PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 28 marks短结构题 · 可用计算器 · 28 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. Always measure angles from the normal, and convert all lengths to metres before substituting. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。角度一律从法线量起，代入前把所有长度化为米。数值答案保留适当的有效数字。
A ray of light in air ($n_1 = 1.00$) strikes the flat top surface of a glass block of refractive index $n_2 = 1.50$ at an angle of incidence of $50^{\circ}$. Take $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$.空气（$n_1 = 1.00$）中一束光以 $50^{\circ}$ 入射角射到折射率 $n_2 = 1.50$ 的玻璃块平整上表面。取 $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$。
(a)State the angle of reflection of the ray at the surface, and the law that gives it.写出该光线在表面的反射角，以及给出它的定律。[2]
(b)Calculate the angle of refraction inside the glass, and state whether the ray bends toward or away from the normal.计算玻璃内的折射角，并说明光线偏向还是偏离法线。[2]
(c)Calculate the speed of light inside the glass.计算玻璃内的光速。[2]
Q2MEDIUMPaper 1superposition and path difference叠加与光程差[4 marks]
Two loudspeakers driven in phase by the same signal generator emit sound of wavelength $0.50\ \mathrm{m}$. At a point $P$ the path length from speaker 1 is $4.00\ \mathrm{m}$ and from speaker 2 is $5.25\ \mathrm{m}$.两个由同一信号发生器同相驱动的喇叭发出波长 $0.50\ \mathrm{m}$ 的声音。在 $P$ 点，到喇叭 1 的路程为 $4.00\ \mathrm{m}$，到喇叭 2 为 $5.25\ \mathrm{m}$。
(a)Using the principle of superposition, determine whether $P$ is a point of constructive or destructive interference. Justify your answer with the path difference in wavelengths.利用叠加原理，判断 $P$ 是相长还是相消干涉点。用以波长表示的光程差说明理由。[3]
(b)State why a steady interference pattern is heard here but would not be heard if the two speakers were driven by two separate, independent generators.说明为何此处能听到稳定的干涉图样，而若两喇叭由两台独立的信号发生器驱动则不能。[1]
A glass block has refractive index $n = 1.50$ and is surrounded by air.一玻璃块折射率 $n = 1.50$，周围为空气。
(a)Calculate the critical angle for the glass-to-air boundary.计算玻璃-空气界面的临界角。[2]
(b)A ray travelling inside the glass meets this boundary at an angle of incidence of $45^{\circ}$. State, with reasons referring to both conditions for total internal reflection, what happens to the ray.玻璃内一束光以 $45^{\circ}$ 入射角到达该界面。结合全内反射的两个条件，说明该光线的去向并解释理由。[2]
(c)Explain how total internal reflection allows an optical fibre to carry a light signal over a long distance.解释全内反射如何使光纤能远距离传输光信号。[2]
Monochromatic light of wavelength $600\ \mathrm{nm}$ passes through a single slit of width $b = 0.040\ \mathrm{mm}$ and forms a diffraction pattern on a screen $D = 2.0\ \mathrm{m}$ away.波长 $600\ \mathrm{nm}$ 的单色光通过缝宽 $b = 0.040\ \mathrm{mm}$ 的单缝，在 $D = 2.0\ \mathrm{m}$ 外的屏上形成衍射图样。
(a)Calculate the angular position of the first diffraction minimum.计算第一衍射极小的角位置。[2]
(b)Hence calculate the width of the central bright maximum on the screen.由此计算屏上中央亮极大的宽度。[2]
(c)State and explain what happens to the width of the central maximum if the slit width is halved.说明并解释当缝宽减半时中央极大的宽度如何变化。[2]
In a Young's double-slit experiment, two narrow slits separated by $d$ are illuminated by monochromatic light and produce fringes of spacing $s$ on a screen a distance $D$ away.在杨氏双缝实验中，相距 $d$ 的两条窄缝被单色光照射，在 $D$ 外的屏上形成间距为 $s$ 的条纹。
(a)State the condition on the path difference from the two slits for a bright fringe to appear.写出两缝光程差满足什么条件时出现亮纹。[1]
(b)The slit separation $d$ is tripled while $\lambda$ and $D$ are unchanged. State and justify the effect on the fringe spacing $s$.在 $\lambda$ 与 $D$ 不变时把缝间距 $d$ 增大为三倍。说明并论证对条纹间距 $s$ 的影响。[2]
(c)The red light is then replaced by blue light in the same apparatus. State the effect on the fringe spacing and explain why, distinguishing the role of the slit separation $d$ from the slit width $b$.随后在同一装置中用蓝光替换红光。说明对条纹间距的影响并解释原因，并区分缝间距 $d$ 与缝宽 $b$ 的不同作用。[3]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分
Graph and Data Questions图像与数据题
These items reward careful reading of gradients and correct handling of uncertainties. Quote uncertainties to one significant figure and round the value to match. Read a gradient from the line or from widely spaced points, never from one data pair divided out.这些题考查对斜率的细致读取与对不确定度的正确处理。不确定度保留 1 位有效数字，并使数值的末位与之对齐。斜率应从直线或相距较远的点读取，切勿用单个数据对相除。
Q6HARDPaper 1BHL ONLYdiffraction grating: $\sin\theta$ vs order衍射光栅：$\sin\theta$ 对级数[11 marks]
A diffraction grating ruled with $300\ \mathrm{lines\,mm^{-1}}$ is illuminated normally with monochromatic light. A student measures the angle $\theta$ of each diffracted order $n$ and tabulates $\sin\theta$ against $n$:一块刻有 $300\ \mathrm{lines\,mm^{-1}}$ 的衍射光栅被单色光垂直照射。学生测量每一级 $n$ 的衍射角 $\theta$，并将 $\sin\theta$ 对 $n$ 列表：
$n$
$1$
$2$
$3$
$\sin\theta$
$0.195$
$0.390$
$0.585$
(a)Starting from the grating equation $d\sin\theta = n\lambda$, show that a graph of $\sin\theta$ against $n$ should be a straight line through the origin, and state what the gradient represents.从光栅方程 $d\sin\theta = n\lambda$ 出发，证明 $\sin\theta$ 对 $n$ 的图应为过原点的直线，并说明斜率代表什么。[3]
(b)Calculate the slit spacing $d$ of the grating, then determine the gradient of the line and hence the wavelength of the light.计算光栅的缝间距 $d$，再求该直线的斜率，由此求光的波长。[3]
(c)Determine the highest order of maximum that can be observed with this grating and wavelength.求用此光栅与此波长能观察到的最高级极大。[3]
(d)State one advantage of using a grating with many slits rather than just two slits to measure a wavelength.说出用多缝光栅而非仅用双缝测量波长的一个优点。[2]
Q7HARDPaper 1Bdouble-slit data + uncertainty双缝数据与不确定度[11 marks]
In a double-slit experiment, slits separated by $d = 0.30\ \mathrm{mm}$ produce fringes on a screen $D = 1.50\ \mathrm{m}$ away. A student measures the distance across $8$ fringe spacings (from the first bright fringe to the ninth) as $(21.6 \pm 0.2)\ \mathrm{mm}$.双缝实验中，缝间距 $d = 0.30\ \mathrm{mm}$，在 $D = 1.50\ \mathrm{m}$ 外的屏上形成条纹。学生测得跨越 $8$ 个条纹间距（从第一条亮纹到第九条）的距离为 $(21.6 \pm 0.2)\ \mathrm{mm}$。
(a)Explain why measuring across many fringe spacings, rather than a single one, reduces the uncertainty in the fringe spacing $s$.解释为何跨越多个条纹间距测量（而非只测一个）能减小条纹间距 $s$ 的不确定度。[3]
(b)Calculate the fringe spacing $s$ and hence the wavelength of the light, using $s = \lambda D / d$.用 $s = \lambda D / d$ 计算条纹间距 $s$，由此求光的波长。[3]
(c)Calculate the percentage uncertainty in the measured distance across the $8$ spacings.计算跨 $8$ 个间距的测量距离的百分比不确定度。[2]
(d)The slit separation $d$ has a percentage uncertainty of $2\%$ and the distance $D$ has a percentage uncertainty of $1\%$. Estimate the percentage uncertainty in the calculated wavelength, and state how it should be quoted.缝间距 $d$ 的百分比不确定度为 $2\%$，距离 $D$ 的为 $1\%$。估算所求波长的百分比不确定度，并说明应如何表述。[3]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 32 marks长结构题 · 可用计算器 · 32 分
Extended Structured Problems长结构问题
Set up each problem with a clear diagram. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer.每题先画清晰的示意图。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。
Q8HARDPaper 2HL ONLYdouble-slit fringes under the single-slit envelope单缝包络下的双缝条纹[12 marks]
Light of wavelength $589\ \mathrm{nm}$ illuminates two slits, each of width $b = 0.050\ \mathrm{mm}$, whose centres are separated by $d = 0.25\ \mathrm{mm}$. The pattern is observed on a screen $D = 1.8\ \mathrm{m}$ away.波长 $589\ \mathrm{nm}$ 的光照射两条缝，每缝缝宽 $b = 0.050\ \mathrm{mm}$，缝中心间距 $d = 0.25\ \mathrm{mm}$。在 $D = 1.8\ \mathrm{m}$ 外的屏上观察图样。
(a)State which length, $b$ or $d$, sets the fine fringe spacing and which sets the broad diffraction envelope.说明 $b$ 与 $d$ 中哪一个决定细密的条纹间距，哪一个决定宽阔的衍射包络。[2]
(b)Calculate the spacing $s$ of the bright interference fringes on the screen.计算屏上亮干涉条纹的间距 $s$。[3]
(c)Calculate the position on the screen of the first minimum of the single-slit diffraction envelope, measured from the centre.计算单缝衍射包络第一极小在屏上（从中心量起）的位置。[3]
(d)By comparing $d$ and $b$, determine how many bright interference fringes lie within the central diffraction envelope, and identify the order of the fringe that is suppressed (a missing order).通过比较 $d$ 与 $b$，求中央衍射包络内有多少条亮干涉条纹，并指出被抑制（缺级）的条纹级数。[2]
(e)Sketch the intensity pattern on the screen, showing the fine fringes modulated by the broad envelope and the suppressed order.画出屏上的强度分布图，显示被宽包络调制的细密条纹以及被抑制的级。[2]
A step-index optical fibre has a core of refractive index $n_1 = 1.50$ and a cladding of refractive index $n_2 = 1.45$.一根阶跃折射率光纤的纤芯折射率 $n_1 = 1.50$，包层折射率 $n_2 = 1.45$。
(a)Light enters the flat end face of the core from air ($n = 1.00$) at an angle of incidence of $30^{\circ}$ to the normal. Calculate the angle of refraction inside the core.光从空气（$n = 1.00$）以与法线成 $30^{\circ}$ 入射到纤芯的平整端面。计算纤芯内的折射角。[2]
(b)Calculate the critical angle for the core-cladding boundary.计算纤芯-包层界面的临界角。[3]
(c)A ray strikes the core-cladding wall at $78^{\circ}$ to the normal. State whether it is guided along the fibre, and justify your answer using the result of (b).一束光以与法线成 $78^{\circ}$ 射到纤芯-包层壁。说明它是否被导引沿光纤传播，并用 (b) 的结果论证。[2]
(d)Unpolarized light of intensity $I_0 = 120\ \mathrm{W\,m^{-2}}$ from the fibre passes through a polarizer and then an analyser whose transmission axis is at $30^{\circ}$ to that of the polarizer. Calculate the intensity transmitted after each component.从光纤射出的强度 $I_0 = 120\ \mathrm{W\,m^{-2}}$ 的非偏振光先通过一个偏振片，再通过一个检偏器，其透光轴与偏振片成 $30^{\circ}$。计算每个元件后透射的强度。[3]
(e)State the angle to which the analyser should be rotated to extinguish the light completely, and explain why this confirms that light is a transverse wave.写出检偏器应旋转到的角度以完全消光，并解释为何这证实光是横波。[2]
Q10HARDPaper 2two-source interference and coherence双源干涉与相干性[8 marks]
Two microwave transmitters, $S_1$ and $S_2$, are driven in phase by the same oscillator at a frequency of $3.0\ \mathrm{GHz}$. A receiver is moved through the region in front of them. Take $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$.两台微波发射器 $S_1$ 与 $S_2$ 由同一振荡器以 $3.0\ \mathrm{GHz}$ 同相驱动。接收器在它们前方区域移动。取 $c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$。
(a)Calculate the wavelength of the microwaves.计算微波的波长。[2]
(b)At a point $Q$ the paths from $S_1$ and $S_2$ are $1.00\ \mathrm{m}$ and $1.25\ \mathrm{m}$. State, with a calculation of the path difference in wavelengths, whether $Q$ is a maximum or a minimum of the received signal.在 $Q$ 点，到 $S_1$ 与 $S_2$ 的路程分别为 $1.00\ \mathrm{m}$ 与 $1.25\ \mathrm{m}$。通过计算以波长表示的光程差，说明 $Q$ 是接收信号的极大还是极小。[3]
(c)The transmitters are now placed a distance $d = 0.20\ \mathrm{m}$ apart and the receiver moves along a line $D = 2.0\ \mathrm{m}$ away, parallel to the line joining the transmitters. Calculate the spacing of adjacent maxima detected by the receiver.现把两发射器相距 $d = 0.20\ \mathrm{m}$ 放置，接收器沿 $D = 2.0\ \mathrm{m}$ 外、平行于两发射器连线的直线移动。计算接收器测得的相邻极大间距。[3]