It is really very easy, and we presume thatwe already know how to do it. However, we shall outline the procedure in somedetail. First, we can, if we are clever with mathematics and know enough aboutcosines and sines, simply work it out. The easiest such case is the one where A1and A2 are equal, let us say they are both equal to A .In those circumstances, for example (we could call this the trigonometricmethod of solving the problem), we have
这确实很容易，我们假定，我们已经知道了如何做此事。然而，我们将用一些细节，概述一下此过程。首先，如果我们精通数学，且知道足够多的余弦和正弦函数，那么，我们就可以简单地得出。最容易的情况就是这种：A1 和 A2是相等的，比如都等于A。在这种情形下，例如（我们可以把此方法，称为解决问题的三角函数方法），我们就有：
R=A[cos(ωt+ϕ1)+cos(ωt+ϕ2)]. (29.9)
Once, in our trigonometry class, we mayhave learned the rule that
在我们的三角函数课中，一旦我们学会了规则：
(29.10)
If we know that, then we can immediatelywrite R as
如果我们知道了这个，那么，我们就可以立即把R写为：
(29.11)
So we find that we have an oscillatory wavewith a new phase and a new amplitude. In general, the result will be anoscillatory wave with a new amplitude AR , which we maycall the resultant amplitude, oscillating at the same frequency but with aphase difference ϕR , called the resultant phase. Inview of this, our particular case has the following result: that the resultantamplitude is
于是，我们发现，我们有了一个震荡波，它有新的相位，和新的振幅。一般来说，结果会是一个震荡波，有新的振幅 AR，这我们可称之为结果振幅，以同样的频率在震荡，但有一个相位差ϕR，被称为结果相位。鉴于这点，我们具体的情况，就有着如下的结果：结果振幅就是：
(29.12)
and the resultant phase is the average ofthe two phases, and we have completely solved our problem.
结果相位，就是两个相位的平均值，我们的问题，已经完全解决。

Now suppose that we cannot remember thatthe sum of two cosines is twice the cosine of half the sum times the cosine ofhalf the difference. Then we may use another method of analysis which is moregeometrical. Any cosine function of ωt can be considered as thehorizontal projection of a rotating vector. Suppose there were avector A1 of length A1rotating with time, so that its angle with the horizontal axis is ωt+ϕ1. (We shall leave out the ωt in a minute, and see that it makes nodifference.) Suppose that we take a snapshot at the time t=0 , although,in fact, the picture is rotating with angular velocity ω(Fig. 29–9).The projection of A1 along the horizontal axis isprecisely A1cos(ωt+ϕ1) . Nowat t=0 the second wave could be represented by another vector, A2, of length A2 and at an angle ϕ2, and also rotating. 现在，假设我们不能记得：两个余弦函数之和，就是2，乘以差的余弦，再乘以和的余弦。那么，我们就可以利用另外一种分析方法，它更具几何意义。任何ωt的余弦函数，都可被考虑为：一个旋转矢量的水平投影。设有矢量 A1，其长度为A1，随时间旋转，于是，其与水平轴的角度，就是ωt+ϕ1（马上我们就会不管ωt了，你将看到，没有什么差别）假设，我们在时间t=0，取了一个快照，虽然事实上。这个图像，是以角速度ω在旋转（图29-9）。A1沿着水平轴的投影，就是 非常精确的A1cos(ωt+ϕ1)。现在，在t=0，第二个波，可被另一个矢量A2来表示，其长度为A2，角度为ϕ2，也在旋转。They are both rotating with the same angular velocity ω, and therefore the relative positions of the two are fixed. The systemgoes around like a rigid body. The horizontal projection of A2is A2cos(ωt+ϕ2) . But we knowfrom the theory of vectors that if we add the two vectors in the ordinary way,by the parallelogram rule, and draw the resultant vector AR, the x -component of the resultant is the sum of the x-components of the other two vectors. That solves our problem. It is easy tocheck that this gives the correct result for the special case we treated above,where A1= A2= A . In this case, wesee from Fig. 29–9 that ARlies midway between A1 and A2and makes an angle (1/2)(ϕ2−ϕ1) witheach. Therefore we see that AR=2Acos(1/2)(ϕ2−ϕ1), as before. Also, as we see from the triangle, the phase of AR, as it goes around, is the average angle of A1and A2 when the two amplitudes are equal.Clearly, we can also solve for the case where the amplitudes are not equal,just as easily. We can call that the geometrical way of solving theproblem.
它们都以同样的加速度ω在旋转，因此，它们之间的相对位置，是固定的。系统像一个刚体一样，在运转。A2的水平投影，就是A2cos(ωt+ϕ2)。但是，从矢量的原理出发，我们知道，如果我们把两个矢量，以普通的方式、即通过平行四边形的规则，来相加，并画出结果矢量 AR，那么，结果的x分量，就是那两个矢量的x分量之和。这就解决了我们的问题。对于我们上面所讨论的特殊情况，即A1= A2=A，很容易验证，这给出了正确的结果。在这种情况下，我们从图29-9看到，AR在A1和 A2的中间，且与每个的夹角，都是 (1/2)(ϕ2−ϕ1){？错，非角平分线也}。因此，我们看到AR=2Acos(1/2)(ϕ2−ϕ1)，正如以前一样。另外，正如我们从三角形中所看到，当AR在转动时，在A1和 A2的振幅相等时，AR的相位，就是A1和 A2的平均角度。很清楚，对于振幅不相等的情况，我们也可以解决，同样容易。我们可以把此，称为解决问题的几何方法。

Fig. 29–9.A geometrical method forcombining two cosine waves. The entire diagram is thought of as rotatingcounterclockwise with angular frequency ω . 图29-9 把两个余弦波组合起来的几何方法。整个示意图，被认为是按逆时针方向，旋转的角频率为 ω。

There is still another way of solving theproblem, and that is the analytical way. That is, instead of havingactually to draw a picture like Fig. 29–9, wecan write something down which says the same thing as the picture: instead ofdrawing the vectors, we write a complex number to represent each of thevectors. The real parts of the complex numbers are the actual physicalquantities. So in our particular case the waves could be written in this way: A1ei(ωt+ϕ1)[the real part of this is A1cos(ωt+ϕ1)] and A2ei(ωt+ϕ2). Now we can add the two:
解决这个问题，还有另一种方式，即分析的方式，也就是说，像图29-9那样，不是画一张图，而是换种方式，我们可以写下某种东西，它所说的，与图像所表达的，一样，即我们不画矢量，而是写一个复数，以代表每一个矢量。此复数的实部，就是实际的物理量。于是，在我们的具体例子中，波就可写作：A1ei(ωt+ϕ1)[其实部，就是A1cos(ωt+ϕ1)]和A2ei(ωt+ϕ2)。现在，我们可以把它们两个相加：
(29.13)
or
或
(29.14)
This solves the problem that we wanted tosolve, because it represents the result as a complex number of magnitude ARand phase ϕR .
这就解决了我们想要解决的问题，因为，它用大小AR和相位ϕR，所组成的一个复数，代表了结果。

To see how this method works, let us findthe amplitude AR which is the “length” of R^. To get the “length” of a complex quantity, we always multiply the quantity byits complex conjugate, which gives the length squared. The complex conjugate isthe same expression, but with the sign of the i ’s reversed. Thus wehave
要看到这个方法是如何起作用的，我们找出振幅AR，它是R^的“长度”。要得到一个复数量的长度，我们总是让这个量，乘以其共轭复数，这会给出长度的平方。共轭复数，就是同样的表达式，但i的符号取反。这样我们就有：
(29.15)
In multiplying this out, we get A21+A22(here the e ’s cancel), and for the cross terms we have
把这个乘出来，我们就得到A21+A22（这里，e被抵消了），对于交叉项，我们有：

Now
现在

That is to say, eiθ+e−iθ=2cosθ. Our final result is therefore
也就是说，eiθ+e−iθ=2cosθ。因此，我们的最终结果就是：
(29.16)
As we see, this agrees with the lengthof AR in Fig. 29–9,using the rules of trigonometry.
正如我们所看到，这与图29中的长度，是一致的，那里，用的是三角函数规则。

Thus the sum of the two effects has theintensity A21 we would get with one of themalone, plus the intensity A22 we would getwith the other one alone, plus a correction. This correction we call the interferenceeffect. It is really only the difference between what we get simply byadding the intensities, and what actually happens. We call it interferencewhether it is positive or negative. (Interference in ordinary language usuallysuggests opposition or hindrance, but in physics we often do not use languagethe way it was originally designed!) If the interference term is positive, wecall that case constructive interference, horrible though it may soundto anybody other than a physicist! The opposite case is called destructiveinterference.
这样，两者效果之和，就是强度A21，加上A22，再加上一个修正，这里A21和A22，我们都可以单独得到。这个修正，我们称之为干涉效应。它实际上是：我们通过简单增加强度所能得到的强度，与实际发生的强度，之差。无论它是正、还是负，我们都称其为干涉。（在通常的语言中，干涉一般表示相反或妨碍，但在物理学中，我们通常并不使用它的这个本意）。如果干涉像是正的，那么，我们称这种情况，为构造干涉，对于一个非物理学家，这听上去，有点恐怖！相反的情况，我们就称为析构干涉。

Fig. 29–10.Two oscillators of equalamplitude, with a phase difference α between them. 图29-10 两个振荡器，振幅相等，相位差为α。
Nowlet us see how to apply our general formula (29.16)for the case of two oscillators to the special situations which we havediscussed qualitatively. To apply this general formula, it is only necessary tofind what phase difference, ϕ2−ϕ1 , existsbetween the signals arriving at a given point. (It depends only on the phasedifference, of course, and not on the phase itself.) So let us consider thecase where the two oscillators, of equal amplitude, are separated by somedistance d and have an intrinsic relative phase α .(When one is at phase zero, the phase of the other is α .) Then weask what the intensity will be in some azimuth direction θ from theE–W line. [Note that this is not the same θ as appears in (29.1).We are torn between using an unconventional symbol like U, or the conventionalsymbol θ (Fig. 29–10).] 对于两个震荡器的情况，我们有普遍公式（29.16），现在，让们看看，如何把此普遍公式，应用到我们曾定性讨论过的特殊情况。要应用这个普遍公式，有必要找到：当信号到达一个给定点时，它们之间所存在的相位差ϕ2−ϕ1（当然，这只依赖于相位差，而非相位本身）。所以，让我们考虑这种情况：两个振荡器，振幅相等，分开的距离为d，有一个内在固有的相关相位α。（即当一个的相位为零时，另一个的为α。）因此，我们要问，在从东--西方向上算的某方位角θ处，强度是什么？[这与（29.1）中所出现的θ，并非同一个。究竟是使用一个非传统的符号、比如U，还是使用一个传统的符号θ，我们很纠结。（图29-10）]

Now let us see what happens in our variouscases. The reason we know, for example, that the intensity is 2 at 30∘ in Fig. 29–5 isthe following: the two oscillators are (1/2)λ apart, so at 30∘ , dsinθ=λ/4 . Thus ϕ2−ϕ1=2πλ/4λ= π/2 , and so the interference term is zero. (Weare adding two vectors at 90∘ .) The result is the hypotenuse of a 45∘ right-angle triangle, which is 21/2 times the unitamplitude; squaring it, we get twice the intensity of one oscillator alone. Allthe other cases can be worked out in this same way.
现在，让我们看看，在不同的情况下，会发生什么。例如，我们知道，在图29-5中的30度处，强度是2，此事的原因为：两个振荡器，相差(1/2)λ，于是，在30度，dsinθ=λ/4。这样，ϕ2−ϕ1= 2πλ/4λ=π/2，于是，干涉项就是零（我们在90度，增加了两个矢量。）结果就是，一个45度的直角三角形的斜边，就是根号2，乘以单位波长；对它取平方，我们就得到：一个振荡器的强度的两倍。{？}其他的情况，也可用这种方式求出。

Chapter30.Diffraction第30章 衍射
30–1The resultant amplitude due to n equal oscillators 30-1 可归于n个相等的振荡器的合成振幅
This chapter is a direct continuation ofthe previous one, although the name has been changed from Interferenceto Diffraction. No one has ever been able to define the differencebetween interference and diffraction satisfactorily. It is just a question ofusage, and there is no specific, important physical difference between them.The best we can do, roughly speaking, is to say that when there are only a fewsources, 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 ismore often used. So, we shall not worry about whether it is interference ordiffraction, but continue directly from where we left off in the middle of thesubject in the last chapter.
这一章，是上一章的直接继续，虽然题目，已经从干涉，变成了衍射。干涉与衍射之间的差别。从来没有人能够令人满意地定义之。它只是一个使用的问题，它们之间，并没有什么具体的、重要的物理差别。概略地说，我们所能做的最好的事情，就是说，当只有少数几个源，比如说两个，在相互干涉时，那么，结果通常被称为干涉，但是，当有大量源时，更经常使用的，就是衍射。所以，对于究竟是叫干涉还是衍射，无需挂怀；上一章的那个话题，没有讲完，现在，我们将直接继续该话题。

Thus we shall now discuss the situationwhere there are n equally spaced oscillators, all of equalamplitude but different from one another in phase, either because they aredriven differently in phase, or because we are looking at them at an angle suchthat there is a difference in time delay. For one reason or another, we have toadd something like this:
这样，我们将讨论下面这种情况：有n个振荡器，在空间中，等距分布，振幅相等，但是，一个与一个之间，相位上有差别，这种相位差，要么是因为，它们被驱动时，有差别，要么是因为，我们看它们的角度所造成的，即在时间延迟上，有差别。由于某种原因，我们必须做如下的加法：
R=A[cosωt+cos(ωt+ϕ)+cos(ωt+2ϕ)+⋯+cos(ωt+(n−1)ϕ)], (30.1)
where ϕ is the phase differencebetween one oscillator and the next one, as seen in a particular direction.Specifically, ϕ=α+2πdsinθ/λ . Now we mustadd all the terms together. We shall do this geometrically. The first one is oflength A , and it has zero phase. The next is also of length Aand it has a phase equal to ϕ . The next one is again oflength A and it has a phase equal to 2ϕ , and so on. Sowe are evidently going around an equiangular polygon with n sides(Fig. 30–1).
这里。ϕ是在一个具体方向看时，一个震荡器与下一个震荡器之间的相位差。具体地说，ϕ=α+2πdsinθ/λ。现在，我们必须把所有的项，加起来。我们将用几何的方式，来做此事。第一项的长度是 A，相位等于零。下一个，长度也是 A，相位等于ϕ。再下一个，长度也是 A，相位等于2ϕ，如此等等。所以，很明显，我们是在绕着一个等角的多边形在走（图31-1）

Fig. 30–1.The resultant amplitude of n=6 equally spaced sources with net successive phase differences ϕ. 图30-1 n =6的合成振幅：六个源，在空间中等距分布，有着连续的相位差 ϕ。

Now the vertices, of course, all lie on acircle, and we can find the net amplitude most easily if we find the radius ofthat circle. Suppose that Q is the center of the circle. Then we knowthat the angle OQS is just a phase angle ϕ . (This isbecause the radius QS bears the same geometrical relation to A2as QO bears to A1 , so they form an angle ϕbetween them.) Therefore the radius r must be such that A=2rsinϕ/2, which fixes r . But the large angle OQT is equalto nϕ , and we thus find that AR=2rsinnϕ/2. Combining these two results to eliminate r , we get
现在，顶点当然全在圆上，如果我们找到此圆半径的话，则净的振幅，就很容易发现。假设Q是圆的中心。因此，我们知道，角度OQS正是一个相位角。（这是因为，半径QS与A2的几何关系，与QO与A1的关系，一样，于是，在它们之间，就形成了一个角度ϕ。）因此，半径r，应该满足A=2rsinϕ/2，这就固定了r。但是，大的角度OQT等于nϕ，这样，我们就发现AR=2rsinnϕ/2。把这两个结果，合在一起，消去r，我们就得到：
(30.2)
The resultant intensity is thus
这样，合成振幅就是：
(30.3)

Now let us analyze this expression andstudy some of its consequences. In the first place, we can check it for n=1. It checks: I=I0 . Next, we check it for n=2 :writing sinϕ=2sinϕ/2cosϕ/2 , we find that AR=2Acosϕ/2, which agrees with (29.12).
现在，让我们分析这个表达式，并研究它的一些后果。首先，我们检查n=1。这就表示：I=I0。接下来，我们检查n=2：写sinϕ=2sinϕ/2cosϕ/2 , 我们发现AR=2Acosϕ/2，这与（29.12）是一致的。

Now the idea that led us to consider theaddition of several sources was that we might get a much stronger intensity inone direction than in another; that the nearby maxima which would have beenpresent if there were only two sources will have gone down in strength. 现在，对于几个源的情况，我们考虑它的想法就是，我们可以在一个方向上，比在另外一个方向上，得到一个强的多的强度；当只有两个源时，附近会有最大值，而在多个源的情况下，这种附近的最大强度，将会减弱。In order to see this effect, we plot the curve that comesfrom (30.3),taking n to be enormously large and plotting the region near ϕ=0. In the first place, if ϕ is exactly 0 , we have 0/0 , but if ϕis infinitesimal, the ratio of the two sines squared is simply n2, since the sine and the angle are approximately equal. Thus the intensity ofthe maximum of the curve is equal to n2 times theintensity of one oscillator. That is easy to see, because if they are all inphase, then the little vectors have no relative angle and all n ofthem add up so the amplitude is n times, and the intensity n2 times, stronger.
为了看到这个效果，我们依据（30.3），来画曲线，把n取得非常大，在 ϕ=0附近画。在第一个地方，如果ϕ准确是零，我们就有0/0，但是，如果ϕ是无限小，那么，两个正弦平方的比率，直接就是n2，由于正弦和角度，大约相等。这样，曲线的强度最大值，就等于：n2乘以一个振荡器的强度。这很容易看到，因为，如果它们全是同相位的，那么，小的矢量，将没有相对角度，它们n个，可全加起来，于是，振幅就是n倍的强，强度就是n2倍的强。

As the phase ϕ increases, theratio of the two sines begins to fall off, and the first time it reaches zerois when nϕ/2=π , because sinπ=0 . In other words, ϕ=2π/ncorresponds to the first minimum in the curve (Fig. 30–2). Interms of what is happening with the arrows in Fig. 30–1, thefirst minimum occurs when all the arrows come back to the starting point; thatmeans that the total accumulated angle in all the arrows, the total phasedifference between the first and last oscillator, must be 2π tocomplete the circle.
随着相位ϕ的增加，两个正弦函数的比率，开始减少，它第一次到零的时间，就是当nϕ/2=π时，因为sinπ=0。换句话说，ϕ=2π/n相应于曲线（图30-2）中的第一个最小值。用图30-1中的箭头这种话语来说，所发生的事情就是，第一个最小值，出现在：所有箭头回到起点之时；这就意味着，要完成这个循环，则由所有箭头所积累的总的角度，即第一个和最后一个震荡器之间的相位差，应该是2π。

Fig. 30–2.The intensity as a function ofphase angle for a large number of oscillators of equal strength. 图30-2 大量的振荡器，强度相等，如图，是它们的‘作为相位角的函数’的强度。

Now we go to the next maximum, and we want to see thatit is really much smaller than the first one, as we had hoped. We shall not goprecisely to the maximum position, because both the numerator and thedenominator 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。