsidiary maxima on the sides It is possible to prove that the area of the whole curve including all the little bumps,is equal to 2nlo, or twice the area of the dotted rectangle in Fig. 30-2 Now let us consider further how we may apply Eq.(30 iferent cir cumstances, 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 n of them, all spaced by a distance d, and we shall suppose that the intrinsic relative phase, one to the next is a. Then if we are observing in a given direction 6 from the normal, there is additional phase 2Td sin e/a because of the time delay between each successive two, which we talked about before. Thus First, we shall take the case a =0. That is, all oscillators are in phase, and ve want to know what the intensity is as a function of the angle 0. In order to iga 3os-d v a iea r array of n equal ind out, we merely have to put o a kd sin e into formula (303) and see what when all the oscillators are in phase there is a strong intensity in the direction 8=0. On the other hand, an interesting question is, where is the first minimum? That occurs when 2T/n. In other words, when 2Td sin B/A= 2T/n,we get the first minumum of the curve. If we get rid of the 2T's so we can look at it little better, it says that the total length L of the array. Referring to Fig 30-3, we see that na sin is Now let us understand physically why we get a minimum at that position. nd L sin 8= 4. What (30.5)says is that when A is equal to one wavelength, we get a minimum. Now why do we get a minimum when 4=x? Because the contribu tions of the various oscillators are then uniformly distributed in phase from 0to
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 = A, we get a minimum. That is the first minimum There is another important feature about formula (30.3), which is that if the angle is increased by any multiple of 2T, it makes no difference to the formula So we will get other strong maxima at 2, 4T, 6r, and so forth. Near each 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? The condition is that 2Tm, where m is any integer. That is, 2d sin 0/ 2Tm. Dividing by 2T, we see that 30.6) This looks like the other formula, (30.5). No, that formula was nd sin 0=x. The difference is that here we have to look at the individual sources, and when we say d sin 8 =mA, that means that we have an angle 0 such that 8= mA. 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 in phase because out of phase by 3600 is the same as being in phase. So they all contribute in phase and produce just as good a maximum as the one for m=0 that we dis cussed before. The subsidiary bumps, the whole shape of the pattern, is just 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 aximum 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. m is called the order of the beam We call attention to the fact that if d is less than A, Eq.(30.6)can have no solution except m =0, so that if the spacing is too small there is only one possible beam, the zero-order one centered at 0=0. (Of course, there is also a beam in the opposite direction. In order to get subsidiary great maxima, we must have the spacing d of the array greater than one wavelength 30-2 The diffraction 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 infinitesimal wires and drive them all with a given phase. But there is a very easy way to do what amounts to the same thing. ing an electric field which arrives at each one of the wires at the same phase (it is so far away that the time delay is the same for all of the wires ).(One can work out cases with curved arrays, but let us take a plane one. Then the external electric field will drive the electrons up and down in each wire. That is, the field which is coming from the original source will shake the electrons up and down, and in moving, these represent new generators. This 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. Therefore 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. If the incidence is normal, the phases will be equal, and we will get exactly the tance we have been di g. Therefore, if the wire in the normal direction, and in certain other directions given by (30battering spacing is greater than the wavelength, we will get a strong intensity of This can also be done with light! Instead of wires, we use a flat piece of glass and make notches in it such that each of the notches scatters a little differently than the rest of the glass. If we then shine light on the glass, each one of the notches 30-3
will represent a source, and if we space the lines very finely, 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 notches! Such objects have actually been made and are in common use-they are calle In one of its forms, a diffraction grating consists of nothing but a plane glass sheet transparent and co with scratches on it. There are often several hundred scratches to the millimeter, very carefully arranged so as to be equally spaced. The effect of such a grating can be seen by arranging a projector so as to 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 another strong patch of light which is olored. This, of course, is the slit image spread out over a wide angular range, because the angle 0 in(30.6) depends upon X, and lights of different colors, as we know, correspond to different frequencies, and therefore different wavelengths The longest visible wavelength is red, and since d sin 6=A, that requires a larger 8. And we do, in fact, find that red is at a greater angle out from the central image There should also be a beam on the other side and indeed we see one on the screen 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 equal! The reason is that the grating has been carefully built to do just this. How? If the grating consists of very fine notches, infinitesimally wide, spaced evenly, then all the intensities would indeed be equal. But, as a matter of fact although we have taken the simplest case, we could also have considered an array of pairs of antennas, in which each member of the pair has a certain strength and some relative phase. In this case, it is possible to get intensities which are different in the different orders. A is often made with little"sawtooth"cuts instead of little symmetrical notches. By carefully arranging the"sawteeth, "more light may be sent into one particular order of spectrum than into the others. In a yould like to have as much light as possible in one orders. This may seem a complicated point to bring in, but it is a very clever thing to do, because it makes the grating more useful So far, we have taken the case where all the phases of the sources are equ But we also have a formula for when the phases differ from one to the next by an angle a. That requires wiring up our antennas with a slight phase shift between each one. Can we do that with light? Yes, we can do it very easily, for suppose that there were a source of light at infinity, at an angle such that the light is coming in at an angle Bin, and let us say that we wish to discuss the scattered dsinθ d sin 8 beam, which is leaving at an angle lout. the bout is the same 0 as we have had before but the Bin is merely a means for arranging that the phase of each source is different: the light coming from the distant driving source first hits one scratch ein out then the next, then the next, and so on, with a phase shift from one to the other, which, as we see, is a = -d sin Bin/. Therefore we have the formula for a grating in which light both comes in and goes out at an angle Fig. 30-4. The path difference for rays scattered from adjacent rulings of a p= 2Td sin Bout/A- 2rd sin ]in/. (307) ing is d sin Bout -d sin Bin. Let us try to find out where we get strong intensity in these circumstances. The condition for strong inter of course that should There are several interesting points to be noted d is less than X; in fact, this is the only solution. In this ca ponds to m=0, where One case of rather great interest is that which corr sin Bin, which means that the light comes out in the same direction as the light which was exciting the grating. We might think that the light"goes right through No, it is diferent light that we are talking about. The light that goes right through is from the original sot talking about is th
enerated by scattering. It turns out that the scattered light is going in the same direction as the original light, in fact it can interfere with it-a feature which we will study later There is another solution for this same case. For a given in, Bout may be the supplement of Bin. So not only do we get a beam in the same direction as the in coming beam but also one in another direction, which, if we consider it carefully, is such that the angle of incidence is equal to the angle of scattering. This we call the reflected beam So we begin to understand the basic machinery of reflection: the light that comes in generates motions of the atoms in the reflector, and the reflector then aerates a new wave, and one of the solutions for the direction of scattering, the only solution if the spacing of the scatterers is small compared with one wavelength, is that the angle at which the light comes out is equal to the angle at which it comes Next, we discuss the special case when d-0. That is, we have just a solid piece of material, so to speak, but of finite length. In addition, we want the phase shift from one scatterer to the next to go to zero. In other words, we put more and more antennas between the other ones, so that each of the phase differences is getting smaller, but the number of antennas is increasing in such a way that the otal phase difference, between one end of the line and the other, is constant. Let us see what happens to(30.)if we keep the difference in phase no from one end to the other constant(say no= D), letting the number go to infinity and the phase shiftφ of each one go to zero. But nowφ is so small that sinφ=中, and if we also recognize n2Io as Im, the maximum intensity at the center of the beam (308) This limiting case is what is shown in Fig. 30-2. In such circumstances we find the same general kind of a picture as for finite spacing with d>A; all the side lobes are practically the same as before, but there are no higher-order maxima. If the scatterers are all in phase, we get a maximum in the direction Bout =0, and a minimum when the distance a is equal to A, just as for finite d and n. So we can even analyze a continuous distribution of scatterers or oscillators, by using integrals instead of summing an example, suppose there were a long line of oscillators, with the charge Fig. 30-5. The intensity pattern of a oscillating along the direction of the line(Fig. 30-5). From such an array the continuous line of oscillators has a single greatest intensity is perpendicular to the line. There is a little bit of intensity up strong maximum and many weak"side nd down from the equatorial plane, but it is very slight. With this result, we can lobes handle a more complicated situation. Suppose we have a set of such lines, each producing a beam only in a plane perpendicular to the line. To find the intensity in various directions from a series of long wires, instead of infinitesimal wires, is the same problem as it was for infinitesimal wires, so long as we are in the central plane perpendicular to the wires; we just add the contribution from each of the long wires. That is why, although we actually analyzed only tiny antennas, we might as well have used a grating with long, narrow slots. Each of the long slots produces an effect only in its own direction, not up and down, but they are all set next to each other horizontally, so they produce interference that way. Thus we can build up more complicated situations by having various distrib tions of scatterers in lines, planes, or in space. The first thing we did was to cor sider scatterers in a line, and we have just extended the analysis to strips; we can work it out by just doing the necessary summations, adding the contributions from the individual scatterers. The principle is always the same 30-3 Resolving power of a grating We are now in a position to understand a number of interesting phenomena For example, consider the use of a grating for separating wavelengths. We noticed that the whole spectrum was spread out on the screen, so a grating can be used an instrument for separating light into its different wavelengths. One of the