📚 AS Physics Unit 4 Mark Scheme Jun19: Formula Derivations Explained | AS 物理 Unit 4 评分方案（2019年6月）公式推导详解
In the June 2019 AS Physics Unit 4 examination, several questions required candidates to derive standard wave and optics formulae from first principles. Understanding the mark scheme logic not only helps you score full marks but also deepens your grasp of key concepts such as interference, diffraction and standing waves. This article reconstructs those derivations step by step, highlighting exactly what examiners looked for in the mark scheme.
在 2019 年 6 月的 AS 物理 Unit 4 考试中，有多道试题要求学生从基本原理出发推导标准的波动与光学公式。理解评分方案的逻辑不仅有助于你获得满分，还能加深你对干涉、衍射和驻波等核心概念的掌握。本文将一步一步重现这些推导过程，并着重指出评分方案中阅卷人所关注的具体要点。
1. Understanding the Mark Scheme Context | 理解评分方案背景
The Unit 4 mark scheme for June 2019 rewarded clear logical steps, correct identification of physical quantities, and the appropriate use of approximations. Simply stating the final formula was never enough; candidates had to show how path difference leads to constructive interference and then use geometry to relate position on the screen to the angle of deviation.
2019 年 6 月 Unit 4 的评分方案鼓励清晰的逻辑步骤、正确识别物理量以及恰当使用近似条件。仅写出最终公式是永远不够的；考生必须展示光程差如何导致相长干涉，然后利用几何关系将屏幕上的位置与偏离角度联系起来。
Examiners typically allocated marks for: drawing a labelled diagram, writing the path difference expression, applying the small angle approximation where required, substituting the appropriate order number, and rearranging the algebra correctly. Missing a diagram or failing to justify the small angle approximation could cost marks even if the final answer was correct.
阅卷人通常会为以下要点分配分数：绘制带标注的示意图、写出光程差表达式、在需要时使用小角度近似、代入正确的干涉级次，以及正确地进行代数变换。即使最终答案正确，缺少示意图或未能对小角度近似做出合理解释也可能导致失分。
2. Path Difference and Constructive Interference | 光程差与相长干涉
derivation is the condition for bright fringes: path difference = mλ, where m = 0, ±1, ±2, … tion for minima uses (m + ½)λ but the June 2019 paper focused on bright fringe separation.
任何双源干涉推导的核心都是亮条纹条件：光程差 = mλ，其中 m = 0, ±1, ±2, … 这是因为同相到达的波会相长叠加，产生强度最大值。暗纹条件则使用 (m + ½)λ，但 2019 年 6 月的试卷侧重于亮纹间距。
lacement x from the central maximum. The path difference δ between waves from the two slits reaching P is δ = S₂P – S₁P ≈ a sinθ, where θ is the angle between the central axis and the line to P.
为了从几何上表达光程差，我们考虑两条相距为 a 的狭缝，屏幕距离为 D，屏幕上一点 P 距离中央极大值的位移为 x。由两狭缝到达 P 点的波的光程差 δ 为 δ = S₂P – S₁P ≈ a sinθ，其中 θ 是中心轴线与到 P 点的连线之间的夹角。
3. Deriving Fringe Spacing for Double Slit | 双缝干涉条纹间距推导
from the geometry tanθ = x/D. Therefore the path difference becomes δ ≈ a (x/D). For the mth bright fringe, we set δ = mλ, giving a(x/D) = mλ.
假设 D 远大于 a 和 x，则角度 θ 很小。在此小角度近似下，sinθ ≈ tanθ，而根据几何关系 tanθ = x/D。因此光程差变为 δ ≈ a(x/D)。对于第 m 级亮条纹，我们令 δ = mλ，得到 a(x/D) = mλ。
Rearranging gives the position of the mth bright fringe as xm = (mλD)/a. The fringe spacing Δx is the distance between adjacent bright fringes: Δx = xm+1 – xm = [(m+1)λD/a] – formula, and the derivation requires candidates to explicitly subtract the two positions.
整理可得第 m 级亮条纹的位置为 xm = (mλD)/a。条纹间距 Δx 是相邻亮条纹之间的距离：Δx = xm+1 – xm = [(m+1)λD/a] – [mλD/a] = λD/a。这就是标准的杨氏双缝公式，推导过程要求考生明确地将两个位置相减。
4. Applying the Small Angle Approximation | 小角度近似应用
wrote Δx = λD/a without justifying the step often lost the approximation mark. In a large number of practical setups, D is at least 1 m and a is a fraction of a millimetre, making the approximation perfectly valid.
评分方案会特别奖励那些说明当 θ 以弧度为单位且小于大约 10° 时 sinθ ≈ θ ≈ tanθ 的考生。那些仅写出 Δx = λD/a 而未为此步骤做出解释的考生通常会失去近似分。在大多数实际装置中，D 至少为 1 米，而 a 不到 1 毫米，因此该近似完全适用。
For questions where the angles were slightly larger, the exam might ask why the fringe pattern deviates from equal spacing. This happens because the small angle approximation breaks down, and the positions are no longer linearly proportional to m. Understanding this boundary gives deeper insight into the derivation’s limitations.
对于角度稍大的问题，考题可能会询问为何条纹图案会偏离等间距。这是因为小角度近似不再成立，位置不再与 m 成线性比例。理解这一边界条件能帮助考生更深入地认识推导的局限性。
5. Diffraction Grating: Deriving d sinθ = nλ | 衍射光栅：推导 d sinθ = nλ
wo. The path difference between waves from adjacent slits is d sinθ. For principal maxima, this must equal an integer number of wavelengths: d sinθ = nλ, where n is the order number (n = 0, 1, 2, …).
对于缝间距为 d 的透射衍射光栅，几何关系类似，但我们现在考虑的是大量狭缝而非只有两个。相邻两狭缝所发出的波的光程差为 d sinθ。为了得到主极大，该光程差必须等于整数倍波长：d sinθ = nλ，其中 n 为级次（n = 0, 1, 2, …）。
ern. The formula works for all angles, and one mark was often assigned for explaining that the central maximum (n=0) corresponds to all waves arriving in phase regardless of wavelength, which is why white light produces a central white fringe.
2019 年 6 月的评分方案要求考生认识到，如果直接通过衍射图样测量 θ，则此处无需使用小角度近似。该公式适用于所有角度，并且通常会有一个分数给予解释：中央主极大（n=0）对应所有波同相到达，与波长无关，这就是白光会产生中央白色条纹的原因。
6. Critical Angle and Total Internal Reflection | 临界角与全内反射推导
Snell’s law states n1 sinθ1 = n2 sinθ2. When light travels from an optically denser medium (refractive index n) into air (refractive index 1), total internal reflection occurs if the angle of incidence exceeds a critical value C. At the critical angle, the angle of refraction is exactly 90°, so sinθ2 = 1.
斯涅尔定律为 n1 sinθ1 = n2 sinθ2。当光从光密介质（折射率 n）射向空气（折射率为 1）时，如果入射角超过临界值 C，就会发生全内反射。在临界角处，折射角恰好为 90°，因此 sinθ2 = 1。
. The mark scheme expected candidates to start from Snell’s law, identify the two media clearly, and then set θ2 = 90° with a brief statement that this represents the onset of total internal reflection.
代入斯涅尔定律：n sin C = 1 × 1，化简得 sin C = 1/n。评分方案期望考生从斯涅尔定律入手，清楚地标明两种介质，然后令 θ2 = 90°，并简要说明这表示全内反射的起始条件。
7. Standing Waves on a String: λ from Length | 弦上驻波：波长与弦长关系
ls an integer number of half-wavelengths. This condition is L = n(λ/2), where n = 1, 2, 3, … corresponds to the harmonic number. Rearranging gives the wavelength for the nth harmonic: λn = 2L/n.
对于两端固定的拉紧的弦，当弦长 L 等于半波长的整数倍时会形成驻波。该条件为 L = n(λ/2)，其中 n = 1, 2, 3, … 对应于谐波数。整理后可得到第 n 次谐波的波长：λn = 2L/n。
e the frequencies of the harmonics using the wave equation v = fλ. Combining v = fλ with λn = 2L/n gives fn s at the open ends or, in the case of a string, nodes at the fixed ends, with a clear diagram.
2019 年 6 月的 Unit 4 试卷可能需要考生结合波速公式 v = fλ 来推导谐波频率。将 v = fλ 与 λn = 2L/n 相结合可得 fn = nv/(2L)。评分方案通常会给分点包括说明基频模式（n=1）在两端具有波节（对于固定端）或波腹，并配有清晰的示意图。
8. Mark Scheme Points: Common Pitfalls | 评分方案要点：常见错误
n step to demonstrate that Δx is independent of m. Simply writing x = mλD/a and claiming that Δx is the coefficient of m was not sufficient.
一个常见错误是将条纹间距 Δx 与某一条纹的位置 x 混淆。评分方案明确要求进行相减步骤，以证明 Δx 与 m 无关。仅写出 x = mλD/a 并声称 Δx 就是 m 的系数是不够的。
stent SI units as an implicit part of the derivation. In the diffraction grating question, forgetting that the number of lines per metre N and d are related by d = 1/N was a common mistake.
另一个陷阱与单位有关：使用纳米为波长单位代入公式而未将其转换为米会导致结果不合理。阅卷人会将使用一致的国际单位制视为推导的隐含部分。在衍射光栅问题中，忘记每米刻线数 N 与 d 之间的关系 d = 1/N 是常见错误。
9. Numerical Examples Using Derivations | 使用推导进行数值示例
ived formula Δx = λD/a yields Δx = (6.00×10⁻⁷ m)(2.00 m) / (5.00×10⁻⁴ m) = 2.40×10⁻³ m = 2.40 mm. Such calculations were often part of follow‑up questions where marks depended on using the correct derived expression.
考虑一个双缝实验，缝间距 a = 0.500 mm，屏距 D = 2.00 m，所用光波长 λ = 6.00 × 10⁻⁷ m。利用推导出的公式 Δx = λD/a 计算得 Δx = (6.00×10⁻⁷ m)(2.00 m) / (5.00×10⁻⁴ m) = 2.40×10⁻³ m = 2.40 mm。这类计算经常是后续题目的一部分，其得分依赖于使用正确的导出表达式。
00×10⁻⁷ m is found from sinθ = nλ/d = 5.00×10⁻⁷ / 3.33×10⁻⁶ = 0.150, giving θ ≈ 8.63°. Demonstrating this step reinforces the derivation’s physical meaning.
对于每毫米 300 线的衍射光栅，缝间距 d = 1/(300×10³ m⁻¹) = 3.33×10⁻⁶ m。对于 λ = 5.00×10⁻⁷ m 的第一级（n=1）条纹，角度由 sinθ = nλ/d = 5.00×10⁻⁷ / 3.33×10⁻⁶ = 0.150 求得，θ ≈ 8.63°。演示这一步骤能强化该推导的物理意义。
10. Summary: Key Steps to Secure Marks | 总结：确保得分的关键步骤
tion (path difference = mλ), apply the geometric relation and the small angle approximation if needed, carry out the algebraic manipulation clearly, and present the final formula with correct units. The derivation is not just an academic exercise; it shows the examiner you understand why the formula works, which is the essence of AS Physics.
要想复现 2019 年 6 月评分方案所期望的成功推导，一定要从带标注的示意图开始，定义所用符号，陈述物理条件（光程差 = mλ），应用几何关系和必要的小角度近似，清晰地进行代数变换，并给出带有正确单位的最终公式。推导并不只是纸上谈兵，它向阅卷人表明你理解公式为何成立，而这正是 AS 物理的精髓所在。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图