Young’s Double-Slit Experiment: Key Exam Points | 杨氏双缝干涉实验考点梳理

📚 Young’s Double-Slit Experiment: Key Exam Points | 杨氏双缝干涉实验考点梳理

Young’s double-slit experiment is one of the most important pieces of evidence for the wave nature of light. In A-Level Physics, it is a popular topic for structured questions, calculations and practical-based reasoning.

杨氏双缝干涉实验是证明光具有波动性的最重要实验之一。在 A-Level 物理中，它是结构化问题、计算题和实验分析题的高频考点。

This article summarises the essential definitions, conditions, formulas and common mistakes you need to master for the CIE examination.

本文系统梳理 CIE 考试中必须掌握的核心概念、条件、公式和常见易错点。

1. Experimental Setup and Purpose | 实验装置与目的

A monochromatic light source, such as a laser, is directed at a narrow single slit. The single slit ensures that light spreading onto the double slits is spatially coherent.

将单色光源（如激光）照射到一个狭窄的单缝上。单缝的作用是使到达双缝的光具有较好的空间相干性。

The light then reaches two closely spaced narrow slits, labelled S₁ and S₂, which act as two coherent sources. These two slits emit overlapping spherical wavefronts.

随后光到达两个相距很近的窄缝 S₁ 和 S₂，它们相当于两个相干光源。两缝发出的光波在空间中叠加。

A screen, or a micrometer eyepiece, is placed at a distance L from the double slits. On the screen, alternating bright and dark fringes are observed.

在距离双缝 L 处放置光屏或测微目镜，屏幕上会出现明暗相间的干涉条纹。

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2. Conditions for Coherent Sources | 相干条件

For a stable interference pattern, the two sources must be coherent. Coherent sources have the same frequency and a constant phase difference.

要形成稳定的干涉图样，两个光源必须相干。相干光源具有相同的频率和恒定的相位差。

In the double-slit arrangement, light from the single slit reaches both S₁ and S₂ with the same phase. Because both slits originate from the same wavefront, they maintain a fixed phase relationship.

在双缝装置中，来自同一单缝的光到达 S₁ 和 S₂ 时相位相同。由于两缝来自同一个波前，它们之间始终保持固定的相位关系。

If the two slits were illuminated by two independent light bulbs, the pattern would change randomly and disappear because the phase difference would not be constant.

如果使用两个独立灯泡分别照亮两缝，由于相位差不恒定，干涉图样会随机变化并消失。

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3. Path Difference and Phase Difference | 光程差与相位差

Consider a point P on the screen. The distances from S₁ and S₂ to P are generally not equal. The path difference is defined as δ = S₂P − S₁P.

考虑屏上某点 P，S₁ 和 S₂ 到 P 的距离通常不相等。光程差定义为 δ = S₂P − S₁P。

If the double-slit separation is a and the angle from the central axis to point P is θ, then for small angles the path difference can be approximated as:

若双缝间距为 a，P 点相对中心轴的角度为 θ，则在小角度近似下，光程差可表示为：

δ = a sin θ

The phase difference Δφ is related to the path difference by the equation:

相位差 Δφ 与光程差的关系为：

Δφ = 2πδ / λ

Here λ is the wavelength of the light. A path difference of one whole wavelength corresponds to a phase difference of 2π radians.

其中 λ 为光的波长。当一个光程差等于一个完整波长时，相位差为 2π 弧度。

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4. Conditions for Bright and Dark Fringes | 明纹与暗纹条件

Bright fringes occur where the waves from S₁ and S₂ arrive in phase, so the path difference is an integer multiple of the wavelength.

当 S₁ 和 S₂ 发出的光在屏上同相叠加时出现亮纹，此时光程差为波长的整数倍。

δ = nλ where n = 0, 1, 2, 3, …

Dark fringes occur where the waves arrive exactly out of phase, so the path difference is a half-integer number of wavelengths.

当两列光在屏上反相叠加时出现暗纹，此时光程差为半波长的奇数倍。

δ = (n + 1/2)λ where n = 0, 1, 2, 3, …

For n = 0, the path difference is zero, so a central bright fringe is formed on the axis directly opposite the midpoint of the double slits.

当 n = 0 时，光程差为零，因此在正对双缝中点的轴上形成中央亮纹。

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5. Derivation of Fringe Spacing Formula | 条纹间距公式推导

Let the distance from the central bright fringe to the nth bright fringe be x. For small θ, the approximation sin θ ≈ tan θ is valid.

设第 n 级亮纹到中央亮纹的距离为 x。在小角度近似下，sin θ ≈ tan θ 成立。

From the geometry, tan θ = x / L, where L is the distance between the double slits and the screen.

根据几何关系，tan θ = x / L，其中 L 为双缝到屏幕的距离。

For a bright fringe, a sin θ = nλ, so using the small-angle approximation:

对亮纹有 a sin θ = nλ，因此利用小角度近似可得：

a x / L = nλ

The distance between two adjacent bright fringes, or the fringe spacing Δx, is therefore:

相邻两条亮纹之间的距离，即条纹间距 Δx，为：

Δx = λL / a

This formula also applies to adjacent dark fringes. It shows that fringes are equally spaced for a given wavelength and geometry.

该公式同样适用于相邻暗纹。它表明在给定波长和装置参数下，条纹是等间距分布的。

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6. Factors Affecting Fringe Spacing | 影响条纹间距的因素

The fringe spacing Δx depends on three factors: wavelength λ, screen distance L, and slit separation a.

条纹间距 Δx 由三个因素决定：波长 λ、屏距 L 和双缝间距 a。

Factor | 因素	Change | 变化	Effect on Δx | 对条纹间距的影响
Wavelength λ | 波长	Increases | 增大	Fringes become wider | 条纹变宽
Screen distance L | 屏距	Increases | 增大	Fringes become wider | 条纹变宽
Slit separation a | 双缝间距	Increases | 增大	Fringes become narrower | 条纹变窄

Increasing the wavelength or the screen distance makes the pattern spread out more. Increasing the slit separation compresses the pattern.

增大波长或屏距会使干涉图样更加展开；增大双缝间距则会使图样更加压缩。

A common mistake is to confuse slit separation with slit width. Slit width affects the intensity envelope through diffraction, but it does not change the fringe spacing in the double-slit formula.

常见错误是混淆双缝间距与缝宽。缝宽通过衍射影响光强包络，但不会改变双缝公式中的条纹间距。

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7. White Light Interference and Coloured Fringes | 白光干涉与彩色条纹

When white light is used, every wavelength in the visible spectrum produces its own fringe pattern. The patterns overlap on the screen.

使用白光时，可见光范围内的每一种波长都会产生各自的一组条纹，这些图样在屏幕上发生重叠。

At the centre, all wavelengths have zero path difference, so the central fringe is bright and white.

在中央处，所有波长的光程差均为零，因此中央亮纹为白色。

Away from the centre, the bright fringes for different wavelengths occur at slightly different positions. Since red light has a longer wavelength, its fringes are more widely spaced than violet fringes.

在远离中心处，不同波长的亮纹位置略有不同。由于红光波长较长，其条纹间距比紫光更大。

As a result, the first-order and higher-order fringes appear coloured: blue/violet is closer to the centre, while red is further outward.

因此，第一级及更高级次的条纹呈现彩色：蓝紫色靠近中央，红色位于外侧。

The white-light pattern is not as sharp as monochromatic interference because each colour has a different fringe spacing, so the maxima wash out at larger angles.

白光干涉图样不如单色光清晰，因为每种颜色的条纹间距不同，在较大角度处极大值会变得模糊。

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8. Measuring Wavelength Using Double Slits | 用双缝实验测量波长

The double-slit experiment can be used to measure the wavelength of monochromatic light using the equation λ = aΔx / L.

利用公式 λ = aΔx / L，可以用双缝实验测量单色光的波长。

In practice, the slit separation a is measured using a travelling microscope. The distance L between the slits and the screen is measured with a metre ruler.

实际测量中，用读数显微镜测量双缝间距 a，用米尺测量双缝到屏幕的距离 L。

To reduce errors, do not measure one fringe spacing directly. Measure the distance across several bright fringes and divide by the number of intervals.

为减小误差，不要直接测量一条条纹间距，而应测量多条亮纹间的总距离，再除以间隔数量。

For example, measure the distance between the central bright fringe and the 5th bright fringe, then divide by 5 to obtain Δx.

例如，测量中央亮纹到第 5 级亮纹的距离，再除以 5 得到 Δx。

Repeat the measurement for different orders and use the average value. This reduces both random errors and the effect of zero-error in the ruler.

对不同级次重复实验并取平均值，这样可以减小随机误差，并且补偿刻度尺的零点误差。

Using a laser is advantageous because it is highly monochromatic and produces a bright, clear pattern that is easy to measure.

使用激光器的优点是单色性好、亮度高、条纹清晰，便于准确测量。

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9. Common Exam Points and Pitfalls | 常见考点与易错点

• Bright fringe condition: path difference = nλ. Make sure you can state this in words and symbols.
• 暗纹条件：光程差 = (n + 1/2)λ。需要同时会用文字和符号表述。

• Dark fringe condition: path difference = (n + 1/2)λ.
• 亮纹条件：光程差 = nλ。

• Fringe spacing formula: Δx = λL / a. Be careful with unit conversions, especially nm to m.
• 条纹间距公式：Δx = λL / a。注意单位换算，尤其是纳米与米之间的转换。

• Coherent sources must have constant phase difference and same frequency. Laser light is coherent; ordinary light is not.
• 相干光源必须具有恒定相位差和相同频率。激光是相干的，普通光不是。

• Do not confuse the interference fringe spacing with the width of the central diffraction maximum caused by a single slit.
• 不要把双缝干涉的条纹间距与单缝衍射中央亮纹宽度混淆。

• For white light, remember the colour order near the centre: blue/violet is inner, red is outer.
• 对于白光，记住中央附近的颜色顺序：蓝紫色在内，红色在外。

• If the screen is moved farther away, the pattern expands but the fringes become dimmer because the same energy spreads over a larger area.
• 屏幕移远时，图样会变得稀疏，但由于能量分布到更大区域，条纹会变暗。

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10. Worked Example | 典型例题解析

A student uses a double-slit setup with slit separation a = 0.50 mm and screen distance L = 1.20 m. The distance between the central bright fringe and the 5th bright fringe is measured as 6.0 mm. Calculate the wavelength of the light.

某同学使用双缝实验装置，双缝间距 a = 0.50 mm，缝屏距离 L = 1.20 m。测得中央亮纹到第 5 级亮纹的距离为 6.0 mm。求该光的波长。

Step 1: Determine the fringe spacing. The 5th bright fringe is 5 intervals from the centre.

第一步：求条纹间距。第 5 级亮纹与中央亮纹之间相隔 5 个间距。

Δx = 6.0 mm / 5 = 1.2 mm = 1.2 × 10⁻³ m

Step 2: Rearrange the fringe spacing formula to find λ.

第二步：改写条纹间距公式求 λ。

λ = aΔx / L

Step 3: Substitute values with all quantities in SI units.

第三步：将所有物理量换算为国际单位并代入。

λ = (0.50 × 10⁻³ × 1.2 × 10⁻³) / 1.20 = 5.0 × 10⁻⁷ m = 500 nm

The wavelength is therefore 500 nm, which corresponds to green light.

因此该光的波长为 500 nm，对应绿光。

Notice that converting millimetres to metres is essential. If you forget the 10⁻³ factors, the answer will be wrong by a factor of 10⁶.

注意必须将毫米换算为米。如果遗漏 10⁻³ 因子，答案将相差 10⁶ 倍。

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