📚 Young's Double-Slit Experiment & Interference of Light | 杨氏双缝实验与光的干涉
The double-slit experiment, first performed by Thomas Young in 1801, is one of the most pivotal demonstrations in physics. It provided conclusive evidence for the wave nature of light and forms the bedrock of wave optics. This article systematically reviews the experiment's setup, underlying principles, key formulas, and typical examination points relevant to A-level and equivalent curricula.
1801年托马斯·杨首次完成的杨氏双缝实验，是物理学史上最重要的演示实验之一。它为光的波动性提供了决定性证据，是波动光学的基石。本文系统梳理该实验的装置、原理、核心公式与典型考点，帮助考生透彻理解这一必考内容。
1. Historical Background & The Wave Nature of Light | 历史背景与光的波动性
In the early 19th century, Newton's corpuscular theory of light dominated scientific thought, suggesting light consisted of tiny particles. However, it failed to satisfactorily explain phenomena such as diffraction and interference. Young's experiment demonstrated that light passing through two closely spaced slits could produce alternate bright and dark bands—an interference pattern—proving that light propagates as waves that can superpose.
19世纪初，牛顿的微粒说主导着科学界，认为光由微小粒子组成。然而，微粒说无法合理解释衍射和干涉现象。杨氏实验表明，光通过两个相距很近的狭缝后会产生明暗相间的条纹——干涉图样——从而证明光以波的形式传播，并且能够叠加。
This experiment decisively supported Huygens' wave theory of light and laid the foundation for the later development of the electromagnetic wave theory by Maxwell.
该实验决定性地支持了惠更斯的波动理论，并为后来麦克斯韦电磁波理论奠定了实验基础。
2. Experimental Setup | 实验装置
A typical setup includes a monochromatic light source (e.g., a laser or sodium lamp), a single narrow slit to ensure spatial coherence, two closely spaced parallel slits (S₁ and S₂), and a screen placed at a distance D from the double slits. The distance between the two slits is denoted by a (often written as d), and the distance from slits to screen is D, with condition D ≫ a.
典型装置包括：单色光源（如激光或钠灯）、一个用于保证空间相干性的单缝、两个相距很近且平行的双缝（S₁和S₂），以及距离双缝为D的光屏。双缝间距记作a（常写作d），缝到屏的距离为D，通常满足D ≫ a。
The single slit ensures that the light reaching S₁ and S₂ is coherent, meaning the two sources maintain a constant phase difference. On the screen, alternating bright and dark fringes are observed, symmetric about the central maximum.
单缝的作用是确保到达S₁和S₂的光相干，即两个光源保持恒定的相位差。屏幕上观察到以中央亮纹为对称中心的明暗相间的条纹。
3. Coherent Sources & Conditions for Interference | 相干光源与干涉条件
To observe sustained (observable) interference patterns, the two light sources must be coherent. Coherent sources have the same frequency (or wavelength) and a constant phase difference. In Young's experiment, the double slits illuminated by a single wavefront automatically satisfy these conditions.
要观察到稳定的干涉图样，两列光源必须相干。相干光源要求频率（或波长）相同且相位差恒定。在杨氏实验中，来自同一波阵面照亮的双缝天然满足这些条件。
- Same frequency / wavelength: The two waves must oscillate at the same frequency to produce a fixed pattern.
- Constant phase difference: The phase relation between the waves does not change with time.
- Sufficient amplitude (intensity): The interfering waves should have comparable amplitudes to achieve high contrast fringes.
- 频率（波长）相同：两列波必须以相同频率振动才能产生稳定图样。
- 相位差恒定：两列波之间的相位关系不随时间变化。
- 振幅（强度）相当：两列相干波的振幅应相近，才能获得高对比度的条纹。
Additionally, the two waves must have a non-zero (but small) path difference so that they overlap on the screen. In practice, the double slits are illuminated by the same wavefront, guaranteeing coherence.
此外，两列波必须具有非零但较小的光程差，以便在屏幕上叠加。实际操作中，双缝由同一波阵面照明，自然保证相干性。
4. Path Difference & Phase Difference | 光程差与相位差
Consider a point P on the screen at a distance x from the central axis. The path difference between the waves arriving from S₁ and S₂ is approximately:
考虑屏上距中心轴线x处的某点P。来自S₁和S₂的两列波到达P点的光程差近似为：
Δ = S₂P − S₁P ≈ (a · x) / D
where a is the slit separation and D is the slit-screen distance. This approximation holds when D ≫ a, and the angle θ is small. The corresponding phase difference δ relates to the path difference through:
其中a是双缝间距，D是缝到屏的距离。当D ≫ a且角度θ很小时该近似成立。对应的相位差δ与光程差的关系为：
δ = (2π / λ) · Δ = (2π / λ) · (a · x / D)
Bright fringes occur when the path difference equals an integer multiple of the wavelength: Δ = nλ (n = 0, 1, 2, …). Dark fringes occur when the path difference equals an odd half-integer multiple of the wavelength: Δ = (n + ½)λ.
亮纹出现在光程差等于波长的整数倍时：Δ = nλ（n = 0, 1, 2, …）。暗纹出现在光程差等于波长的半整数倍时：Δ = (n + ½)λ。
5. Derivation of Fringe Spacing | 条纹间距的推导
From the bright-fringe condition Δ = nλ, the position xn of the n-th bright fringe is:
由亮纹条件Δ = nλ，第n级亮纹的位置xₙ为：
xₙ = (n · λ · D) / a
The separation between two consecutive bright fringes (e.g., n and n+1) is:
相邻两条亮纹（例如n级和n+1级）之间的间距为：
Δx = xₙ₊₁ − xₙ = (λ · D) / a
This quantity Δx is called the fringe width (or fringe spacing). It is constant across the interference pattern for monochromatic light, making the fringes equally spaced.
该量Δx称为条纹宽度（或条纹间距）。单色光干涉图样中各条纹间距恒定，因此条纹等间距分布。
- Fringe width increases with the wavelength λ and the screen distance D.
- Fringe width decreases as the slit separation a increases.
- 条纹间距增大：当波长λ或缝屏距离D增大时。
- 条纹间距减小：当双缝间距a增大时。
6. Intensity Distribution & Interference Pattern | 强度分布与干涉图样
The intensity distribution on the screen follows a cosine-squared relationship. If each slit alone produces intensity I₀, the resultant intensity at a point with phase difference δ is given by:
屏上的光强分布满足余弦平方规律。设每个缝单独到达P点的光强为I₀，则相位差为δ处的合成光强为：
I = 4I₀ cos²(δ / 2)
At constructive interference points (bright fringes), δ = 2nπ, giving I = 4I₀. At destructive interference points (dark fringes), δ = (2n + 1)π, giving I = 0. The central maximum is twice as wide as the other bright fringes because it corresponds to both positive and negative first-order diffraction angles merging.
在相长干涉点（亮纹），δ = 2nπ，光强I = 4I₀。在相消干涉点（暗纹），δ = (2n + 1)π，光强I = 0。中央亮纹的宽度是其他亮纹的两倍，因为它对应正负一级衍射角交汇的区域。
In practice, because the slits have finite width, the interference pattern is modulated by the single-slit diffraction envelope. Larger slit width narrows the diffraction envelope, reducing the number of visible fringes.
实际实验中，由于狭缝具有一定宽度，干涉图样受到单缝衍射包络的调制。缝宽增大，衍射包络变窄，可观察到的可见条纹数目减少。
7. White Light vs Monochromatic Light | 白光与单色光
When a monochromatic light source (such as a laser) is used, the interference pattern consists of equally spaced bright and dark fringes. When white light is used, each wavelength produces its own fringe pattern. The central maximum for all wavelengths coincides at the centre, producing a white bright central fringe.
使用单色光源（如激光）时，干涉图样由等间距的明暗条纹组成。使用白光光源时，每个波长都产生自己的条纹图样。所有波长的中央亮纹在中心重合，因此中央亮纹呈白色。
Away from the centre, the fringes of different wavelengths separate, leading to a spectrum of colours. The shorter wavelengths (blue/violet) produce narrower fringes and appear closer to the centre, while longer wavelengths (red) appear further out. Clear interference fringes are only observed for a few orders because different colours quickly wash each other out.
远离中心处，不同波长的条纹彼此分离，形成彩色光谱。波长较短的（蓝/紫光）条纹间距更小，靠近中心；波长较长的（红光）条纹位于更外侧。由于不同颜色的条纹迅速相互重叠抵消，只有少数几级条纹能够清晰分辨。
8. Applications & Real-World Relevance | 应用与实际意义
The double-slit experiment is not merely a historical demonstration; its principles are used in modern science and industry:
杨氏双缝实验不仅是一个历史性的演示实验，其原理在现代科学和工业中有着广泛应用：
- Measuring wavelength of light: By measuring fringe spacing Δx, the slit separation a, and the screen distance D, one can determine the wavelength λ = (a · Δx) / D.
- Optical interferometry: Interference techniques are used in precision measurements of small displacements, refractive index changes, and surface flatness testing.
- Anti-reflective coatings: Thin-film interference (related in principle) is used to reduce reflections on lenses and glass surfaces.
- Quantum mechanics: The double-slit experiment with single particles (electrons, photons) reveals wave-particle duality—a cornerstone of quantum theory.
- 测量光的波长：通过测量条纹间距Δx、双缝间距a和缝屏距离D，可求出波长λ = (a · Δx) / D。
- 光学干涉测量：干涉技术用于微小位移、折射率变化及表面平整度的高精度测量。
- 增透膜：薄膜干涉（原理相通）被用来减少透镜和玻璃表面的反射。
- 量子力学：单粒子（电子、光子）通过双缝的实验揭示了波粒二象性——量子理论的基石。
9. Common Exam Questions & Pitfalls | 常见考点与易错点
Students should be aware of the following common exam scenarios and misconceptions.
考生应特别注意以下常见考试情境和典型错误。
- Misidentifying conditions: Bright fringe requires Δ = nλ; dark fringe requires Δ = (n + ½)λ. Do not confuse the two.
- Neglecting the approximation: The formula Δx = λD/a assumes a small angle approximation (θ in radians, sin θ ≈ tan θ ≈ θ). For large angles, the approximation breaks down.
- Effect of increasing slit width: Increasing the width of the individual slits (while keeping separation constant) does not change the fringe spacing but reduces the visibility (contrast) due to diffraction envelope narrowing.
- Effect of moving screen closer/farther: Increasing D increases fringe spacing; decreasing D decreases fringe spacing.
- Immersing in water: In a medium of refractive index n, the wavelength becomes λ' = λ/n, so the fringe spacing becomes Δx' = Δx/n, which is smaller than in air.
- 混淆条件：亮纹要求Δ = nλ；暗纹要求Δ = (n + ½)λ。切勿混淆两者。
- 忽略近似条件：公式Δx = λD/a基于小角度近似（θ以弧度计，sin θ ≈ tan θ ≈ θ）。当角度较大时，该近似不再成立。
- 增加缝宽的影响：增大单缝的宽度（保持双缝间距不变）不会改变条纹间距，但由于衍射包络变窄会降低条纹的可见度（对比度）。
- 移动光屏的影响：增大D会使条纹间距增大；减小D会使条纹间距减小。
- 浸入水中：在折射率为n的介质中，波长变为λ' = λ/n，条纹间距变为Δx' = Δx/n，比空气中的间距小。
10. Worked Example | 例题解析
Example: In a Young's double-slit experiment, light of wavelength 600 nm is used. The slit separation is 0.30 mm and the screen is placed 1.2 m from the slits. Calculate the fringe spacing.
例题：在杨氏双缝实验中，使用波长为600 nm的光。双缝间距为0.30 mm，光屏距双缝1.2 m。求条纹间距。
Solution: Use Δx = λD / a. First convert all lengths to metres:
解答：利用Δx = λD / a。先将所有长度换算为米：
λ = 600 nm = 6.00 × 10⁻⁷ m, a = 0.30 mm = 3.0 × 10⁻⁴ m, D = 1.2 m
Substituting into the formula:
代入公式：
Δx = (6.00 × 10⁻⁷ × 1.2) / (3.0 × 10⁻⁴) = 2.4 × 10⁻³ m = 2.4 mm
Thus, each bright fringe is separated by 2.4 mm. If the experiment is repeated under water (n = 1.33), the fringe spacing becomes Δx' = 2.4 mm / 1.33 ≈ 1.8 mm.
因此，相邻亮纹间距为2.4 mm。若将实验置于水中重复（n = 1.33），条纹间距变为Δx' = 2.4 mm / 1.33 ≈ 1.8 mm。
11. Summary of Key Formulas | 关键公式总结
Here are the essential equations that students must master.
以下是考生必须掌握的核心方程。
| Physical Quantity 物理量 |
Formula / Condition 公式 / 条件 |
| Path difference 光程差 |
Δ ≈ (a x) / D |
| Bright fringe condition 亮纹条件 |
Δ = nλ (n = 0, 1, 2, …) |
| Dark fringe condition 暗纹条件 |
Δ = (n + ½)λ (n = 0, 1, 2, …) |
| Position of n-th bright fringe 第n级亮纹位置 |
xₙ = nλD / a |
| Fringe spacing 条纹间距 |
Δx = λD / a |
| Intensity distribution 强度分布 |
I = 4I₀ cos²(δ/2) |
Mastering these relationships and recognising the conditions under which they are valid will help you confidently tackle any Young's double-slit exam question.
熟练掌握这些关系式并认清其适用条件，将帮助您自信应对任何杨氏双缝实验相关的考题。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图