30–5Colored films; crystals 50-5 彩色薄膜；晶体
Theabove, then, are some of the effects of interference obtained by adding thevarious waves. But there are a number of other examples, and even though we donot understand the fundamental mechanism yet, we will some day, and we canunderstand even now how the interference occurs. For example, when a light wavehits a surface of a material with an index n , let us say at normalincidence, some of the light is reflected. The reason for the reflectionwe are not in a position to understand right now; we shall discuss it later.But suppose we know that some of the light is reflected both on entering andleaving a refracting medium. Then, if we look at the reflection of a lightsource in a thin film, we see the sum of two waves; if the thicknesses aresmall enough, these two waves will produce an interference, either constructiveor destructive, depending on the signs of the phases. 因此，以上就是干涉的效果；这些干涉，是通过增加各种波，而得到的。但是，还有若干其他的例子，尽管现在，我们还不理解其基础机制，但是，总有一天，甚至现在，我们都会理解：干涉是如何发生的。例如，当一束光波，指数为 n，打到一个材料表面时，比如说，是法向入射，则光的一部分，会被反射。反射的原因为何，现在还不能完全理解，稍后我们将讨论之。但是，假设我们知道，光在进入和离开折射媒介时，有一部分光，被反射了。因此，如果我们在一个薄的胶片中，看光源的反射，那么，我们会看到：两波之和；如果厚度足够小，那么，这两个波，就会产生干涉，要么是建设性的，要么是拆除性的，这依赖于相位的符号。

It might be, for instance, that for redlight, we get an enhanced reflection, but for blue light, which has a differentwavelength, perhaps we get a destructively interfering reflection, so that wesee a bright red reflection. If we change the thickness, i.e., if we look atanother place where the film is thicker, it may be reversed, the redinterfering and the blue not, so it is bright blue, or green, or yellow, orwhatnot. So we see colors when we look at thin films and the colorschange if we look at different angles, because we can appreciate that thetimings are different at different angles. Thus we suddenly appreciate anotherhundred thousand situations involving the colors that we see on oil films, soapbubbles, etc. at different angles. But the principle is all the same: we areonly adding waves at different phases.
例如，情况可能是这样，对于红光，我们得到一个增强的反射，但是，对于蓝光，由于其波长不同，或许，我们就会得到一个拆除性的干涉反射。如果我们换一个厚度，也就是，在胶片比较厚的地方去看，那么，情况可能就会相反，红光会干涉，而蓝光不会，所以，我们会得到一个明亮的蓝光、或绿光、或黄光、或某种东西。所以，当我们看一个薄的胶片时，我们能看到颜色，如果我们换个角度，则颜色也就会改变，因为，我们可以赞同，角度不同，时间不同。这样，对于油膜、肥皂泡等，我们在不同的角度，会看到不同的颜色，对这种情况，我们突然就可以赞同了。情况万千，原理一样：我们只是在不同的相位上，增加了波而已。

As another important application ofdiffraction, we may mention the following. We used a grating and we saw thediffracted image on the screen. If we had used monochromatic light, it wouldhave been at a certain specific place. 下面我们要提的，是折射的另外一个重要应用。利用光栅，我们可以在屏幕上，看到不同的衍射图像。如果我们使用的是单色光，那么，它将会在一个确定的具体位置上。Then there were various higher-order images also. From the positionsof the images, we could tell how far apart the lines on the grating were, if weknew the wavelength of the light. 因此，也将会有各种高阶的图像。根据图像的位置，且如果我们知道了光的波长，那么，我们就会知道光栅上的线，相距多少。 From the difference in intensity of the various images, we couldfind out the shape of the grating scratches, whether the grating was made ofwires, sawtooth notches, or whatever, without being able to see them.This principle is used to discover the positions of the atoms in a crystal.The only complication is that a crystal is three-dimensional; it is a repeatingthree-dimensional array of atoms. We cannot use ordinary light, because we mustuse something whose wavelength is less than the space between the atoms or weget no effect; so we must use radiation of very short wavelength, i.e., x-rays.各种图像，强度有别，由此可知，光栅刮痕的形状，无需实际地去看它们，我们就可知道，光栅是由线构成的，还是由锯齿形的槽、或其他什么构成的。这个原理，被用来发现晶体中原子的位置。唯一的复杂性就是，晶体是三维的，它是原子的三维重复的数组。我们不能用普通的光，因为，我们用的东西，其波长，要比原子之间的空间，更小，否则，我们就得不到任何效果；所以，我们就应该用波长非常短的放射线，亦即，X射线。So, by shining x-raysinto a crystal and by noticing how intense is the reflection in the variousorders, we can determine the arrangement of the atoms inside without ever beingable to see them with the eye! It is in this way that we know the arrangementof the atoms in various substances, which permitted us to draw those picturesin the first chapter, showing the arrangement of atoms in salt, and so on. Weshall later come back to this subject and discuss it in more detail, andtherefore we say no more about this most remarkable idea at present.
所以，通过用X射线，照射晶体，且注意到，对于不同品级的晶体，反射强度如何不同，我们就可决定，晶体内部的原子排列，而无需用眼睛去看！正是通过这种方式，我们知道了，不同材料中的原子的排列，该方式，允许我们画出第一章中所看到各种图像，以显示，盐等材料中的原子排列。稍后，我们将再回到这个话题，更仔细地讨论它。因此现在，对于这个著名的想法，不再多谈。

30–6Diffraction by opaque screens 30-6 通过不透明屏幕的衍射
Now we come to a very interestingsituation. Suppose that we have an opaque sheet with holes in it, and a lighton one side of it. We want to know what the intensity is on the other side.What most people say is that the light shines through the holes, and producesan effect on the other side. It will turn out that one gets the right answer,to an excellent approximation, if he assumes that there are sources distributedwith uniform density across the open holes, and that the phases of thesesources are the same as they would have been if the opaque material wereabsent. Of course, actually there are no sources at the holes, in factthat is the only place that there are certainly no sources.Nevertheless, we get the correct diffraction patterns by considering the holesto be the only places that there are sources; that is a rather peculiarfact. We shall explain later why this is true, but for now let us just supposethat it is.
现在，我们来到一个非常有趣的情况，假设我们有一个不透明的薄板，其上有洞，一侧有光。我们想知道，另外一侧，光强如何。大多数人会这样说，光会通过洞而照耀，在另一侧，产生效果。结果就是，如果某人假定，源是沿着打开的洞，以均匀强度而分布，且这些源的相位，与不透明材料不在场时，是一样的，那么，他就会得到一个正确的答案，一个近似，非常精彩。当然，实际上，洞中无源，事实上，洞中是唯一没有源的地方。尽管如此，通过把洞，考虑为唯一有源之地，我们还是得到了正确的衍射模式；这个事实，相当独特。稍后，我们将会解释，为什么这是真的，但现在，我们只假定它如此。

In the theory of diffraction there isanother kind of diffraction that we should briefly discuss. It is usually notdiscussed in an elementary course as early as this, only because themathematical formulas involved in adding these little vectors are a littleelaborate. Otherwise it is exactly the same as we have been doing all along.All the interference phenomena are the same; there is nothing very much moreadvanced involved, only the circumstances are more complicated and it is harderto add the vectors together, that is all.
在衍射理论中，还有另外一种衍射，我们应该简明地讨论一下。通常，在这种初级课程中，不会讨论它，只是因为，把那些小的矢量加起来所牵扯到的数学公式，有点精细。否则的话，它与我们一直所做，几乎完全一样。所有的干涉现象，都是一样；没有更高级的东西，被牵扯到了，只是情形，更加复杂，把矢量加在一起，更加困难；就是这些。

Fig. 30–7.A distant light source casts ashadow of an opaque object on a screen. 图30-7 一个远距离的光源，把一个不透明对象的影子，投射到一个屏幕上。
Suppose that we have light coming in frominfinity, casting a shadow of an object. Figure 30–7 showsa screen on which the shadow of an object AB is made by a lightsource very far away compared with one wavelength. Now we would expect thatoutside the shadow, the intensity is all bright, and inside it, it is all dark.As a matter of fact, if we plot the intensity as a function of position nearthe shadow edge, the intensity rises and then overshoots, and wobbles, andoscillates about in a very peculiar manner near this edge (Fig. 30–9). Wenow shall discuss the reason for this. If we use the theorem that we have notyet proved, then we can replace the actual problem by a set of effectivesources uniformly distributed over the open space beyond the object.
假设我们有光，来自无穷远，把一个对象，投影。图30-7显示了一个屏幕，在其上，对象AB的影子，通过一个光源，被形成，光源的距离，与一个波长相比，要大很多。现在，我们可以期待，在影子之外，强度全是亮的，在影子之内，强度全是黑的。事实上，如果我们把强度，当作接近影子边缘处的函数，而画其曲线，那么，强度就会在这个边缘（图30-9），以一种非常独特的方式，升起、超过、摇摆和震荡。现在，我们将讨论其原因。有些定理，我们尚未证明，如果我们使用之，那么，我们就可以把实际问题，替换为：一组有效的源，在超过对象的开放空间中，均匀分布。

We imagine a large number of very closelyspaced antennas, and we want the intensity at some point P . Thatlooks just like what we have been doing. Not quite; because our screen is notat infinity. We do not want the intensity at infinity, but at a finite point.To calculate the intensity at some particular place, we have to add thecontributions from all the antennas. First there is an antenna at D, exactly opposite P ; if we go up a little bit in angle, let ussay a height h , then there is an increase in delay (there is alsoa change in amplitude because of the change in distance, but this is a very smalleffect if we are at all far away, and is much less important than thedifference in the phases). 我们可以想象，有大量天线，在空间中，非常接近，我们想知道，某点P处的强度。这看上去，与我们一直所做，很像。然而，并不完全一样；因为我们的屏幕，并非在无穷远。我们并不想要无穷远处的强度，而是有限点处的。要计算某个具体位置处的强度，我们必须把所有天线的贡献，加在一起。首先，在D处有天线，它正好与P相对；如果我们把角度稍微抬高一点，比如说抬高h，那么，延迟就会增加一点。（振幅也会改变，因为距离变了，但是，如果我们很远的话，这个影响很小，且与相位差相比，远不重要。）Now the path difference EP−DP isapproximately h2/2s , so that the phasedifference is proportional to the square of how far we go from D, while in our previous work s was infinite, and the phase differencewas linearly proportional to h . When the phases arelinearly proportional, each vector adds at a constant angle to the next vector.What we now need is a curve which is made by adding a lot of infinitesimalvectors with the requirement that the angle they make shall increase, notlinearly, but as the square of the length of the curve. To constructthat curve involves slightly advanced mathematics, but we can always constructit by actually drawing the arrows and measuring the angles. In any case, we getthe marvelous curve (called Cornu’s spiral) shown in Fig. 30–8. Nowhow do we use this curve?
现在，路径差EP−DP，约为 h2/2s，于是，相位差就正比于‘我们从D走了多远’的平方，在我们前面的工作中，s是无限的，而相位差，线性地正比于 h。当相位是线性正比的时候，每个矢量，相对于下一个矢量，都是增加了一个常数的角度。现在我们需要的，是一条曲线，它是由很多无限小的矢量，相加而成，这些矢量的角度，不断增加，{此增加}不是线性的，而是作为曲线长度的平方。建造这种曲线，所涉数学，稍微高级点，但是，通过实际地画箭头和测量角度，我们总是可以建造它。不管怎样，我们得到了这条不可思议的曲线（被称为羊角形螺线），如图30-8所示。现在，我们如何使用这条曲线呢？

Fig. 30–8.The addition of amplitudes formany in-phase oscillators whose phase delays vary as the square of the distancefrom point D of the previous figure. 图30-8 很多振荡器，相位相同，它们的相位延迟，随着‘到前图点 D处的距离’的平方，而变化；此图，是这些振荡器的振幅的相加。

If we want the intensity, let us say, at point P, we add a lot of contributions of different phases from point D onup to infinity, and from D down only to point BP. So we start at BP in Fig. 30–8, anddraw a series of arrows of ever-increasing angle. Therefore the totalcontribution above point BP all goes along the spiraling curve. Ifwe were to stop integrating at some place, then the total amplitude would be avector from B to that point; in this particular problem we are going toinfinity, so the total answer is the vector BP∞. Now the position on the curve which corresponds to point BPon the object depends upon where point P is located, sincepoint D , the inflection point, always corresponds to the positionof point P . Thus, depending upon where P is locatedabove B , the beginning point will fall at various positions on thelower left part of the curve, and the resultant vector BP∞ , will have many maxima and minima (Fig. 30–9).
如果我们想得到强度，比如说，点P处的强度，那么，我们从点D往上走，一直到无穷大，把很多不同相位的贡献，相加，然后，从点D往下，也如此相加，但只走到点BP。所以，我们从图30-8中的BP处开始，画一串箭头，其角度，不断增加。因此，在点BP之上的总的贡献，都是沿着羊角形螺线的。如果我们要在某处，停止积分，那么，总的振幅，就应该是从B到该点的一个矢量；在这个具体问题中，我们要去往无穷远，所以，总的答案，就是矢量B P∞。现在，设曲线上，有某点，相应于对象上的点BP，则此某点，依赖于点P位于何处，这是由于，点 D、即弯曲点，总是相应于点P的位置。这样，就依赖于：B在P上方的何处，开始点，将会落在曲线的左边、较低处的不同位置上，而合成矢量B P∞，则将会有很多最大值和最小值（图30-9）。

Fig. 30–9.The intensity near the edge of ashadow. The geometrical shadow edge is at x0 . 图30-9 接近影子边缘处的强度。几何阴影边缘，在x0。

On the other hand, if we are at Q, on the other side of P , then we are using only one end of thespiral curve, and not the other end. In other words, we do not even startat D , but at BQ , so on this side we get anintensity which continuously falls off as Q goes farther into theshadow.
另一方面，如果我们在Q，它在点P的另一侧，那么，我们将只会使用羊角形螺线的一端，而没有使用另一端。换句话说，我们甚至都不是从D开始的，而是从BQ开始的，于是，在这一侧，随着Q不断地往阴影深处走，我们得到的强度，就会不断地下降。

One point that we can immediately calculatewith ease, to show that we really understand it, is the intensity exactlyopposite the edge. The intensity here is 1/4 that of the incident light.Reason: Exactly at the edge (so the endpoint B of the arrow isat D in Fig. 30–8) wehave half the curve that we would have had if we were far into the brightregion. If our point R is far into the light we go from one end ofthe curve to the other, that is, one full unit vector; but if we are at theedge of the shadow, we have only half the amplitude—1/4 the intensity.
为了指出，我们真的理解了，有一点的强度，我们可以立即轻松计出，它就是正好对着边缘的那个点。这里的强度，是入射光强度的1/4。理由：在这个边缘处（于是，箭头的端点B，就在图30-8中的D。），我们所拥有的曲线，是下面曲线的一半，即如果我们深入到光明区域中，所能得到的曲线。如果我们的点R，是在光明区的深处，我们就是从曲线的一端，走到另一端，也就是说，一个完整的单位矢量；但是，如果我们是在阴影的边缘处，则我们只有振幅的一半—即强度的1/4。

In this chapter we have been finding theintensity produced in various directions from various distributions of sources.As a final example we shall derive a formula which we shall need for the nextchapter on the theory of the index of refraction. Up to this point relativeintensities have been sufficient for our purpose, but this time we shall findthe complete formula for the field in the following situation.
在这一章，我们一直在找：由不同的源的分布，在不同方向上所产生的强度。作为最后一个例子，我们将导出一个公式，在下一章，关于折射率的理论中，我们需要该公式。到目前为止，对于我们的目的来说，相对强度，已经很充分了，但是，这一次，对于下面情况中的场，我们将找到完整的公式。

30–7The field of a plane of oscillatingcharges 30-7 布满震荡电荷的平面上的场
Suppose that we have a plane full ofsources, all oscillating together, with their motion in the plane and allhaving the same amplitude and phase. What is the field at a finite, but verylarge, distance away from the plane? (We cannot get very close, of course,because we do not have the right formulas for the field close to the sources.)If we let the plane of the charges be the xy -plane, then we want tofind the field at the point P far out on the z -axis(Fig. 30–10).We suppose that there are η charges per unit area of the plane,and that each one of them has a charge q . All of the charges movewith simple harmonic motion, with the same direction, amplitude, and phase. Welet the motion of each charge, with respect to its own average position,be x0cosωt . Or, using the complex notation andremembering that the real part represents the actual motion, the motion can bedescribed by x0eiωt .
假设我们有一个平面，上面布满了源，且所有的源，都在一起震荡，振幅、相位都相同。那么，在距平面很远、但却有限的距离处，场是什么呢？（当然，我们不可能非常接近，因为，对于接近源的场，我们并没有正确的公式。）如果我们设电荷所在平面，为xy平面，那么，我们想找出的，就是z轴上很远处一点P处的场（图30-10）。我们假定，每单位平面面积上，有η个电荷，每个电荷的电量为q。所有的电荷，都在做余弦加速度运动，方向、振幅、和相位，都相同。对于每个电荷，就其平均位置而言，我们可设其运动为x0cosωt。或者，使用复数标记法，注意，实部代表着实际的运动，故此运动，可用 x0eiωt来描述。

Fig. 30–10.Radiation field of a sheet ofoscillating charges. 图30-10 一个震荡电荷平面的辐射场。

Now we find the field at the point Pfrom all of the charges by finding the field there from each charge q, and then adding the contributions from all the charges. We know that theradiation field is proportional to the acceleration of the charge, which is −ω2x0eiωt(and is the same for every charge). The electric field that we want at thepoint P due to a charge at the point Q is proportional tothe acceleration of the charge q , but we have to remember that thefield at the point P at the instant t is given by theacceleration of the charge at the earlier time t′=t−r/c, where r/c is the time it takes the waves to travel thedistance r from Q to P . Therefore the fieldat P is proportional to
现在，通过找出，每一个电荷q，在点P所形成的场，然后，把其相加，我们就可得到:所有电荷在点P所形成的场。我们知道，辐射场正比于电荷的加速度，即−ω2x0eiωt（对每个电荷都一样）。点Q的一个电荷q，在点P所形成的场，正比于q的加速度，但是，我们应该记住， t时刻，点P的场，是通过较早时刻t′的电荷的加速度，而给出的，这里t′=t−r/c，这里r，是波从Q传播到P，所走过的距离，r/c就是所需时间。因此，点P的场，就正比于：
(30.10)
Using this value for the acceleration asseen from P in our formula for the electric field at large distancesfrom a radiating charge, we get
一个辐射电荷，在远距离处，可以形成场，在我们的公式中，对于此场，可以使用这个值，作为在点P处所看到的加速度，故我们得到：
(30.11)

Now this formula is not quite right,because we should have used not the acceleration of the charge but itscomponent perpendicular to the line QP . We shall suppose,however, that the point P is so far away, compared with thedistance of the point Q from the axis (the distance ρin Fig. 30–10),for those changes that we need to take into account, that we can leave out thecosine factor (which would be nearly equal to 1 anyway).
现在，这个公式，并不完全正确，因为，我们应该使用的，不是电荷的加速度，而是其垂直于线QP的分量。然而，我们将假定，与点Q到轴的距离相比（图30-10中的距离ρ），点P是如此之远，以至于，对于那些我们应该算上的电荷来说，我们可以忽略余弦因子（无论如何，它应该近似地等于1）。

To get the total field at P ,we now add the effects of all the charges in the plane. We should, of course,make a vector sum. But since the direction of the electric field isnearly the same for all the charges, we may, in keeping with the approximationwe have already made, just add the magnitudes of the fields. To ourapproximation the field at P depends only on the distance r, so all charges at the same r produce equal fields. So we add,first, the fields of those charges in a ring of width dρ andradius ρ . Then, by taking the integral over all ρ , wewill obtain the total field.
要得到点P的总场，我们现在，把平面上所有电荷的效果，都加起来。当然，我们应该做一个矢量和。但是，由于对所有电荷来说，电场的方向，几乎一样，且我们已经做了一种近似，为了保持与此近似一致，我们可以只加电场的大小。对于我们的近似来说，点P的场，只依赖于距离r，同样距离的电荷，产生同等的场。设有圆环，宽度为dρ，半径为ρ，我们首先，把环内所有电荷产生的场，全加起来。然后，通过对所有的ρ进行积分，我们就可得到总场。