Interference of Light for IGCSE CIE Physics | 光的干涉：IGCSE CIE 物理考点精讲

📚 Interference of Light for IGCSE CIE Physics | 光的干涉：IGCSE CIE 物理考点精讲

Interference of light is one of the key phenomena that demonstrate the wave nature of light. In the IGCSE CIE Physics syllabus, students are expected to understand how two coherent light sources produce an interference pattern of bright and dark fringes, and how this pattern can be analysed quantitatively. This article provides a comprehensive breakdown of all essential concepts, experiments and formulae you need for your examination.

光的干涉是体现光具有波动性的关键现象之一。在 IGCSE CIE 物理大纲中，学生需要理解两个相干光源如何产生明暗相间的干涉条纹，以及如何定量分析这一图案。本文将全面拆解考试中必需的所有核心概念、实验和公式。

1. What is Interference? | 什么是干涉？

Interference occurs when two or more waves overlap in space. The resultant displacement at any point is the algebraic sum of the displacements due to each individual wave. This is called the principle of superposition. When light waves from coherent sources overlap, they form a stable pattern of alternating bright and dark regions called interference fringes.

当两个或多个波在空间中重叠时，就会发生干涉。任意一点的合位移是各个波单独产生的位移的代数和，这称为叠加原理。当来自相干光源的光波重叠时，会形成明暗交替的稳定图案，称为干涉条纹。

2. Conditions for Observable Interference | 可观察干涉的条件

For a clear and stable interference pattern with light, two conditions must be met. First, the sources must be coherent, meaning they emit waves with a constant phase relationship and the same frequency. Second, the waves must have the same or very similar amplitude, and they should overlap in space.

要想获得清晰稳定的光干涉图样，必须满足两个条件。第一，光源必须是相干光源，即它们发出的波具有恒定的相位关系和相同的频率。第二，这些波必须具有相同或非常相近的振幅，并且要在空间中重叠。

If the sources are not coherent, the phase difference changes randomly over time, and no fixed pattern is observed. Ordinary light sources such as a filament lamp produce incoherent light because the atoms emit light randomly and independently.

如果光源不相干，相位差会随时间随机变化，就无法观察到固定的图案。普通光源（如白炽灯）发出的光是不相干的，因为原子发光是随机、独立的。

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3. Young’s Double-Slit Experiment | 杨氏双缝实验

Thomas Young demonstrated the interference of light in 1801 using a simple but elegant setup. Monochromatic light from a single source is passed through a narrow single slit to ensure the light is coherent and in phase. This light then illuminates two closely spaced parallel slits, which act as two coherent sources. The overlapping light waves produce an interference pattern on a screen placed some distance away.

托马斯·杨在 1801 年用一个简单而精巧的装置演示了光的干涉。单色光从一个光源通过一条狭窄的单缝，以确保光相干且同位相。然后，这束光照射两条靠近的平行双缝，它们就充当了两个相干光源。重叠的光波在远处的屏幕上产生干涉图样。

4. The Concept of Path Difference | 光程差的概念

At a point on the screen, light from the two slits travels slightly different distances. The difference in distance travelled is called the path difference. If the path difference is a whole number of wavelengths, the waves arrive in phase and constructive interference occurs, giving a bright fringe. If the path difference is an odd number of half-wavelengths, the waves arrive out of phase and destructive interference gives a dark fringe.

在屏幕上的某一点，来自两条缝的光传播的距离略有不同。这个传播距离的差值称为光程差。如果光程差是波长的整数倍，两列波到达时同位相，发生相长干涉，形成亮条纹。如果光程差是半波长的奇数倍，两列波到达时反位相，发生相消干涉，形成暗条纹。

Path difference = nλ → bright fringe (constructive)

Path difference = (n + ½)λ → dark fringe (destructive)

光程差 = nλ → 亮条纹（相长）

光程差 = (n + ½)λ → 暗条纹（相消）

Here n is an integer (0, 1, 2, …). The central bright fringe corresponds to n = 0, where the path difference is zero.

这里 n 是整数（0, 1, 2, …）。中央亮条纹对应 n = 0，光程差为零。

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5. Bright and Dark Fringes in Detail | 明暗条纹详解

Bright fringes are formed where crest meets crest or trough meets trough. The amplitudes add up, resulting in maximum intensity. Dark fringes are formed where crest meets trough, causing the displacements to cancel each other out. The pattern consists of equally spaced bright and dark bands parallel to the slits.

亮条纹形成于波峰遇波峰或波谷遇波谷处。振幅相加，产生最大光强。暗条纹形成于波峰遇波谷处，位移相互抵消。干涉图案由等间距且平行于双缝的明暗条纹组成。

The intensity of the bright fringes decreases as you move away from the centre because the light energy spreads out. This is an important experimental observation.

随着远离中心，亮条纹的强度逐渐减弱，因为光能量分散开来。这是一条重要的实验观察结果。

6. The Fringe Separation Formula | 条纹间距公式

The distance between adjacent bright fringes (or adjacent dark fringes) is called the fringe separation or fringe width, denoted by Δx (or w). This quantity is related to the wavelength λ, the distance from the slits to the screen L (or D), and the separation between the two slits d (or a). The formula is:

相邻亮条纹（或相邻暗条纹）之间的距离称为条纹间距，记作 Δx（或 w）。这一物理量与波长 λ、双缝到屏幕的距离 L（或 D）以及双缝间距 d（或 a）有关。公式为：

Δx = λL / d

Δx = λL / d

In some textbooks, Δx is written as w, L as D, and d as a. Always check the symbols used in the exam paper and substitute carefully.

在一些教材中，Δx 写作 w，L 写作 D，d 写作 a。考试时务必看清题目使用的符号，仔细代入。

This formula applies only when the screen is far away compared with the slit separation (L >> d), so that the small-angle approximation sinθ ≈ tanθ ≈ θ is valid.

该公式仅在屏幕距离远大于双缝间距（L >> d）时才成立，因为此时小角近似 sinθ ≈ tanθ ≈ θ 才有效。

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7. Relating Wavelength, Slit Separation and Screen Distance | 波长、缝距和屏距的关系

From Δx = λL / d, we can deduce three proportionalities. First, fringe separation Δx is directly proportional to the wavelength λ: red light gives wider fringes than blue light. Second, Δx is directly proportional to the screen distance L: moving the screen farther away increases fringe spacing. Third, Δx is inversely proportional to the slit separation d: bringing the slits closer together spreads the fringes apart.

由 Δx = λL / d 可以得出三个比例关系。第一，条纹间距 Δx 与波长 λ 成正比：红光的条纹比蓝光更宽。第二，Δx 与屏幕距离 L 成正比：将屏幕移远会增大条纹间距。第三，Δx 与双缝间距 d 成反比：双缝靠得越近，条纹分得越开。

This formula is extremely useful for determining an unknown wavelength experimentally. By measuring Δx, L and d, you can calculate λ. This is a standard required practical in many specifications.

该公式对通过实验测定未知波长极为有用。通过测量 Δx、L 和 d，可以计算出 λ。这是许多考纲中要求的必做实验。

8. White Light Interference | 白光干涉

If a white light source is used instead of monochromatic light, a central white fringe is observed, flanked by coloured fringes. This happens because white light is a mixture of all visible wavelengths. Each wavelength produces its own interference pattern with a slightly different fringe separation. The central fringe is white because all colours undergo constructive interference at the centre (path difference = 0). On either side, the inner edge of the first-order fringe appears violet and the outer edge appears red, because violet has a shorter wavelength and therefore smaller fringe spacing, while red has a longer wavelength and larger spacing.

如果用白光光源代替单色光，会观察到中央白色条纹，两侧是彩色条纹。这是因为白光是所有可见波长的混合。每种波长各自产生干涉图样，条纹间距略有不同。中央条纹为白色，因为所有颜色的光在中心都发生相长干涉（光程差 = 0）。在两侧，一级条纹的内缘呈现紫色，外缘呈现红色，因为紫光波长较短、条纹间距较小，而红光波长较长、条纹间距较大。

After a few orders, the colours overlap so much that the pattern becomes a uniform white again, and distinct fringes are no longer visible. Only the first couple of orders show clear dispersion.

过了若干级后，各种颜色重叠严重，图案重新变成一片均匀的白色，不再能看到清晰的条纹。只有最初几级能显示明显的色散。

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9. Experimental Details and Precautions | 实验细节与注意事项

To obtain a clear interference pattern, the single slit must be narrow and exactly parallel to the double slits. The double slits must be very fine and closely spaced (typically a fraction of a millimetre). The screen should be placed at a distance of at least a metre to give a measurable fringe separation. Measurements of fringe separation are best taken by measuring the distance across several fringes and dividing by the number of spacings.

要获得清晰的干涉图样，单缝必须狭窄并且与双缝严格平行。双缝必须极细且靠近（通常零点几毫米）。屏幕应放置在至少一米之外，以便获得可测量的条纹间距。测量条纹间距的最佳方法是测量若干条条纹的总宽度，再除以间隔数。

It is also crucial to use a dark room or a darkened environment, and to ensure the slits are clean and free from dust. Laser light is often used in modern versions because it is already highly coherent and monochromatic, eliminating the need for a single slit, but careful safety precautions must be observed.

使用暗室或遮光环境也很关键，并需保持双缝洁净无尘。现代实验中常使用激光，因为激光本身就高度相干且单色性好，无需单缝，但务必遵守激光安全规范。

10. Common Mistakes and Misconceptions | 常见错误与误区

Many students think that increasing the intensity of the light source will increase the fringe separation. This is incorrect: intensity affects the brightness of fringes, not their spacing. Another common error is to confuse the distance between slits d with the slit width. Only the separation between the centres of the slits matters in the formula.

许多学生误以为增加光源强度会使条纹间距变大。这是错误的：光强影响条纹的亮度，而不是间距。另一个常见错误是混淆双缝间距 d 和缝宽。公式中只有缝中心之间的距离才重要。

Some may also think that the central fringe in white light interference is a rainbow. However, the central fringe is white because all colours have zero path difference there. Colour separation only appears at higher orders.

还有人可能以为白光干涉的中央条纹是彩虹色。然而，中央条纹是白色的，因为所有颜色的光在此处的光程差均为零。色散只出现在更高级次的条纹中。

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11. Applications of Interference | 干涉的应用

Interference of light is utilised in many technologies. Anti-reflection coatings on lenses rely on destructive interference of a thin film to cancel reflected light. Interferometers are precision instruments that use interference patterns to measure tiny distances or changes in refractive index. The phenomenon also confirms the wave nature of light and is used in holography and optical testing.

光的干涉应用于多种技术。透镜上的抗反射涂层利用薄膜的相消干涉来消除反射光。干涉仪是利用干涉图样精确测量微小距离或折射率变化的精密仪器。这一现象也证实了光的波动性，并用于全息照相和光学检测。

In the IGCSE syllabus, you are mainly expected to understand how interference provides evidence for the wave model of light, and how to apply the fringe separation formula in simple calculations and experiments.

在 IGCSE 大纲中，主要要求学生理解干涉如何为光的波动模型提供证据，以及如何在简单计算和实验中应用条纹间距公式。

12. Summary and Key Equations | 总结与关键公式

Interference of light requires two coherent sources. Young’s double-slit experiment produces bright and dark fringes. The central bright fringe is followed by equally spaced fringes. The fringe separation is given by Δx = λL / d. For white light, the central fringe is white, surrounded by spectra with violet on the inner side and red on the outer side. Remember the conditions for constructive and destructive interference in terms of path difference.

光的干涉需要两个相干光源。杨氏双缝实验产生明暗条纹。中央亮条纹两侧是等间距的条纹。条纹间距由 Δx = λL / d 给出。对于白光，中央条纹为白色，两侧是光谱，内侧为紫色，外侧为红色。记住用光程差表述的相长干涉和相消干涉的条件。

Constructive: path difference = nλ

Destructive: path difference = (n + ½)λ

Δx = λL / d

相长：光程差 = nλ

相消：光程差 = (n + ½)λ

Δx = λL / d

Master these concepts and practise applying the formula in various rearrangements. Past paper questions often ask you to describe the experiment, explain colour formation with white light, or calculate one of the variables from provided data.

掌握这些概念，并练习在不同变式中应用该公式。历年真题常要求描述实验、解释白光的彩色条纹成因，或根据给定数据计算其中一个变量。

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