IGCSE Edexcel Physics: Light Interference Revision Guide | IGCSE Edexcel 物理：光的干涉考点精讲

📚 IGCSE Edexcel Physics: Light Interference Revision Guide | IGCSE Edexcel 物理：光的干涉考点精讲

Interference of light is a fundamental wave phenomenon that provides strong evidence for the wave nature of light. In the IGCSE Edexcel Physics course, you need to understand Young’s double-slit experiment, be able to describe the formation of bright and dark fringes, and apply the fringe spacing equation confidently. This revision guide covers every key point from the specification, with paired English and Chinese explanations to help you master the topic efficiently.

光的干涉是证明光具有波动性的重要现象。在 IGCSE Edexcel 物理课程中，你需要理解杨氏双缝实验，能够描述明暗条纹的形成，并熟练应用条纹间距公式。本篇复习指南涵盖大纲中的每一个考点，搭配中英双语解释，帮助你高效掌握本主题。

1. What is Interference? | 什么是干涉？

Interference occurs when two or more waves superpose and combine to form a resultant wave of greater, lower, or zero amplitude. For light, interference can only be observed when the sources are coherent, meaning they emit waves with a constant phase relationship and the same frequency.

当两个或多个波叠加时，会形成振幅更大、更小或为零的合成波，这一现象称为干涉。对于光来说，只有相干光源才能产生可观测的干涉，相干光源发出的光波必须具有恒定的相位关系和相同的频率。

The principle of superposition states that when two waves meet, the resultant displacement at any point is the vector sum of the individual displacements. Interference fringes are a direct consequence of this principle.

叠加原理指出，当两列波相遇时，任一点的合位移等于各列波单独在该点引起的位移的矢量和。干涉条纹正是这一原理的直接结果。

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2. Conditions for Interference: Coherent Sources | 干涉条件：相干光源

To produce a stable interference pattern with light, the sources must be coherent. Coherence means the waves have a constant phase difference and the same wavelength. Ordinary light bulbs emit photons randomly, so they are incoherent. In the lab, coherence is achieved by using a laser or by splitting light from a single monochromatic source.

要产生稳定的光干涉图样，光源必须是相干的。相干意味着光波具有恒定的相位差和相同的波长。普通灯泡发出的光子是随机的，因此是非相干光源。在实验室中，可以通过使用激光或将来自单一单色光源的光进行分光来实现相干。

Lasers produce highly coherent, monochromatic light, which is ideal for the double-slit experiment. Alternatively, a single-colour filter and a narrow slit can be placed in front of a filament lamp to create a reasonably coherent source, but the fringes are dimmer.

激光可以产生高度相干、单色的光，是双缝实验的理想光源。另一种方法是，在灯丝灯泡前放置一个单色滤光片和一个窄缝，以形成一个较为相干的光源，但产生的条纹亮度较暗。

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

The classic experiment to demonstrate interference of light is Young’s double-slit experiment. Monochromatic light from a laser is directed onto two narrow, parallel slits that are very close together. The slits act as two coherent point sources, emitting waves that spread out (diffract) and overlap on a screen placed some distance away.

证明光干涉现象的经典实验是杨氏双缝实验。将激光发出的单色光照射到两个间距很小的平行窄缝上，这两个狭缝就成为两个相干的点光源，发出的波发生衍射并相互重叠，在远处的屏幕上形成干涉图样。

The distance from the double slit to the screen is denoted as D, the separation between the two slits is a, and the distance between adjacent bright fringes on the screen is called the fringe spacing x. A precise setup allows measurement of x to determine the wavelength of light.

双缝到屏幕的距离记为 D，双缝间距为 a，屏幕上相邻亮条纹之间的距离称为条纹间距 x。精密的实验装置可以测量出 x，从而求出光的波长。

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4. Path Difference and Interference Pattern | 光程差与干涉图样

The pattern observed on the screen consists of a central bright fringe (maximum) with alternating bright and dark fringes on either side. Whether a point on the screen appears bright or dark depends on the path difference between the waves arriving from the two slits.

屏幕上观察到的图样由中央亮条纹（极大）及两侧交替出现的亮条纹和暗条纹组成。屏幕上某点是亮还是暗，取决于从两条狭缝到达该点的光波之间的光程差。

The path difference is the extra distance one wave travels compared to the other. For a bright fringe, the waves arrive in phase and interfere constructively. For a dark fringe, they arrive out of phase and interfere destructively.

光程差是指一列波比另一列波多走的距离。对于亮条纹，两列波到达时同相，发生相长干涉。对于暗条纹，它们到达时反相，发生相消干涉。

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5. Constructive vs Destructive Interference | 相长干涉与相消干涉

Constructive interference occurs when the path difference is a whole number of wavelengths, i.e. path difference = nλ, where n = 0, 1, 2, … At such points, the crests of one wave align with crests of the other, resulting in a bright fringe with maximum amplitude.

当光程差等于波长的整数倍，即光程差 = nλ（n = 0, 1, 2, …）时，发生相长干涉。在这些点，一列波的波峰与另一列波的波峰重合，形成振幅最大的亮条纹。

Destructive interference occurs when the path difference is an odd number of half-wavelengths, i.e. path difference = (n + ½)λ, where n = 0, 1, 2, … The waves arrive completely out of phase, so the resultant amplitude is zero, producing a dark fringe.

当光程差等于半波长的奇数倍，即光程差 = (n + ½)λ（n = 0, 1, 2, …）时，发生相消干涉。此时两列波完全反相到达，合振幅为零，形成暗条纹。

Bright fringe condition: path difference = nλ

亮条纹条件：光程差 = nλ

Dark fringe condition: path difference = (n + ½)λ

暗条纹条件：光程差 = (n + ½)λ

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6. The Fringe Spacing Equation: x = λD / a | 条纹间距公式：x = λD / a

The fringe spacing x, the distance between the centres of two adjacent bright (or dark) fringes, is directly proportional to the wavelength λ and the screen distance D, and inversely proportional to the slit separation a. The relationship is given by the equation:

条纹间距 x（相邻两条亮条纹或暗条纹中心之间的距离）与波长 λ 和屏幕距离 D 成正比，与双缝间距 a 成反比。这一关系由以下公式表示：

x = λD / a

Rearranging for wavelength gives λ = ax / D. This formula allows you to determine the wavelength of light from measurable quantities. All quantities must be in SI units: x and a in metres, D in metres, λ in metres.

将公式变形可得到 λ = ax / D，从而通过可测量的量求出光的波长。所有物理量必须使用国际单位制：x 和 a 的单位为米，D 的单位为米，λ 的单位为米。

If the fringe spacing is very small, you can measure across several fringes and divide by the number of fringes to improve accuracy. For example, measure the distance across 10 bright fringes and divide by 10 to find x.

如果条纹间距很小，可以通过测量多条条纹的总宽度再除以条纹数来提高精度。例如，测量10条亮条纹的总宽度，然后除以10得到 x。

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7. Measuring Wavelength of Light | 测量光的波长

To determine the wavelength of laser light using Young’s interference apparatus, follow these steps: set up the laser, double slit, and screen; measure D with a metre ruler; measure a with a travelling microscope or given data; measure the fringe spacing x by taking the distance across several fringes and dividing.

使用杨氏干涉装置测量激光波长的步骤如下：安装好激光器、双缝和屏幕；用米尺测量 D；用移测显微镜或给出的数据测量 a；通过测量多条条纹的宽度并除以条纹数得到条纹间距 x。

Then use λ = ax / D. A typical set of results might be: a = 0.5 × 10⁻³ m, D = 2.00 m, x = 2.5 × 10⁻³ m, yielding λ = (0.5×10⁻³ × 2.5×10⁻³) / 2.00 = 6.25 × 10⁻⁷ m, which is 625 nm, a typical red laser wavelength.

然后利用公式 λ = ax / D 计算。一组典型数据可能是：a = 0.5 × 10⁻³ m，D = 2.00 m，x = 2.5 × 10⁻³ m，求得 λ = (0.5×10⁻³ × 2.5×10⁻³) / 2.00 = 6.25 × 10⁻⁷ m，即 625 nm，这是典型的红光激光波长。

Safety: never look directly into the laser beam and avoid reflections. Use a low-power Class 2 laser and ensure the laboratory is appropriately lit so that pupils do not dilate excessively.

安全事项：切勿直视激光束，避免反射光。使用低功率的2类激光器，并确保实验室光线适当，防止瞳孔过度放大。

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8. White Light Interference | 白光干涉

If you replace the monochromatic source with a white light source, the interference pattern changes dramatically. White light contains all visible wavelengths. Each wavelength produces its own fringe pattern with a different fringe spacing x, because x is proportional to λ.

如果用白光光源替代单色光源，干涉图样会发生显著变化。白光包含所有可见光波长。由于条纹间距 x 与波长 λ 成正比，不同波长的光产生间距不同的干涉条纹。

At the central maximum, all wavelengths interfere constructively (path difference = 0), so a white central fringe is observed. On either side, the next bright fringe shows colour fringes: blue/violet on the inner edge (smaller λ, smaller x) and red on the outer edge (larger λ, larger x). Higher-order fringes overlap increasingly, and only a few coloured fringes are visible before the pattern washes out to uniform white.

在中央极大处，所有波长的光均发生相长干涉（光程差 = 0），因此观察到白色中央条纹。在中央条纹两侧的第一级亮纹呈现出彩色：内侧为蓝/紫色（λ 较小，x 较小），外侧为红色（λ 较大，x 较大）。更高级次的条纹重合得更加严重，往往只能看到少数几条彩色条纹，之后图样便融合成均匀的白光。

For white light: central fringe is white, outer fringes show spectra with blue inside and red outside.

白光干涉图样：中央条纹为白色，两侧条纹呈现内蓝外红的光谱色。

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9. Effect of Changing Parameters | 改变实验参数的影响

Understanding how the fringe pattern changes when D, a, or λ is varied is a common exam hurdle. Use x = λD / a to reason qualitatively and quantitatively.

理解改变 D、a 或 λ 时干涉图样如何变化是常见的考试难点。应利用公式 x = λD / a 进行定性和定量分析。

• Increasing D (screen distance): x increases, fringes become wider and more widely spaced. 增大屏幕距离 D 会使 x 增大，条纹变宽、间距变大。
• Decreasing a (slit separation): x increases, fringes spread out. 减小双缝间距 a 会使 x 增大，条纹向外扩展。
• Using light with longer wavelength λ: x increases; red light gives wider fringes than blue light. 使用波长更长的光会使 x 增大；红光的条纹比蓝光的更宽。

If the source is replaced by one with a shorter wavelength but everything else remains the same, the fringe spacing will decrease, making the pattern more closely packed.

如果光源换成波长更短的光而其他条件不变，条纹间距将减小，图样会变得更密集。

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10. Common Exam Mistakes and Tips | 常见考试错误与技巧

Don’t confuse diffraction and interference. Diffraction is the spreading of waves through a gap or around an obstacle; interference is the superposition of waves from two or more sources. In the double-slit experiment, both occur: light diffracts at the slits, and the diffracted waves then interfere.

不要混淆衍射和干涉。衍射是波通过狭缝或绕过障碍物时发生的扩展现象；干涉是来自两个或多个光源的波的叠加。在双缝实验中，这两种现象都存在：光在狭缝处发生衍射，衍射后的波再发生干涉。

State the equation in words, not just symbols. Know that fringe spacing is directly proportional to wavelength and screen distance, and inversely proportional to slit separation. The examiner may ask you to explain a change without calculation.

既能写公式，也能用文字表述。要知道条纹间距与波长和屏幕距离成正比，与双缝间距成反比。考官可能会要求你不通过计算，用文字解释变化趋势。

Remember the central fringe is always bright for monochromatic light because the path difference is zero. Label this correctly in diagrams. When measuring x, always measure between centres of fringes, not edges.

记住单色光的中央条纹总是亮条纹，因为光程差为零。在图中要正确标出。测量 x 时，一定测量条纹中心之间的距离，而不能测边缘。

Be precise with units. Convert all lengths to metres before plugging into x = λD / a. For wavelengths, you might need to express the answer in nanometres (1 nm = 1 × 10⁻⁹ m).

注意单位。在代入公式 x = λD / a 之前，必须把所有长度单位换算成米。求出的波长可能需要以纳米（1 nm = 1 × 10⁻⁹ m）表示。

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11. Practice Calculation Example | 计算题示例

In a double-slit experiment, a laser of wavelength 650 nm illuminates two slits separated by 0.40 mm. The screen is 1.50 m from the slits. Calculate the fringe spacing.

在双缝实验中，波长为 650 nm 的激光照射间距为 0.40 mm 的两条狭缝。屏幕距离狭缝 1.50 m。计算条纹间距。

Convert units: λ = 650 nm = 650 × 10⁻⁹ = 6.5 × 10⁻⁷ m; a = 0.40 mm = 4.0 × 10⁻⁴ m; D = 1.50 m.

单位换算：λ = 650 nm = 6.5 × 10⁻⁷ m；a = 0.40 mm = 4.0 × 10⁻⁴ m；D = 1.50 m。

x = λD / a = (6.5 × 10⁻⁷ m × 1.50 m) / (4.0 × 10⁻⁴ m) = 2.44 × 10⁻³ m (or 2.44 mm).

x = λD / a = (6.5 × 10⁻⁷ m × 1.50 m) / (4.0 × 10⁻⁴ m) = 2.44 × 10⁻³ m（即 2.44 mm）。

Always check that your answer is sensible. A fringe spacing of a few millimetres is typical for such experiments.

务必检查答案的合理性。对于此类实验，几毫米的条纹间距是典型的数值。

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

Light interference provides compelling evidence for the wave model. The essential knowledge for the IGCSE Edexcel exam includes the concepts of coherence, constructive and destructive interference, the appearance of the fringe pattern, the equation x = λD / a, and the behaviour with white light.

光的干涉为波动模型提供了有力证据。应对 IGCSE Edexcel 考试必备的知识包括：相干概念、相长与相消干涉、干涉图样的特征、公式 x = λD / a 以及白光干涉的特性。

Quantity 物理量	Symbol 符号	Unit 单位
Fringe spacing 条纹间距	x	metres (m) 米
Wavelength 波长	λ	metres (m) 米
Distance to screen 屏幕距离	D	metres (m) 米
Slit separation 双缝间距	a	metres (m) 米

Memorise the fringe condition rules: bright = nλ, dark = (n + ½)λ. Practise unit conversions and rearrangements. With this foundation, you can confidently tackle any interference question on the IGCSE Edexcel Physics paper.

牢记条纹条件：亮纹对应 nλ，暗纹对应 (n + ½)λ。多练习单位换算和公式变形。有了这些基础，你就能自信应对 IGCSE Edexcel 物理试卷中任何干涉类题目。

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