Edexcel Physics: Interference of Light | Edexcel 物理：光的干涉 考点精讲

📚 Edexcel Physics: Interference of Light | Edexcel 物理：光的干涉 考点精讲

Interference of light is a cornerstone of wave optics in the Edexcel A-level Physics syllabus. Mastering this topic involves understanding coherence, the double-slit experiment, fringe spacing calculations, and thin-film interference. This article breaks down every key concept, formula, and common examination pitfall to help you achieve top marks.

光的干涉是 Edexcel A-level 物理大纲中波动光学的核心内容。掌握这一主题需要理解相干性、双缝实验、条纹间距计算以及薄膜干涉。本文将分解每个关键概念、公式和常见考试陷阱，帮助你获得高分。

1. What Is Interference? | 什么是干涉？

Interference is the superposition of two or more waves arriving from coherent sources, giving rise to regions of reinforcement and cancellation. Where crest meets crest or trough meets trough, constructive interference produces a resultant wave with larger amplitude, while crest meeting trough leads to destructive interference and reduced amplitude. In optics, this manifests as alternating bright and dark fringes on a screen.

干涉是来自相干光源的两个或多个波的叠加，产生增强和抵消的区域。当波峰遇波峰或波谷遇波谷时，相长干涉产生振幅更大的合成波；而波峰遇波谷则导致相消干涉，振幅减小。在光学中，这表现为屏幕上明暗相间的条纹。

Interference patterns can only be observed clearly when the overlapping waves maintain a constant phase relationship over time. This property is called coherence, and it is the very first condition you must check in any interference problem.

只有当重叠的波在时间上保持恒定的相位关系时，才能清晰地观察到干涉图样。这一特性称为相干性，它是任何干涉问题中必须检查的首要条件。

2. Conditions for Interference & Coherence | 干涉条件与相干性

For a stable interference pattern, the two sources must be coherent. This means they must emit waves of the same frequency (or wavelength), with a constant phase difference, and ideally have similar amplitudes. Polarisation must also be considered: the electric field oscillations should not be perpendicular, or no interference occurs.

要获得稳定的干涉图样，两个光源必须是相干的。这意味着它们必须发射相同频率（波长）的波，具有恒定的相位差，并且理想情况下振幅相似。还必须考虑偏振：电场振荡不应相互垂直，否则不会发生干涉。

In the laboratory, it is almost impossible to obtain two independent sources that are highly coherent. Instead, we produce two coherent sources by dividing a single wavefront – either by division of wavefront (Young’s slits) or division of amplitude (thin films). A laser is an excellent coherent source because its output has a long coherence length and near-perfect monochromaticity.

在实验室中，几乎不可能获得两个高度相干的独立光源。相反，我们通过分割单一波前来产生两个相干光源——要么通过波前分割（杨氏双缝），要么通过振幅分割（薄膜）。激光是极好的相干光源，因为其输出具有较长的相干长度和近乎完美的单色性。

If the path difference between waves from the two sources is greater than the coherence length, the phase relation becomes random and interference fringes wash out. This is why sodium light and laser diodes are commonly used; they provide relatively long coherence lengths.

如果两个光源发出的波之间的光程差大于相干长度，相位关系会变得随机，干涉条纹将消失。这就是为什么钠光灯和激光二极管被广泛使用；它们能提供较长的相干长度。

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

Young’s double-slit experiment was the first convincing demonstration of the wave nature of light. Monochromatic light is passed through a single narrow slit to produce a coherent wavefront, which then falls on a pair of closely spaced parallel slits, S₁ and S₂. The two slits act as secondary coherent sources, and the overlapping waves create an interference pattern on a distant screen.

杨氏双缝实验是证明光具有波动性的第一个令人信服的实验。单色光通过一个单缝产生相干波前，然后照射在一对紧密排列的平行狭缝 S₁ 和 S₂ 上。这两个狭缝充当次级相干光源，重叠的波在远处的屏幕上形成干涉图样。

The key geometrical parameters are: slit separation, d (usually a fraction of a millimetre); distance from the slits to the screen, D (typically 1–2 metres); and the wavelength of light, λ. The observation plane is perpendicular to the optical axis, and fringes appear parallel to the slits.

关键的几何参数为：狭缝间距 d（通常为零点几毫米）；狭缝到屏幕的距离 D（一般 1–2 米）；以及光的波长 λ。观察平面垂直于光轴，条纹平行于狭缝出现。

4. Path Difference and the Condition for Bright/Dark Fringes | 光程差及明暗纹条件

Consider a point P on the screen at a distance x from the central maximum. The path difference between the two waves arriving at P from S₂ and S₁ is approximately δ = d sin θ, where θ is the angle between the central axis and the line from the midpoint of the slits to P.

考虑屏幕上距中央极大距离为 x 的一点 P。从 S₂ 和 S₁ 到达 P 的两列波之间的光程差近似为 δ = d sin θ，其中 θ 是中心轴与从双缝中点指向 P 的连线之间的夹角。

δ = d sin θ

Constructive interference (bright fringe) occurs when the path difference is an integer multiple of the wavelength. Destructive interference (dark fringe) occurs when the path difference is an odd multiple of half the wavelength.

当光程差是波长的整数倍时，发生相长干涉（亮纹）。当光程差是半波长的奇数倍时，发生相消干涉（暗纹）。

Bright: d sin θ = mλ, m = 0, ±1, ±2, …

Dark: d sin θ = (m + ½)λ, m = 0, ±1, ±2, …

The integer m is called the order of interference. The central bright fringe (m = 0) is formed when the path difference is zero. All wavelengths in white light constructively interfere here, giving a white central fringe.

整数 m 称为干涉级次。中央亮纹（m = 0）在光程差为零时形成。白光中所有波长在此处相长干涉，产生白色中央条纹。

5. Deriving the Fringe Spacing Formula | 推导条纹间距公式

For small angles, sin θ ≈ tan θ ≈ x/D, where x is the perpendicular distance from the centre of the screen to the fringe of interest. Substituting into the bright fringe condition gives x = (mλ D)/d. The separation between adjacent bright fringes (or adjacent dark fringes) is therefore constant and known as the fringe spacing Δx.

对于小角度，sin θ ≈ tan θ ≈ x/D，其中 x 是从屏幕中心到所关注条纹的垂直距离。代入亮纹条件得 x = (mλ D)/d。因此相邻亮纹（或相邻暗纹）之间的间距是恒定的，称为条纹间距 Δx。

Δx = λD / d

This equation is central to all double-slit calculations in Edexcel exams. It shows that fringe spacing is directly proportional to wavelength and screen distance, and inversely proportional to slit separation. A larger d produces narrower fringes that are harder to resolve; a larger D or λ spreads the pattern out.

该方程是 Edexcel 考试中所有双缝计算的核心。它表明条纹间距与波长和屏幕距离成正比，与狭缝间距成反比。较大的 d 会产生较窄且难以分辨的条纹；较大的 D 或 λ 使图样扩展。

Experimentally, one often measures the total width of several fringes and divides by the number of fringes to obtain a more accurate Δx. Always ensure D is measured to the screen, and d is given or measured with a travelling microscope.

在实验中，通常测量若干条条纹的总宽度，除以条纹数，以获得更精确的 Δx。务必确保 D 测量到屏幕，d 由已知或通过读数显微镜测得。

6. Factors Affecting Fringe Pattern & Experimental Design | 影响条纹图样的因素与实验设计

Changing any of the three variables λ, D, or d directly alters the fringe spacing as predicted by Δx = λD/d. Using red laser (longer λ) instead of blue gives wider fringes. Increasing D also widens the pattern but reduces intensity. Reducing d is a very effective way to increase Δx, which is why Young’s original slits were extremely close together.

改变 λ、D 或 d 三个变量中的任何一个，都会根据 Δx = λD/d 直接改变条纹间距。用红色激光（波长较长）替代蓝色激光会产生更宽的条纹。增大 D 也会使图样变宽，但会降低强度。减小 d 是增大 Δx 极为有效的方法，这也是杨氏最初的双缝间距极小的原因。

To ensure good contrast, the slits must be narrow enough to produce significant diffraction, which allows the wavelets to spread out and overlap. A single slit before the double slits ensures the wavefront arriving at S₁ and S₂ is continuous and in phase. If white light is used, a colour filter can select a narrow wavelength band for clearer fringes.

为确保良好的对比度，狭缝必须足够窄以产生明显的衍射，使子波扩散并重叠。双缝之前的单缝确保到达 S₁ 和 S₂ 的波前连续且同相。如果使用白光，彩色滤光片可以选出窄波长范围，使条纹更清晰。

7. White Light Interference | 白光干涉

When white light is used, the central maximum (m = 0) remains white because all wavelengths interfere constructively at the centre. On either side, however, each colour produces its own fringe pattern with a spacing Δx ∝ λ. As a result, the first-order fringe appears as a continuous spectrum, with violet closest to the centre and red furthest away. Higher orders overlap and produce a wash of colours, making it hard to distinguish individual maxima.

当使用白光时，中央极大（m = 0）保持白色，因为所有波长在此处均相长干涉。但在两侧，每种颜色会按照 Δx ∝ λ 产生各自的条纹图样。因此第一级条纹呈现为连续光谱，紫色最靠近中心，红色最远。更高级次的条纹会重叠并产生混色，难以分辨单个亮纹。

In exam questions, you may be asked to sketch the white-light fringe pattern and label colours, or to explain why only a few orders are visible. The answer lies in the overlapping of spectra: once the red end of the mth order overlaps with the violet of the (m+1)th order, clear fringes disappear.

在考试题中，你可能需要画出白光干涉图样并标出颜色，或者解释为什么只能看到很少的级次。答案在于光谱的重叠：当第 m 级红色与第(m+1)级紫色重叠时，清晰的条纹便消失了。

8. Phase Change on Reflection | 反射时的相位变化

When light reflects from a boundary where the refractive index increases (from lower to higher n), the reflected wave undergoes a phase change of π radians – equivalent to adding or subtracting half a wavelength to the optical path. This is often called the half-wavelength loss. No such phase change occurs when reflection is from a medium of lower refractive index.

当光在折射率增加的界面（从较低 n 到较高 n）反射时，反射波会发生 π 弧度的相位变化——相当于在光程上增加或减去半个波长。这通常称为半波损失。当从折射率较低的介质反射时不发生这种相位变化。

This concept is crucial for thin-film interference. The net effect depends on which reflections undergo the phase change. If only one reflection introduces a π phase shift, the conditions for constructive and destructive interference in reflected light are swapped compared to the simple path-difference analysis.

这一概念对薄膜干涉至关重要。净效应取决于哪些反射引入了 π 相移。如果仅有一次反射引入 π 相移，那么反射光中相长干涉和相消干涉的条件将与简单光程差分析的结果互换。

9. Thin-Film Interference: Soap Bubbles & Air Wedges | 薄膜干涉：肥皂泡与空气劈尖

A thin film, such as a soap bubble or an oil slick, shows vivid colours because of interference between light reflected from the top and bottom surfaces of the film. For a film of thickness t and refractive index n, the optical path difference for near-normal incidence is 2nt, plus any additional contributions from phase changes on reflection.

薄膜（如肥皂泡或油膜）显示出鲜艳的色彩，这源于从薄膜上表面和下表面反射的光之间发生干涉。对于厚度为 t、折射率为 n 的薄膜，接近垂直入射时的光程差为 2nt，再加上反射时任何额外的相位变化。

For a soap film in air, light reflecting from the air–film interface (n_film > n_air) undergoes a π phase change, but at the film–air boundary there is no phase change. Thus the effective path difference includes an extra half wavelength. The conditions become: constructive (bright) for 2nt = (m + ½)λ, and destructive (dark) for 2nt = mλ. This is why soap films appear coloured in reflected light and dark in certain thicknesses.

对于空气中的肥皂膜，光在空气–薄膜界面（n_film > n_air）反射时发生 π 相位变化，而在薄膜–空气界面不发生相位变化。因此等效光程差包含额外的半波长。条件变为：相长（明纹）为 2nt = (m + ½)λ，相消（暗纹）为 2nt = mλ。这就是肥皂膜在反射光中呈现彩色，在某些厚度下呈现暗色的原因。

A similar analysis applies to an air wedge formed between two glass plates. The air film has n ≈ 1, but one reflection (at the lower glass–air interface) has no phase change and the other (at the upper air–glass interface) does, again leading to the same inverted conditions if illumination is from above. The fringe pattern appears as straight, equally spaced bands parallel to the wedge apex.

类似的分析适用于两片玻璃板之间形成的空气劈尖。空气膜的 n ≈ 1，但一个反射（在下玻璃–空气界面）没有相移，另一个反射（在上空气–玻璃界面）有相移，同样导致相同的逆转条件，若从上方照明。条纹图样呈现为平行于劈尖顶端的、等间距的直线条纹。

10. Anti-Reflection Coatings | 减反膜

Anti-reflection coatings on camera lenses and spectacles exploit destructive thin-film interference. A transparent dielectric layer with thickness t and refractive index n_coating is deposited on glass (n_glass). Ideally, n_coating is between n_air and n_glass so that both the air–coating and coating–glass reflections involve a phase change of π. The two reflected waves are then 180° out of phase when the optical path difference 2 n_coating t = λ/2, i.e., t = λ/(4 n_coating). This cancels reflection for that specific wavelength, often chosen in the green part of the spectrum.

相机镜头和眼镜上的减反膜利用了相消薄膜干涉。在玻璃（n_glass）上沉积一层厚度为 t、折射率为 n_coating 的透明介质层。理想情况下 n_coating 介于 n_air 和 n_glass 之间，这样空气–涂层和涂层–玻璃两个反射面都发生 π 相变。当光程差 2 n_coating t = λ/2，即 t = λ/(4 n_coating) 时，两束反射光相位相差 180°，相互抵消。这种消反射针对特定波长，通常选择光谱中绿色部分。

Partial anti-reflection can be achieved for a range of wavelengths by using multi-layer coatings. In exams, you may be asked to calculate the minimum thickness for destructive interference on a given coating, always remembering whether phase changes occur at one or both interfaces.

使用多层薄膜可以在一个波长范围内实现部分减反。在考试中，你可能需要计算给定涂层上实现相消干涉的最小厚度，并始终要考虑在界面处是否发生一次还是两次相位变化。

11. Common Exam Pitfalls & Top Tips | 常见考试陷阱与重要提示

Many students forget to confirm coherence conditions before answering. Always state that the sources must be coherent (same frequency, constant phase) and produced from a single source. When describing Young’s experiment, mention the single slit’s role in ensuring this.

许多学生在答题前忘记确认相干条件。务必说明光源必须相干（相同频率、恒定相位）且由一个单一光源产生。在描述杨氏实验时，要提及单缝的作用在于确保这一点。

Confusing path difference with phase difference is a recurrent error. A path difference of one wavelength corresponds to a phase difference of 2π radians. Using d sinθ = mλ directly when the angle is not small can lead to serious mistakes – estimate sinθ carefully or use the exact geometry if required.

混淆光程差与相位差是反复出现的错误。一个波长的光程差对应 2π 弧度的相位差。当角度不小时直接使用 d sinθ = mλ 可能导致严重错误——仔细估算 sinθ，或在需要时使用精确几何关系。

In fringe-spacing problems, always convert all lengths to metres. Avoid using millimetres for D but centimetres for Δx; keep units consistent. When measuring fringe width, measure across several fringes and divide – this reduces percentage uncertainty. In graphs of x vs. m, the gradient gives λD/d.

在条纹间距问题中，务必将所有长度转换为米。避免对 D 使用毫米、对 Δx 使用厘米；保持单位一致。测量条纹宽度时，跨越多个条纹测量然后除以条纹数——这样可以降低百分比不确定度。在 x 对 m 的图中，斜率给出 λD/d。

For thin films, always check the refractive indices at the boundaries. A missing half-wavelength adjustment can flip the constructive/destructive conditions. Practice with an air wedge, a soap film, and an oil film to internalise when the π phase change occurs on one or both reflections.

对于薄膜，始终检查边界处的折射率。缺少半波调整可能会颠倒相长/相消条件。练习空气劈尖、肥皂膜和油膜的各种情况，内化何时在单次还是两次反射中发生 π 相位变化。

Finally, when white light is used, do not draw equal spacing for all colours. The fringe spacing increases with wavelength, so red is further out than blue. Label colours clearly and note that the central fringe is white.

最后，当使用白光时，不要把所有颜色画成等间距。条纹间距随波长增大而增大，因此红色比蓝色更靠外。清晰地标出颜色，并注明中央条纹是白色的。

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