Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus C3.1 to C3.6考纲 C3.1 至 C3.6PHYSICS HL
Ray in air ($n_1 = 1.00$) hits glass ($n_2 = 1.50$) at $50^{\circ}$. (a) angle of reflection and its law; (b) angle of refraction and bending direction; (c) speed of light in the glass.空气（$n_1 = 1.00$）中光以 $50^{\circ}$ 射到玻璃（$n_2 = 1.50$）。(a) 反射角及其定律；(b) 折射角与偏折方向；(c) 玻璃中的光速。
By the law of reflection the angle of reflection equals the angle of incidence, both measured from the normal. (B1)
So the angle of reflection is $50^{\circ}$. (A1)
Apply Snell's law $n_1 \sin\theta_1 = n_2 \sin\theta_2$: $\sin\theta_2 = \dfrac{(1.00)\sin 50^{\circ}}{1.50}$. (M1)
$$ \sin\theta_2 = \frac{0.766}{1.50} = 0.5107, \qquad \theta_2 = \arcsin(0.5107) \approx 30.7^{\circ}. $$The ray enters a denser medium, so it bends toward the normal ($30.7^{\circ} < 50^{\circ}$). (A1)
From $n = c/v$, rearrange to $v = c/n$: (M1)
$$ v = \frac{3.00 \times 10^{8}}{1.50} = 2.00 \times 10^{8}\ \mathrm{m\,s^{-1}}. $$(A1)
由反射定律，反射角等于入射角，二者均从法线量起。(B1)
故反射角为 $50^{\circ}$。(A1)
用 Snell 定律 $n_1 \sin\theta_1 = n_2 \sin\theta_2$：$\sin\theta_2 = \dfrac{(1.00)\sin 50^{\circ}}{1.50}$。(M1)
$$ \sin\theta_2 = \frac{0.766}{1.50} = 0.5107, \qquad \theta_2 = \arcsin(0.5107) \approx 30.7^{\circ}. $$光进入光密介质，故偏向法线（$30.7^{\circ} < 50^{\circ}$）。(A1)
由 $n = c/v$ 解出 $v = c/n$：(M1)
$$ v = \frac{3.00 \times 10^{8}}{1.50} = 2.00 \times 10^{8}\ \mathrm{m\,s^{-1}}. $$(A1)
Two in-phase speakers, $\lambda = 0.50\ \mathrm{m}$; paths to $P$ are $4.00\ \mathrm{m}$ and $5.25\ \mathrm{m}$. (a) constructive or destructive at $P$; (b) why a steady pattern needs a shared source.两个同相喇叭，$\lambda = 0.50\ \mathrm{m}$；到 $P$ 的路程为 $4.00\ \mathrm{m}$ 与 $5.25\ \mathrm{m}$。(a) $P$ 处相长还是相消；(b) 为何稳定图样需共用一个源。
By superposition the resultant signal at $P$ depends on the path difference of the two coherent waves: $\Delta = 5.25 - 4.00 = 1.25\ \mathrm{m}$. (M1)
Express in wavelengths: $\dfrac{\Delta}{\lambda} = \dfrac{1.25}{0.50} = 2.5 = \left(2 + \tfrac{1}{2}\right)$. (M1)
This is a half-integer number of wavelengths, the destructive condition $\Delta = (n + \tfrac{1}{2})\lambda$ with $n = 2$, so $P$ is a point of destructive interference (a quiet spot). (A1)
Driving both speakers from one generator keeps a constant phase relationship between them: they are coherent, so the pattern of maxima and minima stays fixed. Two independent generators would drift in relative phase, washing the pattern out. (R1)
由叠加原理，$P$ 处合信号取决于两列相干波的光程差：$\Delta = 5.25 - 4.00 = 1.25\ \mathrm{m}$。(M1)
用波长表示：$\dfrac{\Delta}{\lambda} = \dfrac{1.25}{0.50} = 2.5 = \left(2 + \tfrac{1}{2}\right)$。(M1)
这是半整数倍波长，符合相消条件 $\Delta = (n + \tfrac{1}{2})\lambda$（$n = 2$），故 $P$ 为相消干涉点（安静处）。(A1)
用同一台发生器驱动两喇叭，使二者之间保持恒定相位关系：它们相干，故极大与极小的图样保持固定。两台独立发生器之间的相对相位会漂移，使图样消失。(R1)
Glass block $n = 1.50$ in air. (a) critical angle; (b) fate of a ray hitting the boundary at $45^{\circ}$ (both TIR conditions); (c) how TIR carries a signal along a fibre.空气中玻璃块 $n = 1.50$。(a) 临界角；(b) 光以 $45^{\circ}$ 入射界面的去向（两个全反射条件）；(c) 全反射如何沿光纤传输信号。
Glass to air, so $\sin\theta_c = 1/n$: (M1)
$$ \sin\theta_c = \frac{1}{1.50} = 0.6667, \qquad \theta_c = \arcsin(0.6667) \approx 41.8^{\circ}. $$(A1)
Both total-internal-reflection conditions must hold: the ray travels from dense to less-dense (glass to air, satisfied), and the angle of incidence must exceed the critical angle. Here $45^{\circ} > 41.8^{\circ}$. (R1)
Both conditions are met, so the ray undergoes total internal reflection: no light escapes into the air. (A1)
Light launched into the core strikes the core wall at angles greater than the critical angle, so it is totally internally reflected rather than refracted out. (A1)
The light bounces along the fibre by repeated total internal reflection, staying trapped in the core and carrying the signal over long distances with very little loss. (A1)
玻璃到空气，故 $\sin\theta_c = 1/n$：(M1)
$$ \sin\theta_c = \frac{1}{1.50} = 0.6667, \qquad \theta_c = \arcsin(0.6667) \approx 41.8^{\circ}. $$(A1)
全内反射的两个条件须同时满足：光从光密射向光疏（玻璃到空气，满足），且入射角须大于临界角。此处 $45^{\circ} > 41.8^{\circ}$。(R1)
两个条件都满足，故发生全内反射：无光逸入空气。(A1)
射入纤芯的光以大于临界角的角度撞击纤芯壁，故被全内反射而非折射逸出。(A1)
光靠反复全内反射在光纤内来回反弹，被困在纤芯中，以极低损耗远距离传输信号。(A1)
$\lambda = 600\ \mathrm{nm}$, single slit $b = 0.040\ \mathrm{mm}$, screen $D = 2.0\ \mathrm{m}$. (a) angle of first minimum; (b) width of central maximum; (c) effect of halving $b$.$\lambda = 600\ \mathrm{nm}$，单缝 $b = 0.040\ \mathrm{mm}$，屏 $D = 2.0\ \mathrm{m}$。(a) 第一极小角度；(b) 中央极大宽度；(c) 缝宽减半的影响。
Convert to metres: $\lambda = 600 \times 10^{-9}\ \mathrm{m}$, $b = 0.040 \times 10^{-3} = 4.0 \times 10^{-5}\ \mathrm{m}$. Use $\theta = \lambda / b$: (M1)
$$ \theta = \frac{600 \times 10^{-9}}{4.0 \times 10^{-5}} = 1.5 \times 10^{-2}\ \mathrm{rad}. $$(A1)
The central maximum runs from $-\theta$ to $+\theta$, so its full width on the screen is $w = 2 D \theta$: (M1)
$$ w = 2(2.0)(1.5 \times 10^{-2}) = 6.0 \times 10^{-2}\ \mathrm{m} = 60\ \mathrm{mm}. $$(A1)
Since $\theta = \lambda / b$, halving the slit width doubles $\theta$, so the central maximum width $w = 2D\theta$ doubles to $120\ \mathrm{mm}$. (A1)
A narrower slit spreads the wave more, because the central-maximum width is inversely proportional to slit width. (R1)
化为米：$\lambda = 600 \times 10^{-9}\ \mathrm{m}$、$b = 0.040 \times 10^{-3} = 4.0 \times 10^{-5}\ \mathrm{m}$。用 $\theta = \lambda / b$：(M1)
$$ \theta = \frac{600 \times 10^{-9}}{4.0 \times 10^{-5}} = 1.5 \times 10^{-2}\ \mathrm{rad}. $$(A1)
中央极大从 $-\theta$ 到 $+\theta$，故屏上全宽为 $w = 2 D \theta$：(M1)
$$ w = 2(2.0)(1.5 \times 10^{-2}) = 6.0 \times 10^{-2}\ \mathrm{m} = 60\ \mathrm{mm}. $$(A1)
由 $\theta = \lambda / b$，缝宽减半使 $\theta$ 加倍，故中央极大宽度 $w = 2D\theta$ 加倍为 $120\ \mathrm{mm}$。(A1)
缝越窄，波扩散越大，因为中央极大宽度与缝宽成反比。(R1)
Young's double slits, separation $d$, spacing $s$, screen distance $D$. (a) bright-fringe path condition; (b) effect of tripling $d$; (c) effect of switching red to blue, distinguishing $d$ from slit width $b$.杨氏双缝，缝间距 $d$，条纹间距 $s$，屏距 $D$。(a) 亮纹的光程差条件；(b) 把 $d$ 增大三倍的影响；(c) 红光换蓝光的影响，并区分 $d$ 与缝宽 $b$。
A bright fringe occurs where the path difference from the two slits is a whole number of wavelengths: $d\sin\theta = n\lambda$, with $n = 0, 1, 2, \dots$ (A1)
The fringe spacing is $s = \dfrac{\lambda D}{d}$, so $s \propto \dfrac{1}{d}$. (M1)
Tripling $d$ reduces $s$ to one third of its original value: the fringes crowd closer together. (A1)
From $s = \dfrac{\lambda D}{d}$, $s \propto \lambda$. Blue light has a shorter wavelength than red, so the fringe spacing decreases. (A1)
The separation $d$ (the gap between the two slits) is the length in the fringe-spacing formula and sets how far apart the fringes sit. (A1)
The slit width $b$ is a different length: it sets the broad single-slit envelope that modulates the brightness, not the fringe spacing. (R1)
当两缝光程差为整数倍波长时出现亮纹：$d\sin\theta = n\lambda$，$n = 0, 1, 2, \dots$ (A1)
条纹间距 $s = \dfrac{\lambda D}{d}$，故 $s \propto \dfrac{1}{d}$。(M1)
$d$ 增大三倍使 $s$ 减为原来的三分之一：条纹更密。(A1)
由 $s = \dfrac{\lambda D}{d}$，$s \propto \lambda$。蓝光波长比红光短，故条纹间距减小。(A1)
缝间距 $d$（两缝之间的间隙）是条纹间距公式中的长度，决定条纹相距多远。(A1)
缝宽 $b$ 是另一个长度：它决定调制亮度的宽阔单缝包络，而非条纹间距。(R1)
Grating $300\ \mathrm{lines\,mm^{-1}}$, normal incidence; $\sin\theta$ vs $n$ data: $(1,0.195),(2,0.390),(3,0.585)$. (a) show $\sin\theta$ vs $n$ is linear through the origin and state the gradient; (b) $d$, gradient, wavelength; (c) highest order; (d) advantage of many slits.光栅 $300\ \mathrm{lines\,mm^{-1}}$，垂直入射；$\sin\theta$ 对 $n$ 数据：$(1,0.195),(2,0.390),(3,0.585)$。(a) 证明 $\sin\theta$ 对 $n$ 为过原点直线并说明斜率；(b) $d$、斜率、波长；(c) 最高级；(d) 多缝的优点。
Rearrange the grating equation $d\sin\theta = n\lambda$ for $\sin\theta$: (M1)
$$ \sin\theta = \frac{\lambda}{d}\,n. $$This has the form $y = (\text{gradient})\,x$ with no intercept, so $\sin\theta$ against $n$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $\lambda / d$. (A1)
$300$ lines per mm is $300 \times 10^{3}$ lines per metre, so $d = \dfrac{1}{300 \times 10^{3}} = 3.33 \times 10^{-6}\ \mathrm{m}$. (M1)
Read the gradient from the origin to the far point $(3, 0.585)$: $\text{gradient} = \dfrac{0.585}{3} = 0.195$. (A1)
Since gradient $= \lambda / d$: $\lambda = 0.195 \times 3.33 \times 10^{-6} = 6.5 \times 10^{-7}\ \mathrm{m} = 650\ \mathrm{nm}$. (A1)
The largest possible angle is $\theta = 90^{\circ}$, where $\sin\theta = 1$. From $d\sin\theta = n\lambda$, the order cannot exceed $n = d/\lambda$. (M1)
$$ \frac{d}{\lambda} = \frac{3.33 \times 10^{-6}}{6.5 \times 10^{-7}} = 5.13. $$(M1)
The order must be a whole number, so the highest observable order is $n_{\max} = 5$. (A1)
With many slits each maximum is much sharper and brighter than the broad fringes from only two slits. (B1)
The sharp maxima let the diffraction angle, and hence the wavelength, be measured far more precisely. (B1)
把光栅方程 $d\sin\theta = n\lambda$ 解出 $\sin\theta$：(M1)
$$ \sin\theta = \frac{\lambda}{d}\,n. $$此式形如 $y = (\text{斜率})\,x$，无截距，故 $\sin\theta$ 对 $n$ 为过原点的直线。(A1)
与 $y = mx$ 比较，斜率为 $\lambda / d$。(A1)
每毫米 $300$ 线即每米 $300 \times 10^{3}$ 线，故 $d = \dfrac{1}{300 \times 10^{3}} = 3.33 \times 10^{-6}\ \mathrm{m}$。(M1)
从原点到远点 $(3, 0.585)$ 读斜率：$\text{斜率} = \dfrac{0.585}{3} = 0.195$。(A1)
因斜率 $= \lambda / d$：$\lambda = 0.195 \times 3.33 \times 10^{-6} = 6.5 \times 10^{-7}\ \mathrm{m} = 650\ \mathrm{nm}$。(A1)
最大可能角为 $\theta = 90^{\circ}$，此时 $\sin\theta = 1$。由 $d\sin\theta = n\lambda$，级数不能超过 $n = d/\lambda$。(M1)
$$ \frac{d}{\lambda} = \frac{3.33 \times 10^{-6}}{6.5 \times 10^{-7}} = 5.13. $$(M1)
级数须为整数，故可观察的最高级为 $n_{\max} = 5$。(A1)
缝多时每个极大比仅用双缝得到的宽条纹锐利、明亮得多。(B1)
锐利的极大使衍射角（进而是波长）能被远更精确地测量。(B1)
Double slit $d = 0.30\ \mathrm{mm}$, $D = 1.50\ \mathrm{m}$; distance across $8$ spacings $= (21.6 \pm 0.2)\ \mathrm{mm}$. (a) why measure many spacings; (b) fringe spacing and wavelength; (c) percentage uncertainty in the measured distance; (d) percentage uncertainty in $\lambda$ given $d$ at $2\%$ and $D$ at $1\%$.双缝 $d = 0.30\ \mathrm{mm}$，$D = 1.50\ \mathrm{m}$；跨 $8$ 个间距的距离 $= (21.6 \pm 0.2)\ \mathrm{mm}$。(a) 为何测多个间距；(b) 条纹间距与波长；(c) 测量距离的百分比不确定度；(d) 在 $d$ 为 $2\%$、$D$ 为 $1\%$ 时 $\lambda$ 的百分比不确定度。
The single spacing $s$ is found by dividing the total measured distance by the number of spacings. The absolute reading uncertainty ($\pm 0.2\ \mathrm{mm}$) is roughly fixed by the ruler, whatever distance is spanned. (M1)
Spanning $8$ spacings makes the measured distance $8$ times larger, so the same $\pm 0.2\ \mathrm{mm}$ is a much smaller percentage of it. (A1)
Dividing by $8$ at the end shares that small percentage onto $s$, so $s$ carries far less percentage uncertainty than a single-fringe measurement would. (R1)
Fringe spacing: $s = \dfrac{21.6\ \mathrm{mm}}{8} = 2.70\ \mathrm{mm} = 2.70 \times 10^{-3}\ \mathrm{m}$. (M1)
Rearrange $s = \dfrac{\lambda D}{d}$ for $\lambda$: (A1)
$$ \lambda = \frac{s\,d}{D} = \frac{(2.70 \times 10^{-3})(0.30 \times 10^{-3})}{1.50} = 5.4 \times 10^{-7}\ \mathrm{m} = 540\ \mathrm{nm}. $$(A1)
The reading is $(21.6 \pm 0.2)\ \mathrm{mm}$: (M1)
$$ \frac{0.2}{21.6}\times 100\% = 0.93\% \approx 0.9\%. $$(A1)
For a product or quotient, add the percentage uncertainties. The spacing $s$ inherits the $0.9\%$ of the measured distance, and $\lambda = sd/D$ adds the $2\%$ in $d$ and $1\%$ in $D$: (M1)
$$ 0.9\% + 2\% + 1\% \approx 3.9\% \approx 4\%. $$(A1)
So $4\%$ of $540\ \mathrm{nm}$ is about $20\ \mathrm{nm}$, and the result is quoted as $\lambda = (540 \pm 20)\ \mathrm{nm}$. (A1)
单个间距 $s$ 由测得的总距离除以间距数得到。绝对读数不确定度（$\pm 0.2\ \mathrm{mm}$）大体由尺子决定，与所跨距离无关。(M1)
跨 $8$ 个间距使测得距离大 $8$ 倍，故同样的 $\pm 0.2\ \mathrm{mm}$ 占其百分比小得多。(A1)
最后除以 $8$ 把这个小百分比分摊到 $s$ 上，故 $s$ 的百分比不确定度远小于只测单条纹时。(R1)
条纹间距：$s = \dfrac{21.6\ \mathrm{mm}}{8} = 2.70\ \mathrm{mm} = 2.70 \times 10^{-3}\ \mathrm{m}$。(M1)
把 $s = \dfrac{\lambda D}{d}$ 解出 $\lambda$：(A1)
$$ \lambda = \frac{s\,d}{D} = \frac{(2.70 \times 10^{-3})(0.30 \times 10^{-3})}{1.50} = 5.4 \times 10^{-7}\ \mathrm{m} = 540\ \mathrm{nm}. $$(A1)
读数为 $(21.6 \pm 0.2)\ \mathrm{mm}$：(M1)
$$ \frac{0.2}{21.6}\times 100\% = 0.93\% \approx 0.9\%. $$(A1)
对乘积或商，百分比不确定度相加。$s$ 继承测量距离的 $0.9\%$，而 $\lambda = sd/D$ 再加上 $d$ 的 $2\%$ 与 $D$ 的 $1\%$：(M1)
$$ 0.9\% + 2\% + 1\% \approx 3.9\% \approx 4\%. $$(A1)
故 $540\ \mathrm{nm}$ 的 $4\%$ 约为 $20\ \mathrm{nm}$，结果表述为 $\lambda = (540 \pm 20)\ \mathrm{nm}$。(A1)
$\lambda = 589\ \mathrm{nm}$, slit width $b = 0.050\ \mathrm{mm}$, separation $d = 0.25\ \mathrm{mm}$, screen $D = 1.8\ \mathrm{m}$. (a) which length sets fringes vs envelope; (b) fringe spacing $s$; (c) position of the envelope first minimum; (d) number of fringes in the central envelope and the missing order; (e) sketch the intensity pattern.$\lambda = 589\ \mathrm{nm}$，缝宽 $b = 0.050\ \mathrm{mm}$，间距 $d = 0.25\ \mathrm{mm}$，屏 $D = 1.8\ \mathrm{m}$。(a) 哪个长度决定条纹、哪个决定包络；(b) 条纹间距 $s$；(c) 包络第一极小位置；(d) 中央包络内条纹数与缺级；(e) 画强度分布。
The slit separation $d$ sets the fine interference fringe spacing through $s = \lambda D / d$. (A1)
The slit width $b$ sets the broad single-slit diffraction envelope through $\theta = \lambda / b$. (A1)
Convert to metres: $\lambda = 589 \times 10^{-9}\ \mathrm{m}$, $d = 0.25 \times 10^{-3} = 2.5 \times 10^{-4}\ \mathrm{m}$. (M1)
Use $s = \dfrac{\lambda D}{d}$: (M1)
$$ s = \frac{(589 \times 10^{-9})(1.8)}{2.5 \times 10^{-4}} = 4.24 \times 10^{-3}\ \mathrm{m} \approx 4.24\ \mathrm{mm}. $$(A1)
The envelope first minimum is at angle $\theta = \dfrac{\lambda}{b}$ with $b = 0.050 \times 10^{-3} = 5.0 \times 10^{-5}\ \mathrm{m}$: (M1)
$$ \theta = \frac{589 \times 10^{-9}}{5.0 \times 10^{-5}} = 1.178 \times 10^{-2}\ \mathrm{rad}. $$Its position on the screen is $y = D\theta$: (M1)
$$ y = (1.8)(1.178 \times 10^{-2}) = 2.12 \times 10^{-2}\ \mathrm{m} \approx 21.2\ \mathrm{mm}. $$(A1)
The bright fringes lie at $y_n = n s$. The number that fit inside the central envelope (out to $y \approx 21.2\ \mathrm{mm}$) follows from the ratio $\dfrac{d}{b} = \dfrac{0.25}{0.050} = 5$. (M1)
The $n = \pm 5$ fringes would fall exactly on the envelope minimum and are suppressed (a missing order). The central envelope therefore contains the fringes $n = -4$ to $n = +4$, that is $9$ bright fringes, with the $5$th order absent on each side. (A1)
Draw evenly spaced sharp interference fringes (spacing $4.24\ \mathrm{mm}$) whose peak heights are modulated by a broad single-slit envelope that falls to zero at $y = \pm 21.2\ \mathrm{mm}$. (B1)
Show the $5$th fringe on each side missing because it coincides with the envelope minimum, and the central fringe the brightest. (B1)
缝间距 $d$ 通过 $s = \lambda D / d$ 决定细密的干涉条纹间距。(A1)
缝宽 $b$ 通过 $\theta = \lambda / b$ 决定宽阔的单缝衍射包络。(A1)
化为米：$\lambda = 589 \times 10^{-9}\ \mathrm{m}$、$d = 0.25 \times 10^{-3} = 2.5 \times 10^{-4}\ \mathrm{m}$。(M1)
用 $s = \dfrac{\lambda D}{d}$：(M1)
$$ s = \frac{(589 \times 10^{-9})(1.8)}{2.5 \times 10^{-4}} = 4.24 \times 10^{-3}\ \mathrm{m} \approx 4.24\ \mathrm{mm}. $$(A1)
包络第一极小在角度 $\theta = \dfrac{\lambda}{b}$，其中 $b = 0.050 \times 10^{-3} = 5.0 \times 10^{-5}\ \mathrm{m}$：(M1)
$$ \theta = \frac{589 \times 10^{-9}}{5.0 \times 10^{-5}} = 1.178 \times 10^{-2}\ \mathrm{rad}. $$它在屏上的位置为 $y = D\theta$：(M1)
$$ y = (1.8)(1.178 \times 10^{-2}) = 2.12 \times 10^{-2}\ \mathrm{m} \approx 21.2\ \mathrm{mm}. $$(A1)
亮条纹位于 $y_n = n s$。落入中央包络（至 $y \approx 21.2\ \mathrm{mm}$）的条纹数由比值 $\dfrac{d}{b} = \dfrac{0.25}{0.050} = 5$ 给出。(M1)
$n = \pm 5$ 的条纹恰落在包络极小处而被抑制（缺级）。故中央包络内含 $n = -4$ 至 $n = +4$ 的条纹，即 $9$ 条亮纹，两侧各缺第 $5$ 级。(A1)
画出等间距的锐利干涉条纹（间距 $4.24\ \mathrm{mm}$），其峰高被在 $y = \pm 21.2\ \mathrm{mm}$ 处降为零的宽阔单缝包络调制。(B1)
标出两侧第 $5$ 条纹因与包络极小重合而缺失，中央条纹最亮。(B1)
Step-index fibre, core $n_1 = 1.50$, cladding $n_2 = 1.45$. (a) refraction into the core from air at $30^{\circ}$; (b) critical angle of the core-cladding boundary; (c) is a $78^{\circ}$ ray guided; (d) Malus's law with unpolarized $I_0 = 120\ \mathrm{W\,m^{-2}}$, analyser at $30^{\circ}$; (e) extinction angle and what it proves.阶跃光纤，纤芯 $n_1 = 1.50$，包层 $n_2 = 1.45$。(a) 从空气以 $30^{\circ}$ 折射进入纤芯；(b) 纤芯-包层界面临界角；(c) $78^{\circ}$ 的光是否被导引；(d) 非偏振光 $I_0 = 120\ \mathrm{W\,m^{-2}}$、检偏器 $30^{\circ}$ 的马吕斯定律；(e) 消光角及其证明。
Apply Snell's law at the end face, air to core: $n_{\text{air}} \sin 30^{\circ} = n_1 \sin\theta$. (M1)
$$ \sin\theta = \frac{(1.00)\sin 30^{\circ}}{1.50} = \frac{0.500}{1.50} = 0.3333, \qquad \theta \approx 19.5^{\circ}. $$(A1)
Both media are glass, so use the general form $\sin\theta_c = \dfrac{n_2}{n_1}$: (M1)
$$ \sin\theta_c = \frac{1.45}{1.50} = 0.9667. $$(M1)
$$ \theta_c = \arcsin(0.9667) \approx 75.2^{\circ}. $$(A1)
The ray travels from the denser core into the less-dense cladding and strikes at $78^{\circ}$, which exceeds the critical angle $75.2^{\circ}$. (A1)
Both conditions for total internal reflection are met, so the ray is totally internally reflected and stays guided along the fibre. (R1)
Unpolarized light through the first polarizer loses half its intensity: $I_1 = \tfrac{1}{2}(120) = 60\ \mathrm{W\,m^{-2}}$. (M1·A1)
The light is now polarized; apply Malus's law $I = I_1 \cos^2\theta$ with $\theta = 30^{\circ}$: (A1)
$$ I_2 = (60)\cos^2 30^{\circ} = (60)(0.8660)^2 = (60)(0.750) = 45\ \mathrm{W\,m^{-2}}. $$Malus's law gives $I = 0$ when $\cos^2\theta = 0$, i.e. $\theta = 90^{\circ}$: the analyser must be crossed at $90^{\circ}$ to the polarizer. (A1)
That light can be extinguished by rotating a polarizer shows its oscillation has a definite direction perpendicular to travel, which is only possible for a transverse wave; a longitudinal wave could not be blocked this way. (R1)
在端面用 Snell 定律，空气到纤芯：$n_{\text{air}} \sin 30^{\circ} = n_1 \sin\theta$。(M1)
$$ \sin\theta = \frac{(1.00)\sin 30^{\circ}}{1.50} = \frac{0.500}{1.50} = 0.3333, \qquad \theta \approx 19.5^{\circ}. $$(A1)
两介质都是玻璃，故用一般式 $\sin\theta_c = \dfrac{n_2}{n_1}$：(M1)
$$ \sin\theta_c = \frac{1.45}{1.50} = 0.9667. $$(M1)
$$ \theta_c = \arcsin(0.9667) \approx 75.2^{\circ}. $$(A1)
光从光密纤芯射向光疏包层，以 $78^{\circ}$ 入射，超过临界角 $75.2^{\circ}$。(A1)
全内反射的两个条件都满足，故光被全内反射，保持沿光纤被导引。(R1)
非偏振光通过第一个偏振片后强度减半：$I_1 = \tfrac{1}{2}(120) = 60\ \mathrm{W\,m^{-2}}$。(M1·A1)
此时光已偏振；用马吕斯定律 $I = I_1 \cos^2\theta$（$\theta = 30^{\circ}$）：(A1)
$$ I_2 = (60)\cos^2 30^{\circ} = (60)(0.8660)^2 = (60)(0.750) = 45\ \mathrm{W\,m^{-2}}. $$马吕斯定律给出当 $\cos^2\theta = 0$ 即 $\theta = 90^{\circ}$ 时 $I = 0$：检偏器须与偏振片正交（$90^{\circ}$）。(A1)
光能通过旋转偏振片被消光，说明其振动有垂直于传播方向的确定方向，这只有横波才可能；纵波无法被这样阻挡。(R1)
Two in-phase microwave transmitters at $3.0\ \mathrm{GHz}$, $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$. (a) wavelength; (b) at $Q$ paths $1.00\ \mathrm{m}$ and $1.25\ \mathrm{m}$, maximum or minimum; (c) with $d = 0.20\ \mathrm{m}$ and $D = 2.0\ \mathrm{m}$, spacing of adjacent maxima.两台同相微波发射器，$3.0\ \mathrm{GHz}$，$c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$。(a) 波长；(b) 在 $Q$ 路程 $1.00\ \mathrm{m}$ 与 $1.25\ \mathrm{m}$，极大还是极小；(c) 当 $d = 0.20\ \mathrm{m}$、$D = 2.0\ \mathrm{m}$ 时相邻极大间距。
Use $c = f\lambda$, so $\lambda = c / f$: (M1)
$$ \lambda = \frac{3.00 \times 10^{8}}{3.0 \times 10^{9}} = 0.10\ \mathrm{m}. $$(A1)
Path difference: $\Delta = 1.25 - 1.00 = 0.25\ \mathrm{m}$. (M1)
In wavelengths: $\dfrac{\Delta}{\lambda} = \dfrac{0.25}{0.10} = 2.5 = \left(2 + \tfrac{1}{2}\right)$. (A1)
This is a half-integer number of wavelengths, the destructive condition $\Delta = (n + \tfrac{1}{2})\lambda$, so $Q$ is a minimum of the received signal. (A1)
The two-source spacing follows the same fringe formula as the double slit, $s = \dfrac{\lambda D}{d}$, valid here because $D \gg d$. (M1)
Substitute $\lambda = 0.10\ \mathrm{m}$, $D = 2.0\ \mathrm{m}$, $d = 0.20\ \mathrm{m}$: (M1)
$$ s = \frac{(0.10)(2.0)}{0.20} = 1.0\ \mathrm{m}. $$(A1)
用 $c = f\lambda$，故 $\lambda = c / f$：(M1)
$$ \lambda = \frac{3.00 \times 10^{8}}{3.0 \times 10^{9}} = 0.10\ \mathrm{m}. $$(A1)
光程差：$\Delta = 1.25 - 1.00 = 0.25\ \mathrm{m}$。(M1)
用波长表示：$\dfrac{\Delta}{\lambda} = \dfrac{0.25}{0.10} = 2.5 = \left(2 + \tfrac{1}{2}\right)$。(A1)
这是半整数倍波长，符合相消条件 $\Delta = (n + \tfrac{1}{2})\lambda$，故 $Q$ 为接收信号的极小。(A1)
双源间距遵循与双缝相同的条纹公式 $s = \dfrac{\lambda D}{d}$，此处 $D \gg d$ 故成立。(M1)
代入 $\lambda = 0.10\ \mathrm{m}$、$D = 2.0\ \mathrm{m}$、$d = 0.20\ \mathrm{m}$：(M1)
$$ s = \frac{(0.10)(2.0)}{0.20} = 1.0\ \mathrm{m}. $$(A1)