📚 IGCSE WJEC Physics: Interference of Light Exam Essentials | IGCSE WJEC 物理：光的干涉 考点精讲
Interference of light is one of the most compelling pieces of evidence for the wave nature of light. In the IGCSE WJEC Physics syllabus, you are expected to explain the conditions required for interference, describe Young’s double-slit experiment, apply the fringe-spacing formula, and understand how a diffraction grating improves measurements. This article covers all the key concepts, equations, and exam tips you need, with clear explanations in both English and Chinese.
光的干涉是证明光具有波动性的最有力证据之一。在 IGCSE WJEC 物理考试大纲中，你需要解释干涉产生的条件，描述杨氏双缝实验，应用条纹间距公式，并理解衍射光栅如何提高测量精度。本文将用中英双语系统地讲解所有核心概念、公式和应试技巧，助你轻松掌握这一考点。
1. What is Interference? | 什么是干涉？
Interference is the phenomenon that occurs when two or more coherent waves overlap in space. The resultant displacement at any point is the vector sum of the displacements due to the individual waves. Where the waves meet in phase, they reinforce each other, producing a larger amplitude – this is called constructive interference. Where they meet out of phase, they cancel each other out – this is destructive interference. The fact that light can produce interference fringes is a key proof that light behaves as a wave.
干涉是指两个或多个相干波在空间中相遇叠加时所产生的现象。某一点的合位移是各波单独引起位移的矢量和。当波同相相遇时，它们互相增强，产生更大的振幅，这称为相长干涉。当它们反相相遇时，则会互相抵消，这称为相消干涉。光能够产生干涉条纹这一事实，是光具有波动性的重要证据。
2. Conditions for Interference | 干涉产生的条件
To obtain a stable and observable interference pattern, the overlapping light waves must be coherent. Two sources are coherent if they emit waves with the same frequency and a constant phase difference. Ordinary light sources such as a filament lamp emit light in short, random bursts, so they are incoherent. To achieve coherence in the laboratory, we often use a single light source and split its wavefront – for example, by passing the light through a single slit before it reaches a double slit, as in Young’s experiment. The waves must also have the same polarisation, though this is less commonly tested at IGCSE.
要获得稳定且可观察的干涉图样，叠加的光波必须是相干的。如果两个波源发出频率相同且相位差恒定的波，则它们是相干的。普通光源（如白炽灯）发出的光波是短暂且随机的，因而是非相干的。在实验室中实现相干，通常使用单一光源并分割其波前——例如，让光先通过一个单缝再照射到双缝上，这就是杨氏实验的做法。此外，两列波还必须有相同的偏振方向，不过这一点在 IGCSE 阶段较少考查。
3. Young’s Double-Slit Experiment | 杨氏双缝实验
Thomas Young’s famous double-slit experiment, first performed in 1801, provided clear evidence for the wave theory of light. The apparatus consists of a monochromatic light source illuminating a single narrow slit, which acts as a point source of coherent light. The light waves then pass through two closely spaced parallel slits (the double slit) and spread out due to diffraction. Where the two diffracted wavefronts overlap on a screen placed at a distance D away, they interfere, producing a pattern of equally spaced bright and dark fringes.
托马斯·杨在1801年首次完成了著名的双缝实验，为光的波动说提供了明确证据。实验装置使用单色光源照亮一个狭窄的单缝，单缝作为相干点光源。光波随后通过两个相距很近的平行狭缝（双缝），并因衍射而散开。在距离双缝 D 处放置的屏幕上，两个衍射波前重叠并发生干涉，产生一系列等间距的明暗条纹。
4. Path Difference and Phase Difference | 路径差与相位差
Whether interference at a point on the screen is constructive or destructive depends on the path difference between the two waves arriving at that point. The path difference is the extra distance one wave travels compared to the other. If the path difference is a whole number of wavelengths, the waves arrive in phase and constructive interference occurs. If the path difference is an odd number of half-wavelengths, they arrive out of phase and destructive interference occurs. The relationship between path difference p.d. and phase difference Δφ is: Δφ = (2π / λ) × p.d.
屏幕上某点发生的是相长干涉还是相消干涉，取决于到达该点的两列波之间的路程差。路程差指一列波比另一列波多走的距离。如果路程差是波长的整数倍，则两波同相到达，产生相长干涉。如果路程差是半波长的奇数倍，则两波反相到达，产生相消干涉。路程差 p.d. 与相位差 Δφ 之间的关系为：Δφ = (2π / λ) × p.d.
5. Constructive and Destructive Interference Conditions | 相长干涉与相消干涉的条件
For Young’s double slits with slit separation a, consider a point on the screen at an angle θ from the centre. The extra path length from the upper slit is approximately a sin θ. Therefore, the condition for a bright fringe is:
a sin θ = n λ
where n = 0, 1, 2, … is the order number.
对于缝间距为 a 的杨氏双缝，考虑屏幕上与中心夹角为 θ 的一点。从上方狭缝来的额外路径长度近似为 a sin θ。因此，亮纹的条件是：
a sin θ = n λ
其中 n = 0, 1, 2, … 是条纹级数。
The condition for a dark fringe is that the path difference equals an odd number of half-wavelengths:
a sin θ = (n + ½) λ
where n = 0, 1, 2, …
暗纹产生的条件是路程差等于半波长的奇数倍：
a sin θ = (n + ½) λ
其中 n = 0, 1, 2, …
These equations link the geometry of the setup to the wavelength of the light, allowing us to determine λ if we can measure a, θ, and the order n.
这些方程把实验装置的几何参数与光波长联系起来，只要能测量出 a、θ 和级数 n，就可以计算波长 λ。
6. Fringe Pattern and the Spacing Formula | 干涉条纹图样与间距公式
The interference fringes on the screen are equally spaced and parallel to the slits. The central bright fringe (n = 0) is the brightest, and the intensity decreases for higher orders. When the screen is far away compared with the slit separation (D ≫ a), the angle θ is small, so we can use the approximations sin θ ≈ tan θ ≈ θ (in radians). In that case, the fringe separation x – the distance between the centres of two adjacent bright (or dark) fringes – is given by the simple formula:
x = λ D / a
屏幕上的干涉条纹是等间距的，且平行于双缝。中央亮纹（n = 0）最亮，级数越高强度越低。当屏幕到双缝的距离远大于缝间距（D ≫ a）时，角度 θ 很小，我们可以使用近似 sin θ ≈ tan θ ≈ θ（以弧度计）。此时，条纹间距 x（相邻两条亮纹或暗纹中心之间的距离）可用以下简单公式计算：
x = λ D / a
Rearranging, we obtain the version often used to calculate wavelength: λ = a x / D. It is essential that all quantities are in the same units, normally metres.
将此式变形即可得到常用于计算波长的形式：λ = a x / D。必须注意所有物理量使用相同的单位，通常统一用米。
7. Measuring Wavelength Using Young’s Slits | 利用杨氏双缝测波长
To find the wavelength of monochromatic light, you need to measure the slit separation a, the distance D from the slits to the screen, and the fringe separation x. In practice, x is very small (often a fraction of a millimetre), so it is better to measure the distance across several fringes and divide by the number of fringes to get an average value. A travelling microscope or a metre rule with a vernier scale can be used. Care must be taken to minimise errors: ensure the screen is perpendicular to the light path, measure D with a metre rule to the nearest millimetre, and avoid parallax errors when reading x.
要测量单色光的波长，你需要测出缝间距 a、双缝到屏幕的距离 D 以及条纹间距 x。实际操作中 x 通常很小（往往不到一毫米），因此最好测量多个条纹的总宽度再除以条纹数，以获得平均值。可以使用移测显微镜或带游标的米尺进行测量。必须注意减少误差：确保屏幕垂直于光路，用来测量 D 的米尺要精确到毫米，并在读取 x 时避免视差。
Common sources of error include uncertainty in a (often measured with a micrometer) and difficulty in accurately locating the centre of a fringe. Repeating measurements and using a dark room improve reliability.
常见的误差来源包括 a 的不确定度（通常用千分尺测量）以及准确找出条纹中心的困难。重复测量并在暗室中进行实验可以提高结果的可靠性。
8. White Light Interference | 白光的干涉
If the monochromatic source is replaced by white light, a striking pattern appears. White light contains all visible wavelengths, and each wavelength produces its own set of fringes with a slightly different spacing because λ is different. The central fringe (n = 0) is white because all wavelengths arrive in phase at the centre. On either side, however, the bright fringes become coloured, with violet (shorter λ) on the inner edge and red (longer λ) on the outer edge. Higher-order fringes overlap so much that they merge into a uniform white illumination, so only the first one or two orders are clearly seen as spectra.
如果把单色光源换成白光，就会出现引人注目的图样。白光包含所有可见光波长，由于各波长的 λ 不同，每种波长都会形成间距略有差异的条纹。中央条纹（n = 0）是白色的，因为所有波长的光在中央都同相到达。但在中央两侧，亮纹变成彩色，内侧为紫光（波长较短），外侧为红光（波长较长）。高级数条纹因重叠严重而混合成均匀的白光，因此只能清晰地看到前一两级的彩色光谱。
This dispersion of white light by interference reinforces the wave model and is a popular exam topic. Expect questions that ask you to describe the appearance and explain why the central fringe is white while the others are coloured.
白光通过干涉产生的这种色散现象进一步支持了波动模型，也是考试的热点。要准备回答描述条纹外观并解释为什么中央条纹是白色而其他条纹是彩色的问题。
9. Diffraction Grating: An Interference Device | 衍射光栅：干涉器件
A diffraction grating consists of a large number of equally spaced parallel slits. When monochromatic light passes through or reflects from a grating, interference of waves from many slits produces very sharp and bright maxima at specific angles. The grating equation is:
d sin θ = n λ
where d is the grating spacing (the distance between adjacent slits), θ is the angle of the nth-order maximum, and n is the order number.
衍射光栅由大量等间距的平行狭缝组成。当单色光透过光栅或从其表面反射时，来自众多狭缝的波发生干涉，在特定角度上产生非常锐利且明亮的极大。光栅方程为：
d sin θ = n λ
其中 d 是光栅常数（相邻狭缝间的距离），θ 是第 n 级极大对应的角度，n 是级数。
If you know the number of lines per millimetre N, then d = 1 / N (in millimetres, which must be converted to metres). The grating produces maxima that are much sharper than the fringes from a double slit, so the wavelength can be measured more precisely. In the exam you may be asked to compare the two setups or to calculate λ, d, or θ using the grating equation.
如果已知每毫米的刻线数 N，则 d = 1 / N（单位从毫米换算成米）。光栅产生的亮纹比双缝干涉条纹锐利得多，因此可以更精确地测量波长。考试中可能会要求比较这两种装置，或者使用光栅方程计算 λ、d 或 θ。
10. Exam Tips and Common Pitfalls | 考试技巧与常见错误
When answering questions on light interference, always state explicitly that the sources must be coherent. Use the phrase ‘same frequency and constant phase difference’. Never confuse double-slit separation a with fringe spacing x or screen distance D. Rearrange the formula correctly: many students lose marks by mixing up λ = a x / D. Check that you have converted all lengths into metres before substituting into equations.
在回答关于光的干涉问题时，一定要明确说明光源必须是相干的，并使用“频率相同且相位差恒定”这样的表述。千万不要把双缝间距 a、条纹间距 x 和屏幕距离 D 混淆。正确变换公式：许多学生由于搞混 λ = a x / D 而丢分。代入公式前，要核查是否已将所有长度都换算为米。
For the diffraction grating, remember that d = 1/N and that N is often given in lines per mm, so d must be calculated in metres. If the angle θ is very small, you may use the approximation sin θ ≈ θ, but only if the question explicitly states it or if you are asked to compare small-angle and exact results. Also, pay attention to significant figures when recording data and final answers.
对于衍射光栅，要记住 d = 1/N，而 N 经常以每毫米线数给出，因此 d 需要计算并换算成米。如果角度 θ 很小，你可以使用近似 sin θ ≈ θ，但只有在题目明确说明或要求比较小角度结果和精确结果时才能使用。此外，记录数据和写出答案时要注意有效数字。
Finally, in describe-type questions, use scientific terminology: ‘superposition’, ‘path difference’, ‘constructive/destructive interference’, and link the observation directly to the wave model of light. A well-structured answer with a clear diagram reference always scores higher.
最后，在需要描述的问题中，要使用科学术语：“叠加”、“路程差”、“相长/相消干涉”，并将观察到的现象直接与光的波动模型联系起来。答题结构清晰，并配合适当的图示说明，往往能获得更高分数。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply