Now we go to the next maximum, and we wantto see that it is really much smaller than the first one, as we had hoped. Weshall not go precisely to the maximum position, because both the numerator andthe denominator of (30.3)are variant, but sinϕ/2 varies quite slowly compared with sinnϕ/2when n is large, so when sin2nϕ/2=1 we are very closeto the maximum. The next maximum of sin2nϕ/2 comesat nϕ/2=3π/2 , or ϕ=3π/n . Thiscorresponds to the arrows having traversed the circle one and a half times. 现在，我们去到下一个最大值，我们想看看，它是真的比第一个小很多，就像我们所希望的那样。我们将精确地走到最大值的位置，因为，（30.3）中的分子和分母，都是变量，但是，当n很大时，与sinnϕ/2相比，sinϕ/2变的相当慢，所以，当sin2nϕ/2=1时，我们就可以非常接近最大值了。 sin2nϕ/2的下一个最大值，是在nϕ/2=3π/2，或 ϕ=3π/n。这相当于那些箭头，已经走过了1.5个循环的时间。On putting ϕ=3π/ninto the formula to find the size of the maximum, we find that sin23π/2=1in the numerator (because that is why we picked this angle), and in thedenominator we have sin23π/2n . Now if n issufficiently large, then this angle is very small and the sine is equal to theangle; so for all practical purposes, we can put sin3π/2n=3π/2n. Thus we find that the intensity at this maximum is I=I0(4n2/9π2). But n2I0 was the maximum intensity, andso we have 4/9π2 times the maximum intensity, which isabout 0.045 , less than 5 percent, of the maximum intensity! Ofcourse there are decreasing intensities farther out. So we have a very sharpcentral maximum with very weak subsidiary maxima on the sides.
在把ϕ=3π/n代入公式，以找到最大值的大小时，我们发现，在分子中，sin23π/2=1（因为这就是我们为什么选这个角度），在分母中，我们有sin23π/2n。现在，如果n充分大，那么，这个角度就很小，这样，其正弦就等于这个角度；于是，对于所有实践的目的来说，我们可以让sin3π/2n=3π/2n。这样，我们发现，在这个最大值处的强度就是I=I0(4n2/9π2)。但是，n2I0是最大的强度，所以，我们就有：4/9π2乘以最大强度，大约是0.045，比最大强度的5%略小！当然，在更远处，还有不断降低的强度。于是，在中心，我们就有一个非常尖锐的最大值，在边上，还有非常弱的、附属的最大值。

It is possible to prove that the area ofthe whole curve, including all the little bumps, is equal to 2πnI0, or twice the area of the dotted rectangle in Fig. 30–2.
可以证明，整个曲线的面积，包括那些小隆起的，等于 2πnI0，或者，是图30-2中的点线所构成的矩形面积的两倍。

Now let us consider further how we mayapply Eq. (30.3)in different circumstances, and try to understand what is happening. Let usconsider 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 , andwe shall suppose that the intrinsic relative phase, one to the next, is α. Then if we are observing in a given direction θ from the normal,there is an additional phase 2πdsinθ/λ because of thetime delay between each successive two, which we talked about before. Thus
现在，让我们进一步考虑，如何把方程（3.13），应用到不同的情形中，并尝试去理解，究竟发生了什么。我们可以设想，我们的源，都在一条线上，如图30-3所示，共有n个，间距为d，我们将假定，内置的相对相位，即一个对其下一个的，是α。因此，设角度 θ，是与法线的夹角，如果我们沿这个方向去观察，那么，由于时间延迟的原因，在每两个挨着的振荡器之间，就有一个附加相位 2πdsinθ/λ， 这我们以前谈过。这样：
Φ =α+2πdsinθ/λ =α+kdsinθ. (30.4)

First, we shall take the case α=0 .That is, all oscillators are in phase, and we want to know what the intensityis as a function of the angle θ . In order to find out, we merelyhave to put ϕ=kdsinθ into formula (30.3)and see what happens. In the first place, there is a maximum when ϕ=0 .That means that when all the oscillators are in phase there is a strongintensity in the direction θ=0 . On the other hand, an interestingquestion is, where is the first minimum? That occurs when ϕ=2π/n. In other words, when 2πdsinθ/λ=2π/n , weget the first minimum of the curve. If we get rid of the 2π ’s sowe can look at it a little better, it says that
首先，我们取α=0这种情况。也就是说，所有的振荡器，都是同相位的，且我们想知道，强度，作为角度θ的函数，究竟是什么？为了找出这个，我们只是把ϕ=kdsinθ，带入公式（30.3），看看会发生什么。首先，当ϕ=0等于零时，有一个最大值。这就意味着，当所有的振荡器，都是同相位时，在θ=0方向，有一个强的强度。另一方面，一个有趣的问题就是，第一个最小值，在哪里？它出现在ϕ=2π/n时。换句话说，当2πdsinθ/λ=2π/n时，我们就得到了曲线的第一个最小值。如果我们消去2π，我们就可以更好地观察它，它说：
ndsinθ=λ. (30.5)

Now let us understand physically why we geta minimum at that position. nd is the total length L of thearray. Referring to Fig. 30–3, wesee that ndsinθ=Lsinθ=Δ . What (30.5)says is that when Δ is equal to one wavelength, we get a minimum. Nowwhy do we get a minimum when Δ=λ ? Because the contributions of thevarious 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 Δ=λ , we geta minimum. That is the first minimum.
现在，让我们来从物理上来理解，为什么我们在那个位置上，会得到一个最小值。nd是阵列的总长度 L。参考图30-3，我们看到ndsinθ=Lsinθ=Δ。（30.5）所说，就是当Δ等于一个波长时，我们得到的一个最小值。现在，为什么当Δ=λ时，我们却得到了一个最小值？因为，那时候，不同振荡器的贡献，在相位上，是在0到360度之间，均匀分布。（图30-1）中的箭头，将会走一个圆，我们在所有的方向，都增加相等的矢量，这样，总和就是零。于是，当我们有了一个角度，以让Δ=λ时，我们就得到了一个最小值。它是第一个最小值。

There is another important feature aboutformula (30.3),which is that if the angle ϕ is increased by any multiple of 2π, it makes no difference to the formula. So we will get other strong maximaat ϕ=2π , 4π , 6π , and so forth.Near each of these great maxima the pattern of Fig. 30–2 isrepeated. We may ask ourselves, what is the geometrical circumstance that leadsto these other great maxima? The condition is that ϕ=2πm , where mis any integer. That is, 2πdsinθ/λ=2πm . Dividingby 2π , we see that
公式（30.3），还有另外一个重要的特性，那就是，如果角度ϕ的增量，是2π的任何倍数，那么，对此公式来说，将没有任何不同。所以，我们在ϕ=2π , 4π , 6π等处，就会得到其他的、强的最大值。在接近每一个这些最大值的地方，图30-2中的模型，都会被重复。我们可以问我们自己：究竟是什么样的几何模型，才导致了这些其他的最大值呢？条件就是ϕ=2πm，这里，m是任何整数。也就是说，2πdsinθ/λ=2πm。除以 2π，我们看到：
dsinθ=mλ. (30.6)

This looks like the other formula, (30.5).No, that formula was ndsinθ=λ . The difference is thathere we have to look at the individual sources, and when we say dsinθ=mλ, that means that we have an angle θ such that δ=mλ .In other words, each source is now contributing a certain amount, andsuccessive ones are out of phase by a whole multiple of 360∘ , and therefore are contributing in phase, because out ofphase by 360∘ is the same asbeing in phase. So they all contribute in phase and produce just as good amaximum as the one for m=0 that we discussed before. The subsidiarybumps, the whole shape of the pattern, is just like the one near ϕ=0 ,with exactly the same minima on each side, etc. Thus such an array will sendbeams in various directions—each beam having a strong central maximum and acertain number of weak “side lobes.” The various strong beams are referred toas the zero-order beam, the first-order beam, etc., according to the valueof m . m is called the order of the beam.
这看上去，像公式（30.5）。非也，那个公式是ndsinθ=λ。差别就是，这里，我们必须看个别的源，当我们说dsinθ=mλ时，就意味着，我们有一个角度θ，使得 δ=mλ。换句话说，每个源，现在都要贡献一定的量，相继的源，相位相差360度的整数倍，因此，贡献是有相位差的，但因为，当相位差360度时，与同向位是一样的。所以，它们的贡献，就又是同向位的，所产生的最大值，与m=0处产生最大值，同样的好，那个我们已经讨论过了。边上的隆起，模型的整个形状，与ϕ=0附近的类似，在每一侧，都有着几乎同样的最小值等等。这样，这样一个阵列，将会在不同的方向，发送电波—每个电波都有一个强的中心最大值，和若干弱的“边缘叶”。各个强的电波，依据m的值，按第零电波、第一电波等等，来指出，m被称为电波的序号。

We call attention to the fact that if dis less than λ , Eq. (30.6)can have no solution except m=0 , so that if the spacing is too small thereis only one possible beam, the zero-order one centered at θ=0 . (Ofcourse, there is also a beam in the opposite direction.) In order to getsubsidiary great maxima, we must have the spacing d of the arraygreater than one wavelength.
我们要注意一个事实，即如果d小于λ，那么，方程（30.6），除了m=0外，就没有解，所以，如果空间太小，那么，就只有一个可能的电波，即集中在θ=0处的零序电波（当然，在相反的方向，还有电波。）为了得到次要的最大值，我们应该让空间的间距d，大于一个波长。

30–2The diffraction grating 30-2 衍射光栅
In technical work with antennas and wiresit is possible to arrange that all the phases of the little oscillators, orantennas, are equal. The question is whether and how we can do a similar thingwith light. We cannot at the present time literally make littleoptical-frequency radio stations and hook them up with infinitesimal wires anddrive them all with a given phase. But there is a very easy way to do whatamounts to the same thing.
在技术工作中，用不同的天线和线，来安排所有小振荡器的、或天线的相位，是相等的{？}。问题是，对于光，是否能够、及如何做同样的事情？现在，我们不能真正地做光学频率的广播电台，并用无限小的线，把它们钩在一起，及用的给定的相位，来驱动它们。但是，有一个非常容易的方式，来做一件事，该事，可以被算作是同样的事情。{？}

Suppose that we had a lot of parallel wires, equallyspaced at a spacing d , and a radiofrequency source very far away,practically at infinity, which is generating an electric field which arrives ateach one of the wires at the same phase (it is so far away that the time delayis 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 theelectrons up and down in each wire. That is, the field which is coming from theoriginal source will shake the electrons up and down, and in moving, theserepresent new generators. This phenomenon is called scattering: a lightwave from some source can induce a motion of the electrons in a piece ofmaterial, and these motions generate their own waves. 假设我们有很多平行的线，它们在空间中，等间距分布，距离为d，而一个无线电频率的源，在很远处，实际上，可以在无限远，它所产生的电场，以同样的相位，到达每一条线（它是如此之远，以至于，对所有的线，时间延迟，都是一样的。）（人们可以用弯曲的阵列，来做这个实验，但是，我们将用一个平面阵列）因此，此外部的电场，将会驱动电子，在每条线中，上下运动。也就是说，从最初的源所来的场，将会上下摇动电子，使其运动，这代表着一种新的发生器。这个现象，被称为散射：从某个源来的光波，可以在一片材料中，导致电子运动，这些运动，会产生其自己的波。

Therefore all that is necessary is to setup a lot of wires, equally spaced, drive them with a radiofrequency source faraway, and we have the situation that we want, without a whole lot of specialwiring. If the incidence is normal, the phases will be equal, and we will getexactly the circumstance we have been discussing. Therefore, if the wirespacing is greater than the wavelength, we will get a strong intensity ofscattering in the normal direction, and in certain other directions givenby (30.6).
因此，我们所要做的，就是设置很多线，让其等距，用一个很远处的无线电频率的源，来驱动它们，这样，我们想要的情形，就有了，而不用很多特殊的配线{？}。如果入射角是法线方向的，那么，相位就是相等的，我们就会得到我们曾经讨论过的情况。因此，如果线间的距离，比波长大，那么，在法线方向，我们就会得到一个强的散射强度，在某些其他的方向，也会得到一些，通过（30.6）给出{？}。

This can also be done with light! Instead of wires, we use a flat piece of glass and make notches init such that each of the notches scatters a little differently than the rest ofthe glass. If we then shine light on the glass, each one of the notches willrepresent a source, and if we space the lines very finely, but not closer thana wavelength (which is technically almost impossible anyway), then we wouldexpect 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 thespacing of the notches! Such objects have actually been made and are in commonuse—they are called diffraction gratings.
也可以用光，来做此事！不是用线，我们使用一片玻璃，在上面刻出槽来，这样，每个槽的散射，与玻璃其他部分的散射，都略有不同。因此，如果我们用光，照这个玻璃，那么，每个槽，都将会代表一个源，如果我们非常精细地画线{刻槽}，但并不接近于一个波长（技术上讲，几乎不可能），那么，我们将会期待一个奇迹般的现象：光不仅会直接通过，而且，在一个有限的角度上，还会是很强的光波，这依赖槽间的距离。这种对象，实际上已经造出，且很常用，它们被称为散射光栅。

In one of its forms, a diffraction gratingconsists of nothing but a plane glass sheet, transparent and colorless, withscratches on it. There are often several hundred scratches to the millimeter, verycarefully arranged so as to be equally spaced. The effect of such a grating canbe seen by arranging a projector so as to throw a narrow, vertical line oflight (the image of a slit) onto a screen. When we put the grating into thebeam, with its scratches vertical, we see that the line is still there but, inaddition, on each side we have another strong patch of light which is colored.散射光栅，有一种形式，包含着平的玻璃薄片，透明无色，上面有槽。通常在一毫米内，有数百个槽，非常仔细地排列着，空间的间距相等。这种光栅的效果，通过安排一个投影仪，可以看到，即把一个窄的、垂直的光线（缝隙的图像），投到一个屏幕上。当我们把光栅放到光线中，让槽是垂直的，我们看到，线仍在那里，但是另外，在每一边，我们都看到另外一个光斑，它是彩色的。This, of course, is the slit image spread out over a wide angularrange, because the angle θ in (30.6)depends upon λ , and lights of different colors, as we know,correspond to different frequencies, and therefore different wavelengths. Thelongest visible wavelength is red, and since dsinθ=λ ,that requires a larger θ . 这当然是缝隙图像，通过了一个宽的角度范围，因为，（30.6）中的角度 θ，依赖于 λ，而正如我们所知，不同颜色的光 ，相应于不同的频率，因此，就是相应于不同的波长。最长的光波是红色，由于dsinθ=λ，这就要求一个较大的θ。And we do, in fact, findthat red is at a greater angle out from the central image! There should also bea beam on the other side, and indeed we see one on the screen. Then, theremight be another solution of (30.6)when m=2 . We do see that there is something vaguely there—very weak—andthere are even other beams beyond.
事实上，我们确实发现了，从中心图像中，红色，是以一个较大的角度而射出！在另一侧，还应该有一个光束，我们在屏幕上，也确实看到了。因此，对于（30.6），当m =2时，还应该有另一个解。我们确实看到，那里有一些模糊的东西，非常弱，甚至，在更远处，还有一些其他的光束。

We have just argued that all these beamsought to be of the same strength, but we see that they actually are not and, infact, not even the first ones on the right and left are equal! The reason isthat the grating has been carefully built to do just this. How? If the gratingconsists of very fine notches, infinitesimally wide, spaced evenly, then allthe intensities would indeed be equal. But, as a matter of fact, although wehave taken the simplest case, we could also have considered an array of pairsof antennas, in which each member of the pair has a certain strength and somerelative phase. 我们已经论证了，所有这些光束，强度都应相同，但是，我们看到，实际上并非如此，而事实上，甚至，在右边和左边，都不是同样的强度{？}！原因是这样，光栅被仔细地制造出来，就是为做此事。怎么做的呢？如果光栅包含着非常精细的槽，槽宽无限小，均匀分布，那么，所有强度，确实就会是相等的。但事实上，虽然我们所取情况，是最简单的，但我们还是可以考虑一个天线对的阵列，在其中，对的每一个成员，都有一个确定的强度，和某个相对的相位。 In this case, it is possible to get intensities which are differentin the different orders. A grating is often made with little “sawtooth” cutsinstead of little symmetrical notches. By carefully arranging the “sawteeth,”more light may be sent into one particular order of spectrum than into theothers. In a practical grating, we would like to have as much light as possiblein one of the orders. This may seem a complicated point to bring in, but it isa very clever thing to do, because it makes the grating more useful.
在这种情况下，得到不同序的强度，就是可能的。造光栅时，通常会做一些小的“锯齿”，而不是小的对称的槽。通过仔细地安排这些“锯齿”切口，更多的光，可以被发送到光谱的某个具体的序中，而不是其他地方。在一个实际的光栅中，对于某个序，我们希望有尽可能多的光。这一点，非常复杂，但引入它，非常聪明，因为，它会使得光栅更有用。

So far, we have taken the case where allthe phases of the sources are equal. But we also have a formula for ϕwhen the phases differ from one to the next by an angle α . Thatrequires wiring up our antennas with a slight phase shift between each one. Canwe do that with light? Yes, we can do it very easily, for suppose that therewere a source of light at infinity, at an angle such that the light iscoming in at an angle θin , and let us say that we wish to discussthe scattered beam, which is leaving at an angle θout (Fig. 30–4). Theθout is the same θ as we have had before, but the θin ismerely a means for arranging that the phase of each source is different: thelight coming from the distant driving source first hits one scratch, then thenext, then the next, and so on, with a phase shift from one to the other,which, as we see, is α=−2πdsinθin/λ .Therefore we have the formula for a grating in which light both comes in andgoes out at an angle:
目前为止，我们所选的案例中，所有源的相位，都相等。但是，当一个源的相位，与另一个，相差角度α时，对于ϕ，我们还有一个公式。这就要求，在对我们的天线布线时，在两个源之间，要有一个轻微的相位偏移。我们用光，也能做到此事吗？是的，能做，且非常容易，因为，假设有一个光源，在无穷远处，有一定的角度，这样，光就是以角度θin到来，让我们说，我们希望讨论：散射出去的光束，其角度为θout（图30-4）。θout与我们以前的θ，是同样的，但是，θin只是一种方式，该方式的目的，就是要安排每个源的相位，使其不同：光从远处的一个驱动性的源而来，它首先打到一个槽上，然后，是下一个，然后，再是下一个，如此等等，一个于一个之间，有一个相位的偏移，我们看到，它就是 α=−2πdsinθin/λ。因此，对于一个光栅，光进来出去，各有角度，所以我们就有一个公式：
(30.7)