两道缝为什么会出现明暗条纹？——托马斯·杨与光的叠加

两支手电照向同一面墙，我们通常期待重叠处更亮。如果添上一束光，某些地方反而变暗，直觉便遇到了麻烦。关键不只是来了多少光，还在于两束光的振动怎样配合；普通手电缺少稳定的配合，通常看不见固定条纹。[3]

Aim two torches at the same wall and we normally expect their overlapping beams to make it brighter. If adding light instead darkens some places, intuition runs into trouble. What matters is how the oscillations of the two beams relate, as well as how much light arrives. Ordinary torches lack a stable relationship and generally produce no stationary fringes. [3]

1801年11月12日，托马斯·杨在英国皇家学会宣读《论光与颜色的理论》，论文于1802年刊出。他没有把自己写成推倒一切的英雄，开篇反而称赞牛顿的光学观察，想用简明的共同原理，重新连接那些看似分散的事实。[1]

On 12 November 1801, Thomas Young read On the Theory of Light and Colours to the Royal Society; the paper appeared in 1802. He did not present himself as a hero overturning everything. He praised Newton's optical observations near the outset and sought simple common principles that could connect apparently disparate facts. [1]

在论文的第八个命题中，他提出，两列波相遇时，合成运动可以加强，也可以削弱。他把研究声音时熟悉的叠加思路带进光学：同一套规则，或许既能说明亮处，也能说明暗处。困难是怎样让光把这种配合显露出来。[1]

In his eighth proposition, Young argued that when two trains of waves meet, their combined motion can strengthen or weaken. He brought into optics a principle of superposition familiar from his work on sound: one rule might account for darkness as well as brightness. The challenge was to make light reveal this relationship. [1]

另一次讲演把证据推到眼前：杨于1803年11月24日宣读《有关物理光学的实验与计算》，次年发表。他在开头强调，接下来描述的实验只需身边的器材，有阳光时便容易重复；这给读者留下了亲手核对的入口。[2]

A later lecture brought the evidence into view. Young read Experiments and Calculations Relative to Physical Optics on 24 November 1803, and it appeared the following year. He began by stressing that the experiments required readily available apparatus and could be repeated easily when the sun shone, inviting readers to check the evidence themselves. [2]

他用窗板外的小镜子反射阳光，让光近乎水平穿过覆盖窗孔的厚纸针孔，送过桌面。他把一条窄卡片放进光束，观察墙面或纸屏上的影子。出人意料的是，影子内部也出现平行的明暗纹，中间仍然明亮。[2]

A small mirror outside the shutter reflected sunlight almost horizontally through a needle hole in thick paper covering the shutter opening, directing it across a table. Young placed a narrow strip of card in the beam and examined its shadow on a wall or paper screen. Parallel fringes appeared within the shadow itself, with a bright centre. [2]

杨又插入一块小屏，挡住从卡片一侧进入影区的光，内部那组条纹随即消失。这个动作比单看条纹更有说服力：它检验了两侧光的共同作用。不能因此说一侧光毫无结构；这里消失的是原先影区中的干涉纹。[2]

Young then inserted a small screen to intercept light entering the shadow from one side of the card. The internal fringes disappeared. This intervention was more telling than the pattern alone: it tested the joint contribution of light from both sides. It does not establish that one-sided light has no structure; the disappearing pattern was the interference inside the shadow. [2]

因此，“杨的双缝实验”不宜被压缩成1801年一次与现代图画完全相同的演示。下面采用两道狭缝和远处光屏，是为了把两路光及其路程差分清楚的现代教学模型；它与窄卡片实验共享叠加的核心问题。[1,2,3,4]

Young's double-slit experiment should therefore not be compressed into a single 1801 demonstration identical to a modern diagram. We will use two narrow slits and a distant screen as a modern teaching model that separates the two contributions and their path difference. It shares the card experiment's central question about superposition. [1,2,3,4]

先选近似单色的光，让同一光源对称照亮两缝，使两路频率相同、相位差稳定；再取相同偏振方向。这里的“相干”指振动保持确定的相位关系，不要求它们到达光屏每一点时都同相。[3]

Choose approximately monochromatic light and illuminate both slits symmetrically from one source, giving the contributions the same frequency and a stable phase difference. Take the same polarization direction too. Coherence means a definite phase relationship; it does not require the waves to arrive in phase at every point on the screen. [3]

用现代电磁波语言，某点的电场是E=E₁+E₂，正负表示电场方向。探测器读出的光强与合成电场平方的时间平均有关。因此不能先把两路各自的亮度相加；必须先让有方向的振动相加，再取平方。[3,5]

In modern electromagnetic language, the electric field at a point is E = E₁ + E₂, with signs indicating field direction. Measured intensity depends on the time average of the square of the combined field. We must therefore add the directed oscillations before squaring, rather than first adding the intensities of the separate contributions. [3,5]

设每一路单独在该点产生的光强都是I₀。同相到达时振幅加倍，光强便是4I₀；相差半个周期时，等振幅的电场完全抵消，理想光强为零。两路光合起来的局部范围是零到4I₀，最亮处不是2I₀。[3,5]

Let each contribution alone produce intensity I₀ at the chosen point. Arrival in phase doubles the amplitude and gives 4I₀. A half-cycle offset cancels equal field amplitudes and gives an ideal zero. The local intensity can range from zero to 4I₀; its maximum is not 2I₀. [3,5]

若相位差在探测时间内均匀随机变化，忽强忽弱的交叉项平均为零，读数才成为2I₀。稳定条纹要求相干，振幅相等才可能全暗。这里使用的是后来的波动表述与计算，不能把电场符号和现代公式原样归给杨。[1,3,5]

If the relative phase samples its cycle uniformly at random during detection, the fluctuating interference term averages to zero and the reading becomes 2I₀. Stable fringes require coherence, while complete darkness requires equal amplitudes. This is a later wave formulation and calculation; its electric-field notation and modern equations should not be attributed directly to Young. [1,3,5]

同相出发的两路光，到光屏某点走过的距离可能不同。多走整整一个波长λ，仍像两人合着节拍；多走半个波长，就错开半个周期。移动观察位置，改变的是路程差，所以明与暗会依次交替。[3,4]

Two contributions leaving in phase may travel different distances to a point on the screen. An extra full wavelength λ keeps them in step; an extra half wavelength puts them half a cycle apart. Moving across the screen changes this path difference, producing successive bright and dark fringes. [3,4]

两缝中心相距d，光屏距离为L；在L远大于d且观察角度很小时，距中央y处的路程差近似为ΔL≈dy/L。这个式子把细小的路程差放大成屏上可量的位置，近似条件也规定了这把尺能在哪里使用。[4]

Let the slit centres be separated by d and the screen be a distance L away. When L is much greater than d and the observation angle is small, the path difference at displacement y from the centre is approximately ΔL ≈ dy/L. This converts a tiny path difference into a measurable screen position, within the stated approximations. [4]

路程差为整数个波长时出现亮纹，为半整数个波长时出现暗纹。于是相邻亮纹间距Δy≈λL/d：拉远光屏，条纹展开；增大缝距，条纹变密。公式描述空间位置，不是说光在亮纹之间停止传播。[3,4]

Integer-wavelength path differences give bright fringes; half-integer differences give dark ones. Adjacent bright fringes are therefore separated by Δy ≈ λL/d. Moving the screen farther away spreads the pattern, while increasing slit separation packs it more closely. This equation locates fringes; it does not imply that light stops propagating between bright fringes. [3,4]

例如取λ=550纳米、d=0.25毫米、L=1米，算得亮纹间距约2.2毫米。保持其余条件，把波长换成600纳米，间距变为2.4毫米。这是教学计算，显示我们怎样由屏上的毫米间隔，反推肉眼分不出的波长。[4]

For a teaching example, take λ = 550 nm, d = 0.25 mm and L = 1 m. The bright-fringe spacing is about 2.2 mm. Changing the wavelength to 600 nm while holding the other quantities fixed gives 2.4 mm. Measurable millimetre intervals can thus reveal wavelengths far too small to distinguish by eye. [4]

暗处的能量去了哪里？在完整、无损的光学系统中，干涉改变能量的空间分布：有些位置减少，另一些位置增加。局部光强为零，并不等于能量被消灭；同样，某一点达到4I₀，也不能推出整张屏的总功率翻了四倍。[5]

Where has the energy at a dark fringe gone? In a complete lossless optical system, interference changes its spatial distribution: some places receive less and others more. Zero local intensity does not mean that energy has been destroyed. Nor does 4I₀ at one point imply four times the total power across the screen. [5]

真实狭缝还有宽度，各缝自身的衍射会给条纹罩上明暗变化的包络。前面的等振幅模型忽略了这个包络，只适合解释局部条纹；不能把它无限延伸后积分，冒充整个装置的能量账。模型越简洁，越要记得它省略了什么。[6]

Real slits have finite width. Diffraction from each slit gives the fringes a broader envelope of varying brightness. Our equal-amplitude model neglects that envelope and explains the local pattern; extending it without limit and integrating would not provide the energy account of the whole apparatus. A simple model requires attention to what it leaves out. [6]

再看杨挡住一侧光的动作：他没有只问“这里亮不亮”，而是追问“改变哪一部分，条纹就会消失”。一条卡片与一块小屏，让看似不合常理的暗纹成为可重复的证据。发现的力量，就藏在这个把解释交给实验检验的动作中。[2]

Return to Young's act of blocking light from one side. He asked which contribution could be changed to make the fringes disappear, moving beyond a judgment of brightness alone. A narrow card and a small screen turned seemingly puzzling darkness into repeatable evidence. The discovery gains its force from this act of putting an explanation to an experimental test. [2]