📚 IB Physics: Single-Slit Diffraction Patterns and Fringe Rules | IB物理：单缝衍射图样与条纹规律
When monochromatic light passes through a narrow slit, it spreads out and produces a characteristic pattern of bright and dark bands on a screen. This phenomenon, called single-slit diffraction, is a core topic in IB Physics because it directly demonstrates the wave nature of light.
当单色光通过一条狭窄的狭缝时，光会向外扩展，并在屏幕上产生明暗相间的特征图样。这种现象称为单缝衍射，是 IB 物理的核心内容，因为它直接体现了光的波动性。
1. What Is Single-Slit Diffraction? | 什么是单缝衍射？
Diffraction is the bending and spreading of waves when they pass through an opening or around an obstacle. In single-slit diffraction, the opening is much wider than the wavelength of light, but still narrow enough (typically a fraction of a millimetre) that noticeable spreading occurs.
衍射是波通过开口或绕过障碍物时发生的弯曲和扩展现象。在单缝衍射中，开口比光的波长大得多，但仍然足够窄（通常为零点几毫米量级），从而产生明显的扩展。
The resulting pattern is not simply a sharp image of the slit. Instead, it consists of a broad, intense central maximum flanked by narrower, dimmer secondary maxima, separated by dark minima.
产生的图样并不是狭缝的清晰影像，而是由一个宽阔而明亮的中央极大以及两侧较窄、较暗的次极大组成，它们之间由暗纹（极小值）隔开。
2. Huygens' Principle and Wavefront Construction | 惠更斯原理与波前构建
Single-slit diffraction can be explained using Huygens' principle: every point on a wavefront acts as a source of secondary spherical wavelets. These wavelets propagate in all directions and interfere with one another.
单缝衍射可以用惠更斯原理来解释：波前上的每一点都可以看作一个发出次级球面波子波的波源。这些子波向各个方向传播，并彼此干涉。
For a slit of width a, each point across the slit produces a wavelet. At a distant point on the screen, these wavelets arrive with different phases because they have travelled different distances. The superposition of all these wavelets determines the observed intensity.
对于宽度为 a 的狭缝，缝上每一点都会产生子波。在屏幕上较远的一点处，这些子波因传播距离不同而具有不同的相位。所有这些子波的叠加决定了观察到的强度。
3. Experimental Setup for Observing Single-Slit Diffraction | 观察单缝衍射的实验装置
A typical arrangement uses a laser or a monochromatic light source, a single narrow slit, and a screen placed at a large distance D from the slit. The slit width a is typically between 0.01 mm and 0.5 mm.
典型的实验装置包括激光或单色光源、一条狭窄单缝，以及距离狭缝较远（距离为 D）的屏幕。缝宽 a 通常在 0.01 mm 到 0.5 mm 之间。
The light source must be coherent and monochromatic for a clear pattern.
光源必须具有相干性和单色性，才能获得清晰的图样。
The slit must be straight-edged and uniform in width.
狭缝的边缘必须平直，且宽度均匀。
The screen distance D should be large so that the small-angle approximation is valid.
屏幕距离 D 应足够大，以便使用小角度近似。
4. General Shape of the Diffraction Pattern | 衍射图样的总体形状
The first key feature of the single-slit diffraction pattern is its central maximum: a bright, wide band at the centre of the screen. Its angular width is roughly twice that of the secondary maxima.
单缝衍射图样的第一个关键特征是中央极大：屏幕上中央的一条明亮宽条纹。其角宽度大约是次极大角宽度的两倍。
On both sides of the central maximum, there are alternating dark and bright bands. The bright bands (secondary maxima) are much dimmer than the central maximum, typically less than 5% of its peak intensity.
在中央极大两侧，明暗条纹交替排列。次级极大远暗于中央极大，其峰值强度通常不到中央极大的 5%。
The pattern is symmetric about the centre, because the slit geometry is symmetric. The intensity falls off rapidly with increasing distance from the centre.
由于狭缝的几何结构具有对称性，图样关于中心对称。随着离中心距离增大，强度迅速下降。
5. Condition for Dark Fringes (Minima) | 暗纹（极小值）条件
For destructive interference to occur at a point on the screen, the waves from different parts of the slit must cancel each other out. The condition for the n-th dark fringe is:
为了在屏幕上某一点发生相消干涉，来自狭缝不同部分的光波必须相互抵消。第 n 级暗纹的条件是：
a sin θ = n λ, n = 1, 2, 3, ...
where a is the slit width, θ is the angle from the centre of the slit to the point on the screen, λ is the wavelength, and n is the order of the minimum.
其中 a 是缝宽，θ 是从狭缝中心到屏幕上某点的角度，λ 是波长，n 是暗纹的级数。
n, not a dark fringe. Therefore the first dark fringe on either side corresponds to n = 1:
注意 n = 0 会给出 θ = 0，这是图样中心，而不是暗纹。因此两侧第一暗纹对应 n = 1：
sin θ₁ = λ / a
6. Positions of Bright Fringes | 明纹位置
Bright fringes in single-slit diffraction occur approximately halfway between adjacent dark fringes. This gives the approximate condition for the n-th bright fringe (secondary maximum):
单缝衍射中的明纹近似出现在相邻暗纹的中间位置。由此可得第 n 级明纹（次极大）的近似条件：
a sin θ ≈ (n + ½) λ, n = 1, 2, 3, ...
ximum is not included in this formula because it is the main bright band centred at θ = 0.
其中 n = 1 对应第一级次极大，n = 2 对应第二级次极大，以此类推。中央极大不包括在该公式中，因为它是以 θ = 0 为中心的中央亮带。
It is important to remember that this is an approximation. The exact maxima of the single-slit diffraction pattern occur at slightly different angles, but IB Physics normally accepts the halfway approximation.
需要记住这是一个近似结果。单缝衍射图样的实际极大值出现在略微不同的角度，但 IB 物理通常接受这种“中间位置”近似。
7. Width of the Central Maximum | 中央极大宽度
rst dark minimum on the other side. Using the small-angle approximation sin θ ≈ θ ≈ tan θ:
中央极大从一侧第一暗纹延伸到另一侧第一暗纹。利用小角度近似 sin θ ≈ θ ≈ tan θ：
θ₁ ≈ λ / a
Therefore the full angular width of the central maximum is:
因此中央极大的总角宽度为：
Δθ = 2λ / a
If the screen is at distance D from the slit, the linear width of the central maximum is:
如果屏幕与狭缝距离为 D，则中央极大的线宽度为：
Δy = 2D tan θ₁ ≈ 2D λ / a
8. Effect of Slit Width on the Pattern | 缝宽对图样的影响
θ₁ = λ/a increases, so the pattern becomes wider and the central maximum becomes broader.
缝宽 a 直接控制衍射图样的扩展程度。如果 a 减小，θ₁ = λ/a 增大，因此图样变宽，中央极大变得更加宽阔。
Conversely, if the slit is made wider, the diffraction pattern narrows; when a becomes much larger than λ, the spreading becomes negligible and the light behaves almost geometrically.
相反，如果狭缝加宽，衍射图样会收窄；当 a 远大于 λ 时，扩展变得可以忽略，光的行为几乎接近几何光学。
r slit → wider and dimmer diffraction pattern; wider slit → narrower and brighter pattern.
一个重要规律：缝越窄，衍射图样越宽且越暗；缝越宽，图样越窄且越亮。
9. Effect of Wavelength on the Pattern | 波长对图样的影响
he same slit width. Thus red light (λ ≈ 700 nm) spreads more than blue light (λ ≈ 450 nm).
由于暗纹条件为 a sin θ = n λ，在同一缝宽下，波长越长，衍射角越大。因此红光（λ ≈ 700 nm）比蓝光（λ ≈ 450 nm）扩展得更明显。
If white light is used, each wavelength produces its own diffraction pattern. The central maximum appears white because all wavelengths overlap at the centre, but the edges of the central maximum and the secondary maxima show coloured fringes, with blue closer to the centre and red further out.
如果使用白光，每个波长都会产生各自的衍射图样。中央极大因所有波长在中心重叠而呈现白色，但中央极大边缘和次级极大会出现彩色条纹：蓝色靠近中心，红色远离中心。
10. Intensity Distribution and Relative Brightness | 强度分布与相对亮度
The intensity of a single-slit diffraction pattern is not uniform. The central maximum is very intense, while the secondary maxima are much weaker and decrease in intensity as their distance from the centre increases.
单缝衍射图样的强度并不均匀。中央极大非常明亮，而次级极大暗得多，并且随着离中心距离增加，强度进一步降低。
The principal maxima intensities, relative to the central maximum, are approximately:
各级极大相对于中央极大的强度约为：
| Order | Relative Intensity | 级数 | 相对强度 |
| Central maximum | 1.00 | 中央极大 | 1.00 |
| First secondary maximum | ≈ 0.047 | 第一级次极大 | ≈ 0.047 |
| Second secondary maximum | ≈ 0.017 | 第二级次极大 | ≈ 0.017 |
| Third secondary maximum | ≈ 0.008 | 第三级次极大 | ≈ 0.008 |
11. Single-Slit Diffraction vs Double-Slit Interference | 单缝衍射与双缝干涉的比较
A common source of confusion is the difference between single-slit diffraction and Young's double-slit interference. They are both wave phenomena, but their patterns differ in key ways.
一个常见的混淆点是比较单缝衍射和杨氏双缝干涉。两者都是波动现象，但图样有明显差异。
| Feature | Single-Slit Diffraction | Double-Slit Interference |
| Bright fringe condition | a sin θ ≈ (n + ½) λ (secondary maxima) | d sin θ = n λ |
| Dark fringe condition | a sin θ = n λ | d sin θ = (n + ½) λ |
| Central maximum | Very broad and bright (twice the width of secondary maxima) | Similar width and brightness to other fringes |
| Fringe spacing | Not uniform; central maximum is widest | Uniform (for small angles) |
| Intensity variation | Fast fall-off in secondary maxima | Fringes nearly equally bright (ideal case) |
In real double-slit experiments, each slit itself produces diffraction, so the overall pattern is a double-slit interference pattern modulated by a single-slit diffraction envelope.
在真实双缝实验中，每条缝本身也会产生衍射，因此整体图样是双缝干涉图样被单缝衍射包络调制的结果。
12. Worked Example and Exam Tips | 例题与考试技巧
Worked Example: A monochromatic light of wavelength 600 nm passes through a slit of width 0.10 mm. A screen is placed 2.0 m away. Calculate the width of the central maximum.
例题：波长为 600 nm 的单色光通过宽度为 0.10 mm 的狭缝，屏幕放置在 2.0 m 远处。求中央极大的宽度。
Solution: First, find the angle of the first dark minimum:
解答：首先求第一级暗纹的角度：
sin θ₁ = λ / a = (600 × 10⁻⁹) / (1.0 × 10⁻⁴) = 6.0 × 10⁻³
Since θ₁ is small, y₁ ≈ D θ₁ = 2.0 × 6.0 × 10⁻³ = 0.012 m = 12 mm.
因为 θ₁ 很小，y₁ ≈ D θ₁ = 2.0 × 6.0 × 10⁻³ = 0.012 m = 12 mm。
The central maximum extends from y = -12 mm to y = +12 mm, so its full width is 24 mm.
中央极大从 y = -12 mm 延伸到 y = +12 mm，因此总宽度为 24 mm。
Exam tips:
考试技巧：
the dark fringe condition as a sin θ = n λ, and clearly list what each symbol represents.
始终写明暗纹条件 a sin θ = n λ，并清楚地解释每个符号的含义。
Remember that the central maximum is between the first minima on either side, so its width is 2Dλ/a, not Dλ/a.
记住中央极大位于两侧第一暗纹之间，因此其宽度为 2Dλ/a，而不是 Dλ/a。
Check units: slit width and wavelength must both be in metres.
检查单位：缝宽和波长都必须换算成米。
For small angles, use sin θ ≈ tan θ ≈ θ (in radians), but only when θ is small.
在小角度情况下，可使用 sin θ ≈ tan θ ≈ θ（以弧度为单位），但仅当 θ 很小时才可如此近似。
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