Light Diffraction in A-Level CIE Physics | A-Level CIE 物理：光的衍射考点精讲

📚 Light Diffraction in A-Level CIE Physics | A-Level CIE 物理：光的衍射考点精讲

Diffraction is a fundamental wave phenomenon that explains how light spreads when it encounters an obstacle or a narrow opening. In A-Level CIE Physics, mastering the principles of diffraction is essential for solving problems involving single slits, diffraction gratings, and optical resolution. This article breaks down the core concepts, derivations, and exam techniques you need to excel in the topic.

衍射是波动的基本现象，它解释了光遇到障碍物或狭缝时如何发生弯曲扩展。在 A-Level CIE 物理中，掌握衍射原理对于解决涉及单缝、衍射光栅和光学分辨率的题目至关重要。本文拆解核心概念、推导过程和应试技巧，助你轻松攻克这一考点。

1. What is Diffraction? | 什么是衍射？

Diffraction is the spreading of waves as they pass through a gap or around an obstacle. It is a property common to all types of waves, including light, sound, and water waves. The amount of diffraction depends on the relative size of the wavelength compared to the aperture or obstacle.

衍射是指波在通过缝隙或绕过障碍物时发生的扩展现象。它是所有波动（包括光波、声波和水波）的共性。衍射的显著程度取决于波长与缝隙或障碍物尺寸的相对大小。

For light, diffraction becomes noticeable only when the size of the opening or obstacle is comparable to the wavelength of light (around 10⁻⁷ m). In everyday life, we do not observe light diffracting around large objects because the wavelength is far too small relative to the object size.

对于光波，只有当缝隙或障碍物的尺寸与光波长（约 10⁻⁷ m）相当时，衍射才会变得明显。日常生活中我们看不到光绕过大物体衍射，正是因为波长相对物体尺寸来说实在太小。

2. Conditions for Observable Diffraction | 明显衍射的条件

Significant diffraction occurs when the size of the aperture or obstacle, a, is roughly equal to or smaller than the wavelength, λ. The degree of spreading increases as the ratio λ/a increases. If a is much larger than λ, diffraction effects are negligible and light travels in straight lines, consistent with geometric optics.

当缝隙或障碍物的尺寸 a 与波长 λ 近似相等或更小时，衍射现象显著。扩展程度随 λ/a 增大而增大。若 a 远大于 λ，衍射效应可忽略，光沿直线传播，这与几何光学一致。

In a single-slit experiment, a narrow slit (a ~ 0.1 mm) illuminated by a laser (λ ~ 600 nm) produces a broad diffraction pattern because a ≫ λ is not satisfied; the slit width must be decreased further to see pronounced fringes.

在单缝实验中，用激光（λ ~ 600 nm）照射一窄缝（a ~ 0.1 mm）会产生较宽的衍射图样，因为此时 a 并未远大于 λ；要看到更明显的条纹，需进一步减小缝宽。

3. Single-Slit Diffraction | 单缝衍射

When monochromatic light passes through a single narrow slit, the wavefronts emerging from different points across the slit interfere. This produces a central bright maximum flanked by a series of dark and bright fringes on a screen. The central maximum is twice as wide as the subsidiary maxima and is much brighter.

当单色光通过一个单狭缝时，从缝上不同点发出的子波相互干涉。在屏幕上形成中央亮纹，两侧分布着一系列暗纹和亮纹。中央明纹的宽度是次级明纹的两倍，且亮度最高。

The intensity of secondary maxima drops rapidly: the first side maximum has roughly 4.7% of the central peak intensity, the next even less. This distinctive intensity distribution is a key feature of single-slit Fraunhofer diffraction.

次级明纹的强度迅速下降：第一侧明纹强度约为中央主极大的 4.7%，后续的更弱。这种独特的强度分布是单缝夫琅禾费衍射的关键特征。

4. Intensity Distribution in Single-Slit Pattern | 单缝衍射图样的强度分布

The intensity I at an angle θ from the centre is given by:

I = I₀ [sin(β) / β]² where β = (π a sinθ) / λ

在偏离中心 θ 角处的强度 I 由下式给出：

I = I₀ [sin(β) / β]²，其中 β = (π a sinθ) / λ

Here I₀ is the intensity at the centre, a is the slit width, and λ is the wavelength. Minima (dark fringes) occur when sin(β)=0 but β≠0, leading to the condition a sinθ = nλ, with n = ±1, ±2, ±3,…

式中 I₀ 是中心的强度，a 是缝宽，λ 是波长。当 sin(β)=0 且 β≠0 时出现暗纹，由此导出条件 a sinθ = nλ，n = ±1, ±2, ±3,…

The central maximum lies between n = -1 and n = +1 minima, giving an angular width of approximately 2λ/a. Subsidiary maxima occur roughly when a sinθ = (m+½)λ, where m = ±1, ±2,…

中央明纹位于 n = -1 和 n = +1 暗纹之间，角宽度约为 2λ/a。次级明纹大致出现在 a sinθ = (m+½)λ 处，m = ±1, ±2,…。

5. Derivation of Single-Slit Minima Condition | 单缝衍射极小条件推导

Divide the slit into two equal halves. For the first minimum (n=1), rays from one half must be exactly out of phase by π with rays from the corresponding points in the other half. This happens when the path difference between a ray from the top of the slit and one from the centre is (λ/2) sinθ = λ/2, giving a sinθ = λ.

将狭缝分成相等的两半。对于第一级暗纹（n=1），来自上半部分的射线与来自下半部分对应点的射线必须相位差 π。当从缝顶和缝中心出发的两条射线之间的光程差为 (a/2) sinθ = λ/2 时满足条件，从而得出 a sinθ = λ。

Generalising by dividing the slit into 2n equal parts yields a sinθ = nλ for the nth minimum. This simple derivation is often examined in A-Level CIE Physics, requiring students to explain the principle using phasor or wavefront division.

将缝等分为 2n 份，推广可得第 n 级暗纹条件 a sinθ = nλ。这一简洁推导常出现在 CIE 物理考试中，要求考生用旋转矢量或波前分割原理解释。

6. Diffraction Grating | 衍射光栅

A diffraction grating consists of a large number of equally spaced, parallel slits (or grooves) ruled on a glass or metal surface. The grating constant d (or spacing) is the distance between adjacent slits. Typically, gratings have several hundred to thousands of lines per millimetre.

衍射光栅由大量等间距、平行的刻线（狭缝或沟槽）组成，刻在玻璃或金属表面。光栅常量 d（光栅间距）是相邻狭缝间的距离。常见光栅每毫米有数百至数千条刻线。

When light passes through or reflects from a grating, each slit acts as a coherent source. The multiple beams interfere to produce very sharp, well-defined maxima (principal maxima) at certain angles, separated by wide dark regions.

当光通过光栅或从光栅反射时，每条狭缝都充当一个相干光源。多光束干涉会在特定角度产生非常锐利、清晰的主极大，主极大之间被宽阔的暗区隔开。

7. The Grating Equation | 光栅方程

The angular positions of the principal maxima are given by the grating equation:

d sinθ = nλ, n = 0, ±1, ±2, ±3, …

主极大的角位置由光栅方程给出：

d sinθ = nλ，n = 0, ±1, ±2, ±3, …

Here d is the slit spacing (grating constant), θ is the angle of diffraction measured from the normal to the grating, λ is the wavelength, and n is the order number. The central maximum (n=0) is the zeroth order; the others are first order, second order, etc.

式中 d 是狭缝间距（光栅常量），θ 是从光栅法线量起的衍射角，λ 是波长，n 是级次。中央主极大（n=0）是零级；其余分别为一级、二级等。

It is vital to remember that the equation only gives the angles of the principal maxima. Minima and secondary maxima occur between them, but their positions are not required for standard CIE calculations.

务必记住，该方程只给出主极大的角度。极小和次极大出现在它们之间，但在标准 CIE 计算中不要求确定其位置。

8. Spectra and Orders | 光谱和级次

When white light is used with a diffraction grating, each wavelength component is diffracted at a slightly different angle according to d sinθ = nλ. This separates the light into a spectrum. In each order (except n=0), a continuous rainbow spectrum is observed, with violet closest to the centre and red furthest away.

当白光照射衍射光栅时，根据 d sinθ = nλ，各个波长成分以略微不同的角度衍射，从而将光分成光谱。在每一级（n=0 除外），都能观察到连续的彩虹光谱，紫光靠近中心，红光最远。

Higher orders can overlap: for example, the blue end of the 3rd-order spectrum may overlap with the red end of the 2nd-order spectrum. The maximum possible order n_max is limited by sinθ ≤ 1, so n_max ≤ d/λ.

高级次光谱可能重叠：例如，三级光谱的蓝端可能与二级光谱的红端重叠。最大可能级次 n_max 受 sinθ ≤ 1 限制，因此 n_max ≤ d/λ。

9. Resolvance of a Grating (Rayleigh Criterion) | 光栅的分辨率（瑞利判据）

The ability of a grating to separate two closely spaced wavelengths is described by its resolving power R:

R = λ / Δλ = Nn

这里 Δλ 为恰好能分辨的最小波长差，N 为光栅被照明的刻线总数，n 为级次。

This formula, derived from the Rayleigh criterion, states that two wavelengths are just resolved when the principal maximum of one coincides with the first minimum of the other. Greater N and higher order n yield better resolution.

该公式由瑞利判据导出：当一波长的主极大恰好落在另一波长的第一极小上时，两波长恰能被分辨。N 越大、级次 n 越高，分辨率越高。

In practice, a grating with 500 lines/mm illuminated over 2 cm width gives N = 10,000 lines, offering very high resolution suitable for precision spectroscopy.

实际应用中，每毫米 500 线的光栅在 2 cm 宽照明下 N = 10000 条刻线，可提供极高分辨率，适用于精密光谱学。

10. Applications of Diffraction Gratings | 衍射光栅的应用

Diffraction gratings are widely used in spectrometers to analyse the spectral composition of light from stars, flames, and discharge tubes. By measuring the angles θ for each wavelength, the line spectrum of an element can be determined, allowing identification of chemical species.

衍射光栅广泛应用于光谱仪，用于分析恒星、火焰和放电管发出的光的频谱成分。通过测量每个波长的衍射角 θ，可以确定元素的线状光谱，从而鉴定化学物质。

They also feature in laser tuning, optical communication (wavelength division multiplexing) and structural analysis of materials using X-ray diffraction, where the crystal lattice acts as a three-dimensional grating.

光栅还用于激光调谐、光通信（波分复用）以及利用 X 射线衍射分析材料结构，此时晶格充当三维光栅。

11. Comparison: Single Slit vs. Double Slit vs. Grating | 比较：单缝、双缝与光栅

Feature	Single Slit	Double Slit	Diffraction Grating
Fringe pattern	Broad central max, weak secondary maxima	Equally spaced, similar intensity fringes	Very sharp principal maxima, wide dark gaps
Intensity	Decreases rapidly from centre	Nearly uniform (modulated by single-slit envelope)	Very bright peaks, almost zero elsewhere
Key equation	a sinθ = nλ (min)	d sinθ = nλ (max)	d sinθ = nλ (max)

In double-slit interference, the observed pattern is actually a combination of double-slit interference and single-slit diffraction, leading to missing orders when an interference maximum coincides with a diffraction minimum.

在双缝干涉中，观察到的图样实际上是双缝干涉与单缝衍射的结合，当干涉极大与衍射极小重合时会出现缺级现象。

12. Common Misconceptions & Exam Tips | 常见误区与应试技巧

Misconception 1: ‘Diffraction always makes fringes.’ | 误区 1：“衍射总会产生条纹。”

Diffraction itself is simply the spreading of waves. Fringes appear only when waves from different sources or different parts of the same wavefront overlap and interfere. A single slit produces a diffraction pattern that is an interference pattern arising from the superposition of wavelets across the slit.

衍射本身只是波的扩展。只有不同波源或同一波前不同部分发出的波重叠并干涉时，才会出现条纹。单缝产生的衍射图样其实是来自缝上各处子波叠加的干涉图样。

Misconception 2: ‘The slit width is not important for a grating.’ | 误区 2：“缝宽对光栅不重要。”

While the grating equation uses the spacing d, the individual slit width a affects the intensity envelope via single-slit diffraction. This envelope can suppress certain orders, causing them to be missing.

尽管光栅方程使用间距 d，单个狭缝宽度 a 会通过单缝衍射影响强度包络。该包络可能抑制某些级次，导致缺级。

Exam tips: Always convert distance units to metres before calculations. For grating problems you are often given lines per millimetre — find d by taking the reciprocal and converting: d (m) = 1 / (lines per m). Check that sinθ does not exceed 1; if the calculated sinθ > 1, that order does not exist. When comparing two wavelengths, use the grating equation to find the angular separation or use resolving power to check if they can be distinguished.

应试技巧：计算前务必将距离单位转换为米。光栅题目常给出每毫米线数——取倒数后转换成米得到 d（m）= 1 /（每米线数）。检查 sinθ 是否超过 1；若计算出的 sinθ > 1，则该级次不存在。比较两个波长时，使用光栅方程求角分离，或利用分辨本领判断它们是否能被区分。

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Light Diffraction in A-Level CIE Physics | A-Level CIE 物理：光的衍射 考点精讲