Superposition
A-Level Physics Topic 8 19:48 English narration · English + 中文 subtitles burned in
Chapters
Transcript
Put on a pair of noise-cancelling headphones, and the roar of a plane engine fades to silence.
戴上一副降噪耳机，飞机引擎的轰鸣就渐渐归于寂静。
But here is the strange part: they do not block the sound — they add more sound.
但奇怪的地方在于： 它们并不是挡住声音——它们是加入了更多声音。
A tiny microphone listens to the noise coming in.
一个微小的麦克风倾听传进来的噪声。
The headphones then play a second wave, an exact upside-down copy of it.
耳机随即播放第二个波，一个和它完全上下颠倒的副本。
Where the noise pushes up, the copy pushes down.
噪声向上推的地方，副本就向下推。
The two cancel, and you hear quiet.
两者相互抵消，你听到的就是安静。
This is superposition, the key to this whole topic.
这就是叠加，也是整个专题的钥匙。
When waves meet, they add.
当波相遇时，它们相加。
This simple idea builds stationary waves, interference patterns, and the beautiful physics of light through slits.
这个简单的想法造就了驻波、干涉图样， 以及光穿过狭缝的美妙物理。
Today: superposition, standing waves, diffraction, and interference.
今天：叠加、驻波、衍射和干涉。
Let's begin.
让我们开始吧。
Here is the rule.
规则在这里。
When two waves overlap, the displacement at each point is the vector sum of the two — the principle of superposition.
当两个波重叠时，每一点的位移就是两者之和——这就是叠加原理。
If two crests meet, in phase, they build a bigger wave: constructive interference.
如果两个波峰相遇、同相，它们叠成一个更大的波：相长干涉。
If a crest meets a trough, exactly out of phase, they cancel: destructive interference.
如果波峰遇到波谷、恰好反相，它们相互抵消：相消干涉。
And remember, intensity goes as amplitude squared — so two waves adding in phase give not double, but four times the intensity.
记住，强度与振幅的平方成正比——所以两个同相相加的波，给出的不是两倍，而是四倍的强度。
The principle, in the words the scheme prints: when two or more waves meet at a point, the resultant displacement is the sum of the displacements of the individual waves.
这条原理，用评分标准印出来的说法是： 当两列或更多的波在某一点相遇时，合位移等于各列波位移的和。
Displacements, not intensities, and they add with their signs — which is why two waves can superpose to give nothing at all.
是位移，不是强度，而且是带符号相加的—— 这正是两列波叠加之后可以什么都不剩的原因。
Look at the same idea drawn carefully.
仔细看同一想法的画法。
On the left, two identical waves arrive in phase — crest on crest.
左边，两个完全相同的波同相到达——波峰叠在波峰上。
Their resultant has twice the amplitude: constructive interference.
它们的合成波振幅是原来的两倍：相长干涉。
On the right, the waves arrive exactly out of phase, crest on trough, with a phase difference of pi.
右边，两波恰好反相到达，波峰叠在波谷上， 相位差为圆周率。
The resultant is a flat line of zero amplitude: complete destructive interference.
合成波是一条振幅为零的直线：完全相消干涉。
Between these extremes, any other phase difference gives an amplitude somewhere in the middle.
在这两个极端之间， 任何其他相位差都会给出介于中间的振幅。
And after they meet, the waves pass through each other and come out unchanged — superposition never destroys the original waves.
而且相遇之后，波彼此穿过并原样出来—— 叠加从不毁掉原来的波。
Send two identical waves toward each other, in opposite directions — for example, a wave on a string and its own reflection.
让两个完全相同的波朝相反方向彼此靠近——比如，绳子上的波和它自己的反射波。
Where they overlap, something surprising happens: the pattern stops moving.
在它们重叠的地方，一件令人惊讶的事发生了：图样停止移动。
This is a stationary wave.
这就是驻波。
Certain points never move at all, while others swing with the largest amplitude.
某些点完全不动，而另一些点以最大的振幅摆动。
The wave no longer travels — it just stands there, vibrating in place.
这个波不再传播——它只是站在那里，原地振动。
Here are five snapshots of the same setup, at zero, a quarter period, half a period, three quarters, and a full period.
这里是同一装置的五张快照：零、四分之一周期、半周期、四分之三周期和整周期。
Two progressive waves travel in opposite directions; their sum is the middle curve.
两个行波朝相反方向传播；它们的和是中间那条曲线。
Watch the points labelled N: they stay on the axis every time — those are nodes.
注意标着节的点：它们始终在轴上—— 那些是波节。
The points labelled A swing with the largest amplitude — antinodes.
标着腹的点以最大振幅摆动——波腹。
The envelope stays put while the progressive waves keep racing through.
包络线固定不动，而行波仍在穿过。
Examples are everywhere: a wave on a string reflected from a fixed end, sound in an air column reflected from a closed end, and microwaves bouncing between an emitter and a metal sheet.
例子随处可见：绳子上从固定端反射回来的波，空气柱中从闭端反射的声波， 以及在发射器和金属板之间来回的微波。
Explain how the stationary wave is formed is worth three or four marks, and the same answer serves a string, an air column and microwaves.
「解释驻波是怎么形成的」值得三到四分， 而且同一个答案对弦、气柱和微波都适用。
The wave from the source travels to the far end — the wall, the closed end, the metal plate — and is reflected.
从波源出发的波传到远端——墙、闭口端、金属板——并被反射。
The incident and reflected waves have the same frequency and speed and travel in opposite directions, so they superpose: where they always arrive in phase there is an antinode, and where they always arrive antiphase there is a node.
入射波和反射波频率相同、速度相同、传播方向相反，于是它们叠加： 永远同相到达的地方是波腹，永远反相到达的地方是波节。
Those special points have names.
那些特殊的点有名字。
A node is a point that never moves — the two waves always cancel there.
波节是一个永不移动的点——两个波在那里总是相互抵消。
An antinode is a point of maximum swing.
波腹是摆动最大的点。
Next-door nodes are half a wavelength apart, and so are next-door antinodes — a handy way to measure the wavelength.
相邻的波节相距半个波长，相邻的波腹也是——这是测量波长的好办法。
On a string fixed at both ends, the simplest pattern is one loop, so the string length is half a wavelength.
在两端固定的绳子上，最简单的图样是一个环，所以绳长等于半个波长。
A point-eight metre string, with a wave speed of two hundred and forty, gives a fundamental frequency of one hundred and fifty hertz.
一根零点八米的绳子，波速为二百四十，给出一百五十赫兹的基频。
The fundamental mode on a stretched string is one loop: a node at each fixed end, and a single antinode in the middle.
张紧绳子上的基频模式是一个环：每个固定端一个波节，中间一个波腹。
Length L equals half a wavelength.
绳长等于半个波长。
A node and the next antinode are a quarter of a wavelength apart.
波节与相邻波腹相距四分之一波长。
Particles between two nodes all oscillate in phase with each other, though with different amplitudes — largest at the antinode, zero at the nodes.
两个波节之间的质点彼此同相振动， 只是振幅不同——波腹最大，波节为零。
Particles on opposite sides of a node oscillate in antiphase: their phase difference is pi.
波节两侧的质点反相振动：相位差是圆周率。
Higher modes squeeze more loops into the same length, each loop still half a wavelength long.
更高的模式在同样长度里塞进更多的环，每个环仍是半个波长。
How does a stationary wave differ from a progressive wave?
驻波与行波有何不同？
A stationary wave does not carry energy along its length — the pattern just stores energy, rocking in place.
驻波并不沿其长度传输能量——图样只是储存能量，原地摇晃。
The pattern itself does not move along, and the nodes stay fixed.
图样本身不沿长度移动，波节固定不动。
A progressive wave has the same amplitude everywhere and carries energy from one place to another.
行波处处振幅相同，并把能量从一处带到另一处。
That contrast is a classic exam mark: energy stored versus energy transferred.
这一对比是经典得分点：能量被储存，还是被传输。
Drive a string with a vibrator at frequency f until a stationary pattern appears.
用振动器以频率艾夫驱动绳子，直到出现驻波图样。
Measure the distance between two well-separated nodes and divide by the number of half-wavelengths between them. That gives lambda.
量出两个相隔较远的波节之间的距离， 再除以它们之间半波长的个数，就得到波长。
Then the wave speed is frequency times wavelength — v equals f lambda.
然后波速等于频率乘以波长——维等于艾夫兰姆达。
This is how a sonometer or Melde's experiment finds the speed of waves on a string.
这就是弦音计或梅尔德实验求绳子上波速的方法。
The same ideas apply to air columns.
同样的想法也适用于空气柱。
For a tube closed at one end and open at the other — a resonance tube — the closed end is a displacement node and the open end is a displacement antinode.
一端封闭、一端开口的管子——共鸣管——闭端是位移波节， 开端是位移波腹。
The fundamental has length L equal to a quarter of a wavelength.
基频时长度エル等于四分之一波长。
The next resonance is at three quarters of a wavelength, then five quarters, and so on.
下一个共鸣在四分之三波长处， 然后是四分之五，以此类推。
For a tube open at both ends, both ends are antinodes, so the fundamental is L equal to half a wavelength — just like a string fixed at both ends, but with antinodes where the string had nodes.
两端都开口的管子，两端都是波腹，所以基频是长度等于半波长—— 就像两端固定的绳子，只是波腹出现在绳子上是波节的地方。
A third experiment the syllabus names is the dust tube: fine powder lying along a tube is shaken into little heaps at the nodes, where the air is still, and cleared from the antinodes, so the heap spacing is half a wavelength and you can measure it with a rule.
考纲点名的第三个实验是尘埃管： 铺在管子里的细粉末被振动堆成一小堆一小堆，堆在波节处——那里的空气是静止的—— 波腹处的粉末则被扫开， 所以两堆之间的间距是半个波长，用尺子就能量出来。
Waves do not only travel in straight lines — they bend around obstacles and spread out through gaps.
波不只沿直线传播——它们会绕过障碍物，并穿过缝隙散开。
This is diffraction.
这就是衍射。
How much they spread depends on the gap — on the ratio of wavelength to gap width.
它们散开多少，取决于缝隙。
If the gap is much wider than the wavelength, the wave barely bends.
如果缝隙比波长宽得多，波几乎不弯折。
But when the gap is about the same size as the wavelength, the wave spreads out in a wide arc.
但当缝隙与波长大小相当时，波就散开成一个大大的弧形。
That is why you can hear around a corner, but not see around it.
这就是为什么你能听到拐角后的声音，却看不见拐角后的东西。
The definition to give: diffraction is the spreading of a wave as it passes through a gap — an aperture — or around the edge of an obstacle.
要给出的定义是： 衍射，是波通过一个缝隙——也就是孔径——或者绕过障碍物边缘时发生的扩展。
The spreading is greatest when the gap is about the same size as the wavelength.
当缝隙的尺寸和波长差不多时，扩展最明显。
Show this with water waves in a ripple tank.
用水波槽里的水波来展示这一点。
Straight waves meet a barrier with a gap.
直波遇到带缝的挡板。
Through a wide gap, labelled a, they pass almost straight on with very little spreading.
通过宽缝，标为甲， 它们几乎直行通过，几乎不散开。
Through a narrow gap, labelled b, they fan out in curved wavefronts — strong diffraction.
通过窄缝，标为乙，它们呈弯曲波前散开——强衍射。
If the gap is even smaller than the wavelength, the spreading is stronger still, and the gap acts almost like a point source.
如果缝比波长还小，散开更强，缝几乎像一个点源。
Speech has a wavelength near one metre, close to doorway size, so sound diffracts around corners.
语音波长约一米，接近门缝大小， 所以声音会绕过拐角。
Visible light has a wavelength of about five hundred nanometres — far smaller than any doorway — so light barely diffracts and you cannot see around the corner.
可见光波长大约五百纳米——远小于任何门缝——所以光几乎不衍射， 你看不见拐角后的东西。
All waves diffract: water, sound, light, and microwaves.
所有波都会衍射：水波、声波、光波和微波。
Now send light through two narrow slits, close together.
现在让光穿过两条靠得很近的狭缝。
Each slit spreads the light by diffraction, and the two spreading waves overlap.
每条缝都通过衍射把光散开，两个散开的波相互重叠。
Where crests meet crests, the light adds — a bright band.
波峰遇到波峰的地方，光相加——一条亮带。
Where crests meet troughs, the light cancels — a dark band.
波峰遇到波谷的地方，光相消——一条暗带。
The result is a striped pattern of bright and dark fringes, printed on the screen by nothing but waves adding and cancelling.
结果是一排明暗相间的条纹，完全由波的相加和相消印在屏幕上。
Interference is the superposition of two coherent waves to give a steady pattern of high-amplitude regions — constructive — and low-amplitude regions — destructive.
干涉是两个相干波的叠加，形成稳定的高振幅区域——相长——和低振幅区域——相消。
The shifting colours on a soap bubble come from light interfering with itself after reflecting from the front and back of the film.
肥皂泡上变幻的颜色，来自光在薄膜前后表面反射后与自身干涉。
In the diagram, two coherent point sources send out circular wavefronts.
图中，两个相干点源 发出圆形波前。
Blue dots mark where crest meets crest; lines of maximum displacement fan out from between the sources.
蓝色圆点标出波峰与波峰相遇处；最大位移的线从两源之间向外散开。
Those are the constructive ridges of a two-source interference pattern.
那些就是双源干涉图样中的相长脊线。
The same effect in a real ripple tank: two side-by-side sources send out circular water waves that overlap.
真实水波槽里的同一效应：两个并排的源发出圆形水波并相互重叠。
Bright bands of strong ripples fan out where crests reinforce; calm lines of cancellation fan out between them.
波峰相长的地方 亮带状强波纹向外散开；相消的平静线夹在它们之间。
Because the sources are driven together, they stay coherent and the pattern holds steady — not a flicker, but a fixed map of constructive and destructive regions.
因为两源一起驱动，它们保持相干， 图样稳定——不是闪烁，而是一幅固定的相长与相消区域图。
For a clear, steady pattern, the two sources must be coherent — they must keep a constant phase difference, never drifting.
要得到清晰、稳定的图样，两个源必须相干——它们必须保持恒定的相位差，绝不漂移。
That is why we use one light source split into two, not two separate lamps.
这就是为什么我们用一个光源分成两束，而不是两盏独立的灯。
Two conditions matter: the sources must be coherent, and they should have roughly equal amplitude, so the dark fringes are truly dark.
两个条件很重要： 两个源必须相干，而且它们的振幅应大致相等，这样暗纹才是真正的暗。
Get these right, and stable fringes appear.
把这些做对，稳定的条纹就出现了。
Spell out the full list for two-source fringes.
把双源条纹的完整条件写清楚。
You need two coherent sources — constant phase difference, and therefore the same frequency.
你需要两个相干源——恒定相位差，因而频率相同。
You need roughly equal amplitudes, or the dark regions never go truly dark.
振幅要大致相等，否则暗区永远不会真正变暗。
The waves must overlap where you look.
波必须在你观察的地方重叠。
And for light, a transverse wave, they need the same plane of polarisation.
对于光这种横波，它们还需要同一偏振面。
Two separate lamps fail the first test: their phases wander randomly, so any pattern flickers too fast to see and you get only an average.
两盏独立的灯过不了第一条： 它们的相位随机漂移，任何图样闪得太快看不见，你只能得到平均效果。
For two coherent sources, what happens at a point depends on the path difference — how much further one wave has travelled than the other.
对两个相干源，某点发生什么取决于路程差——一列波比另一列多走了多少。
Constructive interference when the path difference is a whole number of wavelengths: zero, one, two, and so on, times lambda.
相长干涉出现在路程差为波长整数倍时：零、一、二等等乘以兰姆达。
Destructive interference when the path difference is a half-integer number of wavelengths: one half, three halves, five halves, times lambda.
相消干涉出现在路程差为半整数倍波长时：二分之一、二分之三、二分之五乘以兰姆达。
Memorise both: path difference equals n lambda for bright, and n plus a half times lambda for dark.
两条都要背熟：路程差等于恩兰姆达是亮，恩加二分之一乘以兰姆达是暗。
The double-slit pattern follows a simple equation.
双缝图样遵循一个简单的方程。
The fringe spacing equals the wavelength, times the distance to the screen, divided by the slit separation.
条纹间距等于波长，乘以到屏幕的距离，再除以缝的间隔。
Rearranged, the wavelength is the slit spacing, times the fringe spacing, divided by the screen distance.
变形一下，波长等于缝的间距，乘以条纹间距，再除以屏幕距离。
Measure the fringes, and you have measured the wavelength of light — a length smaller than a thousandth of a millimetre.
量出条纹， 你就测出了光的波长——一个比千分之一毫米还小的长度。
Young's experiment starts with monochromatic light through a single slit, then a double slit a distance a apart.
杨氏实验先让单色光穿过单缝，再穿过间距为诶的双缝。
The single slit makes the two slits coherent sources, because both are lit by the same wavefront.
单缝让两条缝成为相干源， 因为它们被同一波前照亮。
Diffracted light from each slit overlaps on a screen a distance D away, forming fringes of spacing x.
每条缝衍射的光在距离为迪的屏幕上重叠，形成间距为艾克斯的条纹。
Bright fringes — maxima — sit where the path difference is a whole number of wavelengths.
亮纹——极大——出现在路程差为波长整数倍处。
Dark fringes — minima — sit where it is n plus a half times lambda.
暗纹——极小——出现在恩加二分之一倍波长处。
The fringes are equally spaced.
条纹等间距。
The equation is lambda equals a x over D, or rearranged, x equals lambda D over a.
方程是兰姆达等于诶艾克斯除以迪，或变形为艾克斯等于兰姆达迪除以诶。
Worked example.
例题。
In a double-slit experiment the slits are zero point five zero millimetres apart and lit by light of wavelength six hundred nanometres.
双缝实验中，缝间距为零点五零毫米，所用光波长为六百纳米。
The screen is two point zero metres away.
屏幕在两点零米处。
Find the fringe spacing.
求条纹间距。
Use x equals lambda D over a.
用艾克斯等于兰姆达迪除以诶。
Put the numbers in with SI units: lambda is six hundred times ten to the minus nine, D is two point zero, and a is zero point five zero times ten to the minus three.
用国际单位代入：兰姆达是六百乘以十的负九次方， 迪是两点零，诶是零点五零乘以十的负三次方。
That gives two point four times ten to the minus three metres — two point four millimetres.
结果是二点四乘以十的负三次方米——二点四毫米。
Explain how the pattern of bright and dark fringes is formed: light from the two slits is coherent, so the two waves arriving at a point on the screen have a constant phase difference set by their path difference.
「解释明暗条纹的图样是怎么形成的」： 从两条缝出来的光是相干的，所以到达屏上某一点的两列波之间， 有一个由它们的路程差决定的恒定相位差。
Where the path difference is a whole number of wavelengths they arrive in phase and superpose constructively, giving a bright fringe; where it is an odd number of half wavelengths they arrive antiphase and cancel, giving a dark one.
路程差是波长的整数倍的地方，两波同相到达、相长叠加，形成亮纹； 路程差是半波长的奇数倍的地方，两波反相到达、相互抵消，形成暗纹。
From x equals lambda D over a, you can control the pattern.
由艾克斯等于兰姆达迪除以诶，你可以控制图样。
To make the fringe spacing larger: increase lambda, increase D, or decrease a — bring the slits closer.
要让条纹间距变大：增大波长、增大迪， 或减小诶——把缝靠得更近。
To make the fringe spacing smaller: increase a, the slits further apart; reduce D, the screen closer; or use a shorter wavelength — bluer light.
要让条纹间距变小：增大诶，缝离得更远；减小迪，屏幕更近； 或用更短的波长——更蓝的光。
Exam questions love asking which change spreads the fringes out.
考题很爱问哪种改动会把条纹拉开。
For a sharper measurement, use a diffraction grating — not two slits, but thousands, packed close together.
要得到更精确的测量，就用衍射光栅——不是两条缝，而是成千上万条紧密排列的缝。
The many slits make the bright fringes razor-sharp and far apart.
这许多条缝让亮纹变得极其锐利、彼此远离。
Each bright fringe appears where the path difference between neighbouring slits is a whole number of wavelengths.
每一条亮纹都出现在相邻缝之间的光程差 恰好是波长的整数倍处。
That is the grating equation: the slit spacing, times the sine of the angle, equals a whole number, times the wavelength.
这就是光栅方程：缝的间距，乘以角度的正弦，等于整数乘以波长。
From the angle, you get the wavelength, very precisely.
从这个角度，你就能非常精确地得到波长。
A parallel beam of monochromatic light strikes the grating and splits into several sharp beams that reach a screen at different angles — the orders.
一束平行单色光打在光栅上，分裂成几束锐利的光，以不同角度到达屏幕——这就是各级次。
The central beam is n equals zero.
中央光束是恩等于零。
First order is n equals one on each side, then second order, and so on.
一级是两侧各恩等于一，然后是二级，以此类推。
Compared with the double slit, a grating gives much sharper maxima, because more slits add together — every other direction is cancelled by many slits at once.
与双缝相比， 光栅给出的极大尖锐得多，因为更多的缝一起相加——其他方向同时被许多缝相消掉。
Worked example.
例题。
A diffraction grating has five hundred lines per millimetre.
一块衍射光栅每毫米有五百条线。
Light of wavelength six hundred nanometres is shone normally on it.
波长六百纳米的光正入射。
Find the angle of the first-order maximum, n equals one.
求一级极大的角度，恩等于一。
First find the slit spacing: d equals one over five hundred millimetres, which is two point zero times ten to the minus six metres.
先求缝间距：迪等于一除以五百毫米，即二点零乘以十的负六次方米。
Then sine theta equals n lambda over d — six hundred times ten to the minus nine, over two point zero times ten to the minus six — so sine theta is zero point three zero.
然后正弦西塔等于恩兰姆达除以迪—— 六百乘以十的负九次方，除以二点零乘以十的负六次方——所以正弦西塔是零点三零。
Therefore theta is about seventeen degrees.
因此西塔大约十七度。
Three more grating facts.
再记三条光栅事实。
If a grating has N lines per millimetre, then d equals one over N millimetres, or ten to the minus three over N metres.
若光栅每毫米有恩条线，则迪等于一除以恩毫米，或十的负三次方除以恩米。
For four hundred and fifty lines per millimetre, d is about two point two two micrometres.
对每毫米四百五十条线，迪约二点二二微米。
Second: sine theta cannot exceed one, so the highest order you can see is the integer part of d over lambda.
第二：正弦西塔不能超过一，所以你能看到的最高级次 是迪除以兰姆达的整数部分。
If d over lambda is three point two seven, orders up to n equals three exist; n equals four needs sine greater than one and is not seen.
若迪除以兰姆达是三点二七，则存在直到恩等于三的级次； 恩等于四需要正弦大于一，看不见。
Third: to find an unknown wavelength, shine parallel light normally on the grating, measure the angle of the first-order maximum from the centre, and use lambda equals d sine theta.
第三：求未知波长时，让平行光正入射光栅， 从中心量出一级极大的角度，用兰姆达等于迪正弦西塔。
Repeat for higher orders and average to cut error.
对更高级次重复并取平均以减小误差。
If you are asked to describe the diffraction at the grating rather than to calculate with it: each slit diffracts the light so that it spreads across the whole screen, and the light from all the slits then superposes.
如果题目要你描述光栅上的衍射，而不是用它做计算： 每一条缝都让光发生衍射，使它散布到整个屏上， 然后来自所有缝的光再叠加。
A maximum appears only in the directions where the path difference between adjacent slits is a whole number of wavelengths, so with many slits the maxima are far sharper than with two.
只有在相邻两缝的路程差是波长整数倍的那些方向上才会出现极大， 所以缝数很多时，极大比双缝时锐利得多。
Three marks to secure.
三个要拿稳的分。
First, superposition just means add the displacements — constructive when in phase, destructive when out of phase.
第一，叠加就是把位移相加——同相时相长，反相时相消。
Second, in a stationary wave, adjacent nodes are half a wavelength apart.
第二，在驻波中，相邻的波节相距半个波长。
Third, coherent sources are essential for stable interference fringes.
第三，相干的源对稳定的干涉条纹至关重要。
Master these, and superposition is yours.
掌握这些，叠加就是你的了。
Four equation marks.
四个方程分。
For two-source interference: path difference equals n lambda for constructive, and n plus a half times lambda for destructive — and the sources must be coherent.
双源干涉：相长时路程差等于恩兰姆达，相消时等于恩加二分之一乘以兰姆达—— 且源必须相干。
Double slit: lambda equals a x over D — know every symbol.
双缝：兰姆达等于诶艾克斯除以迪——每个符号都要认得。
Diffraction grating: d sine theta equals n lambda.
衍射光栅： 迪正弦西塔等于恩兰姆达。
On a stationary wave, mark nodes and antinodes; adjacent nodes are half a wavelength apart; it stores energy but does not transfer it.
在驻波上标出波节和波腹；相邻波节相距半波长； 它储存能量但不传输能量。
A stationary wave needs two waves of the same frequency travelling in opposite directions.
驻波需要两列同频率、相向传播的波。
The fixed-wording definitions, one answer only.
固定措辞的定义，只给一个答案。
Principle of superposition: when two or more waves meet at a point, the resultant displacement is the sum of the displacements of the individual waves.
叠加原理：两列或更多的波在某点相遇时，合位移等于各列波位移之和。
A stationary wave: the pattern formed when two progressive waves of the same frequency and speed travel in opposite directions and superpose, with nodes and antinodes that do not move.
驻波：两列频率和速度相同、传播方向相反的行波叠加所形成的图样， 其波节和波腹的位置不移动。
A node: displacement always zero.
波节：位移始终为零的点。
An antinode: amplitude a maximum.
波腹：振幅最大的点。
Diffraction: the spreading of a wave through a gap or around an obstacle.
衍射：波通过缝隙或者绕过障碍物时的扩展。
Interference: the superposition of waves from coherent sources giving a steady pattern of maxima and minima.
干涉：来自相干波源的波叠加，形成稳定的极大极小图样。
Coherence: a constant phase difference, and so the same frequency.
相干：相位差恒定，因而频率相同。
The traps.
陷阱。
Displacements add in superposition, not amplitudes or intensities.
叠加时相加的是位移，不是振幅，也不是强度。
Adjacent nodes are HALF a wavelength apart, and a node to the next antinode is a quarter.
相邻两个波节相距半个波长，波节到相邻波腹是四分之一。
In d sine theta equals n lambda, d is a spacing — invert lines per millimetre to get it.
在 d sinθ 等于 nλ 里，d 是间距——把每毫米的刻线数取倒数才能得到它。
Measure theta from the normal, and remember the angle between the two first-order beams is twice theta one.
θ 要从法线量起，而且要记得两条一级光束之间的夹角是 2θ₁。
Coherent means a constant phase difference, not "in phase".
相干指的是相位差恒定，不是「同相」。
And a brighter source or a narrower slit changes the contrast and how many fringes you can see, not the fringe spacing, which is lambda D over a.
另外，光源更亮或者缝更窄，改变的是对比度和能看到几条条纹， 而不是条纹间距，间距是 λD 除以 a。