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Solutions详解

Light and Geometric Optics · Solutions光与几何光学 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 20 marksAP 选择题 + 安/卑省考短答 · 共 20 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 EM spectrum电磁波谱 · HS-PS4-3 [3 marks][3 分]

Order the EM spectrum regions from lowest to highest photon energy.把电磁波谱区域按光子能量从低到高排列。

Answer:答案：  (A)  radio, infrared, visible, ultraviolet, gamma无线电、红外、可见光、紫外、伽马

(a) Use $E = hf$: photon energy rises with frequency用 $E = hf$：光子能量随频率升高 M1·A1·A1

Photon energy is $E = hf$, so ordering by energy is the same as ordering by frequency (and the reverse of ordering by wavelength). From low frequency to high frequency:光子能量为 $E = hf$，故按能量排序与按频率排序相同（与按波长排序相反）。从低频到高频： $$ \text{radio} \to \text{microwave} \to \text{infrared} \to \text{visible} \to \text{ultraviolet} \to \text{X-ray} \to \text{gamma}. $$ Dropping the regions not listed, the requested order is (A): radio, infrared, visible, ultraviolet, gamma.略去未列出的区域，所求顺序为 (A)：无线电、红外、可见光、紫外、伽马。
Why the distractors fail.干扰项分析。
(B): this is highest-to-lowest energy, the reverse of what was asked.这是能量从高到低，与所求方向相反。
(C), (D): place radio (lowest energy) after visible, scrambling the order.把无线电（能量最低）排在可见光之后，顺序错乱。
One spectrum, three equivalent rankings.同一波谱，三种等价排序。 Because $c = f\lambda$ (constant $c$) and $E = hf$, frequency, photon energy, and the inverse of wavelength all increase together across the spectrum. Memorise one order (radio at the low-energy, long-wavelength end; gamma at the high-energy, short-wavelength end) and you can answer any "rank these" question. Visible light sits in the middle, roughly $400$ to $700$ nm. A frequent error is to confuse "longest wavelength" with "highest energy"; they are opposite ends.由于 $c = f\lambda$（$c$ 恒定）且 $E = hf$，频率、光子能量与波长的倒数在整个波谱中同步增大。只需记住一种顺序（无线电在低能量、长波长端；伽马在高能量、短波长端），便能回答任何"排序"问题。可见光位于中段，约 $400$ 至 $700$ nm。常见错误是把"波长最长"误当作"能量最高"，二者恰好相反。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Law of reflection反射定律 · HS-PS4-3 [3 marks][3 分]

Ray strikes a flat mirror at $25^{\circ}$ to the mirror surface. Angle of reflection from the normal?光线以与镜面成 $25^{\circ}$ 射到平面镜上。从法线量的反射角？

Answer:答案：  (B)  $65^{\circ}$

(a) Convert the surface angle to the angle from the normal, then apply $\theta_i = \theta_r$把镜面角换算为从法线量的角，再用 $\theta_i = \theta_r$ M1·A1·A1

The normal is perpendicular to the surface, so the angle of incidence measured from the normal is the complement of the $25^{\circ}$ surface angle:法线垂直于镜面，故从法线量起的入射角是 $25^{\circ}$ 镜面角的余角： $$ \theta_i \;=\; 90^{\circ} - 25^{\circ} \;=\; 65^{\circ}. $$ By the law of reflection $\theta_r = \theta_i = 65^{\circ}$, option (B).由反射定律 $\theta_r = \theta_i = 65^{\circ}$，选 (B)。
Why the distractors fail.干扰项分析。
(A) $25^{\circ}$: uses the angle from the surface directly, forgetting to convert to the normal.直接用从镜面量的角，忘记换算到法线。
(C) $50^{\circ}$: doubles the surface angle.把镜面角加倍。
(D) $115^{\circ}$: adds $90^{\circ}$ instead of subtracting.把 $90^{\circ}$ 相加而非相减。
Every angle in reflection and refraction is measured from the normal.反射与折射中所有角度都从法线量起。 This is the single most common slip in geometric optics. Both the law of reflection ($\theta_i = \theta_r$) and Snell's law use the angle between the ray and the line perpendicular to the surface, not the surface itself. When a problem states the angle "to the surface" or "from the mirror," subtract it from $90^{\circ}$ first. Drawing the normal as a dashed line at the point of incidence before reading off any angle eliminates this error.这是几何光学中最常见的失误。反射定律（$\theta_i = \theta_r$）与斯涅尔定律使用的都是光线与垂直于界面之直线（法线）的夹角，而非与界面本身的夹角。当题目给出"与界面成"或"与镜面成"的角时，先用 $90^{\circ}$ 减去它。在读取任何角度前，先在入射点把法线画成虚线，即可消除此类错误。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §1 Wave-particle nature波粒二象性 · SPH3U [5 marks][5 分]

Blue light, $\lambda = 480$ nm in vacuum. (a) Frequency. (b) Photon energy. (c) Which model each step uses. $c = 3.00\times10^8$, $h = 6.63\times10^{-34}$.蓝光，真空中 $\lambda = 480$ nm。(a) 频率。(b) 光子能量。(c) 各步所用模型。$c = 3.00\times10^8$，$h = 6.63\times10^{-34}$。

Answer:答案：  (a) $f = 6.25\times10^{14}\ \text{Hz}$  ·  (b) $E = 4.14\times10^{-19}\ \text{J}$  ·  (c) (a) wave model, (b) particle model(a) 波动模型，(b) 粒子模型

(a) Frequency from $c = f\lambda$由 $c = f\lambda$ 求频率 M1·A1

Convert nm to m: $480\ \text{nm} = 480 \times 10^{-9}\ \text{m}$.把 nm 换算为 m：$480\ \text{nm} = 480 \times 10^{-9}\ \text{m}$。 $$ f \;=\; \frac{c}{\lambda} \;=\; \frac{3.00 \times 10^8}{480 \times 10^{-9}} \;=\; 6.25 \times 10^{14}\ \text{Hz.} $$

(b) Photon energy from $E = hf$由 $E = hf$ 求光子能量 M1·A1

$$ E \;=\; hf \;=\; (6.63 \times 10^{-34})(6.25 \times 10^{14}) \;=\; 4.14 \times 10^{-19}\ \text{J.} $$

(c) Which model each step relies on各步依赖哪种模型 A1

Part (a) treats light as a wave ($c = f\lambda$ relates wave speed, frequency, wavelength). Part (b) treats light as a stream of particles, with each photon carrying a discrete energy quantum $E = hf$.(a) 把光当作波处理（$c = f\lambda$ 联系波速、频率、波长）。(b) 把光当作粒子流处理，每个光子携带离散能量量子 $E = hf$。
Light is described by whichever model fits the question: wave for $c = f\lambda$, particle for $E = hf$.用最契合问题的模型描述光：$c = f\lambda$ 用波动模型，$E = hf$ 用粒子模型。 Frequency is the bridge between the two models: it appears in both the wave relation $c = f\lambda$ and the photon relation $E = hf$. The wave model explains diffraction, interference, and refraction; the particle model explains the photoelectric effect and per-photon energy. NGSS HS-PS4-3 asks students to evaluate exactly this: that EMR can be described by either model and one is sometimes more useful. Note the unit chain: Hz $=$ s$^{-1}$, and J $=$ (J·s)(s$^{-1}$), so the units cross-check the arithmetic.频率是连接两种模型的桥梁：它既出现在波动关系 $c = f\lambda$ 中，也出现在光子关系 $E = hf$ 中。波动模型解释衍射、干涉与折射；粒子模型解释光电效应与单光子能量。NGSS HS-PS4-3 正是要求学生评估这一点：电磁辐射可用任一模型描述，且某一模型有时更有用。注意单位链：Hz $=$ s$^{-1}$，J $=$ (J·s)(s$^{-1}$)，单位可交叉核验算术。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §4 Snell's law斯涅尔定律 · HS-PS4-3 [3 marks][3 分]

Air ($n=1.00$) to water ($n=1.33$), $\theta_1 = 45^{\circ}$. Angle of refraction?空气（$n=1.00$）到水（$n=1.33$），$\theta_1 = 45^{\circ}$。折射角？

Answer:答案：  (A)  $\approx 32^{\circ}$

(a) Apply Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$应用斯涅尔定律 $n_1\sin\theta_1 = n_2\sin\theta_2$ M1·A1·A1

$$ \sin\theta_2 \;=\; \frac{n_1\sin\theta_1}{n_2} \;=\; \frac{1.00 \times \sin 45^{\circ}}{1.33} \;=\; \frac{0.707}{1.33} \;=\; 0.532. $$ $$ \theta_2 \;=\; \arcsin(0.532) \;\approx\; 32^{\circ}. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $45^{\circ}$: assumes the ray does not bend, ignoring the index change.误以为光线不弯折，忽略折射率变化。
(C) $60^{\circ}$: bends the ray away from the normal, the wrong direction for entering a denser medium.让光线偏离法线，进入光密介质时方向错误。
(D) $34^{\circ}$: multiplies by $n_2$ instead of dividing ($\sin\theta_2 = 1.33 \times \sin45^{\circ}$ is impossible $>1$ here, but a mis-step lands near this value).用乘以 $n_2$ 代替除以（此处会得到 $>1$ 的不可能值，但错步会落在此附近）。
Into a denser medium, light bends toward the normal ($\theta_2 < \theta_1$).进入光密介质时，光偏向法线（$\theta_2 < \theta_1$）。 Water ($n = 1.33$) is optically denser than air ($n = 1.00$), so $\theta_2 = 32^{\circ}$ is smaller than the $45^{\circ}$ incidence, exactly as expected. A quick sanity check: rearranging Snell's law, $\sin\theta_2 = (n_1/n_2)\sin\theta_1$; since $n_1/n_2 < 1$, $\sin\theta_2 < \sin\theta_1$, forcing a smaller refraction angle. If you ever compute $\sin\theta_2 > 1$, that signals total internal reflection (only possible going dense to less dense). Always confirm the bending direction against the density change before trusting the number.水（$n = 1.33$）的光学密度大于空气（$n = 1.00$），故 $\theta_2 = 32^{\circ}$ 小于 $45^{\circ}$ 入射角，与预期一致。快速核验：改写斯涅尔定律 $\sin\theta_2 = (n_1/n_2)\sin\theta_1$；因 $n_1/n_2 < 1$，故 $\sin\theta_2 < \sin\theta_1$，折射角必更小。若算出 $\sin\theta_2 > 1$，则提示发生全内反射（只可能由光密射向光疏）。在相信数值前，先用密度变化核对弯折方向。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Curved mirrors曲面镜 · Physics 11 [6 marks][6 分]

Concave mirror, $f = 10$ cm, object at $d_o = 15$ cm. (a) $d_i$. (b) $m$. (c) Nature of image.凹镜，$f = 10$ cm，物在 $d_o = 15$ cm。(a) $d_i$。(b) $m$。(c) 像的性质。

Answer:答案：  (a) $d_i = +30\ \text{cm}$  ·  (b) $m = -2.0$  ·  (c) real, inverted, enlarged实像、倒立、放大

(a) Image distance from the mirror equation由镜方程求像距 M1·A1·A1

Concave mirror: $f = +10$ cm, $d_o = +15$ cm.凹镜：$f = +10$ cm，$d_o = +15$ cm。 $$ \frac{1}{d_i} \;=\; \frac{1}{f} - \frac{1}{d_o} \;=\; \frac{1}{10} - \frac{1}{15} \;=\; \frac{3}{30} - \frac{2}{30} \;=\; \frac{1}{30}. $$ $$ d_i \;=\; +30\ \text{cm} \quad (\text{positive: real image, in front of the mirror}). $$

(b) Magnification放大率 A1

$$ m \;=\; -\frac{d_i}{d_o} \;=\; -\frac{30}{15} \;=\; -2.0. $$

(c) Nature of the image像的性质 A1·A1

$d_i > 0$ means the image is real. $m < 0$ means inverted. $|m| = 2.0 > 1$ means enlarged (twice the object's height).$d_i > 0$ 表示实像。$m < 0$ 表示倒立。$|m| = 2.0 > 1$ 表示放大（为物高的两倍）。
For a concave mirror with the object between $f$ and $C$, the image is real, inverted, and enlarged.凹镜中物体位于 $f$ 与 $C$ 之间时，像为实像、倒立、放大。 Here $f = 10$ cm so the centre of curvature is $C = 2f = 20$ cm; the object at $15$ cm lies between $F$ and $C$, the regime that always gives a real, inverted, magnified image. The single mirror equation $\tfrac1f = \tfrac1{d_o}+\tfrac1{d_i}$ together with the sign convention ($f > 0$ for concave; $d_i > 0$ real) handles every case. The magnification sign and magnitude then deliver orientation and size in one step: negative for inverted, $|m|>1$ for enlarged. A ray diagram (parallel ray through $F$, focal ray to parallel) confirms the same picture.此处 $f = 10$ cm，故曲率中心 $C = 2f = 20$ cm；物体在 $15$ cm 处，位于 $F$ 与 $C$ 之间，这一区间总给出实像、倒立、放大的像。单一镜方程 $\tfrac1f = \tfrac1{d_o}+\tfrac1{d_i}$ 配合符号约定（凹镜 $f > 0$；实像 $d_i > 0$）即可处理所有情形。放大率的符号与大小随后一步给出取向与尺寸：负号即倒立，$|m|>1$ 即放大。光路图（平行光线过 $F$、焦点光线出射后平行）也确认同一结论。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 34 marksAP 衔接简答题 + 荣誉级 · 共 34 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 Converging lens会聚透镜 · HS-PS4-5 [8 marks][8 分]

Converging lens, $f = 12$ cm. (a) Object at $20$ cm: find $d_i$. (b) Magnification and image nature. (c) Object moved to $8$ cm: find $d_i$. (d) Describe image and name the instrument.会聚透镜，$f = 12$ cm。(a) 物在 $20$ cm：求 $d_i$。(b) 放大率与像的性质。(c) 物移至 $8$ cm：求 $d_i$。(d) 描述像并说出仪器名称。

Answer:答案：  (a) $d_i = +30\ \text{cm}$  ·  (b) $m = -1.5$ (real, inverted)（实、倒立）  ·  (c) $d_i = -24\ \text{cm}$  ·  (d) virtual, upright, enlarged: magnifying glass虚像、正立、放大：放大镜

(a) Image distance with object beyond $f$物在 $f$ 外时的像距 M1·A1

$$ \frac{1}{d_i} \;=\; \frac{1}{f} - \frac{1}{d_o} \;=\; \frac{1}{12} - \frac{1}{20} \;=\; \frac{5}{60} - \frac{3}{60} \;=\; \frac{2}{60} \;=\; \frac{1}{30} \;\Longrightarrow\; d_i = +30\ \text{cm.} $$

(b) Magnification and image nature放大率与像的性质 M1·A1

$$ m \;=\; -\frac{d_i}{d_o} \;=\; -\frac{30}{20} \;=\; -1.5. $$ $d_i > 0$ is a real image; $m < 0$ is inverted; $|m| = 1.5 > 1$ is enlarged.$d_i > 0$ 为实像；$m < 0$ 为倒立；$|m| = 1.5 > 1$ 为放大。

(c) Object now inside $f$ ($d_o = 8 < 12$)物现移入 $f$ 内（$d_o = 8 < 12$） M1·A1

$$ \frac{1}{d_i} \;=\; \frac{1}{12} - \frac{1}{8} \;=\; \frac{2}{24} - \frac{3}{24} \;=\; -\frac{1}{24} \;\Longrightarrow\; d_i = -24\ \text{cm.} $$

(d) Describe the image and name the instrument描述像并说出仪器名称 A1·A1

$d_i = -24$ cm is negative, so the image is virtual, on the same side as the object. $m = -d_i/d_o = -(-24)/8 = +3$: upright and enlarged. This is the magnifying glass (simple microscope) configuration.$d_i = -24$ cm 为负，故像是虚像，与物同侧。$m = -d_i/d_o = -(-24)/8 = +3$：正立且放大。这就是放大镜（简单显微镜）的配置。
A converging lens flips behaviour at the focal point: object beyond $f$ gives a real inverted image; object inside $f$ gives a virtual upright magnified image.会聚透镜在焦点处行为反转：物在 $f$ 外成实像、倒立；物在 $f$ 内成虚像、正立、放大。 The same thin-lens equation governs both parts; only the sign of $d_i$ changes. Crossing the focal point is the watershed: at exactly $d_o = f$ the emerging rays are parallel and no image forms (a $1/d_i = 0$ result). Inside $f$ the rays diverge and the eye traces them back to a virtual image, exactly how a magnifying glass works. Recognising "object inside $f$ of a converging lens" as the magnifying-glass case lets you predict the answer before computing. Carrying the negative sign of $d_i$ through to $m$ is what flips orientation from inverted to upright.两部分受同一薄透镜方程支配，只是 $d_i$ 的符号改变。越过焦点是分水岭：恰好 $d_o = f$ 时出射光线平行、不成像（$1/d_i = 0$）。在 $f$ 内光线发散，眼睛沿其反向延长得到虚像，这正是放大镜的原理。把"会聚透镜物在 $f$ 内"识别为放大镜情形，便能在计算前预判答案。将 $d_i$ 的负号一路代入 $m$，正是取向从倒立翻转为正立的关键。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Refraction in glass玻璃中的折射 · SPH3U [8 marks][8 分]

Air ($n=1.00$) to glass ($n=1.50$), $\theta_1 = 50^{\circ}$, $\lambda_{\text{air}} = 600$ nm, $c = 3.00\times10^8$. (a) Refraction angle. (b) Speed in glass. (c) Wavelength in glass. (d) Bending direction.空气（$n=1.00$）到玻璃（$n=1.50$），$\theta_1 = 50^{\circ}$，$\lambda_{\text{air}} = 600$ nm，$c = 3.00\times10^8$。(a) 折射角。(b) 玻璃中速度。(c) 玻璃中波长。(d) 弯折方向。

Answer:答案：  (a) $\theta_2 \approx 30.7^{\circ}$  ·  (b) $v = 2.00\times10^8\ \text{m/s}$  ·  (c) $\lambda_{\text{glass}} = 400\ \text{nm}$  ·  (d) toward the normal偏向法线

(a) Refraction angle from Snell's law由斯涅尔定律求折射角 M1·A1·A1

$$ \sin\theta_2 \;=\; \frac{n_1\sin\theta_1}{n_2} \;=\; \frac{1.00 \times \sin 50^{\circ}}{1.50} \;=\; \frac{0.766}{1.50} \;=\; 0.511. $$ $$ \theta_2 \;=\; \arcsin(0.511) \;\approx\; 30.7^{\circ}. $$

(b) Speed of light in the glass, $n = c/v$玻璃中光速，$n = c/v$ M1·A1

$$ v \;=\; \frac{c}{n} \;=\; \frac{3.00 \times 10^8}{1.50} \;=\; 2.00 \times 10^8\ \text{m/s.} $$

(c) Wavelength in the glass, $\lambda_{\text{glass}} = \lambda_{\text{air}}/n$玻璃中波长，$\lambda_{\text{glass}} = \lambda_{\text{air}}/n$ M1·A1

$$ \lambda_{\text{glass}} \;=\; \frac{\lambda_{\text{air}}}{n} \;=\; \frac{600\ \text{nm}}{1.50} \;=\; 400\ \text{nm.} $$

(d) Bending direction弯折方向 A1

The ray bends toward the normal ($30.7^{\circ} < 50^{\circ}$) because glass is optically denser ($n_2 > n_1$): light slows, so it bends toward the normal on entering.光线偏向法线（$30.7^{\circ} < 50^{\circ}$），因为玻璃光学密度更大（$n_2 > n_1$）：光减速，故进入时偏向法线。
Frequency is conserved across a boundary; speed and wavelength both drop by the factor $n$.频率在界面处守恒；速度与波长都按因子 $n$ 减小。 When light enters a denser medium, the source sets the frequency, so $f$ cannot change. Since $v = f\lambda$ and $v$ falls to $c/n$, the wavelength must fall to $\lambda_{\text{air}}/n$ by the same factor. Here both drop by $1.50$: $v$ from $3.00\times10^8$ to $2.00\times10^8$ m/s, and $\lambda$ from $600$ to $400$ nm. The colour we perceive depends on frequency, which is why the light still looks the same colour underwater even though its wavelength has shortened. A common error is to "shrink" the frequency instead of the wavelength.光进入光密介质时，频率由光源决定，故 $f$ 不能改变。由于 $v = f\lambda$ 且 $v$ 降为 $c/n$，波长必按同一因子降为 $\lambda_{\text{air}}/n$。此处两者都按 $1.50$ 减小：$v$ 从 $3.00\times10^8$ 降到 $2.00\times10^8$ m/s，$\lambda$ 从 $600$ 降到 $400$ nm。我们感知的颜色取决于频率，这就是为何光在水下看起来颜色不变，尽管波长已缩短。常见错误是把"频率"而非"波长"缩小。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §6 Diverging lens & power发散透镜与光焦度 · Physics 11 [8 marks][8 分]

Diverging lens, $f = -15$ cm, object at $30$ cm. (a) $d_i$. (b) Magnification and image nature. (c) Power in dioptres. (d) Why no real image of a real object.发散透镜，$f = -15$ cm，物在 $30$ cm。(a) $d_i$。(b) 放大率与像的性质。(c) 屈光度光焦度。(d) 为何不能对实物体成实像。

Answer:答案：  (a) $d_i = -10\ \text{cm}$  ·  (b) $m = +\tfrac{1}{3}$ (virtual, upright, diminished)（虚、正立、缩小）  ·  (c) $P = -6.7\ \text{D}$  ·  (d) diverging rays never converge发散光线永不会聚

(a) Image distance from the thin-lens equation由薄透镜方程求像距 M1·A1

$$ \frac{1}{d_i} \;=\; \frac{1}{f} - \frac{1}{d_o} \;=\; \frac{1}{-15} - \frac{1}{30} \;=\; -\frac{2}{30} - \frac{1}{30} \;=\; -\frac{3}{30} \;=\; -\frac{1}{10} \;\Longrightarrow\; d_i = -10\ \text{cm.} $$

(b) Magnification and image nature放大率与像的性质 M1·A1

$$ m \;=\; -\frac{d_i}{d_o} \;=\; -\frac{-10}{30} \;=\; +\frac{1}{3} \;\approx\; +0.33. $$ $d_i < 0$: virtual (same side as object). $m > 0$: upright. $|m| < 1$: diminished.$d_i < 0$：虚像（与物同侧）。$m > 0$：正立。$|m| < 1$：缩小。

(c) Power of the lens, $P = 1/f$ (f in metres)透镜光焦度，$P = 1/f$（f 以米计） M1·A1

$$ P \;=\; \frac{1}{f} \;=\; \frac{1}{-0.15\ \text{m}} \;=\; -6.7\ \text{D.} $$

(d) Why a diverging lens cannot form a real image of a real object为何发散透镜不能对实物体成实像 A1·A1

A diverging lens always spreads incoming rays apart, so the refracted rays never actually converge to a point on the far side. The eye traces them backward to a virtual image on the same side as the object. Algebraically, with $f < 0$ and $d_o > 0$, $\tfrac1{d_i} = \tfrac1f - \tfrac1{d_o}$ is always negative, so $d_i < 0$ for every real object position.发散透镜总使入射光线分散，故折射光线在远侧从不真正会聚到一点。眼睛沿其反向延长得到与物同侧的虚像。从代数看，$f < 0$ 且 $d_o > 0$ 时，$\tfrac1{d_i} = \tfrac1f - \tfrac1{d_o}$ 恒为负，故任何实物体位置都有 $d_i < 0$。
A diverging lens has negative power and is the corrective lens for short-sightedness.发散透镜光焦度为负，是矫正近视的镜片。 Power $P = 1/f$ (in dioptres, D) is how optometrists specify lenses: a positive D converges, a negative D diverges. The $-6.7$ D here is strong. Because a diverging lens always gives a virtual, upright, diminished image regardless of object distance, it is predictable and safe to use in spectacles for myopia, where it spreads incoming rays just enough to move the focus back onto the retina. When two thin lenses touch, their powers add ($P_{\text{total}} = P_1 + P_2$), which is why a single dioptre number summarises a prescription.光焦度 $P = 1/f$（单位屈光度 D）是验光师标注透镜的方式：正 D 会聚，负 D 发散。此处 $-6.7$ D 较强。由于发散透镜无论物距如何总成虚像、正立、缩小，其行为可预测，适合用作近视眼镜片，使入射光线恰好散开，把焦点移回视网膜。两块薄透镜紧贴时光焦度相加（$P_{\text{total}} = P_1 + P_2$），这就是为何一个屈光度数即可概括一份验光处方。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 Total internal reflection & fibre optics全内反射与光纤 · HS-PS4-5 [10 marks][10 分]

Glass fibre core $n = 1.50$; $n_{\text{air}} = 1.00$, $n_{\text{water}} = 1.33$, $n_{\text{diamond}} = 2.42$. (a) Two conditions for TIR. (b) Critical angle glass-air. (c) Ray at $45^{\circ}$: TIR? (d) Diamond critical angle and sparkle. (e) Fibre in water: new critical angle and confinement at $45^{\circ}$.玻璃纤芯 $n = 1.50$；$n_{\text{air}} = 1.00$，$n_{\text{water}} = 1.33$，$n_{\text{diamond}} = 2.42$。(a) 全内反射的两个条件。(b) 玻璃-空气临界角。(c) $45^{\circ}$ 光线是否全内反射？(d) 钻石临界角与闪烁。(e) 浸入水中：新临界角与 $45^{\circ}$ 时的约束。

Answer:答案：  (a) dense→less dense; angle $>$ critical angle光密→光疏；角度 $>$ 临界角  ·  (b) $\theta_c \approx 41.8^{\circ}$  ·  (c) yes ($45^{\circ} > 41.8^{\circ}$)会（$45^{\circ} > 41.8^{\circ}$）  ·  (d) $\theta_c \approx 24.4^{\circ}$  ·  (e) $\theta_c \approx 62.5^{\circ}$, no longer confined不再约束

(a) Two conditions required for total internal reflection全内反射所需的两个条件 A1·A1

(1) Light must travel from an optically denser medium toward a less dense one ($n_1 > n_2$). (2) The angle of incidence must exceed the critical angle ($\theta_1 > \theta_c$). Only then is there no refracted ray and all the light reflects back.(1) 光必须从光密介质射向光疏介质（$n_1 > n_2$）。(2) 入射角必须大于临界角（$\theta_1 > \theta_c$）。两者同时满足时才没有折射光线，全部光线反射回原介质。

(b) Critical angle for the glass-air interface, $\sin\theta_c = n_2/n_1$玻璃-空气界面临界角，$\sin\theta_c = n_2/n_1$ M1·A1

$$ \sin\theta_c \;=\; \frac{n_{\text{air}}}{n_{\text{glass}}} \;=\; \frac{1.00}{1.50} \;=\; 0.667 \;\Longrightarrow\; \theta_c \;=\; \arcsin(0.667) \;\approx\; 41.8^{\circ}. $$

(c) Ray at $45^{\circ}$ to the normal与法线成 $45^{\circ}$ 的光线 M1·A1

Compare with the critical angle: $45^{\circ} > 41.8^{\circ}$, so the ray does undergo total internal reflection and stays inside the fibre.与临界角比较：$45^{\circ} > 41.8^{\circ}$，故该光线会发生全内反射，留在纤芯内。

(d) Critical angle for diamond in air, and why diamonds sparkle钻石在空气中的临界角，及钻石为何闪烁 M1·A1

$$ \sin\theta_c \;=\; \frac{1.00}{2.42} \;=\; 0.413 \;\Longrightarrow\; \theta_c \;=\; \arcsin(0.413) \;\approx\; 24.4^{\circ}. $$ The very small critical angle means light entering a cut diamond strikes most internal faces above $24.4^{\circ}$, so it bounces many times by total internal reflection before exiting, producing the intense sparkle.极小的临界角意味着进入切割钻石的光在多数内表面的入射角都超过 $24.4^{\circ}$，故在射出前经多次全内反射，产生强烈的闪烁。

(e) Fibre immersed in water, $\sin\theta_c = n_{\text{water}}/n_{\text{glass}}$光纤浸入水中，$\sin\theta_c = n_{\text{water}}/n_{\text{glass}}$ M1·A1

$$ \sin\theta_c \;=\; \frac{1.33}{1.50} \;=\; 0.887 \;\Longrightarrow\; \theta_c \;=\; \arcsin(0.887) \;\approx\; 62.5^{\circ}. $$ Now $45^{\circ} < 62.5^{\circ}$, so the ray no longer exceeds the critical angle: it refracts out into the water and the fibre no longer confines light at $45^{\circ}$.此时 $45^{\circ} < 62.5^{\circ}$，光线不再超过临界角：它折射进入水中，故光纤在 $45^{\circ}$ 时不再约束光。
A smaller index contrast raises the critical angle, so surrounding a fibre with a higher-index medium can break total internal reflection.折射率反差越小，临界角越大，故用更高折射率的介质包围光纤会破坏全内反射。 The critical angle obeys $\sin\theta_c = n_2/n_1$, so as the outside index $n_2$ climbs toward the core index $n_1$, the ratio approaches 1 and $\theta_c$ approaches $90^{\circ}$. Glass-air ($1.00/1.50$) gives $41.8^{\circ}$, comfortably below the $45^{\circ}$ ray; glass-water ($1.33/1.50$) lifts it to $62.5^{\circ}$, above $45^{\circ}$, so light leaks out. This is exactly why real optical fibres use a low-index cladding glass rather than relying on an air gap: it keeps the critical angle small and tolerant of bends. Diamond is the opposite extreme: its huge index makes $\theta_c$ tiny, trapping light through many internal bounces, which is the optical origin of its brilliance.临界角满足 $\sin\theta_c = n_2/n_1$，故当外侧折射率 $n_2$ 趋近纤芯折射率 $n_1$ 时，比值趋于 1，$\theta_c$ 趋于 $90^{\circ}$。玻璃-空气（$1.00/1.50$）得 $41.8^{\circ}$，明显小于 $45^{\circ}$ 光线；玻璃-水（$1.33/1.50$）把它抬到 $62.5^{\circ}$，超过 $45^{\circ}$，故光泄漏出去。这正是真实光纤采用低折射率包层玻璃而非依赖空气间隙的原因：它使临界角保持较小，对弯折更宽容。钻石是相反的极端：其极高的折射率使 $\theta_c$ 极小，通过多次内反射困住光，这正是其璀璨的光学根源。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §4 + §5 Refraction (applied)折射（应用） · 30-C1.11k [9 marks][9 分]

Sunlight strikes a pool surface at $40^{\circ}$ from the normal; water $n = 1.33$, $n_{\text{air}} = 1.00$, $c = 3.00\times10^8$. (a) Refraction angle in water. (b) Speed in water. (c) Apparent depth of a fish at $2.0$ m. (d) Critical angle water-air and what the fish sees beyond it.阳光以与法线成 $40^{\circ}$ 射到水面；水 $n = 1.33$，$n_{\text{air}} = 1.00$，$c = 3.00\times10^8$。(a) 水中折射角。(b) 水中速度。(c) $2.0$ m 处鱼的表观深度。(d) 水-空气临界角及超过该角时鱼看到什么。

Answer:答案：  (a) $\theta_2 \approx 28.9^{\circ}$  ·  (b) $v = 2.26\times10^8\ \text{m/s}$  ·  (c) $1.5\ \text{m}$  ·  (d) $\theta_c \approx 48.8^{\circ}$, total internal reflection全内反射

(a) Refraction angle from Snell's law由斯涅尔定律求折射角 M1·A1·A1

$$ \sin\theta_2 \;=\; \frac{n_{\text{air}}\sin\theta_1}{n_{\text{water}}} \;=\; \frac{1.00 \times \sin 40^{\circ}}{1.33} \;=\; \frac{0.643}{1.33} \;=\; 0.483. $$ $$ \theta_2 \;=\; \arcsin(0.483) \;\approx\; 28.9^{\circ}. $$

(b) Speed of light in the water, $n = c/v$水中光速，$n = c/v$ M1·A1

$$ v \;=\; \frac{c}{n} \;=\; \frac{3.00 \times 10^8}{1.33} \;=\; 2.26 \times 10^8\ \text{m/s.} $$

(c) Apparent depth $=$ real depth $/\,n$表观深度 $=$ 实际深度 $/\,n$ M1·A1

$$ d_{\text{apparent}} \;=\; \frac{2.0}{1.33} \;\approx\; 1.5\ \text{m.} $$ The fish looks shallower than it really is.鱼看起来比实际位置更浅。

(d) Critical angle for the water-air boundary, $\sin\theta_c = n_{\text{air}}/n_{\text{water}}$水-空气界面临界角，$\sin\theta_c = n_{\text{air}}/n_{\text{water}}$ M1·A1

$$ \sin\theta_c \;=\; \frac{1.00}{1.33} \;=\; 0.752 \;\Longrightarrow\; \theta_c \;=\; \arcsin(0.752) \;\approx\; 48.8^{\circ}. $$ Looking up beyond $48.8^{\circ}$ from the normal, the fish sees no sky: the surface acts as a mirror by total internal reflection, showing a reflected view of the pool bottom.从法线量起超过 $48.8^{\circ}$ 向上看，鱼看不到天空：水面因全内反射如同镜面，呈现池底的反射像。
A single index $n$ controls bending angle, light speed, and apparent depth at a water surface.单一折射率 $n$ 同时支配水面的弯折角、光速与表观深度。 Refraction at the air-water boundary is one physical effect with several everyday consequences. Snell's law bends the entering sunlight toward the normal ($28.9^{\circ} < 40^{\circ}$); the same $n = c/v$ slows the light to $2.26\times10^8$ m/s; and the same $n$ divides the real depth to give the apparent depth ($2.0/1.33 = 1.5$ m), which is why a fish or a pool floor always looks closer to the surface than it is. Going the other way, light leaving water can totally internally reflect once the angle exceeds $\theta_c = 48.8^{\circ}$, so a fish sees the entire sky squeezed into a bright cone overhead (Snell's window) and a mirror everywhere outside it. Notice $\theta_c$ here is exactly the angle whose sine is $1/n$, the same ratio that set the apparent-depth factor.空气-水界面的折射是同一物理效应，却带来多种日常后果。斯涅尔定律使入射阳光偏向法线（$28.9^{\circ} < 40^{\circ}$）；同一 $n = c/v$ 把光减速到 $2.26\times10^8$ m/s；同一 $n$ 把实际深度除得表观深度（$2.0/1.33 = 1.5$ m），这就是为何鱼或池底总看起来比实际更靠近水面。反方向上，光离开水时一旦角度超过 $\theta_c = 48.8^{\circ}$ 便全内反射，故鱼看到整片天空被压缩成头顶上方一个明亮的圆锥（斯涅尔窗），锥外处处如镜。注意此处 $\theta_c$ 恰是正弦为 $1/n$ 的角，与设定表观深度因子的比值相同。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §3 + §7 Curved mirrors (applied)曲面镜（应用） · 30-C1.7k [9 marks][9 分]

Concave makeup mirror $f = 15$ cm, face at $10$ cm. (a) $d_i$. (b) Magnification, image, why it makes a good makeup mirror. (c) Convex security mirror $f = -40$ cm, shopper at $200$ cm: $d_i$ and $m$. (d) Why convex is preferred for store security.凹面化妆镜 $f = 15$ cm，脸在 $10$ cm。(a) $d_i$。(b) 放大率、像、为何适合作化妆镜。(c) 凸面安防镜 $f = -40$ cm，顾客在 $200$ cm：$d_i$ 与 $m$。(d) 安防为何选用凸面镜。

Answer:答案：  (a) $d_i = -30\ \text{cm}$  ·  (b) $m = +3.0$ (virtual, upright, enlarged)（虚、正立、放大）  ·  (c) $d_i \approx -33.3\ \text{cm}$, $m \approx +0.17$  ·  (d) wide field of view视野宽

(a) Image distance from the mirror equation (concave, $f = +15$)由镜方程求像距（凹镜，$f = +15$） M1·A1

$$ \frac{1}{d_i} \;=\; \frac{1}{f} - \frac{1}{d_o} \;=\; \frac{1}{15} - \frac{1}{10} \;=\; \frac{2}{30} - \frac{3}{30} \;=\; -\frac{1}{30} \;\Longrightarrow\; d_i = -30\ \text{cm.} $$

(b) Magnification and why it works as a makeup mirror放大率及为何适合作化妆镜 M1·A1·A1

$$ m \;=\; -\frac{d_i}{d_o} \;=\; -\frac{-30}{10} \;=\; +3.0. $$ $d_i < 0$: virtual (behind the mirror). $m = +3.0$: upright and enlarged $3\times$. The object sits inside the focal length, giving an upright, magnified image of the face, ideal for close work.$d_i < 0$：虚像（在镜后）。$m = +3.0$：正立且放大 $3$ 倍。物体位于焦距以内，给出脸部正立放大的像，适合近距精细操作。

(c) Convex security mirror ($f = -40$, $d_o = 200$)凸面安防镜（$f = -40$，$d_o = 200$） M1·A1·A1

$$ \frac{1}{d_i} \;=\; \frac{1}{-40} - \frac{1}{200} \;=\; -\frac{5}{200} - \frac{1}{200} \;=\; -\frac{6}{200} \;\Longrightarrow\; d_i \approx -33.3\ \text{cm.} $$ $$ m \;=\; -\frac{d_i}{d_o} \;=\; -\frac{-33.3}{200} \;\approx\; +0.17. $$

(d) Why convex is preferred for security安防为何选用凸面镜 A1

A convex mirror always gives an upright, diminished image, so it captures a much wider field of view, letting one mirror watch a large area of the store.凸面镜总成正立、缩小的像，故视野宽得多，一面镜子即可监视商店的大片区域。
The same mirror equation gives a magnifying virtual image for a concave mirror inside its focus and a wide-angle diminished image for a convex mirror.同一镜方程：凹镜在焦距内给出放大虚像，凸镜给出广角缩小像。 Both parts use $\tfrac1f = \tfrac1{d_o}+\tfrac1{d_i}$ with the sign convention $f > 0$ for concave and $f < 0$ for convex. A concave mirror with the object closer than its focal length ($d_o = 10 < f = 15$) gives $d_i < 0$, a virtual, upright, enlarged image, exactly what a makeup or shaving mirror needs. A convex mirror always gives a virtual, upright, diminished image for any real object ($d_i < 0$, $|m| < 1$), trading magnification for a wide field of view, which is why they appear on cars, at blind corners, and in shops. Reading off the signs ($d_i$ for real/virtual, $m$ for orientation and size) classifies the image without redrawing rays.两部分都用 $\tfrac1f = \tfrac1{d_o}+\tfrac1{d_i}$，并采用符号约定：凹镜 $f > 0$，凸镜 $f < 0$。凹镜物距小于焦距（$d_o = 10 < f = 15$）时 $d_i < 0$，成虚像、正立、放大，正是化妆镜或剃须镜所需。凸镜对任何实物体总成虚像、正立、缩小（$d_i < 0$，$|m| < 1$），以放大率换取宽视野，故见于汽车、盲角与商店。读取符号（$d_i$ 定实/虚，$m$ 定取向与尺寸）即可判定像的性质，无需重画光路。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 + §7 Optical instruments光学仪器 · HS-PS4-5 [10 marks][10 分]

Slide projector lens $f = 5.0$ cm, slide at $6.0$ cm. (a) Image distance (screen distance). (b) Magnification and orientation. (c) Telescope $f_o = 60$ cm, $f_e = 2.0$ cm: angular magnification. (d) Objective $6.0$ cm vs pupil $0.60$ cm: light-gathering factor. (e) Corrective lens for a myopic student.幻灯机透镜 $f = 5.0$ cm，幻灯片在 $6.0$ cm。(a) 像距（屏距）。(b) 放大率与取向。(c) 望远镜 $f_o = 60$ cm，$f_e = 2.0$ cm：角放大率。(d) 物镜 $6.0$ cm 对瞳孔 $0.60$ cm：集光倍数。(e) 近视学生的矫正透镜。

Answer:答案：  (a) $d_i = +30\ \text{cm}$  ·  (b) $m = -5.0$ (inverted)（倒立）  ·  (c) $M = 30\times$  ·  (d) $100\times$  ·  (e) diverging (concave) lens发散（凹）透镜

(a) Image distance from the thin-lens equation由薄透镜方程求像距 M1·A1

$$ \frac{1}{d_i} \;=\; \frac{1}{f} - \frac{1}{d_o} \;=\; \frac{1}{5.0} - \frac{1}{6.0} \;=\; \frac{6}{30} - \frac{5}{30} \;=\; \frac{1}{30} \;\Longrightarrow\; d_i = +30\ \text{cm.} $$ The screen must be $30$ cm from the lens.屏幕须距透镜 $30$ cm。

(b) Magnification and orientation放大率与取向 M1·A1

$$ m \;=\; -\frac{d_i}{d_o} \;=\; -\frac{30}{6.0} \;=\; -5.0. $$ $m < 0$ means the projected image is inverted (and $5\times$ enlarged), so the slide must be loaded upside-down to appear upright on the screen.$m < 0$ 表示投影像倒立（且放大 $5$ 倍），故幻灯片须倒置装入才能在屏上正立显示。

(c) Angular magnification of the telescope, $M = f_o/f_e$望远镜角放大率，$M = f_o/f_e$ M1·A1

$$ M \;=\; \frac{f_o}{f_e} \;=\; \frac{60}{2.0} \;=\; 30\times. $$

(d) Light-gathering factor (area $\propto$ diameter$^2$)集光倍数（面积 $\propto$ 直径$^2$） M1·A1

$$ \frac{A_{\text{obj}}}{A_{\text{pupil}}} \;=\; \left(\frac{6.0}{0.60}\right)^2 \;=\; 10^2 \;=\; 100\times. $$

(e) Corrective lens for short-sightedness近视的矫正透镜 A1·A1

A myopic eye is too strong: it focuses distant light in front of the retina. A diverging (concave) lens spreads the incoming rays first, moving the focal point back onto the retina, so distant stars come into focus.近视眼屈光过强：把远处的光聚焦在视网膜之前。发散（凹）透镜先让入射光线散开，把焦点后移到视网膜上，故远处的星星得以聚焦。
Every optical instrument is the thin-lens equation plus one design ratio: magnification, angular magnification, or collecting area.每件光学仪器都是薄透镜方程加一个设计比值：放大率、角放大率或集光面积。 A projector places the object just outside $f$ ($d_o = 6 > f = 5$) to throw a real, inverted, enlarged image far away on a screen, which is why slides load upside-down. A telescope instead chains two lenses and its angular magnification is purely the focal-length ratio $f_o/f_e = 30$, while its light grasp scales with collecting area, not diameter, so a $10\times$ wider objective gathers $10^2 = 100\times$ more light, the real reason big telescopes see faint stars. Vision correction is the same physics run backwards: a myopic eye over-converges, so a diverging lens of negative power pre-spreads the light to land the image on the retina. One equation, three instruments, plus the squared-diameter rule for brightness.幻灯机把物体置于 $f$ 稍外（$d_o = 6 > f = 5$），在远处屏幕上投出实像、倒立、放大，故幻灯片须倒装。望远镜则串联两块透镜，其角放大率纯为焦距比 $f_o/f_e = 30$，而集光能力随集光面积而非直径增长，故物镜宽 $10$ 倍便多集 $10^2 = 100$ 倍的光，这正是大望远镜能看到暗星的真正原因。视力矫正是同一物理的逆用：近视眼会聚过强，故用负光焦度的发散透镜预先散光，使像落到视网膜上。一个方程，三件仪器，外加亮度的直径平方律。