📚 AS Physics: PH02 Unit 2 Formula Derivations and Example Responses | AS物理：PH02单元2公式推导与示例解答
The International AS Physics Unit 2 (PH02) paper requires students not only to recall key formulas but also to derive them from fundamental principles and apply these derivations in exam-style questions. This article provides detailed step-by-step derivations for the most important equations covered in the ‘Physics at Work’ specification, including waves, optics, electricity, and quantum phenomena. Each section pairs an English explanation with a Chinese translation and is followed by a worked example response to illustrate typical mark-scoring approaches.
国际AS物理单元2（PH02）考试不仅要求背诵关键公式，更要求能从基本原理出发进行推导，并将推导过程应用于考试题型中。本文详细解析了“物理学在工作”考纲中最重要的公式推导步骤，涵盖波动、光学、电学和量子现象。每个小节先给出英文解释，紧接着提供中文配对翻译，并附上典型例题的解答范例，帮助理解得分要点。
1. Wave Speed Equation | 波速公式推导
Derivation: A wave travels one full wavelength λ in a time equal to one period T. Therefore, the speed v is distance over time: v = λ / T. Since frequency f = 1/T, substituting gives v = fλ.
推导：波在一个周期 T 内前进一个完整的波长 λ。因此波速 v 等于距离除以时间：v = λ / T。由于频率 f = 1/T，代入即得 v = fλ。
Example response: Calculate the speed of water waves with wavelength 0.40 m and frequency 2.5 Hz. v = fλ = 2.5 × 0.40 = 1.0 m s-1.
示例解答：计算波长为 0.40 m、频率为 2.5 Hz 的水波波速。v = fλ = 2.5 × 0.40 = 1.0 m s-1。
v = fλ
2. Snell’s Law and Refractive Index | 斯涅耳定律与折射率
Derivation: Consider a wavefront changing speed as it crosses a boundary. The absolute refractive index n is defined as n = c / v, where c is the speed of light in vacuum. Using the geometry of wavefronts and the constancy of frequency, we obtain n1 sin θ1 = n2 sin θ2. For light travelling from medium 1 into medium 2, the angles are measured from the normal.
推导：考虑波前在穿过界面时速度发生变化。绝对折射率 n 定义为 n = c / v，其中 c 是真空中光速。利用波前几何关系和频率不变性，可以得到 n1 sin θ1 = n2 sin θ2。光从介质1进入介质2时，角度均从法线量起。
Example response: Light passes from air (n = 1.00) into glass (n = 1.50). If the angle of incidence is 30°, find the angle of refraction. Using 1.00 sin 30° = 1.50 sin θ2, sin θ2 = 0.50 / 1.50 = 1/3, so θ2 = 19.5°.
示例解答：光从空气（n = 1.00）射入玻璃（n = 1.50），入射角为 30°，求折射角。由 1.00 sin 30° = 1.50 sin θ2 得 sin θ2 = 0.50 / 1.50 = 1/3，故 θ2 = 19.5°。
n1 sin θ1 = n2 sin θ2 n = c / v
3. Diffraction Grating Equation | 衍射光栅方程
Derivation: When monochromatic light is incident normally on a diffraction grating, secondary wavelets from adjacent slits interfere constructively when the path difference is an integer number of wavelengths. From geometry, the path difference for two adjacent slits is d sin θ. For constructive interference, d sin θ = nλ, where d is the slit spacing, θ is the angle to the nth maximum, and n is the order number (n = 0, ±1, ±2 …).
推导：单色光垂直入射到衍射光栅时，相邻狭缝发出的子波在光程差等于整数倍波长时发生相长干涉。由几何关系，相邻两狭缝的光程差为 d sin θ。形成亮纹的条件为 d sin θ = nλ，其中 d 为光栅常数，θ 为第 n 级条纹的衍射角，n 为级次（n = 0, ±1, ±2 …）。
Example response: A grating with 500 lines per mm produces a first-order maximum at an angle of 20°. Determine the wavelength of the light. d = 1 / (500 × 103) = 2.0 × 10-6 m. Using d sin θ = nλ with n = 1, λ = 2.0 × 10-6 × sin 20° = 6.8 × 10-7 m.
示例解答：每毫米 500 条刻线的光栅，在 20° 处出现第一级明纹。求光波长。d = 1 / (500 × 103) = 2.0 × 10-6 m。由 d sin θ = nλ ，取 n = 1，得 λ = 2.0 × 10-6 × sin 20° = 6.8 × 10-7 m。
d sin θ = nλ
4. Photon Energy and the Photoelectric Effect | 光子能量与光电效应方程
Derivation: Each photon carries energy E = hf, where h is Planck’s constant. In the photoelectric effect, a photon is absorbed by an electron in the metal; the electron needs a minimum energy φ (the work function) to escape. The excess energy becomes the electron’s maximum kinetic energy: hf = φ + Kmax. The stopping potential Vs relates to Kmax by Kmax = eVs.
推导：每个光子携带能量 E = hf，其中 h 为普朗克常数。在光电效应中，金属中的电子吸收一个光子，需要至少 φ（逸出功）的能量才能逃逸。多余的能量转化为电子的最大动能：hf = φ + Kmax。遏止电压 Vs 与最大动能的关系为 Kmax = eVs。
Example response: Light of frequency 6.0 × 1014 Hz ejects electrons from a metal with a maximum kinetic energy of 0.40 eV. Calculate the work function in joules. hf = 6.63 × 10-34 × 6.0 × 1014 = 3.978 × 10-19 J. Kmax = 0.40 eV = 0.40 × 1.60 × 10-19 = 6.4 × 10-20 J. Then φ = hf – Kmax = 3.34 × 10-19 J.
示例解答：频率为 6.0 × 1014 Hz 的光照射某金属，产生最大动能 0.40 eV 的光电子。以焦耳为单位计算逸出功。hf = 6.63 × 10-34 × 6.0 × 1014 = 3.978 × 10-19 J。Kmax = 0.40 eV = 0.40 × 1.60 × 10-19 = 6.4 × 10-20 J。所以 φ = hf – Kmax = 3.34 × 10-19 J。
E = hf hf = φ + Kmax
5. Ohm’s Law and Electrical Power | 欧姆定律与电功率
Derivation: For an ohmic conductor at constant temperature, the current I through it is directly proportional to the potential difference V across it: V = IR, where R is resistance. Power P is the rate of energy transfer: P = IV. Substituting V = IR gives P = I2R, and using I = V/R gives P = V2/R.
推导：对于恒定温度下的欧姆导体，通过它的电流 I 与两端电压 V 成正比：V = IR，其中 R 为电阻。功率 P 是能量转化的速率：P = IV。代入 V = IR 得到 P = I2R，利用 I = V/R 得到 P = V2/R。
Example response: A 12 V battery supplies a 24 W lamp. Calculate the lamp’s resistance and the current. P = V2/R → R = V2 / P = 144 / 24 = 6.0 Ω. Then I = V/R = 12 / 6.0 = 2.0 A.
示例解答：12 V 电池为一个 24 W 的灯泡供电。计算灯泡的电阻和电流。P = V2/R → R = V2 / P = 144 / 24 = 6.0 Ω。然后 I = V/R = 12 / 6.0 = 2.0 A。
V = IR P = IV = I2R = V2/R
6. Resistors in Series and Parallel | 串联与并联电阻
Derivation for series: The same current I flows through each resistor. The total p.d. is Vtotal = V1 + V2 = IR1 + IR2. Using Vtotal = I Rtotal, we get Rtotal = R1 + R2 (+ R3 …).
串联推导：同一电流 I 流过各个电阻，总电压 Vtotal = V1 + V2 = IR1 + IR2。由 Vtotal = I Rtotal，得到 Rtotal = R1 + R2（+ R3 …）。
Derivation for parallel: The p.d. across each branch is the same, V. The total current splits: Itotal = I1 + I2 = V/R1 + V/R2. Using Itotal = V / Rtotal, we obtain 1 / Rtotal = 1/R1 + 1/R2.
并联推导：各支路两端电压相同，均为 V。总电流分流：Itotal = I1 + I2 = V/R1 + V/R2。由 Itotal = V / Rtotal，可得 1 / Rtotal = 1/R1 + 1/R2。
Example response: Two resistors, 3.0 Ω and 6.0 Ω, are connected in parallel. Find the total resistance. 1/Rtotal = 1/3.0 + 1/6.0 = 0.5, so Rtotal = 2.0 Ω.
示例解答：两个电阻 3.0 Ω 和 6.0 Ω 并联，求总电阻。1/Rtotal = 1/3.0 + 1/6.0 = 0.5，故 Rtotal = 2.0 Ω。
Series: Rtotal = R1 + R2 Parallel: 1/Rtotal = 1/R1 + 1/R2
7. Electromotive Force and Internal Resistance | 电动势与内阻
Derivation: A real cell has internal resistance r. The electromotive force (emf) E is the energy per unit charge supplied by the cell. In a complete circuit, some energy is dissipated inside the cell: the terminal p.d. V = E – Ir, where I is the current drawn. Rearranging, E = I (R + r), giving I = E / (R + r).
推导：实际的电池具有内阻 r。电动势 E 是电池单位电荷所提供的能量。在闭合电路中，一部分能量消耗在电池内部：端电压 V = E – Ir，其中 I 为电路中的电流。整理可得 E = I (R + r)，以及 I = E / (R + r)。
Example response: A cell of emf 1.50 V and internal resistance 0.20 Ω is connected to a 2.80 Ω resistor. Find the terminal p.d. Total resistance Rtotal = 2.80 + 0.20 = 3.00 Ω. Current I = 1.50 / 3.00 = 0.50 A. Then V = E – Ir = 1.50 – (0.50 × 0.20) = 1.40 V.
示例解答：一电池的电动势为 1.50 V、内阻为 0.20 Ω，与 2.80 Ω 电阻相连。求端电压。总电阻 Rtotal = 2.80 + 0.20 = 3.00 Ω。电流 I = 1.50 / 3.00 = 0.50 A。端电压 V = E – Ir = 1.50 – (0.50 × 0.20) = 1.40 V。
V = E – Ir E = I (R + r)
8. Potential Divider Formula | 分压器公式
Derivation: In a series circuit with two resistors R1 and R2 connected to a supply of voltage Vin, the same current I = Vin / (R1 + R2) flows. The voltage across R2 is Vout = I R2 = [Vin / (R1 + R2)] × R2. This is the potential divider equation: Vout = Vin × R2 / (R1 + R2).
推导：在两个电阻 R1 和 R2 串联并接到输入电压 Vin 的电路中，电流相同 I = Vin / (R1 + R2)。R2 两端的电压为 Vout = I R2 = [Vin / (R1 + R2)] × R2。此即分压公式：Vout = Vin × R2 / (R1 + R2)。
Example response: A sensor of resistance 400 Ω is placed in series with a fixed 600 Ω resistor connected to a 5.0 V supply, with the output taken across the sensor. Calculate the output voltage. Vout = 5.0 × 400 / (600 + 400) = 5.0 × 0.40 = 2.0 V.
示例解答：一个阻值 400 Ω 的传感器与 600 Ω 固定电阻串联接到 5.0 V 电源，输出电压取自传感器两端。计算输出电压。Vout = 5.0 × 400 / (600 + 400) = 5.0 × 0.40 = 2.0 V。
Vout = Vin × R2 / (R1 + R2)
9. Young’s Double-Slit Fringe Spacing | 杨氏双缝条纹间距
Derivation: Monochromatic light of wavelength λ passes through two narrow slits separated by distance d. A screen is placed at distance D from the slits, where D >> d. For the bright fringe at distance y from the centre, the path difference is approximately d sin θ. Using the small angle approximation sin θ ≈ tan θ = y / D, the condition for constructive interference d (y/D) = nλ gives y = nλD / d. The distance between adjacent bright fringes (fringe spacing) is Δy = λD / d.
推导：波长为 λ 的单色光通过间距为 d 的两条狭缝，屏幕距狭缝 D（D >> d）。距中心 y 处的亮纹满足光程差近似为 d sin θ。小角度近似 sin θ ≈ tan θ = y / D，干涉加强条件 d (y/D) = nλ，得 y = nλD / d。相邻亮纹间距 Δy = λD / d。
Example response: In a Young’s double-slit experiment, λ = 6.0 × 10-7 m, D = 1.5 m, and the fringe spacing is 0.45 mm. Find the slit separation. d = λD / Δy = (6.0 × 10-7 × 1.5) / (4.5 × 10-4) = 2.0 × 10-3 m (2.0 mm).
示例解答：杨氏双缝实验中 λ = 6.0 × 10-7 m，D = 1.5 m，条纹间距为 0.45 mm。求双缝间距。d = λD / Δy = (6.0 × 10-7 × 1.5) / (4.5 × 10-4) = 2.0 × 10-3 m (2.0 mm)。
Δy = λD / d
10. Critical Angle and Total Internal Reflection | 临界角与全内反射
Derivation: When light travels from a medium with refractive index n1 into one with n2 (where n1 > n2), the angle of refraction increases with angle of incidence. At the critical angle θc, the angle of refraction is 90°. Applying
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