📚 Key Formula Derivations in IAL Physics Unit 2: Insights from the Jan 2021 Examiner's Report | IAL物理Unit 2关键公式推导：2021年1月考官报告解析
The January 2021 examiner's report for International A-Level Physics Unit 2 highlighted that many candidates struggled to secure full marks on questions requiring step-by-step formula derivation or a deep understanding of underlying assumptions. Common pitfalls included poor handling of small-angle approximations, unit inconsistencies, and confusion between standing wave harmonic patterns for different boundary conditions. This article revisits the essential derivations, clarifies the physics behind them, and draws directly on the examiners' observations to help you avoid the most frequent mistakes.
2021年1月国际A-Level物理Unit 2考官报告指出，许多学生在需要分步推导公式或深刻理解潜在假设的题目上难以拿到满分。常见错误包括对小角度近似处理不当、单位不一致，以及混淆不同边界条件下的驻波谐波模式。本文重新梳理关键推导，阐明其背后的物理意义，并直接引用考官的观察，帮助你避开最常见的失分点。
1. Young's Double-Slit Fringe Spacing | 杨氏双缝条纹间距推导
oximation sinθ ≈ tanθ = y / D (with D the slit-to-screen distance), we obtain Δ ≈ a y / D.
推导的核心在于将两缝光到达屏幕某点的光程差与几何关系联系起来。对于距离中央极大竖直距离为 y 的点 P，光程差 Δ = S₂P - S₁P ≈ a sinθ，其中 a 为缝间距。利用小角度近似 sinθ ≈ tanθ = y / D（D 为缝屏间距），得到 Δ ≈ a y / D。
n (distance between adjacent bright fringes) is then Δy = yₙ₊₁ - yₙ = λD / a. The examiner' e approximation explicitly or by mixing up y (distance from centre) and Δy (fringe width).
相长干涉条件为 Δ = nλ，故 a y / D = nλ。因此第 n 级亮纹的位置为 yₙ = nλD / a。相邻亮纹间距则为 Δy = yₙ₊₁ - yₙ = λD / a。考官报告指出，很多考生因为未明确写出小角度近似，或混淆了 y（距中心的距离）与 Δy（条纹宽度）而丢分。
Δy = λD / a
2. Diffraction Grating Equation | 衍射光栅方程推导
ath difference is an integer multiple of the wavelength: d sinθ = nλ, where n = 0, ±1, ±2, … This relation governs the angular positions of bright fringes.
对于线间距为 d 的衍射光栅，相邻缝发出的波之间的光程差为 d sinθ。当该光程差等于波长的整数倍时，出现相长干涉（主极大）：d sinθ = nλ，其中 n = 0, ±1, ±2, … 该关系决定了亮纹的角位置。
The examiner' metres, leading to gross errors. Always express d in metres (d = 1/N, where N is lines/m).
考官报告强调，学生常将光栅当作双缝处理，忽视了光栅的极大非常尖锐，且小角度近似 sinθ ≈ θ（弧度）仅当 θ 很小时才可使用。此外，许多人未能将线间距从每毫米线数换算为米，导致严重错误。始终以米为单位表达 d（d = 1/N，N 为每米线数）。
d sinθ = nλ
3. Standing Waves on a Stretched String | 张紧弦上的驻波频率推导
n its length L equals an integer number of half-wavelengths: L = n(λ/2), with n = 1, 2, 3, … g the wave speed equation v = fλ, we obtain the resonant frequencies f = v/λ = n v / (2L).
两端固定的张紧弦，当其长度 L 等于半波长的整数倍时，可形成驻波：L = n(λ/2)，n = 1, 2, 3, … 故 λ = 2L/n。利用波速公式 v = fλ，得到共振频率 f = v/λ = n v / (2L)。
al frequency (first harmonic) is f₁ = v/(2L). The n-th harmonic is fₙ = n f₁. The examiner' la for a string by mistake. Remember: string fixed at both ends → all harmonics (n = 1,2,3…) are possible.
基频（一次谐波）为 f₁ = v/(2L)。第 n 次谐波为 fₙ = n f₁。考官报告显示，许多考生错误识别了图中所显示的波腹数量，计数有误，或者错误地将管中的公式用于弦。请记住：两端固定的弦 → 所有谐波（n = 1,2,3…）都可能存在。
fₙ = n v / (2L) where n = 1, 2, 3, …
4. Standing Waves in Pipes | 管中的驻波频率推导
leading to the same harmonic sequence as a string: L = n(λ/2), so fₙ = n v/(2L), n = 1,2,3… s: L = (2n-1)λ/4, so λ = 4L/(2n-1) and fₙ = (2n-1) v/(4L). Only odd harmonics are present.
对于两端开口的管，两端均为波腹，其谐波序列与弦相同：L = n(λ/2)，故 fₙ = n v/(2L)，n = 1,2,3… 对于一端封闭的管，闭端为波节，开口端为波腹。长度 L 对应于奇数倍的四分之一波长：L = (2n-1)λ/4，因此 λ = 4L/(2n-1)，fₙ = (2n-1) v/(4L)。只有奇次谐波存在。
A common mistake flagged in the examiner' dium (e.g., using the speed of sound in air but forgetting the temperature effect when given). Always use the v specified or recall v ≈ 340 m s⁻¹ for sound in air at room temperature.
考官报告指出的一个常见错误是混淆开管与闭管的边界条件，尤其是当题目给出驻波示意图时。学生常将弦公式套用到闭管上，忽略了 n 必须为奇数。另一个易错点是使用了未根据介质调整的波速 v（例如，使用空气中的声速但忘记题目给出的温度效应）。始终使用题目指定的 v，或记住室温下空气中声速约为 340 m s⁻¹。
Open pipe: fₙ = n v/(2L) ; Closed pipe: fₙ = (2n-1) v/(4L)
5. Snell's Law from Huygens' Principle | 由惠更斯原理推导斯涅尔定律
Huygens' ce v₂t in medium 2, while the wavelet from B travels v₁t in medium 1. From the geometry, sinθ₁ = (v₁t) / AB and sinθ₂ = (v₂t) / AB. Dividing gives sinθ₁ / sinθ₂ = v₁ / v₂.
惠更斯原理指出，波前上的每一点都可视为子波的波源。当平面波前以一定角度遇到界面时，第二种介质中的子波以不同速度传播。考虑入射到界面的波前 AB。在时间 t 内，来自 A 的子波在介质 2 中传播距离 v₂t，而来自 B 的子波在介质 1 中传播 v₁t。由几何关系，sinθ₁ = (v₁t) / AB，sinθ₂ = (v₂t) / AB。两式相除得 sinθ₁ / sinθ₂ = v₁ / v₂。
m, we can write v₁/v₂ = n₂/n₁. Substituting this gives n₁ sinθ₁ = n₂ sinθ₂, which is Snell's law. The examiner' to the ratio of refractive indices, or they confused which index belonged to which medium.
由于折射率 n = c / v（c 为真空中光速），可写为 v₁/v₂ = n₂/n₁。代入后得到 n₁ sinθ₁ = n₂ sinθ₂，即斯涅尔定律。考官报告注意到，当要求概述此推导时，考生常遗漏将速度比与折射率比联系起来的关键步骤，或混淆了哪个折射率对应哪个介质。
n₁ sinθ₁ = n₂ sinθ₂
6. Resistivity and Resistance | 电阻率与电阻公式推导
y proportional to its length L and inversely proportional to its cross-sectional area A: R ∝ mental to understanding how material property, geometry and temperature affect resistance.
均匀导线的电阻 R 与其长度 L 成正比，与其横截面积 A 成反比：R ∝ L / A。引入比例常数 ρ（电阻率），得到 R = ρL / A。这一关系对于理解材料属性、几何形状和温度如何影响电阻至关重要。
lve the diameter to get the radius or using the diameter directly in A = πr². The examiner' nd gave ρ in Ω cm rather than Ω m. Always keep units consistent: L in metres, A in m², R in ohms gives ρ in Ω m.
在2021年1月的考试中，许多学生由直径计算截面积时出错，忘记将直径除以二得到半径，或在 A = πr² 中直接使用直径。考官报告强调面积必须以 m² 为单位。此外，在变形公式求电阻率时，考生有时混淆单位，给出的 ρ 单位为 Ω cm 而非 Ω m。请始终保持单位一致：L 以米计，A 以 m² 计，R 以欧姆计，得到 ρ 的单位为 Ω m。
R = ρL / A
7. Series Resistors and the Potential Divider | 串联电阻与分压器公式推导
For resistors in series, the same current I flows through each. Ohm's law gives V₁ = I R₁, V₂ = I R₂, … and the total p.d. V = I(R₁ + R₂ + …). Hence the total resistance Rtotal = R₁ + R₂ + … io of voltages is V₁ / V₂ = R₁ / R₂. This is the foundation of the potential divider rule.
对于串联电阻，流过每个电阻的电流 I 相同。由欧姆定律得 V₁ = I R₁，V₂ = I R₂，… 总电压 V = I(R₁ + R₂ + …)。因此总电阻 Rtotal = R₁ + R₂ + … 。更重要的是，电压之比为 V₁ / V₂ = R₁ / R₂。这是分压器规则的基础。
The potential divider equation for two resistors R₁ and R₂ in series across a supply voltage Vin is Vout = Vin × R₂/(R₁ + R₂), where Vout is measured across R₂. A frequent error, noted by examiners, was placing R₁ in the numerator, leading to the wrong output voltage. Another mistake was assuming the current through the divider remained constant when a load was added; actually the effective resistance changes. Always check whether the circuit is unloaded or loaded.
对于两个电阻 R₁ 和 R₂ 串联后连接至电源电压 Vin，分压器公式为 Vout = Vin × R₂/(R₁ + R₂)，其中 Vout 是跨接在 R₂ 两端的电压。考官们注意到一个常见错误是将 R₁ 放在了分子上，导致输出电压算反。另一个错误是当接入负载后仍假设流过原分压器的电流不变；实际上有效电阻会改变。务必检查电路是空载还是带载。
Vout = Vin × (R₂ / (R₁ + R₂))
8. EMF and Internal Resistance | 电动势和内阻的推导与图像
tion shows that plotting V against I yields a straight line with gradient -r and y-intercept ε.
对于具有电动势 ε 的实际电源，当有电流 I 流出时，由于内阻 r 的存在，路端电压 V 会小于 ε。由能量守恒得 ε = I(R + r)，其中 R 为外接负载。变形得：V = ε - I r。这一线性关系表明，以 V 对 I 作图得到一条直线，斜率为 -r，y 轴截距为 ε。
e gradient becomes -1/r and the intercept ε/r. Be very careful with graph axes. Another classic error is using the nominal emf of a cell as exactly the y-intercept when the line does not clearly reach I = 0; interpolation may be needed.
考官报告指出，许多考生难以从 V-I 图中正确提取 r，常常忽略负号而直接将斜率当作 r。他们也会混淆截距，特别是当图像以 I-V（电流在 y 轴）的形式呈现时。此时斜率变为 -1/r，截距为 ε/r。务必留意坐标轴变量。另一个经典错误是当直线并未清晰延伸到 I = 0 时，仍直接把标称电动势当作 y 截距；可能需要进行外推。
V = ε - I r
9. Photoelectric Effect Equation and Graph Analysis | 光电效应方程与图像分析
The energy of a photon is E = hf, where h is Planck' oton strikes a metal surface, the maximum kinetic energy of the emitted photoelectron is Ek,max is the work function of the metal. This can also be written using the stopping potential Vs: e Vs = hf - φ, so Vs = (h/e) f - (φ/e).
光子能量为 E = hf，其中 h 为普朗克常数。当光子撞击金属表面时，逸出光电子的最大动能为 Ek,max = hf - φ，φ 为金属的逸出功。也可用遏止电势 Vs 表达：e Vs = hf - φ，即 Vs = (h/e) f - (φ/e)。
The graph of Vs φ/h. The Jan 2021 report noted that many candidates could not link the gradient to Planck's constant correctly, especially when the graph' efore calculating the gradient and use e = 1.60 × 10⁻¹⁹ C. Another pitfall was confusing Ek,max with the energy of the photon itself; make it clear that not all photon energy goes into kinetic energy — some is used to overcome the work function.
Vs 对 f 作图是一条直线，斜率为 h/e，x 截距为阈频率 f₀ = φ/h。2021年1月的报告指出，许多考生无法正确将斜率与普朗克常数关联，尤其当图像的单位非标准时（如频率以 10¹⁴ Hz 为单位）。在计算斜率前务必转换为 Hz，并使用 e = 1.60 × 10⁻¹⁹ C。另一个易错点是混淆 Ek,max 与光子本身的能量；须说明并非全部光子能量都转化为动能——一部分用于克服逸出功。
e Vs = hf - φ
10. Common Pitfalls and Unit Conversion Mastery | 常见失误与单位换算精通
Across all derivations, the examiner's report repeatedly highlighted the importance of consistent SI units. For instance, using millimetres for slit separation or centimetres for screen distance in the Young's slits formula will yield a fringe spacing off by factors of 10³ or 10². Always convert lengths to metres and frequencies to hertz before substituting into equations. Keep powers of ten clear and use standard form to avoid arithmetic slips.
在所有推导中，考官报告反复强调了统一的国际单位制的重要性。例如，在杨氏双缝公式中使用毫米表示缝距或厘米表示屏距，会使条纹间距计算结果产生 10³ 或 10² 倍的偏差。代公式前务必将长度换算为米、频率换算为赫兹。清晰使用幂指数和标准形式以避免运算失误。
Other generic mistakes included incomplete derivations— ). Students also lost marks for not clearly stating assumptions such as the small-angle approximation or the medium being uniform. When a question asks "derive" or "show that", each logical step must be presented. Simply writing the known result is insufficient.
其他普遍性错误包括推导不完整——直接跳到最终公式而未写出条件（如相长干涉：Δ = nλ）。学生也因未明确陈述假设（如小角度近似或介质均匀）而丢分。当题目要求“推导”或“证明”时，必须呈现每一个逻辑步骤。只写下已知结果是不够的。
The examiners also pointed out that candidates often confused the symbols for slit separation (a) and grating spacing (d), or wavelength (λ) and distance (D). Using a clear, labelled diagram in your answer can help demonstrate your understanding and avoid symbol mix-ups. Remember: a well-drawn sketch with annotations can earn marks even if the algebra goes slightly awry.
考官还指出，考生常混淆缝距 (a) 与光栅间距 (d) 的符号，或波长 (λ) 与距离 (D) 的符号。在答案中使用清晰标注的示意图有助于展示理解并避免符号混淆。请记住：一个带有注释的清晰草图即使代数推导略有偏差也能赚取分数。
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