Mock Exam Analysis: AQA Physics Unit 2 – Waves and Optics | AQA 物理单元测试模拟卷解析：波与光学

📚 Mock Exam Analysis: AQA Physics Unit 2 – Waves and Optics | AQA 物理单元测试模拟卷解析：波与光学

This article provides a comprehensive breakdown of a full mock exam for the AQA Pre-U Physics unit on waves and optics. For each question type, we present model solutions, highlight common mistakes, and reinforce the key concepts you must master for top marks.

本文全面解析一份AQA大学预科物理单元模拟卷（涵盖波与光学），针对每类题型给出标准解答、指出常见错误并强化核心概念，助你稳拿高分。

1. Progressive Waves and the Wave Equation | 行波与波动方程

Question 1: A sound wave of frequency 512 Hz travels through air at 340 m s⁻¹. Calculate its wavelength.

题目1：一频率为512 Hz的声波在空气中以340 m s⁻¹传播，求其波长。

Solution: Use the wave equation v = f λ. Rearranging gives λ = v / f = 340 / 512 = 0.664 m (or 66.4 cm).

解答：运用波动方程 v = f λ，变形得 λ = v / f = 340 / 512 = 0.664 m（即 66.4 cm）。

Many students mix up frequency and wavelength or forget that the speed of sound in air is given and need not be calculated. Always write the equation first before substituting numbers.

许多学生混淆频率与波长，或者忘记给出的是空气中的声速无需另行计算。务必先列出方程再代入数值。

Question 2: An oscilloscope displays a waveform with time-base 2 ms/cm; two complete cycles occupy 5 cm. Determine the frequency.

题目2：示波器显示波形，时基为 2 ms/cm，两个完整周期占据5 cm，求频率。

Solution: Period T = (total distance / number of cycles) × time-base = (5 cm / 2) × 2 ms/cm = 5 ms = 0.005 s. Hence f = 1/T = 200 Hz.

解答：周期 T =（总距离 / 周期数）× 时基 = (5 cm / 2) × 2 ms/cm = 5 ms = 0.005 s，故 f = 1/T = 200 Hz。

Ensure the time-base unit is consistent: convert ms to s before calculating frequency. Some students forget to divide the total length by the number of cycles first.

注意时基单位统一：计算频率前将 ms 转换为 s。部分学生忘记先将总长度除以周期个数。

---

2. Phase Difference and Path Difference | 相位差与波程差

Question: Two points on a progressive wave are 1.2 m apart. The wavelength is 3.0 m. Calculate the phase difference (a) in radians, (b) in degrees.

题目：行波上两点相距1.2 m，波长为3.0 m，求相位差（a）以弧度表示，（b）以度数表示。

Solution: Phase difference Δφ = (2π / λ) × path difference. Δφ = (2π / 3.0) × 1.2 = 0.8π rad ≈ 2.51 rad. In degrees: 0.8π × (180°/π) = 144°.

解答：相位差 Δφ = (2π / λ) × 波程差。Δφ = (2π / 3.0) × 1.2 = 0.8π rad ≈ 2.51 rad。转为度数：0.8π × (180°/π) = 144°。

Do not forget to use the correct relation: 2π radians = 360°. Many errors arise from mixing path difference with separation between two sources, which is a different concept.

切勿忘记2π弧度 = 360°。很多错误源于将波程差与两波源间距混淆，这是不同概念。

A quick check: if path difference = λ, phase difference = 2π rad; if ½λ, phase difference = π rad. This helps verify your answer.

快速验证：若波程差 = λ，相位差 = 2π rad；若½λ，相位差 = π rad，可由此检验答案。

---

3. Superposition and Interference Conditions | 叠加与干涉条件

Concept check: State the necessary conditions for observable interference fringes from two coherent sources.

概念检查：说明两相干光源产生可观察干涉条纹的必要条件。

Model answer: The sources must be coherent (same frequency, constant phase difference), have a similar amplitude, and emit waves of the same type. For light, this usually requires a single source split into two, as in Young’s double-slit.

标准答案：光源必须相干（同频率、恒定相位差）、振幅相近、发射同类型波。对于光，这通常需将单个光源分为两束，如杨氏双缝实验。

A typical exam trap: stating “same frequency and same phase” instead of “constant phase difference”. They need not be in phase; the key is a fixed relationship.

典型考试陷阱：表述为“同频率同相位”而非“恒定相位差”。它们不必同相；关键是相位关系恒定。

Path difference for constructive interference: nλ (n = 0, 1, 2…); for destructive: (n+½)λ. Always link to phase difference: constructive = 0, 2π, 4π…; destructive = π, 3π…

相长干涉波程差：nλ（n=0,1,2…）；相消干涉：(n+½)λ。务必与相位差关联：相长对应0、2π、4π…；相消对应π、3π…

---

4. Young’s Double-Slit Experiment | 杨氏双缝实验

Question: In a Young’s double-slit arrangement, slit separation a = 0.50 mm, screen distance D = 1.5 m. The fringe width Δx is measured as 1.8 mm. Calculate the wavelength λ of the light used.

题目：杨氏双缝实验中，缝间距 a = 0.50 mm，屏距 D = 1.5 m，条纹间距 Δx 测得1.8 mm，求所用光的波长 λ。

Solution: λ = (a Δx) / D. Convert all to metres: a = 5.0 × 10⁻⁴ m, Δx = 1.8 × 10⁻³ m. Then λ = (5.0×10⁻⁴ × 1.8×10⁻³) / 1.5 = 6.0 × 10⁻⁷ m = 600 nm.

解答：λ = (a Δx) / D。全部化为米：a = 5.0 × 10⁻⁴ m，Δx = 1.8 × 10⁻³ m，得 λ = (5.0×10⁻⁴ × 1.8×10⁻³) / 1.5 = 6.0 × 10⁻⁷ m = 600 nm。

Always convert to base SI units. Many candidates lose marks by using mm directly in the formula without scaling.

务必转换为基本国际单位。很多考生直接以 mm 代入公式而未转换比例，导致失分。

The fringe width Δx is the distance between adjacent bright (or dark) fringes. It is constant only for small angles; the approximation sinθ ≈ tanθ ≈ θ holds well for typical set-ups.

条纹间距 Δx 是相邻亮纹（或暗纹）中心间的距离。它仅在小角度下恒定；常规装置中 sinθ ≈ tanθ ≈ θ 近似成立。

---

5. Diffraction Grating | 衍射光栅

Question: A diffraction grating has 500 lines per mm. Light of wavelength 630 nm is incident normally. Calculate the angle θ for the first-order maximum (n=1).

题目：衍射光栅每毫米500条刻线，波长630 nm的光垂直入射，求第一级明纹（n=1）的角位置 θ。

Solution: Grating spacing d = 1 / (500 × 10³ lines/m) = 2.0 × 10⁻⁶ m. Grating equation: d sinθ = nλ. For n=1: sinθ = λ / d = (6.30×10⁻⁷) / (2.0×10⁻⁶) = 0.315. Hence θ = arcsin(0.315) ≈ 18.4°.

解答：光栅常数 d = 1 / (500 × 10³ 条/米) = 2.0 × 10⁻⁶ m。光栅方程 d sinθ = nλ。代入 n=1：sinθ = λ / d = (6.30×10⁻⁷) / (2.0×10⁻⁶) = 0.315，故 θ = arcsin(0.315) ≈ 18.4°。

It is crucial to first find d correctly. If given lines per mm, convert to lines per metre by multiplying by 10³, then take the reciprocal.

准确求得 d 是关键。若给出每毫米线数，乘以10³换算为每米线数，再取倒数。

Note: The grating produces sharp maxima at angles satisfying the equation. Between maxima, the intensity is negligible. This is why grating spectra are used for precise wavelength measurement.

注意：光栅在满足方程的角度产生尖锐极大值，极大值之间光强可忽略，因此光栅光谱常用于精确测波长。

---

6. Stationary Waves on Strings | 弦上的驻波

Question: A stretched string of length 0.80 m is fixed at both ends. It vibrates in its third harmonic (n=3). Sketch the standing wave pattern and state the number of nodes and antinodes.

题目：一根长0.80 m的弦两端固定，以第三谐频（n=3）振动，画出驻波图案并指出波节和波腹的数目。

Solution: For the third harmonic, there are 3 half-wavelengths fitting in the length: L = 3λ/2 → λ = 2L/3 = 0.533 m. The pattern has 4 nodes (including both ends) and 3 antinodes.

解答：第三谐频下，3个半波长恰好等于弦长：L = 3λ/2 → λ = 2L/3 = 0.533 m。图案中共有4个波节（含两端）和3个波腹。

A common mistake is to count the ends as one node each but then miscount the antinodes. Remember: for the nth harmonic, there are n antinodes and n+1 nodes.

常见错误是将两端各计为一个波节却数错波腹数量。记住：第 n 谐频含有 n 个波腹和 n+1 个波节。

The frequency is given by f = n(v/2L) where v is wave speed. If fundamental f₀ = v/2L, then f3 = 3f₀. This relation is often tested.

频率公式为 f = n(v/2L)，其中 v 为波速。若基频 f₀ = v/2L，则 f₃ = 3f₀，此关系常被考查。

---

7. Refraction and Snell’s Law | 折射与斯涅尔定律

Question: Light travels from air (n₁ = 1.00) into glass (n₂ = 1.50). The angle of incidence is 35°. Calculate the angle of refraction inside the glass.

题目：光从空气（n₁ = 1.00）射入玻璃（n₂ = 1.50），入射角为35°，求玻璃中的折射角。

Solution: Snell’s law: n₁ sinθ₁ = n₂ sinθ₂. Thus sinθ₂ = (1.00 × sin35°) / 1.50 ≈ 0.574 / 1.50 = 0.3827. θ₂ = arcsin(0.3827) ≈ 22.5°.

解答：斯涅尔定律：n₁ sinθ₁ = n₂ sinθ₂。于是 sinθ₂ = (1.00 × sin35°) / 1.50 ≈ 0.574 / 1.50 = 0.3827，θ₂ = arcsin(0.3827) ≈ 22.5°。

Always check that the refracted angle is smaller when light enters a denser medium, and larger when emerging. If your result contradicts this, you may have inverted the indices.

务必检查：光进入光密介质时折射角应较小，离开时较大。若结果与此矛盾，可能将折射率倒置了。

In many problems, you might need to find the critical angle when going from glass to air: sin c = n₂/n₁ = 1.00/1.50 → c ≈ 41.8°. Total internal reflection occurs for angles greater than c.

许多问题中需计算从玻璃到空气的临界角：sin c = n₂/n₁ = 1.00/1.50 → c ≈ 41.8°。入射角大于 c 即发生全内反射。

---

8. Total Internal Reflection and Optical Fibres | 全内反射与光纤

Question: An optical fibre has a core of refractive index 1.62 and a cladding of index 1.52. Calculate the critical angle at the core-cladding interface and explain why this is necessary for signal transmission.

题目：某光纤纤芯折射率为1.62，包层折射率为1.52，计算芯-包层界面的临界角，并解释为何这对信号传输是必需的。

Solution: sin c = n_cladding / n_core = 1.52 / 1.62 ≈ 0.9383. c = arcsin(0.9383) ≈ 69.8°. This critical angle ensures that light rays hitting the interface at angles greater than 69.8° undergo total internal reflection, guiding the light along the fibre without loss.

解答：sin c = n包层 / n纤芯 = 1.52 / 1.62 ≈ 0.9383，c = arcsin(0.9383) ≈ 69.8°。该临界角保证入射角大于69.8°的光线发生全内反射，使光无损地沿光纤传导。

Students often misapply the formula by swapping indices. Remember: critical angle exists only when light goes from higher to lower refractive index. The cladding has a lower index to enable TIR at the boundary.

学生常因调换折射率而误用公式。记住：临界角仅存在于光从光密至光疏介质的情形。包层折射率较低正是为了使界面发生全内反射。

The acceptance cone of a fibre is related to the critical angle. A larger core-cladding index difference gives a wider acceptance angle, making coupling of light easier.

光纤的接收角与临界角相关。纤芯与包层折射率差越大，接收角越宽，更易耦合光线。

---

9. Polarisation of Light | 光的偏振

Question: Unpolarised light of intensity I₀ passes through a polariser. The transmitted light then encounters a second polariser (analyser) whose transmission axis is at 60° to the first. Calculate the final intensity.

题目：强度为 I₀ 的非偏振光通过一个起偏器，而后透射光遇到第二个偏振片（检偏器），其透光轴与第一个成60°夹角，求最终光强。

Solution: After the first polariser, intensity becomes I₁ = I₀/2 (since unpolarised light halves). Applying Malus’s law: I₂ = I₁ cos²θ = (I₀/2) × cos²60° = (I₀/2) × (0.5)² = I₀ × 0.125 = 0.125 I₀.

解答：经第一个偏振片后，强度变为 I₁ = I₀/2（非偏振光减半）。应用马吕斯定律：I₂ = I₁ cos²θ = (I₀/2) × cos²60° = (I₀/2) × (0.5)² = 0.125 I₀。

Do not forget the initial halving step for unpolarised light. If the first filter was already placed and the incident light is already polarised, just use Malus’s law directly.

别忘了非偏振光初始减半的步骤。若入射光已是偏振光，则直接使用马吕斯定律即可。

Polarisation is a key evidence for the transverse nature of light waves, since longitudinal waves cannot be polarised.

偏振是证明光波为横波的关键证据，因为纵波无法被偏振。

---

10. Summary of Key Formulas and Common Mistakes | 关键公式与常见错误总结

Key equations you must recall:

必备公式：

v = fλ

λ = a Δx / D

d sinθ = nλ

n₁ sinθ₁ = n₂ sinθ₂

sin c = n₂ / n₁

I = I₀ cos²θ

Consistent unit conversion is vital: convert mm to m, ms to s, and lines per mm to lines per m before substitution. Losing a factor of 1000 is the single most common mistake in the exam.

统一的单位换算至关重要：代入前务必将 mm 化为 m，ms 化为 s，每 mm 线数化为每 m 线数。丢失1000因子是考试中最常见的错误。

For interference and diffraction, clearly distinguish between path difference and phase difference. Constructive interference: path diff = nλ, phase diff = 2nπ. Destructive: path diff = (n+½)λ, phase diff = (2n+1)π.

干涉与衍射问题中，清晰区分波程差与相位差。相长干涉：波程差 = nλ，相位差 = 2nπ；相消干涉：波程差 = (n+½)λ，相位差 = (2n+1)π。

When dealing with stationary waves, always relate the number of antinodes to the harmonic number. Check that the string boundary conditions match the diagram you draw.

处理驻波时，始终将波腹数与谐频序数关联。检查所画图示是否符合弦的边界条件。

Keep a formula sheet with the conditions for each formula (e.g., n₁ > n₂ for critical angle, normal incidence for grating equation). Practise with past papers under timed conditions to internalise these applications.

准备一张公式表，注明各式的适用条件（如临界角要求 n₁ > n₂，光栅方程需垂直入射）。通过限时刷真题内化这些应用。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）