IB Physics HL · 鼎睿学苑

Unit C.3: Wave phenomena单元 C.3：波动现象

The wave-behaviour heart of Theme C. Reflection and refraction at a boundary, the refractive index and Snell's law, total internal reflection and its use in fibre optics, single-slit diffraction, the superposition principle and two-source interference, and Young's double-slit fringe spacing. The HL extension adds the diffraction grating and the polarization of light through Malus's law. These phenomena are the experimental fingerprint that light is a wave, and they reappear across optics, sound, and modern physics.主题 C 中波动行为的核心。界面处的反射与折射、折射率与 Snell 定律、全反射及其在光纤中的应用、单缝衍射、叠加原理与双源干涉，以及杨氏双缝的条纹间距。HL 扩展加入衍射光栅，以及通过马吕斯定律描述的光的偏振。这些现象是"光是波"的实验指纹，并贯穿于光学、声学与近代物理。

IB Physics · Theme C.3 · First Assessment 2025 Papers 1 · 2 6 Topics · SL + HL mix6 个核心专题 · SL + HL 混合
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How to use this guide本指南使用说明

C.3 is the "evidence that light is a wave" unit. The algebra is light — most marks come from picking the right boundary equation (Snell, critical angle, path difference, fringe spacing) and substituting carefully with angles in degrees and wavelengths in metres. The conceptual traps are the dangerous part: which angle is measured from the normal, whether a bright fringe needs a whole or half wavelength of path difference, and what each variable in $s = \lambda D / d$ actually is. Train the geometry alongside the formulas.C.3 是"光是波的证据"单元。代数很轻——分数主要来自选对界面公式（Snell、临界角、光程差、条纹间距）并细心代入（角度用度、波长用米）。真正危险的是概念陷阱：角度是否从法线量起、亮纹需要整数还是半整数波长的光程差，以及 $s = \lambda D / d$ 中每个量到底代表什么。几何与公式要一起练。

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If you are cramming如果你在临阵磨枪

Memorise four equations: Snell's law $n_1 \sin\theta_1 = n_2 \sin\theta_2$, refractive index $n = c/v$, critical angle $\sin\theta_c = 1/n$, and double-slit fringe spacing $s = \lambda D / d$. Always measure angles from the normal, never from the surface. Constructive interference needs path difference $= n\lambda$; destructive needs $(n + \tfrac{1}{2})\lambda$.

背熟四条公式：Snell 定律 $n_1 \sin\theta_1 = n_2 \sin\theta_2$、折射率 $n = c/v$、临界角 $\sin\theta_c = 1/n$、双缝条纹间距 $s = \lambda D / d$。角度永远从法线量起，不要从表面量。相长干涉需光程差 $= n\lambda$；相消需 $(n + \tfrac{1}{2})\lambda$。

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If you are going for a 7如果你目标是 7 分

Be able to derive $s = \lambda D / d$ from the path-difference geometry, and explain why TIR requires light going from dense to less-dense medium. Know that the single-slit first minimum $\theta = \lambda / b$ HL sets the envelope that modulates double-slit fringes. Master the grating $d \sin\theta = n\lambda$ HL and Malus's law $I = I_0 \cos^2\theta$ HL, including the unpolarized-to-polarized halving.

能从光程差几何推出 $s = \lambda D / d$，并解释为何全反射要求光从光密介质射向光疏介质。理解单缝第一极小 $\theta = \lambda / b$ HL 决定了调制双缝条纹的包络。掌握光栅 $d \sin\theta = n\lambda$ HL 与马吕斯定律 $I = I_0 \cos^2\theta$ HL，包括非偏振光通过偏振片后强度减半。

HL flagHL 标记说明 The single-slit first-minimum formula $\theta = \lambda / b$ (in C3.3), the diffraction grating, and polarization with Malus's law (in C3.6) are HL extension content. SL students may safely skip the HL-flagged blocks; the qualitative description of single-slit diffraction is SL.单缝第一极小公式 $\theta = \lambda / b$（C3.3 中）、衍射光栅，以及偏振与马吕斯定律（C3.6 中）为 HL 扩展内容。SL 学生可跳过带 HL 标记的段落；单缝衍射的定性描述属于 SL。
Topic C3.1 · Reflection and RefractionTopic C3.1 · 反射与折射

Reflection, Refraction and Snell's Law反射、折射与 Snell 定律 C.3 SL+HL

Law of reflection. The angle of incidence equals the angle of reflection, both measured from the normal: $\theta_i = \theta_r$. Incident ray, reflected ray and normal lie in one plane.

Refractive index. n = c/v compares the speed of light in vacuum $c$ to its speed $v$ in the medium: $$ n = \frac{c}{v} \;\;(\ge 1). $$ Snell's law (from the data booklet). n₁ sin θ₁ = n₂ sin θ₂ $$ n_1 \sin\theta_1 = n_2 \sin\theta_2. $$ Reading it. Light entering a denser medium ($n_2 > n_1$) bends toward the normal; entering a less-dense medium, it bends away. Frequency $f$ is unchanged across the boundary; speed and wavelength change together ($v = f\lambda$).
反射定律。入射角等于反射角，二者均从法线量起：$\theta_i = \theta_r$。入射线、反射线与法线共面。

折射率（refractive index）。n = c/v 比较真空中光速 $c$ 与介质中光速 $v$： $$ n = \frac{c}{v} \;\;(\ge 1). $$ Snell 定律（数据手册）。n₁ sin θ₁ = n₂ sin θ₂ $$ n_1 \sin\theta_1 = n_2 \sin\theta_2. $$ 如何读。光进入光密介质（$n_2 > n_1$）时偏向法线；进入光疏介质时偏离法线。频率 $f$ 跨界面不变；速度与波长一同改变（$v = f\lambda$）。
Worked Example C3.1 (air into glass)C3.1 例题（空气进玻璃）

A ray of light in air ($n_1 = 1.00$) strikes the flat surface of glass ($n_2 = 1.50$) at an angle of incidence of $40^{\circ}$. Find the angle of refraction and the speed of light in the glass. Take $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$.空气（$n_1 = 1.00$）中一束光以 $40^{\circ}$ 入射角射到玻璃（$n_2 = 1.50$）平面。求折射角与玻璃中的光速。取 $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$。

Identify. Known: $n_1 = 1.00$, $\theta_1 = 40^{\circ}$, $n_2 = 1.50$. Measure angles from the normal.

识别。已知：$n_1 = 1.00$、$\theta_1 = 40^{\circ}$、$n_2 = 1.50$。角度从法线量起。

Set up. Apply Snell's law $n_1 \sin\theta_1 = n_2 \sin\theta_2$:

列式。用 Snell 定律 $n_1 \sin\theta_1 = n_2 \sin\theta_2$：

$$ \sin\theta_2 = \frac{n_1 \sin\theta_1}{n_2} = \frac{(1.00)\sin 40^{\circ}}{1.50} = \frac{0.643}{1.50} = 0.4285. $$ $$ \theta_2 = \arcsin(0.4285) \approx 25.4^{\circ}. $$

Speed in glass. From $n = c/v$, $v = c/n = (3.00 \times 10^{8}) / 1.50 = 2.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$.

玻璃中光速。由 $n = c/v$，$v = c/n = (3.00 \times 10^{8}) / 1.50 = 2.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$。

Evaluate. The ray bent toward the normal ($25.4^{\circ} < 40^{\circ}$), consistent with entering a denser medium. The speed dropped to two-thirds of $c$, exactly the inverse of the index ratio.

评估。光线偏向法线（$25.4^{\circ} < 40^{\circ}$），与进入光密介质一致。光速降为 $c$ 的三分之二，恰为折射率的倒数比。

Going deeper: why wavelength shrinks but frequency stays fixed深入：为何波长缩短而频率不变

At a boundary the wave fronts must stay continuous — crests cannot pile up or vanish — so the same number of crests per second arrive on both sides. Hence the frequency $f$ is conserved across the interface.

在界面处波前必须连续——波峰不能堆积或消失——故两侧每秒到达的波峰数相同，因此频率 $f$ 跨界面守恒。

Since $v = f\lambda$ and $v$ falls in a denser medium, the wavelength must fall in proportion:

由于 $v = f\lambda$ 且光密介质中 $v$ 减小，波长必按比例减小：

$$ \frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} = \frac{n_1}{n_2}. $$

This is why Snell's law can equivalently be written $n_1 \lambda_1 = n_2 \lambda_2$ in terms of wavelengths, and why the colour (set by $f$) of light does not change when it enters glass.

这就是为何 Snell 定律也可用波长写成 $n_1 \lambda_1 = n_2 \lambda_2$，以及为何光进入玻璃时颜色（由 $f$ 决定）不变。

Light of speed $2.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$ travels in a transparent medium. Its refractive index is (take $c = 3.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$):光在某透明介质中速度为 $2.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$。其折射率为（取 $c = 3.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$）：
C3.1 · Q1
$0.67$
$1.5$
$2.0$
$6.0 \times 10^{16}$
$n = c/v = (3.0 \times 10^{8})/(2.0 \times 10^{8}) = 1.5$. The index is always $\ge 1$ since light slows in matter.$n = c/v = (3.0 \times 10^{8})/(2.0 \times 10^{8}) = 1.5$。光在介质中变慢，故折射率 $\ge 1$。
Use $n = c/v$ with $c$ on top. A refractive index is dimensionless and always $\ge 1$.用 $n = c/v$（$c$ 在分子）。折射率无量纲且 $\ge 1$。
A ray passes from glass ($n = 1.5$) into water ($n = 1.33$). Compared with the incident ray, the refracted ray bends:光线从玻璃（$n = 1.5$）射入水（$n = 1.33$）。与入射线相比，折射线：
C3.1 · Q2
Toward the normal偏向法线
Stays exactly along the incident direction沿原入射方向不变
Away from the normal偏离法线
Back into the glass (always)总是返回玻璃
Going from higher to lower index ($1.5 \to 1.33$), Snell's law forces $\sin\theta_2 > \sin\theta_1$, so $\theta_2 > \theta_1$: the ray bends away from the normal (until the critical angle, where TIR begins).从高折射率到低折射率（$1.5 \to 1.33$），Snell 定律给出 $\sin\theta_2 > \sin\theta_1$，故 $\theta_2 > \theta_1$：光线偏离法线（直到临界角处开始全反射）。
Dense-to-less-dense ($n_1 > n_2$) means the ray bends away from the normal. Bending toward the normal happens for the reverse direction.光密到光疏（$n_1 > n_2$）时光线偏离法线。偏向法线发生在相反方向。
Topic C3.2 · Total Internal ReflectionTopic C3.2 · 全反射

Total Internal Reflection and Fibre Optics全反射与光纤 C.3 SL+HL

Two conditions for total internal reflection (TIR).
  • Light must travel from a denser to a less-dense medium ($n_1 > n_2$).
  • The angle of incidence must exceed the critical angle $\theta_c$.
Critical angle (from the data booklet). Set $\theta_2 = 90^{\circ}$ in Snell's law. For a medium of index $n$ against vacuum/air: sin θ_c = 1/n $$ \sin\theta_c = \frac{1}{n}, \qquad \text{or generally} \quad \sin\theta_c = \frac{n_2}{n_1}. $$ At the critical angle the refracted ray grazes the surface ($\theta_2 = 90^{\circ}$). Beyond it, no light refracts out: 100% reflects internally.

Fibre optics. A glass core (higher $n$) clad in lower-$n$ glass traps light by repeated TIR, carrying signals with very low loss over long distances.
全反射（total internal reflection, TIR）的两个条件。
  • 光必须从光密介质射向光疏介质（$n_1 > n_2$）。
  • 入射角必须大于临界角 $\theta_c$。
临界角（数据手册）。在 Snell 定律中令 $\theta_2 = 90^{\circ}$。折射率为 $n$ 的介质对真空/空气：sin θ_c = 1/n $$ \sin\theta_c = \frac{1}{n}, \qquad \text{一般地} \quad \sin\theta_c = \frac{n_2}{n_1}. $$ 在临界角处折射线沿表面掠过（$\theta_2 = 90^{\circ}$）。超过临界角后无光折射出去：100% 内部反射。

光纤。高折射率玻璃纤芯外包低折射率包层，靠反复全反射困住光，以极低损耗远距离传输信号。
Worked Example C3.2 (critical angle of glass)C3.2 例题（玻璃的临界角）

A glass block has refractive index $n = 1.52$. (a) Find the critical angle for a glass-to-air boundary. (b) A ray inside the glass hits the surface at $45^{\circ}$. Does it escape or totally reflect?玻璃块折射率 $n = 1.52$。(a) 求玻璃-空气界面的临界角。(b) 玻璃内一束光以 $45^{\circ}$ 入射表面，它会逸出还是全反射？

(a) Critical angle. Glass to air, so $\sin\theta_c = 1/n$:

(a) 临界角。玻璃到空气，故 $\sin\theta_c = 1/n$：

$$ \sin\theta_c = \frac{1}{1.52} = 0.6579, \qquad \theta_c = \arcsin(0.6579) \approx 41.1^{\circ}. $$

(b) Compare. The incidence angle $45^{\circ}$ exceeds $\theta_c = 41.1^{\circ}$, and the ray goes from dense to less-dense. Both TIR conditions are met, so the ray totally internally reflects — no light escapes into the air.

(b) 比较。入射角 $45^{\circ}$ 大于 $\theta_c = 41.1^{\circ}$，且光从光密射向光疏。两个全反射条件都满足，故发生全反射——无光逸入空气。

Evaluate. Had the ray hit at $35^{\circ}$ (below $\theta_c$), it would have partly refracted out. The critical angle is the sharp boundary between "some light escapes" and "all light reflects".

评估。若以 $35^{\circ}$（小于 $\theta_c$）入射，则部分折射逸出。临界角是"部分逸出"与"全部反射"之间的明确分界。

Direction matters方向很关键 TIR is impossible when light travels from a less-dense into a denser medium (air into glass), because then it bends toward the normal and always finds a refracted ray. Always check the ray is going dense-to-less-dense before testing the critical angle.当光从光疏射向光密（空气进玻璃）时不可能发生全反射，因为它偏向法线、总能找到折射线。检验临界角之前务必先确认光是从光密射向光疏。
Going deeper: why $\sin\theta_c = n_2/n_1$ comes straight from Snell深入：$\sin\theta_c = n_2/n_1$ 如何直接由 Snell 得出

The critical angle is the incidence angle at which the refracted ray just grazes the boundary, i.e. $\theta_2 = 90^{\circ}$. Substitute into Snell's law $n_1 \sin\theta_1 = n_2 \sin\theta_2$:

临界角是折射线恰好沿界面掠过时的入射角，即 $\theta_2 = 90^{\circ}$。代入 Snell 定律 $n_1 \sin\theta_1 = n_2 \sin\theta_2$：

$$ n_1 \sin\theta_c = n_2 \sin 90^{\circ} = n_2 \quad\Rightarrow\quad \sin\theta_c = \frac{n_2}{n_1}. $$

When the second medium is air ($n_2 \approx 1$) this reduces to $\sin\theta_c = 1/n_1$. Because $\sin\theta_c \le 1$ requires $n_2 \le n_1$, the formula automatically encodes the "dense-to-less-dense" condition: if $n_2 > n_1$ there is no real critical angle and TIR cannot occur.

当第二介质为空气（$n_2 \approx 1$）时化为 $\sin\theta_c = 1/n_1$。由于 $\sin\theta_c \le 1$ 要求 $n_2 \le n_1$，该式自动包含"光密到光疏"条件：若 $n_2 > n_1$ 则无实临界角，不可能全反射。

A transparent material has critical angle $48.8^{\circ}$ against air. Its refractive index is approximately:某透明材料对空气的临界角为 $48.8^{\circ}$。其折射率约为：
C3.2 · Q1
$1.33$
$0.75$
$1.52$
$2.0$
$n = 1/\sin\theta_c = 1/\sin 48.8^{\circ} = 1/0.752 \approx 1.33$ (this is water).$n = 1/\sin\theta_c = 1/\sin 48.8^{\circ} = 1/0.752 \approx 1.33$（即水）。
Rearrange $\sin\theta_c = 1/n$ to $n = 1/\sin\theta_c$. The index must exceed 1, ruling out $0.75$.由 $\sin\theta_c = 1/n$ 得 $n = 1/\sin\theta_c$。折射率须大于 1，排除 $0.75$。
Which condition is required for total internal reflection to occur?发生全反射需要满足哪个条件？
C3.2 · Q2
Light travels from less-dense to denser, at any angle光从光疏到光密，任意角度
Incidence angle is below the critical angle入射角小于临界角
The two media have equal refractive index两介质折射率相等
Light travels from denser to less-dense, above the critical angle光从光密到光疏，且大于临界角
TIR needs both: $n_1 > n_2$ (dense to less-dense) and $\theta_1 > \theta_c$. Below $\theta_c$ the ray partly refracts out; from less-dense to denser there is no critical angle at all.全反射需同时满足：$n_1 > n_2$（光密到光疏）且 $\theta_1 > \theta_c$。小于 $\theta_c$ 时部分折射逸出；从光疏到光密则根本无临界角。
Both conditions are mandatory: dense-to-less-dense direction AND incidence above the critical angle.两个条件缺一不可：光密到光疏方向，且入射角大于临界角。
Topic C3.3 · Single-Slit DiffractionTopic C3.3 · 单缝衍射

Diffraction at a Single Slit or Aperture单缝或孔径处的衍射 C.3 SL+HL

Diffraction (qualitative, SL). When a wave passes through an aperture or past an edge it spreads out. The spreading is most pronounced when the slit width $b$ is comparable to the wavelength $\lambda$. Large slit ($b \gg \lambda$): little spread. Narrow slit ($b \sim \lambda$): broad spread.

Single-slit pattern. A bright central maximum (twice as wide as the side maxima) flanked by progressively dimmer side maxima, separated by dark minima.

First minimum (HL). HL For a slit of width $b$, the first diffraction minimum appears at angle $$ \theta = \frac{\lambda}{b} \qquad \text{(small-angle, } \theta \text{ in radians)}. $$ Wider slit or shorter wavelength ⇒ narrower central maximum.
衍射（定性，SL）。波经过孔径或绕过边缘时会扩散。当缝宽 $b$ 与波长 $\lambda$ 相当时扩散最明显。宽缝（$b \gg \lambda$）：扩散小。窄缝（$b \sim \lambda$）：扩散大。

单缝图样。一个明亮的中央极大（宽度为旁极大的两倍），两侧是逐渐变暗的旁极大，中间被暗极小分隔。

第一极小（HL）。HL 对缝宽 $b$，第一衍射极小出现在角度 $$ \theta = \frac{\lambda}{b} \qquad \text{（小角近似，} \theta \text{ 取弧度）}. $$ 缝越宽或波长越短 ⇒ 中央极大越窄。
Worked Example C3.3 (central maximum width) HLC3.3 例题（中央极大宽度）HL

Light of wavelength $600\ \mathrm{nm}$ passes through a single slit of width $0.12\ \mathrm{mm}$ and falls on a screen $2.5\ \mathrm{m}$ away. Find the angular position of the first minimum and the width of the central bright fringe on the screen.波长 $600\ \mathrm{nm}$ 的光通过缝宽 $0.12\ \mathrm{mm}$ 的单缝，照到 $2.5\ \mathrm{m}$ 外的屏上。求第一极小的角位置与屏上中央亮纹的宽度。

Identify. $\lambda = 600 \times 10^{-9}\ \mathrm{m}$, $b = 0.12 \times 10^{-3}\ \mathrm{m}$, $D = 2.5\ \mathrm{m}$.

识别。$\lambda = 600 \times 10^{-9}\ \mathrm{m}$、$b = 0.12 \times 10^{-3}\ \mathrm{m}$、$D = 2.5\ \mathrm{m}$。

First minimum angle. Use $\theta = \lambda / b$:

第一极小角度。用 $\theta = \lambda / b$：

$$ \theta = \frac{600 \times 10^{-9}}{0.12 \times 10^{-3}} = 5.0 \times 10^{-3}\ \mathrm{rad}. $$

Width of central maximum. The central maximum spans from $-\theta$ to $+\theta$, so its half-width on the screen is $D\theta$ and its full width is $2 D \theta$:

中央极大宽度。中央极大从 $-\theta$ 到 $+\theta$，故屏上半宽为 $D\theta$，全宽为 $2 D \theta$：

$$ w = 2 D \theta = 2 (2.5)(5.0 \times 10^{-3}) = 2.5 \times 10^{-2}\ \mathrm{m} = 25\ \mathrm{mm}. $$

Evaluate. Halving the wavelength or doubling the slit width would halve this width — the central maximum sharpens as $b$ grows relative to $\lambda$.

评估。把波长减半或缝宽加倍都会把该宽度减半——当 $b$ 相对 $\lambda$ 增大时中央极大变窄。

Going deeper: where the first minimum comes from HL深入：第一极小的由来 HL

Divide the slit of width $b$ into two halves. Pair each ray in the top half with the ray a distance $b/2$ below it. The path difference between a pair, at angle $\theta$, is $\tfrac{b}{2}\sin\theta$.

把缝宽 $b$ 分成两半。把上半部每条光线与其下方 $b/2$ 处的光线配对。在角度 $\theta$ 处，每对光线的光程差为 $\tfrac{b}{2}\sin\theta$。

Every pair cancels (destructive) when this path difference equals $\lambda/2$:

当此光程差等于 $\lambda/2$ 时每对都相消（相消干涉）：

$$ \frac{b}{2}\sin\theta = \frac{\lambda}{2} \;\Rightarrow\; b\sin\theta = \lambda \;\Rightarrow\; \sin\theta = \frac{\lambda}{b}. $$

For small angles $\sin\theta \approx \theta$, giving the data-booklet result $\theta = \lambda/b$. This is the envelope that modulates the brightness of double-slit fringes in C3.5.

小角时 $\sin\theta \approx \theta$，得到数据手册结果 $\theta = \lambda/b$。这正是 C3.5 中调制双缝条纹亮度的包络。

A single slit is made narrower. The central diffraction maximum on the screen:把单缝变窄，屏上中央衍射极大：
C3.3 · Q1
Becomes narrower变窄
Becomes wider变宽
Stays the same width宽度不变
Splits into two maxima分裂为两个极大
The first minimum sits at $\theta = \lambda/b$. Decreasing $b$ increases $\theta$, so the central maximum spreads wider. Narrower slit, broader pattern.第一极小位于 $\theta = \lambda/b$。减小 $b$ 使 $\theta$ 增大，故中央极大变宽。缝越窄，图样越宽。
From $\theta = \lambda/b$, the central maximum width is inversely proportional to slit width. Narrower slit means a wider spread.由 $\theta = \lambda/b$，中央极大宽度与缝宽成反比。缝越窄，扩散越宽。
HL Light of $500\ \mathrm{nm}$ passes a slit of width $0.10\ \mathrm{mm}$. The angle to the first minimum is:HL $500\ \mathrm{nm}$ 的光通过缝宽 $0.10\ \mathrm{mm}$ 的单缝。第一极小的角度为：
C3.3 · Q2
$5.0 \times 10^{-6}\ \mathrm{rad}$
$2.0 \times 10^{-4}\ \mathrm{rad}$
$5.0 \times 10^{-3}\ \mathrm{rad}$
$0.50\ \mathrm{rad}$
$\theta = \lambda/b = (500 \times 10^{-9})/(0.10 \times 10^{-3}) = 5.0 \times 10^{-3}\ \mathrm{rad}$.$\theta = \lambda/b = (500 \times 10^{-9})/(0.10 \times 10^{-3}) = 5.0 \times 10^{-3}\ \mathrm{rad}$。
Convert both to metres: $\lambda = 5.0 \times 10^{-7}\ \mathrm{m}$, $b = 1.0 \times 10^{-4}\ \mathrm{m}$, then $\theta = \lambda/b$.都化为米：$\lambda = 5.0 \times 10^{-7}\ \mathrm{m}$、$b = 1.0 \times 10^{-4}\ \mathrm{m}$，再用 $\theta = \lambda/b$。
Topic C3.4 · Superposition and InterferenceTopic C3.4 · 叠加与干涉

The Principle of Superposition and Interference叠加原理与干涉 C.3 SL+HL

Principle of superposition. When two or more waves overlap, the resultant displacement at each point is the vector sum of the individual displacements. After overlapping, the waves continue unchanged.

Constructive vs destructive.
  • Constructive: crests meet crests; amplitudes add. Path difference $= n\lambda$.
  • Destructive: crest meets trough; amplitudes subtract (full cancellation if equal). Path difference $= (n + \tfrac{1}{2})\lambda$.
Path difference rules (coherent, in-phase sources). $$ \text{constructive: } \Delta = n\lambda, \qquad \text{destructive: } \Delta = \left(n + \tfrac{1}{2}\right)\lambda, \qquad n = 0, 1, 2, \dots $$ Coherence. A clear, stable interference pattern needs coherent sources: constant phase relationship and (usually) the same frequency.
叠加原理（principle of superposition）。两列或多列波叠加时，各点的合位移为各波位移的矢量和。叠加之后各波继续不受影响地传播。

相长与相消。
  • 相长：峰遇峰；振幅相加。光程差 $= n\lambda$。
  • 相消：峰遇谷；振幅相减（等幅则完全抵消）。光程差 $= (n + \tfrac{1}{2})\lambda$。
光程差规则（相干、同相源）。 $$ \text{相长：} \Delta = n\lambda, \qquad \text{相消：} \Delta = \left(n + \tfrac{1}{2}\right)\lambda, \qquad n = 0, 1, 2, \dots $$ 相干性。清晰稳定的干涉图样需要相干源：相位关系恒定且（通常）频率相同。
Worked Example C3.4 (two-speaker interference)C3.4 例题（双喇叭干涉）

Two loudspeakers driven in phase emit sound of wavelength $0.40\ \mathrm{m}$. At a point $P$, the path from speaker 1 is $3.00\ \mathrm{m}$ and from speaker 2 is $3.60\ \mathrm{m}$. Is $P$ a point of constructive or destructive interference?两个同相驱动的喇叭发出波长 $0.40\ \mathrm{m}$ 的声音。在 $P$ 点，到喇叭 1 的路程为 $3.00\ \mathrm{m}$，到喇叭 2 为 $3.60\ \mathrm{m}$。$P$ 是相长还是相消？

Find the path difference. $\Delta = 3.60 - 3.00 = 0.60\ \mathrm{m}$.

求光程差。$\Delta = 3.60 - 3.00 = 0.60\ \mathrm{m}$。

Express in wavelengths. $\Delta / \lambda = 0.60 / 0.40 = 1.5 = (1 + \tfrac{1}{2})$.

用波长表示。$\Delta / \lambda = 0.60 / 0.40 = 1.5 = (1 + \tfrac{1}{2})$。

Classify. The path difference is a half-integer number of wavelengths, $(n + \tfrac{1}{2})\lambda$ with $n = 1$. The waves arrive exactly out of phase, so $P$ is a point of destructive interference (a quiet spot).

判断。光程差为半整数倍波长，$(n + \tfrac{1}{2})\lambda$（$n = 1$）。两波恰好反相到达，故 $P$ 为相消干涉点（安静处）。

Evaluate. A path difference of exactly $1.0\ \mathrm{m}$ ($2.5\lambda$) would also be destructive; $0.80\ \mathrm{m}$ ($2\lambda$) would be constructive. Only the fractional part of $\Delta/\lambda$ matters.

评估。光程差恰为 $1.0\ \mathrm{m}$（$2.5\lambda$）也会相消；$0.80\ \mathrm{m}$（$2\lambda$）则相长。只有 $\Delta/\lambda$ 的小数部分起作用。

Going deeper: why coherence and a phase offset matter深入：相干性与相位差为何重要

The path-difference rules above assume the two sources start in phase. If the sources have a built-in phase offset of half a cycle (anti-phase), the rules swap: constructive interference then requires $\Delta = (n + \tfrac{1}{2})\lambda$ and destructive requires $\Delta = n\lambda$.

上述光程差规则假设两源同相起振。若两源本身有半周期相位差（反相），规则互换：此时相长要求 $\Delta = (n + \tfrac{1}{2})\lambda$，相消要求 $\Delta = n\lambda$。

Two independent light bulbs never produce a steady interference pattern because their relative phase jitters randomly billions of times per second — they are incoherent. This is why Young split a single beam into two (C3.5): the two slits are then guaranteed coherent.

两个独立灯泡永远不会产生稳定干涉图样，因为其相对相位每秒随机抖动数十亿次——它们不相干。这正是杨氏把单束光分成两束（C3.5）的原因：两缝由此必然相干。

Two coherent in-phase sources of wavelength $\lambda$ reach a point with path difference $2\lambda$. The interference there is:两个相干同相、波长为 $\lambda$ 的源到达某点，光程差为 $2\lambda$。该处干涉为：
C3.4 · Q1
Destructive (cancellation)相消（抵消）
Partial, half amplitude部分相消，半振幅
Constructive (maximum)相长（极大）
Always zero regardless of phase无论相位总为零
$\Delta = 2\lambda = n\lambda$ with $n = 2$ is a whole number of wavelengths, so the waves arrive in phase: constructive interference, a bright/loud maximum.$\Delta = 2\lambda = n\lambda$（$n = 2$）为整数倍波长，两波同相到达：相长干涉，是亮/响的极大。
A whole number of wavelengths ($n\lambda$) is the constructive condition for in-phase sources.整数倍波长（$n\lambda$）是同相源的相长条件。
The principle of superposition states that the resultant displacement of overlapping waves is:叠加原理指出，叠加波的合位移是：
C3.4 · Q2
The vector sum of the individual displacements各波位移的矢量和
The product of the individual displacements各波位移之积
Always the larger of the two displacements总是取较大的位移
Zero everywhere they overlap叠加处处处为零
Superposition: at each instant and point, add the displacements as signed vectors. Crest+crest reinforces; crest+trough cancels. After overlapping, each wave continues unchanged.叠加：每一时刻、每一点把位移作为带符号矢量相加。峰+峰增强；峰+谷抵消。叠加后各波继续不变传播。
Superposition adds displacements (as vectors), it does not multiply them or pick the larger one.叠加是把位移（作为矢量）相加，不是相乘也不是取较大者。
Topic C3.5 · Young's Double-SlitTopic C3.5 · 杨氏双缝

Young's Double-Slit Experiment杨氏双缝实验 C.3 SL+HL

The set-up. Monochromatic, coherent light passes through two narrow slits a distance $d$ apart and forms equally-spaced bright and dark fringes on a screen a distance $D$ away.

Fringe spacing (from the data booklet). s = λD/d $$ s = \frac{\lambda D}{d}, $$ where $s$ = spacing between adjacent bright (or dark) fringes, $\lambda$ = wavelength, $D$ = slit-to-screen distance, $d$ = slit separation.

Bright fringe condition. Path difference $d\sin\theta = n\lambda$ gives a maximum; small-angle $\sin\theta \approx \theta \approx y/D$ leads directly to $s = \lambda D/d$.

Watch the variables. $d$ is the gap between slits, not the slit width $b$; $b$ controls the single-slit envelope (C3.3), $d$ controls the fringe spacing.
装置。单色相干光通过相距 $d$ 的两条窄缝，在 $D$ 外的屏上形成等间距的明暗条纹。

条纹间距（数据手册）。s = λD/d $$ s = \frac{\lambda D}{d}, $$ 其中 $s$ 为相邻亮纹（或暗纹）间距，$\lambda$ 为波长，$D$ 为缝到屏的距离，$d$ 为缝间距。

亮纹条件。光程差 $d\sin\theta = n\lambda$ 给出极大；小角 $\sin\theta \approx \theta \approx y/D$ 直接导出 $s = \lambda D/d$。

注意变量。$d$ 是两缝之间的间距，不是缝宽 $b$；$b$ 控制单缝包络（C3.3），$d$ 控制条纹间距。
Worked Example C3.5 (measuring wavelength)C3.5 例题（测量波长）

In a double-slit experiment, slits separated by $d = 0.25\ \mathrm{mm}$ produce fringes on a screen $D = 1.8\ \mathrm{m}$ away. Ten bright fringes span $4.32\ \mathrm{cm}$. Find the wavelength of the light.双缝实验中，缝间距 $d = 0.25\ \mathrm{mm}$，屏距 $D = 1.8\ \mathrm{m}$。十条亮纹横跨 $4.32\ \mathrm{cm}$。求光的波长。

Find the fringe spacing. Ten fringes span nine gaps, but "10 bright fringes span $4.32\ \mathrm{cm}$" is conventionally the distance across 9 spacings; to keep the arithmetic clean here we take the spacing of adjacent fringes as $s = 4.32\ \mathrm{cm}/9 = 0.48\ \mathrm{cm} = 4.8 \times 10^{-3}\ \mathrm{m}$.

求条纹间距。十条亮纹之间有九个间隔；"10 条亮纹跨 $4.32\ \mathrm{cm}$"通常指跨 9 个间距，故相邻条纹间距 $s = 4.32\ \mathrm{cm}/9 = 0.48\ \mathrm{cm} = 4.8 \times 10^{-3}\ \mathrm{m}$。

Rearrange $s = \lambda D/d$ for $\lambda$.

由 $s = \lambda D/d$ 解出 $\lambda$。

$$ \lambda = \frac{s\, d}{D} = \frac{(4.8 \times 10^{-3})(0.25 \times 10^{-3})}{1.8} = 6.7 \times 10^{-7}\ \mathrm{m} = 670\ \mathrm{nm}. $$

Evaluate. $670\ \mathrm{nm}$ is red light, a physically reasonable visible wavelength. Note how all lengths were converted to metres before substituting.

评估。$670\ \mathrm{nm}$ 为红光，是合理的可见波长。注意代入前所有长度都化为米。

Going deeper: deriving $s = \lambda D/d$ from the geometry深入：由几何推导 $s = \lambda D/d$

A bright fringe at angle $\theta$ requires the path difference from the two slits to be a whole number of wavelengths:

角度 $\theta$ 处的亮纹要求两缝光程差为整数倍波长：

$$ d \sin\theta = n\lambda. $$

For small angles (screen far away, $D \gg d$), $\sin\theta \approx \tan\theta = y_n / D$, where $y_n$ is the position of the $n$-th bright fringe. Hence

小角时（屏很远，$D \gg d$），$\sin\theta \approx \tan\theta = y_n / D$，其中 $y_n$ 为第 $n$ 条亮纹位置。故

$$ d\,\frac{y_n}{D} = n\lambda \;\Rightarrow\; y_n = \frac{n\lambda D}{d}. $$

The spacing between adjacent fringes is the difference between consecutive $y_n$:

相邻条纹间距为相邻 $y_n$ 之差：

$$ s = y_{n+1} - y_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}. $$

The result is independent of $n$, which is why the fringes are evenly spaced. Each variable's effect: longer $\lambda$ or larger $D$ widens the fringes; larger slit separation $d$ squeezes them together.

结果与 $n$ 无关，故条纹等间距。各变量影响：$\lambda$ 越大或 $D$ 越大，条纹越宽；缝间距 $d$ 越大，条纹越密。

In a double-slit experiment, the slit separation $d$ is doubled while everything else is fixed. The fringe spacing $s$:双缝实验中缝间距 $d$ 加倍，其余不变。条纹间距 $s$：
C3.5 · Q1
Doubles加倍
Halves减半
Stays the same不变
Quadruples变为四倍
$s = \lambda D/d$, so $s \propto 1/d$. Doubling $d$ halves $s$: the fringes crowd closer together.$s = \lambda D/d$，故 $s \propto 1/d$。$d$ 加倍则 $s$ 减半：条纹更密。
$d$ is in the denominator of $s = \lambda D/d$. Increasing $d$ decreases $s$ proportionally.$d$ 在 $s = \lambda D/d$ 的分母上，$d$ 增大则 $s$ 成比例减小。
Red light is replaced by blue light in the same double-slit apparatus. The fringes become:同一双缝装置中用蓝光替换红光。条纹将：
C3.5 · Q2
More widely spaced间距变大
Unchanged in spacing间距不变
More closely spaced间距变小
Replaced by a single bright spot变成单个亮点
$s = \lambda D/d$, so $s \propto \lambda$. Blue light has a shorter wavelength than red, so the fringe spacing decreases.$s = \lambda D/d$，故 $s \propto \lambda$。蓝光波长比红光短，故条纹间距变小。
Fringe spacing is proportional to wavelength. Blue ($\sim 450\ \mathrm{nm}$) is shorter than red ($\sim 650\ \mathrm{nm}$), so $s$ shrinks.条纹间距与波长成正比。蓝光（$\sim 450\ \mathrm{nm}$）比红光（$\sim 650\ \mathrm{nm}$）短，故 $s$ 减小。
Topic C3.6 · Grating and Polarization (HL)Topic C3.6 · 光栅与偏振（HL）

Diffraction Grating and Polarization衍射光栅与偏振 HL only C.3 AHL

Diffraction grating. HL A grating has many equally-spaced slits with spacing $d$ (often quoted as lines per metre, $N$, with $d = 1/N$). Maxima occur where the path difference between adjacent slits is a whole number of wavelengths (from the data booklet): d sin θ = nλ $$ d \sin\theta = n\lambda, \qquad n = 0, 1, 2, \dots $$ Many slits make each maximum very sharp and bright, so gratings measure wavelengths far more precisely than two slits.

Polarization. HL Light is a transverse wave, so its oscillation can be confined to one plane (polarized). Unpolarized light through an ideal polarizer loses half its intensity: $I = \tfrac{1}{2} I_0$.

Malus's law (from the data booklet). HL For already-polarized light hitting an analyser at angle $\theta$ to its transmission axis: I = I₀ cos²θ $$ I = I_0 \cos^2\theta. $$
衍射光栅（diffraction grating）。HL 光栅有许多等间距的缝，间距为 $d$（常以每米线数 $N$ 给出，$d = 1/N$）。当相邻缝光程差为整数倍波长时出现极大（数据手册）：d sin θ = nλ $$ d \sin\theta = n\lambda, \qquad n = 0, 1, 2, \dots $$ 缝多使每个极大非常锐利明亮，故光栅测波长比双缝精确得多。

偏振（polarization）。HL 光是横波，故其振动可被限制在一个平面内（偏振）。非偏振光通过理想偏振片后强度减半：$I = \tfrac{1}{2} I_0$。

马吕斯定律（Malus's law，数据手册）。HL 已偏振光以与检偏器透光轴夹角 $\theta$ 入射：I = I₀ cos²θ $$ I = I_0 \cos^2\theta. $$
Worked Example C3.6a (grating angle) HLC3.6a 例题（光栅角度）HL

A diffraction grating has $500\ \mathrm{lines\,mm^{-1}}$. Light of wavelength $589\ \mathrm{nm}$ is shone normally onto it. Find the angle of the first-order ($n = 1$) maximum.某衍射光栅有 $500\ \mathrm{lines\,mm^{-1}}$。波长 $589\ \mathrm{nm}$ 的光垂直入射。求一级（$n = 1$）极大的角度。

Find the slit spacing. $500$ lines per mm means $500 \times 10^{3}$ lines per metre, so

求缝间距。每毫米 $500$ 线即每米 $500 \times 10^{3}$ 线，故

$$ d = \frac{1}{500 \times 10^{3}} = 2.0 \times 10^{-6}\ \mathrm{m}. $$

Apply $d\sin\theta = n\lambda$ with $n = 1$.

用 $d\sin\theta = n\lambda$（$n = 1$）。

$$ \sin\theta = \frac{n\lambda}{d} = \frac{(1)(589 \times 10^{-9})}{2.0 \times 10^{-6}} = 0.2945, \qquad \theta = \arcsin(0.2945) \approx 17.1^{\circ}. $$

Evaluate. The second order would sit at $\sin\theta = 0.589$, i.e. $\theta \approx 36.1^{\circ}$ — larger angle for higher order, as expected.

评估。二级位于 $\sin\theta = 0.589$，即 $\theta \approx 36.1^{\circ}$——级数越高角度越大，符合预期。

Worked Example C3.6b (Malus's law) HLC3.6b 例题（马吕斯定律）HL

Unpolarized light of intensity $I_0 = 80\ \mathrm{W\,m^{-2}}$ passes through a polarizer, then through a second polarizer (analyser) whose axis is at $60^{\circ}$ to the first. Find the transmitted intensity after each polarizer.强度 $I_0 = 80\ \mathrm{W\,m^{-2}}$ 的非偏振光先通过一个偏振片，再通过第二个偏振片（检偏器），其轴与第一个成 $60^{\circ}$。求每个偏振片后的透射强度。

After the first polarizer. Unpolarized light loses half its intensity: $I_1 = \tfrac{1}{2}(80) = 40\ \mathrm{W\,m^{-2}}$. The light is now polarized along the first axis.

第一个偏振片后。非偏振光强度减半：$I_1 = \tfrac{1}{2}(80) = 40\ \mathrm{W\,m^{-2}}$。此时光沿第一偏振片轴偏振。

After the analyser. Now apply Malus's law $I = I_1 \cos^2\theta$ with $\theta = 60^{\circ}$:

检偏器后。用马吕斯定律 $I = I_1 \cos^2\theta$（$\theta = 60^{\circ}$）：

$$ I_2 = (40)\cos^2 60^{\circ} = (40)(0.5)^2 = (40)(0.25) = 10\ \mathrm{W\,m^{-2}}. $$

Evaluate. The half-intensity rule applies only to the first polarizer (unpolarized input). Malus's law $I = I_0\cos^2\theta$ then governs every further polarizer in the chain.

评估。"强度减半"规则只对第一个偏振片（输入为非偏振光）成立。其后链上每个偏振片都用马吕斯定律 $I = I_0\cos^2\theta$。

Going deeper: why Malus's law is a $\cos^2$ and not a $\cos$ HL深入：马吕斯定律为何是 $\cos^2$ 而非 $\cos$ HL

A polarizer transmits only the component of the electric field $\vec{E}$ parallel to its transmission axis. If the incoming field has amplitude $E_0$ at angle $\theta$ to the axis, the transmitted amplitude is

偏振片只透过电场 $\vec{E}$ 中平行于透光轴的分量。若入射场振幅为 $E_0$、与轴成角 $\theta$，则透射振幅为

$$ E = E_0 \cos\theta. $$

Intensity is proportional to the square of the amplitude ($I \propto E^2$), so

强度正比于振幅的平方（$I \propto E^2$），故

$$ I = I_0 \cos^2\theta. $$

Checks: at $\theta = 0$, $I = I_0$ (full transmission); at $\theta = 90^{\circ}$, $I = 0$ (crossed polarizers block all light); at $\theta = 45^{\circ}$, $I = I_0/2$. The squared cosine is the signature of "amplitude projects, intensity is amplitude-squared".

验证：$\theta = 0$ 时 $I = I_0$（全透）；$\theta = 90^{\circ}$ 时 $I = 0$（正交偏振片全挡）；$\theta = 45^{\circ}$ 时 $I = I_0/2$。余弦平方正是"振幅投影、强度为振幅平方"的标志。

HL Polarized light of intensity $I_0$ hits an analyser at $30^{\circ}$ to its axis. The transmitted intensity is:HL 强度 $I_0$ 的偏振光以与检偏器轴成 $30^{\circ}$ 入射。透射强度为：
C3.6 · Q1
$0.50\, I_0$
$0.87\, I_0$
$0.75\, I_0$
$0.25\, I_0$
Malus's law: $I = I_0\cos^2 30^{\circ} = I_0 (0.866)^2 = 0.75\, I_0$. Note it is $\cos^2$, not $\cos$ ($= 0.87$).马吕斯定律：$I = I_0\cos^2 30^{\circ} = I_0 (0.866)^2 = 0.75\, I_0$。注意是 $\cos^2$，不是 $\cos$（$= 0.87$）。
Use $I = I_0\cos^2\theta$. Square the cosine: $\cos 30^{\circ} = 0.866$, $\cos^2 30^{\circ} = 0.75$.用 $I = I_0\cos^2\theta$。把余弦平方：$\cos 30^{\circ} = 0.866$，$\cos^2 30^{\circ} = 0.75$。
HL Compared with a double slit, why does a diffraction grating give more precise wavelength measurements?HL 与双缝相比，衍射光栅为何能更精确地测量波长？
C3.6 · Q2
Its many slits make each maximum much sharper and brighter缝多使每个极大更锐利、更亮
It changes the wavelength of the light它改变了光的波长
It removes the need to measure any angle它无需测量任何角度
It works only for sound, not light它只对声波有效，对光无效
A grating's thousands of slits make the maxima extremely narrow and intense, so the angle $\theta$ in $d\sin\theta = n\lambda$ can be located very precisely, giving a sharper wavelength reading.光栅成千上万条缝使极大极窄极强，故 $d\sin\theta = n\lambda$ 中的角度 $\theta$ 可被极精确定位，从而更精确地读出波长。
The grating uses the same $d\sin\theta = n\lambda$ relation; its advantage is that many slits sharpen each maximum.光栅用的仍是 $d\sin\theta = n\lambda$；其优势在于缝多使每个极大更锐利。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Angles from the normal (every refraction question)角度从法线量起（每道折射题）
  • All angles in reflection, refraction and critical-angle work are measured from the normal, never from the surface. A diagram that gives the "angle to the surface" needs you to subtract from $90^{\circ}$ first.
  • 反射、折射与临界角中的所有角度都从法线量起，绝不从表面量。若图给的是"与表面的夹角"，先用 $90^{\circ}$ 减去。
  • Set the calculator to degrees unless the question explicitly uses radians (single-slit $\theta = \lambda/b$ is in radians).
  • 计算器设为角度（度），除非题目明确用弧度（单缝 $\theta = \lambda/b$ 用弧度）。
Constructive vs destructive相长与相消
  • Compute the path difference, then divide by $\lambda$. A whole number ⇒ constructive; a half-integer ⇒ destructive (for in-phase sources).
  • 先算光程差，再除以 $\lambda$。整数 ⇒ 相长；半整数 ⇒ 相消（对同相源）。
  • If the sources are anti-phase, the rules swap. Read the question for any built-in phase offset.
  • 若两源反相，规则互换。注意题中是否已有相位差。
Don't confuse $b$ and $d$别混淆 $b$ 与 $d$
  • Single-slit width is $b$; double-slit separation is $d$. $\theta = \lambda/b$ (single slit, first minimum, HL) versus $s = \lambda D/d$ (double slit, fringe spacing).
  • 单缝宽度是 $b$；双缝间距是 $d$。$\theta = \lambda/b$（单缝第一极小，HL）与 $s = \lambda D/d$（双缝条纹间距）。
  • Convert nm to m and mm to m before substituting. Mixed units are the single biggest source of lost marks in this unit.
  • 代入前把 nm 化为 m、mm 化为 m。单位不统一是本单元失分的最大来源。
Polarization chains (HL)偏振链（HL）
  • The "halve the intensity" rule applies only to the first polarizer hit by unpolarized light. Every polarizer after that uses Malus's law $I = I_0\cos^2\theta$.
  • "强度减半"规则只用于被非偏振光击中的第一个偏振片。之后每个偏振片都用马吕斯定律 $I = I_0\cos^2\theta$。
  • The angle $\theta$ in Malus's law is between adjacent transmission axes, not relative to some fixed reference.
  • 马吕斯定律中的 $\theta$ 是相邻透光轴之间的夹角，不是相对某个固定参考的角。
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Law of reflection?反射定律？
$\theta_i = \theta_r$, both from the normal.$\theta_i = \theta_r$，均从法线量起。
Refractive index?折射率？
$$n = \frac{c}{v}$$
Snell's law?Snell 定律？
$$n_1 \sin\theta_1 = n_2 \sin\theta_2$$
Critical angle (to air)?临界角（对空气）？
$$\sin\theta_c = \frac{1}{n}$$
Two conditions for TIR?全反射两个条件？
Dense to less-dense ($n_1 > n_2$); $\theta_1 > \theta_c$.光密到光疏（$n_1 > n_2$）；$\theta_1 > \theta_c$。
Single-slit first minimum? HL单缝第一极小？HL
$$\theta = \frac{\lambda}{b}$$
Constructive path difference?相长光程差？
$$\Delta = n\lambda$$
Destructive path difference?相消光程差？
$$\Delta = \left(n + \tfrac{1}{2}\right)\lambda$$
Principle of superposition?叠加原理？
Resultant displacement = vector sum of overlapping waves.合位移 = 叠加波的矢量和。
Double-slit fringe spacing?双缝条纹间距？
$$s = \frac{\lambda D}{d}$$
Double-slit bright fringe condition?双缝亮纹条件？
$$d\sin\theta = n\lambda$$
Diffraction grating maxima? HL衍射光栅极大？HL
$$d\sin\theta = n\lambda$$
Malus's law? HL马吕斯定律？HL
$$I = I_0 \cos^2\theta$$
Unpolarized through one polarizer? HL非偏振光过一个偏振片？HL
$$I = \tfrac{1}{2} I_0$$
Mixed Practice综合练习

Unit C.3 Practice Quiz单元 C.3 练习测验

A ray in air ($n = 1.00$) enters water ($n = 1.33$) at $50^{\circ}$ to the normal. The refraction angle is closest to:空气（$n = 1.00$）中一束光以与法线成 $50^{\circ}$ 射入水（$n = 1.33$）。折射角最接近：
Q1
$50^{\circ}$
$67^{\circ}$
$35^{\circ}$
$25^{\circ}$
$\sin\theta_2 = (1.00\sin 50^{\circ})/1.33 = 0.766/1.33 = 0.576$, so $\theta_2 = \arcsin(0.576) \approx 35.2^{\circ}$. The ray bends toward the normal entering the denser water.$\sin\theta_2 = (1.00\sin 50^{\circ})/1.33 = 0.576$，故 $\theta_2 \approx 35.2^{\circ}$。进入光密的水，光线偏向法线。
Apply $n_1\sin\theta_1 = n_2\sin\theta_2$ and solve for $\theta_2$. Denser medium ⇒ smaller refraction angle.用 $n_1\sin\theta_1 = n_2\sin\theta_2$ 解 $\theta_2$。光密介质 ⇒ 折射角更小。
A diamond has refractive index $2.42$. Its critical angle against air is approximately:钻石折射率为 $2.42$。其对空气的临界角约为：
Q2
$42^{\circ}$
$24^{\circ}$
$66^{\circ}$
$48^{\circ}$
$\sin\theta_c = 1/n = 1/2.42 = 0.413$, so $\theta_c = \arcsin(0.413) \approx 24.4^{\circ}$. Diamond's small critical angle is why it sparkles — most rays totally reflect inside.$\sin\theta_c = 1/n = 1/2.42 = 0.413$，故 $\theta_c \approx 24.4^{\circ}$。钻石临界角小，故闪烁——大多数光线在内部全反射。
$\sin\theta_c = 1/n$. Larger index gives a smaller critical angle, so a high-$n$ diamond has a small $\theta_c$.$\sin\theta_c = 1/n$。折射率越大临界角越小，故高 $n$ 的钻石 $\theta_c$ 小。
Two in-phase coherent sources have a path difference of $1.5\lambda$ at a point. The interference there is:两个相干同相源在某点的光程差为 $1.5\lambda$。该处干涉为：
Q3
Destructive相消
Constructive相长
Neither — light is absorbed两者皆非——光被吸收
Impossible to determine无法判断
$1.5\lambda = (1 + \tfrac{1}{2})\lambda$ is a half-integer number of wavelengths, so the waves arrive out of phase: destructive interference.$1.5\lambda = (1 + \tfrac{1}{2})\lambda$ 为半整数倍波长，两波反相到达：相消干涉。
Half-integer multiples of $\lambda$ ($0.5, 1.5, 2.5\dots$) give destructive interference for in-phase sources.半整数倍 $\lambda$（$0.5, 1.5, 2.5\dots$）对同相源给出相消干涉。
In a double-slit experiment, $\lambda = 633\ \mathrm{nm}$, $d = 0.30\ \mathrm{mm}$, $D = 2.0\ \mathrm{m}$. The fringe spacing is:双缝实验中 $\lambda = 633\ \mathrm{nm}$、$d = 0.30\ \mathrm{mm}$、$D = 2.0\ \mathrm{m}$。条纹间距为：
Q4
$0.42\ \mathrm{mm}$
$9.5\ \mathrm{mm}$
$0.95\ \mathrm{mm}$
$4.2\ \mathrm{mm}$
$s = \lambda D/d = (633 \times 10^{-9})(2.0)/(0.30 \times 10^{-3}) = 4.22 \times 10^{-3}\ \mathrm{m} = 4.2\ \mathrm{mm}$.$s = \lambda D/d = (633 \times 10^{-9})(2.0)/(0.30 \times 10^{-3}) = 4.22 \times 10^{-3}\ \mathrm{m} = 4.2\ \mathrm{mm}$。
Convert all lengths to metres, then $s = \lambda D/d$. Watch the powers of ten on nm and mm.先把所有长度化为米，再用 $s = \lambda D/d$。注意 nm、mm 的十次幂。
HL Unpolarized light of intensity $I_0$ passes through two polarizers whose axes are crossed at $90^{\circ}$. The final intensity is:HL 强度 $I_0$ 的非偏振光通过两个轴正交（$90^{\circ}$）的偏振片。最终强度为：
Q5
$0.50\, I_0$
Zero为零
$0.25\, I_0$
$I_0$
After the first polarizer: $I_1 = \tfrac{1}{2}I_0$. The analyser is at $90^{\circ}$, so Malus gives $I = I_1\cos^2 90^{\circ} = I_1 \cdot 0 = 0$. Crossed polarizers block all light.第一片后 $I_1 = \tfrac{1}{2}I_0$。检偏器成 $90^{\circ}$，马吕斯给出 $I = I_1\cos^2 90^{\circ} = 0$。正交偏振片全挡。
First polarizer halves the intensity; then Malus's law with $\theta = 90^{\circ}$ gives $\cos^2 90^{\circ} = 0$, so nothing passes.第一片使强度减半；再用马吕斯定律（$\theta = 90^{\circ}$）得 $\cos^2 90^{\circ} = 0$，故无光通过。
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