Let us try to find out where we get strongintensity in these circumstances. The condition for strong intensities is, ofcourse, that ϕ should be a multiple of 2π . There areseveral interesting points to be noted.
现在，让我们找出，在这些情形下，我们在哪里，能够得到强的强度。强的强度的条件，当然就是，ϕ应该是2π的整数倍。有几个感兴趣的点，需要注意。

One case of rather great interest is thatwhich corresponds to m=0 , where d is less than λ ;in fact, this is the only solution. In this case we see that sinθout=sinθin, which may mean that θout is the supplement of θin so thelight comes out in the same direction as the light which was excitingthe grating. We might think that the light “goes right through.” No, it is differentlight that we are talking about. The light that goes right through is fromthe original source; what we are talking about is the new light which isgenerated by scattering. It turns out that the scattered light is going inthe same direction as the original light, in fact it can interfere with it—afeature which we will study later.
有一种情况，相当有趣，它相当于于m=0，这里d小于λ；事实上，这是唯一的解。在这种情况下，我们看到，sinθout=sinθin，它就意味着θout是θin补偿{？}，这样，光出去的方向，就与激发光栅的光的方向，是一样的。我们可能会想，光是“直接通过的。”不，我们所谈的，是不同的光。直接通过的光，是从最初的源来的；而我们正在谈论的，则是新的光，它是由散射产生的。结果就是，散射出来的光，与最初的光，所走方向，是一样的，事实上，它可以与它，发生干涉--这个特性，我们稍后将研究。

There is another solution for this samecase: θout may equal θin . So not only do we get a beam inthe same direction as the incoming beam but also one in another direction, suchthat the angle of incidence is equal to the angle of scattering. This wecall the reflected beam.
对于这种情况，还有另外一个解：θout 可以等于 θin。所以，我们不仅可以得到一个光束，它与进入的光束，方向相同，而且，在另外一个方向，还可得到一个光束，这样，入射角等于散射角。这我们可以称为反射束。

So we begin to understand the basicmachinery of reflection: the light that comes in generates motions of the atomsin the reflector, and the reflector then regenerates a new wave, and oneof the solutions for the direction of scattering, the only solution ifthe spacing of the scatterers is small compared with one wavelength, is thatthe angle at which the light comes out is equal to the angle at which it comesin!
于是，我们开始理解反射的基本机制了：进入的光，会让反射器中的原子，产生运动，因此，反射器会产生一个新的波，散射的方向的诸解之一，也是散射器的空间间距，与一个波长相比，较小时的唯一解，就是：光出去的角度，与光进入的角度，是相等的。

Next, we discuss the special case when d→0. That is, we have just a solid piece of material, so to speak, but of finitelength. In addition, we want the phase shift from one scatterer to the next togo to zero. In other words, we put more and more antennas between the otherones, so that each of the phase differences is getting smaller, but the numberof antennas is increasing in such a way that the total phase difference,between one end of the line and the other, is constant. Let us see what happensto (30.3)if we keep the difference in phase nϕ from one end to the otherconstant (say nϕ=Φ ), letting the number go to infinity and the phaseshift ϕ of each one go to zero. But now ϕ is so small that sinϕ=ϕ, and if we also recognize n2I0 as Im, the maximum intensity at the center of the beam, we find
下面，我们讨论d→0时这种特殊情况。也就是说，我们有一个固体的材料片，这么说吧，一个有限长度的固体材料片。另外，我们想让：从一个散射器到另一个散射器之间的相位偏移，趋向于零。换句话说，我们在一对天线之间，放入越来越多的天线，这样，每对天线之间的相位差，越变越小，但是，总的天线的数目，是以这种方式增加：总的相位差，即在线的这一端和那一端之间的差，是一个常数。让我们看看，如果我们保持从一端到另一端的相位差 nϕ，为常数（比如nϕ=Φ），设天线数目n，趋于无穷，且每个相位偏移ϕ，趋向于零，那么，对于（30.3），会发生什么。但是，现在ϕ，是如此之小，以至于sinϕ=ϕ，且如果我们把n2I0，认作是Im，即在光束中心的最大强度，那么，我们发现
(30.8)
This limiting case is what is shown inFig. 30–2.
这种有限的情况，就是图30-2所示的情况。

In such circumstances we find the samegeneral kind of a picture as for finite spacing with d<λ ; allthe side lobes are practically the same as before, but there are nohigher-order maxima. If the scatterers are all in phase, we get a maximum inthe direction θout=0 , and a minimum when the distance Δ is equalto λ , just as for finite d and n . So we caneven analyze a continuous distribution of scatterers or oscillators, byusing integrals instead of summing.
在这种情形下，我们所发现的图像，与空间无限且d<λ时的图像，具有同样的普遍性；与以前的一样，所有的边缘叶，基本上同样的，但是，没有较高序的极大值了。如果散射器都是同相位的，那么，我们在方向θout=0，会得到一个最大值，当距离Δ 等于 λ 时，会得到一个最小值，正如有限的d和 n一样。所以，用积分代替求和，我们甚至可以分析：散热器或天线的连续分布。

Fig. 30–5.The intensity pattern of acontinuous line of oscillators has a single strong maximum and many weak “sidelobes.” 图 30-5 一个振荡器的连续的线，其强度模型，有一个单独的强的最大值，和很多弱的“边缘叶”。
As an example, suppose there were a longline of oscillators, with the charge oscillating along the direction of theline (Fig. 30–5).From such an array the greatest intensity is perpendicular to the line. Thereis a little bit of intensity up and down from the equatorial plane, but it isvery slight. With this result, we can handle a more complicated situation.Suppose we have a set of such lines, each producing a beam only in a planeperpendicular to the line. To find the intensity in various directions from aseries of long wires, instead of infinitesimal wires, is the same problem as itwas for infinitesimal wires, so long as we are in the central planeperpendicular to the wires; we just add the contribution from each of the longwires. That is why, although we actually analyzed only tiny antennas, we mightas well have used a grating with long, narrow slots. Each of the long slotsproduces an effect only in its own direction, not up and down, but they are allset next to each other horizontally, so they produce interference that way.
作为一个例子，假设有一长队的振荡器，电荷震荡，是沿着此队的方向（图30-5）。这样一个阵列，所产生的最大强度，是垂直于此队的。从赤道面来的，在上下方，还有一点小的强度，{？}但是，它很轻微。利用这个结果，我们可以掌控更复杂的情况。假设我们有一组这种队列，每一个队，都会产生一个波束，波束所在平面，垂直于该队。一系列长线、而不是无限小的线，会在不同方向，产生强度，要找出此强度，与找出无限小的线所产生的强度，是同样的问题，只要我们所在的中心平面，是垂直于这些队线的；我们刚刚增加了：来自每条长线的贡献。这就是为什么，虽然，我们实际上只分析了微小的天线，但是，我们也可以利用带有细长槽的光栅。每一个长槽，只会在其自己的方向上，产生效果，而不是在其上方或下方，但是，由于它们是水平地并排排列，所以，它们会以那种方式，产生干涉。

Thus we can build up more complicatedsituations by having various distributions of scatterers in lines, planes, orin space. The first thing we did was to consider scatterers in a line, and wehave just extended the analysis to strips; we can work it out by just doing thenecessary summations, adding the contributions from the individual scatterers.The principle is always the same.
这样，通过在不同的线、面、或空间中，布置散射器，我们就可以造出更复杂的情况。我们所做的第一件事，就是考虑一条线中的散射器，我们已经把分析，扩展到了队列{从点到线}；通过做必要的总结，增加个别散射器所做的贡献，我们就可以完成此工作。原理总是一样的。

Suppose that for light of a certain colorwe happen to have the maximum of the diffracted beam occurring at a certainangle. If we vary the wavelength the phase 2πdsinθ/λis different, so of course the maximum occurs at a different angle. That is whythe red and blue are spread out. How different in angle must it be in order forus to be able to see it? If the two maxima are exactly on top of each other, ofcourse we cannot see them. If the maximum of one is far enough away from theother, then we can see that there is a double bump in the distribution oflight. In order to be able to just make out the double bump, the followingsimple criterion, called Rayleigh’s criterion, is usually used(Fig. 30–6).It is that the first minimum from one bump should sit at the maximum of theother. Now it is very easy to calculate, when one minimum sits on the othermaximum, how much the difference in wavelength is. The best way to do it isgeometrically.
假设对于某确定颜色的光，我们刚好，在某确定的角度，会有衍射波束的最大值。如果能改变波长，则相位2πdsinθ/λ就会不同，于是，当然，最大值就会出现在一个不同的角度。这就是为什么，红和蓝是展开的。为了让我们能够看到这种分开，角度应该差多少呢？如果两个最大值，刚好准确叠加，那么当然，我们看不到它们。如果一个波束的最大值，与另一个的最大值，相距足够远，那么，在光的分布中，我们就可以看到两个隆起。为了刚好能够做出双隆起，经常用到的，就是下面的简单判据，被称为瑞利判据（图30-6）。那就是，一个隆起的第一个最小值，应该与另外一个的最大值，相叠加。现在，当一个的最小值与另外一个的最大值叠加时，计算波长的差是多少，就很容易了。做此事的最好方式，是几何方式。

Fig. 30–6.Illustration of the Rayleighcriterion. The maximum of one pattern falls on the first minimum of the other. 图30-6 瑞利判据示意图。一个模式的最大值，落在另外一个模式的第一个最小值处。

In order to have a maximum forwavelength λ′ , the distance Δ (Fig. 30–3) mustbe nλ′ , and if we are looking at the m th-order beam, itis mnλ′ . In other words, 2πdsinθ/λ′=2πm, so ndsinθ , which is Δ , is mλ′ times n, or mnλ′ . For the other beam, of wavelength λ , we want tohave a minimum at this angle. That is, we want Δ to be exactly onewavelength λ more than mnλ . That is, Δ=mnλ+λ=mnλ′. Thus if λ′=λ+Δλ , we find
为了让波长λ′有一个最大值 , 距离 Δ (图 30–3)应为nλ′ ，如果我们是在看第m个波束，那么，它就是mnλ′ 。换句话说，2πdsinθ/λ′=2πm，于是ndsinθ , 亦即 Δ , 就是mλ′ 乘以n , 或 mnλ′ 。对于其它波长为λ的波束，我们想在这个角度，有一个最大值。也就是说，我们想让Δ ，比mnλ正好多一个波长λ。也就是, Δ=mnλ+λ=mnλ′ 。这样，如果λ′=λ+Δλ ，我们发现：
Δλ/λ=1/mn. (30.9)
The ratio λ/Δλ is calledthe resolving power of a grating; we see that it is equal to the totalnumber of lines in the grating, times the order. It is not hard to prove thatthis formula is equivalent to the formula that the error in frequency is equalto the reciprocal time difference between extreme paths that are allowed tointerfere:1
比率 λ/Δλ，被称为光栅的分辨率；我们看到，它等于光栅上的线的总数，乘以序号。不难证明，这个公式，与下式等价，下式所表示的，就是两个波束，要被允许发生干涉，则各有路径，这两条路径，会形成一个时间差，所以，频率中的错误，就等于此路径取极端值时的时间差的导数{？}：
Δν=1/T.
In fact, that is the best way to rememberit, because the general formula works not only for gratings, but for any otherinstrument whatsoever, while the special formula (30.9)depends on the fact that we are using a grating.
事实上，这是记忆它的最好方式，因为，一般公式，不仅对光栅起作用，而且，对任何其他的仪器，也起作用，而特殊的公式（30.9），则依赖于：我们正在使用着一个光栅这一事实。

30–4The parabolic antenna 30-4 抛物面天线
Now let us consider another problem inresolving power. This has to do with the antenna of a radio telescope, used fordetermining the position of radio sources in the sky, i.e., how large they arein angle. Of course if we use any old antenna and find signals, we would not knowfrom what direction they came. We are very interested to know whether thesource is in one place or another. One way we can find out is to lay out awhole series of equally spaced dipole wires on the Australian landscape. Thenwe take all the wires from these antennas and feed them into the same receiver,in such a way that all the delays in the feed lines are equal. Thus thereceiver receives signals from all of the dipoles in phase. That is, it addsall the waves from every one of the dipoles in the same phase. Now whathappens? If the source is directly above the array, at infinity or nearly so,then its radiowaves will excite all the antennas in the same phase, so they allfeed the receiver together.
现在，让我们考虑分辨率中的另外一个问题。这与无线电望远镜的天线有关，天空中，有无线电波，无线电望远镜，是用来的探测其源的位置的，亦即，源在角度上有多大。当然，如果我们使用任何老的天线，可以发现信号，但我们并不知道，信号来自何方。源在何处这个问题，我们很感兴趣。解决办法之一，就是我们可以在澳大利亚的大地上，布置一系列的偶极子导线，在空间中，等距分布。然后，我们把所有这些天线的导线，都喂给同一个接收器，以这种方式，所有的延迟，在喂入线中，都是相等的。这样，接收器就是从所有同相位的偶极子中，接收信号。也就是说，它把从每个偶极子中接收到的波，都加到同一个相位上。现在会发生什么呢？如果源是在阵列的正上方，距离有限或比较接近，那么，其无线电波，就将会以同样的相位，激发所有的天线{？}，所以，它们会被一起喂给接收器。

Now suppose that the radio source is at aslight angle θ from the vertical. Then the various antennas arereceiving signals a little out of phase. The receiver adds all these out-of-phasesignals together, and so we get nothing, if the angle θ is too big.How big may the angle be? Answer: we get zero if the angle Δ/L=θ(Fig. 30–3)corresponds to a 360∘ phaseshift, that is, if Δ is the wavelength λ . 现在，假设无线电源，与垂直方向，有一个微小的夹角θ。那么，不同天线所接收到的信号，就有小的相位差。接收器会把这些不同相位的信号，加在一起，并且，如果角度θ太大，我们就什么也得不到。这个角度要多大呢？答案：如果角度Δ/L=θ (图30–3)相当于一个360度的相位偏移，那么，我们得到的是零。This isbecause the vector contributions form together a complete polygon with zeroresultant. The smallest angle that can be resolved by an antenna array oflength L is θ=λ/L . 这是因为，矢量的贡献，一起形成了一个完整的多边形，此形的后果为零。一个天线阵列，长度为L，它能解的最小的角度，就是θ=λ/L{？}。Notice that the receiving pattern of an antenna such as this isexactly the same as the intensity distribution we would get if we turned thereceiver around and made it into a transmitter. This is an example of what iscalled a reciprocity principle. It turns out, in fact, to be generally true for anyarrangement of antennas, angles, and so on, that if we first work out what therelative intensities would be in various directions if the receiver were atransmitter instead, then the relative directional sensitivity of a receiverwith the same external wiring, the same array of antennas, is the same as therelative intensity of emission would be if it were a transmitter.
注意，下面两者，几乎一样：一、一个像这样的天线的接收模式；二、如果我们回转接收器，让它变成一个发射器时，它的强度分布。这个例子，就是我们称为互逆原理的例子。结果就是，对于任何天线、角度等，无论怎么安排，这一点，事实上一般都为真，即如果接受器是一个发射器，我们首先得出，在不同的方向上，相对强度是什么样子，那么，对于一个发射器，它有同样的外部线状分布，即同样的天线阵列，则它的相对方向上的强度，与‘它作为一个发射器时’的‘发射的相对强度’，是一样的。{？}

Some radio antennas are made in a different way.Instead of having a whole lot of dipoles in a long line, with a lot of feedwires, we may arrange them not in a line but in a curve, and put the receiverat a certain point where it can detect the scattered waves. This curve iscleverly designed so that if the radiowaves are coming down from above, and thewires scatter, making a new wave, the wires are so arranged that the scatteredwaves reach the receiver all at the same time (Fig. 26–12). In other words, the curve is a parabola,and when the source is exactly on its axis, we get a very strong intensity atthe focus. In this case we understand very clearly what the resolving power ofsuch an instrument is. 有些无线电的天线，是以不同的方式制造的。即所有的偶极子，并不是在一个条长线上，且带有很多喂入线，而是把它们安排在一个曲面上，把接受器，放在某个点上，在那里，它可以探测到散射的波。这个曲面，设计地很聪明，这样，如果无线电波，是从上方来，通过线的散射，会产生一个新的波，这些线的安排，会使得被散射的波，在同一时间，到达接收器（图26-12）。换句话说，此曲面就是一个抛物面，当源刚好，在其轴上，则我们在焦点上，就会得到一个非常强的强度。在这种情况下，我们就能非常清楚地理解：这样一种设备的分辨率，是什么了。The arranging of theantennas on a parabolic curve is not an essential point. It is only aconvenient way to get all the signals to the same point with no relative delayand without feed wires. The angle such an instrument can resolve is still θ=λ/L, where L is the separation of the first and last antennas. It does notdepend on the spacing of the antennas and they may be very close together or infact be all one piece of metal. Now we are describing a telescope mirror, ofcourse. We have found the resolving power of a telescope! 把天线安排在一个抛物曲面上，并非要旨。它只是一种方便的方式，为的是让所有的信号，都到达同一个点，没有相对延迟，没有喂入线。这种仪器可以分辨的角度，仍是θ=λ/L，这里L，是第一个天线，与最后一个天线之间的距离。它并不依赖于天线的分布，天线可以紧密地排在一起，或者，完全就是同一块金属板。现在，当然，我们是在描述望远镜。我们已经找到了望远镜的分辨率。

(Sometimes the resolving power is written θ=1.22λ/L, where L is the diameter of the telescope. The reason that it is notexactly λ/L is this: when we worked out that θ=λ/L, we assumed that all the lines of dipoles were equal in strength, but when wehave a circular telescope, which is the way we usually arrange a telescope, notas much signal comes from the outside edges, because it is not like a square,where we get the same intensity all along a side. We get somewhat less becausewe are using only part of the telescope there; thus we can appreciate that theeffective diameter is a little shorter than the true diameter, and that is whatthe 1.22 factor tells us. In any case, it seems a little pedantic to putsuch precision into the resolving power formula.2)
（有时，分辨率也被写作θ=1.22λ/L，这里L是望远镜的直径。它不是λ/L，乃是因为：当我们得出θ=λ/L时，我们认定了，所有的偶极子的队列，强度都是相等的，但是，通常我们的望远镜，都是圆形的，所以此时，从外边缘所来的信号，并不是同样的多，因为，圆形与四方形，并不一样，在四方形中，沿着每一边，我们得到的强度，是一样的。我们得到有点少，乃是因为，我们只用了望远镜的一部分；这样，我们就可以认同{？}：有效直径，比真正的直径，要稍微短点；因子1.22告诉我们的，正是此事。在任何情况下，把这样一个精度，放到分辨率公式中，似乎都有点迂腐、或学究气{？}。（脚注2））