📚 A-Level Physics Mark Scheme Unit 2 Jan 21: Formula Derivations | A-Level 物理评分方案 Unit 2 2021年1月：公式推导
This article focuses on the key formula derivations assessed in a typical A-Level Physics Unit 2 January 2021 mark scheme, especially for topics like waves, electricity, and quantum physics. Understanding these derivations not only helps you answer structured questions but also deepens your grasp of the underlying physical principles. We will go through step‑by‑step reasoning for each major equation, linking them directly to mark scheme expectations.
本文聚焦于典型的 A‑Level 物理 Unit 2 2021 年 1 月评分方案中所考查的关键公式推导，特别是波、电学与量子物理等主题。理解这些推导不仅有助于回答结构化问题，还能加深对基本物理原理的掌握。我们将逐一梳理每个重要公式的推导步骤，并将它们直接对应到评分方案的要求。
1. Deriving the Double‑Slit Fringe Spacing Formula | 双缝干涉条纹间距公式推导
In Young’s double‑slit experiment, the path difference between waves from the two slits arriving at a point on a distant screen is approximately s sin θ, where s is the slit separation and θ is the angle from the central axis. For bright fringes, constructive interference requires a path difference of nλ, so s sin θ = nλ. For small angles, sin θ ≈ tan θ = y/D, where y is the fringe displacement and D is the slit‑screen distance. Hence, s(y/D) = λ for the first‑order fringe (n=1), giving the fringe spacing Δy = λD / s between adjacent bright fringes.
在杨氏双缝实验中，从两条缝到达远处屏幕上某一点的波程差近似为 s sin θ，其中 s 是缝间距，θ 是与中心轴的夹角。为了形成亮条纹，需要波程差等于整数倍波长，即 s sin θ = nλ。当角度很小时，sin θ ≈ tan θ = y/D，其中 y 是条纹位移，D 是缝到屏幕的距离。因此对于第一级明纹 (n＝1) 有 s(y/D) = λ，由此可得相邻亮条纹的间距 Δy = λD / s。
Δy = λD / s
2. Deriving the Diffraction Grating Equation | 衍射光栅方程推导
For a transmission diffraction grating with line spacing d, the path difference between waves from adjacent slits is d sin θ. Constructive interference occurs when this equals nλ. This gives the grating equation d sin θ = nλ, where n is the order number. The derivation from the mark scheme often requires stating that each slit acts as a coherent source and that maxima are observed when the waves are in phase. No small‑angle approximation is used here because θ can be large.
对于刻线间距为 d 的透射式衍射光栅，相邻狭缝发出的波的波程差为 d sin θ。当该波程差等于 nλ 时发生相长干涉，这就得到光栅方程 d sin θ = nλ，其中 n 为级数。评分方案常要求说明每条缝都可视为相干光源，且当波同相时观察到极大值。这里不采用小角度近似，因为 θ 可能很大。
d sin θ = nλ
3. Deriving Einstein’s Photoelectric Equation | 爱因斯坦光电效应方程推导
Einstein proposed that light consists of photons, each with energy E = hf. When a photon strikes a metal surface, its energy is used to overcome the work function φ (the minimum energy to release an electron) and to provide the electron with kinetic energy. Conservation of energy gives hf = φ + ½mv²ₘₐₓ. Many mark schemes award marks for clearly defining φ and stating that the maximum kinetic energy is independent of intensity. This derivation is fundamental to explaining the photoelectric effect.
爱因斯坦提出光由光子组成，每个光子的能量为 E = hf。当光子撞击金属表面时，其能量一部分用于克服逸出功 φ（释放电子所需的最小能量），剩余部分转化为电子的动能。由能量守恒得到 hf = φ + ½mv²ₘₐₓ。很多评分方案会针对明确给出 φ 的定义并指出最大动能与光强无关而给分。这一推导是解释光电效应的基础。
hf = φ + ½mv²ₘₐₓ
4. Deriving Stopping Potential and Maximum Kinetic Energy | 推导截止电压与最大动能
When a reverse potential Vₛ is applied to a photoelectric cell, the kinetic energy of the fastest electrons is converted into electrical potential energy: ½mv²ₘₐₓ = e Vₛ. Combining this with Einstein’s equation yields e Vₛ = hf – φ. This is often plotted as Vₛ against f; the gradient gives h/e and the intercept on the f‑axis gives the threshold frequency f₀ = φ/h. In a Jan 21 mark scheme, such a graphical derivation may be asked for explicitly.
当对光电管施加反向电压 Vₛ 时，最快电子的动能转化为电势能：½mv²ₘₐₓ = e Vₛ。将其与爱因斯坦方程结合可得 e Vₛ = hf – φ。通常把 Vₛ 对 f 作图，直线的斜率为 h/e，在频率轴上的截距给出阈频率 f₀ = φ/h。在 2021 年 1 月的评分方案中，可能会明确要求进行这种图像推导。
e Vₛ = hf – φ
5. Deriving the Resistivity Equation | 电阻率公式推导
For a uniform wire of length L and cross‑sectional area A, the resistance R is found to be directly proportional to L and inversely proportional to A. Introducing the resistivity ρ as the constant of proportionality gives R = ρL / A. Mark schemes often require a justification using the concept that a longer wire provides more obstacles to electron flow, while a larger area allows more parallel paths. Derivation may be extended by combining with Ohm’s law V = IR to give V = I(ρL/A).
对于长度为 L、横截面积为 A 的均匀导线，电阻 R 与 L 成正比，与 A 成反比。引入比例常数电阻率 ρ，即得 R = ρL / A。评分方案常要求说明：更长的导线对电子流动产生更多阻碍，而更大的横截面积则提供了更多的并联路径。推导还可结合欧姆定律 V = IR 得到 V = I(ρL/A)。
R = ρL / A
6. Deriving Internal Resistance and Terminal p.d. | 内阻与端电压推导
A real power source can be modelled as an ideal emf ε in series with an internal resistance r. When a current I flows, the terminal potential difference V is less than ε because of the voltage drop across r. From energy conservation (or Kirchhoff’s second law): ε = V + Ir, thus V = ε – Ir. The mark scheme frequently tests the derivation using the equation of a straight line: V = –r I + ε, so a graph of V against I gives a slope of –r and an intercept of ε.
实际电源可以等效为一个理想电动势 ε 与内阻 r 串联。当有电流 I 流过时，由于在 r 上产生压降，端电压 V 小于 ε。根据能量守恒（或基尔霍夫第二定律）：ε = V + Ir，因此 V = ε – Ir。评分方案经常考查利用直线方程进行推导：V = –r I + ε，因此 V‑I 图的斜率为 –r，纵截距为 ε。
V = ε – Ir
7. Deriving Ohm’s Law for a Conductor | 导体欧姆定律推导
For an ohmic conductor at constant temperature, the current I is directly proportional to the potential difference V. Starting from the definition of resistance R = V/I, it follows that V = IR. A deeper derivation at the microscopic level uses the drift velocity v of electrons: I = nAve, where n is the charge carrier density and e is the elementary charge. Combined with V = E L and the electric field E, and using an expression for resistivity, one can show that V ∝ I. Mark schemes may ask for the link between micro‑ and macroscopic quantities.
对于恒温下的欧姆导体，电流 I 与电势差 V 成正比。从电阻的定义 R = V/I 出发，可得 V = IR。在微观层面更深层的推导则利用电子的飘移速度 v：I = nAve，其中 n 为载流子密度，e 为元电荷。结合 V = E L 与电场 E，并利用电阻率表达式，能够证明 V ∝ I。评分方案可能会要求将微观量与宏观量联系起来。
V = IR and I = nAve
8. Deriving Conditions for Standing Waves on a String | 弦上驻波形成条件推导
A standing wave forms when two progressive waves of the same frequency and amplitude travel in opposite directions. For a string fixed at both ends, boundary conditions require nodes at the ends. The simplest standing wave (fundamental) has a length L = λ/2. In general, L = nλ/2, where n = 1,2,3… . Using v = fλ, the resonant frequencies are fₙ = nv/(2L). Jan 21 mark schemes may require deriving f = (1/2L)√(T/μ) by substituting v = √(T/μ) for a stretched string with tension T and mass per unit length μ.
当两列频率和振幅相同但传播方向相反的波叠加时，形成驻波。对于两端固定的弦，边界条件要求两端为波节。最简单的驻波（基频）满足长度 L = λ/2。推广到一般情况，L = nλ/2，其中 n = 1,2,3…。利用 v = fλ，可得共振频率 fₙ = nv/(2L)。2021 年 1 月的评分方案可能会要求通过代入 v = √(T/μ)（T 为张力，μ 为单位长度质量）进一步推导 f = (1/2L)√(T/μ)。
fₙ = nv/(2L) and f = (1/2L)√(T/μ)
9. Deriving the Potential Divider Formula | 分压器公式推导
In a series circuit, the current is the same through all components. For two resistors R₁ and R₂ in series connected to a supply Vₛ, the total resistance is Rₜₒₜ = R₁ + R₂. The current I = Vₛ/(R₁+R₂). The voltage across R₁ is Vₒᵤₜ = I R₁ = Vₛ R₁/(R₁+R₂). This potential divider equation appears frequently in mark schemes, and students are expected to derive it from first principles, showing clearly how current and Ohm’s law are used.
在串联电路中，流过所有元件的电流相同。对于两个电阻 R₁ 与 R₂ 串联后接入电源 Vₛ，总电阻 Rₜₒₜ = R₁ + R₂。电流 I = Vₛ/(R₁+R₂)。R₁ 两端的电压为 Vₒᵤₜ = I R₁ = Vₛ R₁/(R₁+R₂)。该分压公式在评分方案中频繁出现，学生应能从基本原理推导，并清楚表明是如何应用电流和欧姆定律的。
Vₒᵤₜ = Vₛ × R₁/(R₁+R₂)
10. Deriving Kirchhoff’s Laws from Conservation Principles | 由守恒定律推导基尔霍夫定律
Kirchhoff’s first law (junction rule) states that the sum of currents entering a junction equals the sum leaving it: Σ Iᵢₙ = Σ Iₒᵤₜ. This follows directly from the conservation of electric charge – charge cannot accumulate at a junction. Kirchhoff’s second law (loop rule) states that around any closed loop, Σ ε = Σ IR. This arises from the conservation of energy; the total energy gained per unit charge from sources equals the total energy dissipated in the resistors. In Jan 21 derivations, marks are given for explicitly linking these laws to conservation laws.
基尔霍夫第一定律（节点定律）指出，流入节点的电流之和等于流出节点的电流之和：Σ Iᵢₙ = Σ Iₒᵤₜ。这直接源于电荷守恒——电荷不能在节点处积累。基尔霍夫第二定律（回路定律）指出，在任一闭合回路中，Σ ε = Σ IR。这源于能量守恒；每单位电荷从电源获得的总能量等于电阻上消耗的总能量。在 2021 年 1 月的推导中，凡明确将这两条定律与守恒定律联系起来均会得到分数。
Σ Iᵢₙ = Σ Iₒᵤₜ and Σ ε = Σ IR
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