หน่วย 2: คลื่นและไฟฟ้า
Pearson Edexcel · International A-Level · Physics หัวข้อ 2 19:22 การบรรยายภาษาอังกฤษ · คำบรรยายภาษาอังกฤษ + 中文 ลอยตัวบนภาพ
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Transcript
A phone connects the two halves of this unit. Waves bring information, and a battery supplies electrical energy.
手机把本单元的两个部分联系起来：波传递信息，电池提供电能。
Start each problem with the physical model.
每道题先确定物理模型。
The reference explains every requirement, while the sheets develop methods and the authentic sets test transfer.
讲义解释全部大纲要求，练习逐步建立方法，真题检验迁移能力。
A sample of papers does not replace the syllabus.
抽样的试卷不能代替大纲。
Look at the axes before looking at the curve.
先读坐标轴，再看曲线。
A distance graph is a snapshot of many particles at one instant. A time graph follows one particle.
距离图是在同一时刻观察许多粒子的快照，时间图则跟踪一个粒子。
The same curve shape can therefore give either wavelength or period.
因此，同样的曲线形状可能表示波长，也可能表示周期。
The vertical axis must also say whether it represents displacement or pressure.
纵轴也要说明表示位移还是压强。
In sound, air molecules move backwards and forwards around resting positions.
声波中，空气分子在平衡位置附近来回运动。
A compression travels onwards even though each molecule stays nearby.
密部向前传播，而每个分子仍留在附近。
A pressure maximum marks crowding, not maximum molecular displacement.
压强最大处表示分子最密集，并非分子位移最大处。
For a sinusoidal travelling sound wave, pressure and displacement patterns are a quarter wavelength apart.
对于正弦行进声波，压强与位移图样在空间上相差四分之一波长。
Use the cycle count to find frequency. Then take its reciprocal for period. Finally use speed equals frequency times wavelength.
用振动次数求频率，再取倒数求周期，最后用波速等于频率乘波长。
Keep units beside the quantities.
保留各物理量的单位。
Pause and write the three values before the solution appears.
继续之前，先写出这三个值。
Twelve divided by zero point zero three gives four hundred hertz.
十二除以零点零三得到四百赫兹。
One divided by four hundred gives a period of two point five milliseconds.
频率的倒数给出二点五毫秒的周期。
Dividing three hundred forty metres per second by four hundred hertz gives zero point eight five metres.
三百四十米每秒除以四百赫兹得到零点八五米。
Multiplying period by frequency gives one, which checks the pair.
周期乘频率等于一，可以用来核对。
Coherence does not require zero phase difference. It requires a stable phase relationship and equal frequency.
相干并不要求相位差为零，而是要求频率相同、相位关系稳定。
For sources initially in phase, one whole wavelength of extra path adds a full cycle and reinforces the other wave. An odd half wavelength reverses the phase.
初始同相的波，多走一个完整波长后仍同相；多走奇数个半波长后反相。
Complete cancellation additionally requires equal arriving amplitudes.
完全抵消还要求到达时的振幅相等。
Study the two snapshots.
观察两个相反时刻的快照。
Nodes remain at zero displacement, while antinodes reach the largest amplitude.
波节位移始终为零，波腹振幅最大。
Each loop moves together, and neighbouring loops move oppositely.
同一波节之间的粒子同相运动，相邻波段反相运动。
The pattern results from opposite travelling waves of the same frequency. It does not represent particles moving along the drawn curve.
这个图样由同频率、反向传播的波叠加形成，并不是粒子沿着曲线运动。
A fixed string has displacement nodes at both ends. Its fundamental fits half a wavelength.
两端固定的弦在两端都是位移波节，基频时容纳半个波长。
In a tube closed at one end, the simplest molecular displacement pattern has a node at the closed end and an antinode near the open end, so it fits a quarter wavelength.
一端封闭的管中，最简单的分子位移图样在封闭端有波节，在开口附近有波腹，因此容纳四分之一波长。
The pressure pattern is reversed.
压强图样与位移图样相反。
Use an end correction if provided.
题目提供端部修正时应使用。
The generator drives the speaker and supplies an oscilloscope reference.
信号发生器驱动扬声器，并给示波器提供参考信号。
Move the microphone along a measured line and compare its signal with that reference.
沿测量线移动麦克风，把其信号与参考信号比较。
Measure between several repeated in-phase positions and divide by the number of wavelengths. Measure several time cycles too.
测量跨越多个同相位置的距离，再除以波长个数；时间测量也跨越多个周期。
Reducing wall reflections improves the signal comparison.
减少墙面反射有助于比较信号。
For a string, wave speed depends on tension divided by mass per unit length.
弦上的波速取决于张力与线密度之比，基频还取决于振动长度。
Fundamental frequency also depends on vibrating length. Investigate one variable at a time. The selected mode must stay fixed, because changing to another harmonic can mimic a change caused by tension or length.
每次只研究一个变量，保持振动模式不变，否则改变谐波可能被误认为张力或长度的影响。
Keep the hanging masses secure.
悬挂的砝码应固定牢靠。
Pause and calculate. A solar cell has area twenty-five square centimetres, intensity eight hundred watts per square metre, and efficiency twenty percent. Can it supply zero point five zero watts?
太阳能电池面积二十五平方厘米，入射强度每平方米八百瓦，效率百分之二十。
Show the area conversion and compare useful power.
它能否提供零点五零瓦？ 先计算再作比较。
At eight hundred watts per square metre over twenty-five square centimetres, incident power is two watts.
强度为每平方米八百瓦、面积为二十五平方厘米时，入射功率为两瓦。
With twenty percent efficiency, useful power is only zero point four watts.
效率为百分之二十时，有用功率只有零点四瓦。
Compare the useful output with the required output.
应将有用输出与所需输出比较。
A correct formula with an unconverted area can still be wrong by ten thousand.
面积未换算时，即使公式正确，结果也可能相差一万倍。
Find the normal first. Measure both ray angles from that line, never from the surface.
先找法线，两边的光线角度都从法线量起，不从界面量起。
When light enters a lower-index medium it speeds up and bends away from the normal.
光进入较低折射率的介质时，速度增加并偏离法线。
Frequency remains fixed by the source, while wavelength changes with speed.
频率由光源决定而保持不变，波长随波速改变。
A ray along the normal can change speed without changing direction.
沿法线入射的光可以改变速度而不改变方向。
For glass of index one point five entering air, the critical angle is about forty-one point eight degrees.
折射率一点五的玻璃向空气传播时，临界角约为四十一点八度。
A forty-five degree incidence therefore gives total internal reflection.
因此四十五度入射会发生全反射。
The same forty-five degrees entering glass from air does not.
若光从空气进入玻璃，同样的四十五度则不会。
Increasing a fibre cladding index increases its critical angle and can prevent trapping.
提高光纤包层的折射率会增大临界角，可能使原先被限制的光线透出。
For a refractive-index experiment, trace the block, mark separated ray points and reconstruct the internal ray after removing the block.
测折射率时，先描出玻璃砖轮廓，标记间距较大的光线点，移开玻璃砖后重建内部光线。
Repeated angle pairs give a graph whose gradient is the index relative to air.
重复测量角度，作图所得斜率为相对空气的折射率。
Polarisation asks a different question: which transverse vibration direction passes through? The light still propagates forwards.
偏振研究的是哪一个横向振动方向能通过，光仍向前传播。
The gap does not change the wave frequency. It restricts which wavefront points supply the outgoing wavelets.
狭缝不会改变波的频率，而是限制哪些波阵面上的点能产生出射子波。
Their envelope spreads into the region behind the opening.
子波的包络面向开口后的区域扩展。
A very wide opening still has edge diffraction, but the central region spreads much less.
很宽的开口边缘仍有衍射，但中央区域扩展较少。
Use this construction to explain, rather than simply name, diffraction.
用这种作图解释衍射，不要只写出名称。
Convert the grating line density to lines per metre, then take its reciprocal for slit spacing.
先将光栅线密度换算成每米的线数，再取倒数得到缝距。
In the practical, average left and right spot displacements, use screen distance to find the angle, and then apply the grating equation.
实验中取左右光斑位移的平均值，利用屏幕距离求角度，然后使用光栅方程。
Check that sine cannot exceed one when deciding which orders are possible.
判断可见级次时，要检查正弦值不超过一。
Keep the laser below eye level.
激光保持在眼睛高度以下。
An electron diffraction pattern is experimental evidence. The de Broglie equation predicts the wavelength associated with the electron momentum.
电子衍射图样是实验证据，德布罗意方程则预测与电子动量对应的波长。
Stating the equation alone does not explain the evidence: discuss the diffraction pattern and why simple straight-path particles cannot account for it.
仅写方程并不能解释证据；需要说明衍射图样以及简单直线运动的粒子模型为何不能解释它。
An electron still also has particle properties.
电子同时仍具有粒子性质。
At five thousand metres per second, an echo delay of twenty microseconds corresponds to a depth of five centimetres.
波速为每秒五千米、回波延迟为二十微秒时，深度为五厘米，总路程为十厘米。
The total path is ten centimetres. Two reflectors four millimetres apart give echoes only one point six microseconds apart, so a two-microsecond pulse overlaps them.
相距四毫米的两个反射面，其回波只相差一点六微秒，因此两微秒脉冲会使它们重叠。
Use a shorter pulse for timing separation.
缩短脉冲可以改善时间上的分辨。
Wave models explained many observations, but a simple continuous-energy model could not explain the photoelectric threshold and prompt emission at low intensity.
波动模型解释了许多观察，但简单的连续能量模型无法解释光电效应的阈值，以及低强度下仍立即发生的发射。
The photon model links one absorption event to a packet of energy.
光子模型把一次吸收事件与一个能量包联系起来。
It complements rather than deletes the wave evidence.
它补充而不是抹去波动证据。
Choose the description that explains the measurement being discussed.
根据讨论的测量选择合适的描述。
A photon supplies energy to one electron.
一个光子向一个电子提供能量。
Some energy is needed to escape from the surface, and the remainder is kinetic energy.
电子离开表面需要能量，剩余部分成为动能。
Electrons that lose more energy inside the material escape more slowly.
在材料内部损失更多能量的电子，逸出时速度较小。
If the photon has less energy than the work function, a negative subtraction means no emission, not negative kinetic energy.
如果光子能量小于逸出功，计算出的负差值表示不发射，而不是出现负动能。
Above threshold, raising intensity at fixed frequency can increase emitted electron count without increasing maximum kinetic energy.
超过阈值后，固定频率而增大强度，可增加发射电子数，但不增加最大动能。
Raising frequency increases maximum kinetic energy.
提高频率会增加最大动能。
At fixed incident power, however, more energy per photon means fewer photons per second.
不过，在入射功率不变时，每个光子的能量增加意味着每秒光子数减少。
Do not promise a photocurrent ratio without considering collection and material response.
还要考虑收集条件与材料响应，不能直接保证光电流的比例。
First put both photon energy and work function into joules. Subtract the work function from photon energy.
先把光子能量和逸出功都换算成焦耳，再用光子能量减去逸出功。
Then divide the maximum kinetic energy by electron charge to find the stopping potential.
最后将最大动能除以电子电荷量，求遏止电压。
State whether the result confirms emission.
说明结果是否支持发生发射。
Try this before continuing.
继续之前先自己尝试。
Two electronvolts becomes three point two times ten to the minus nineteen joules.
二电子伏特等于三点二乘十的负十九次方焦耳。
The photon has five point three zero four times ten to the minus nineteen joules, leaving about two point one times ten to the minus nineteen.
光子能量为五点三零四乘十的负十九次方焦耳，剩余约二点一乘十的负十九次方焦耳。
Dividing by electron charge gives one point three volts.
除以电子电荷量得到约一点三伏。
The positive kinetic energy confirms emission.
正的动能表明可以发生发射。
The atom can occupy allowed discrete energies.
原子只能占据允许的分立能量。
A downward transition emits a photon with the energy difference, not the value of either level alone.
向下跃迁发射的光子能量等于能级差，而不是某一个能级的数值。
From minus two to minus five electronvolts, the photon has three electronvolts.
从负二电子伏特到负五电子伏特，光子能量为三电子伏特。
Count the possible transitions and populated starting levels carefully when predicting spectral lines.
预测谱线时，应仔细检查可能的跃迁及哪些初始能级被占据。
Current measures charge flow, while voltage measures energy per unit charge.
电流描述电荷流动速率，电压描述每单位电荷转移的能量。
Resistance is their ratio at an operating point.
电阻是某一工作点处电压与电流之比。
Ohm’s law is a special linear relationship at constant temperature and other physical conditions.
欧姆定律是在温度等物理条件不变时成立的线性关系。
A lamp still has a resistance even though it does not obey a constant-resistance model as it heats.
灯泡加热时不满足恒定电阻模型，但它仍然有电阻。
Charge conservation gives the junction rule. Energy conservation gives the loop voltage rule.
电荷守恒给出节点电流规律，能量守恒给出回路电压规律。
In series, substitute each voltage drop as current times resistance to derive resistance addition.
串联时，将各电压降写成电流乘电阻，可推导电阻相加。
In parallel, substitute each branch current as voltage divided by resistance.
并联时，将各支路电流写成电压除以电阻。
Derive before memorising so that you can recognise a new arrangement.
先推导再记忆，才能识别新的电路组合。
Pause and calculate. Six ohms is in parallel with three ohms, then in series with four ohms across twelve volts. Find total current and both branch currents.
六欧姆与三欧姆并联，再与四欧姆串联，接在十二伏电源两端。
Check that the branch currents add to the total.
求总电流以及两个支路电流。
The parallel section is two ohms and the whole circuit is six ohms.
六欧与三欧并联得到二欧，整个电路为六欧。
A twelve-volt supply gives two amperes overall, but only four volts across the parallel section.
十二伏电源使总电流为两安，但并联部分只有四伏。
The branch currents are two thirds and four thirds of an ampere.
两支路电流分别为三分之二安和三分之四安。
Use the branch voltage, not the supply voltage, when calculating branch power.
计算支路功率时，使用支路电压，不是电源电压。
A constant-temperature ohmic conductor has a straight graph.
恒温欧姆导体的图像是直线。
A heating filament lamp flattens as resistance rises.
灯丝加热后电阻增加，曲线变平。
A self-heating negative-temperature-coefficient thermistor can steepen as resistance falls.
负温度系数热敏电阻若因电流发热，电阻下降，曲线可能变陡。
A diode mainly conducts forward.
二极管主要沿正向导电。
For a curved graph, the tangent describes a change ratio, not generally the reciprocal of operating-point resistance.
对于曲线，切线表示变化量之比，一般不等于工作点电阻的倒数。
Measure voltage and current for several wire lengths between the actual contacts.
对几个实际接触点间距，分别测量电压和电流。
A resistance-length graph separates the wire gradient from a fixed contact-resistance intercept.
电阻对长度图可以把导线的斜率与固定接触电阻的截距区分开。
Check the micrometer zero and measure diameter at several positions.
检查千分尺零点，并在多个位置测直径。
Diameter error matters strongly because area depends on its square.
面积取决于直径平方，因此直径误差影响较大。
Small current and switching off limit temperature changes.
小电流和间歇断电有助于控制温度。
The symbol n means mobile carrier number per unit volume, not every electron in the material.
这里的n是单位体积中的可移动载流子数，不是材料中的全部电子数。
Drift is the small directed average superposed on random motion.
漂移速度是叠加在随机运动上的微小定向平均速度。
For the same current, halving cross-sectional area doubles drift speed if carrier density is unchanged.
在载流子密度不变且电流相同时，横截面积减半会使漂移速度加倍。
Material resistivity also depends on scattering, not carrier density alone.
材料电阻率还取决于散射，不能只由载流子密度解释。
Along a uniform wire at constant temperature, resistance grows linearly with length, so voltage drop does too.
恒温均匀导线的电阻与长度成正比，因此电压降也与长度成正比。
A two-resistor divider uses the same series-current argument.
两个电阻的分压器使用同样的串联电流关系。
The simple fraction assumes negligible output current.
简单分压比例假设输出电流可忽略。
If a load is attached, combine it in parallel with the output resistor before calculating the fraction.
接上负载后，应先将负载与输出电阻并联，再计算分压比例。
The output is across the lower LDR.
输出跨接在下方光敏电阻两端。
In the dark, four kiloohms out of a total six gives four volts.
暗处总电阻六千欧，光敏电阻占四千欧，输出四伏。
In bright light, one kiloohm out of three gives two volts.
亮处总电阻三千欧，光敏电阻占一千欧，输出两伏。
A significant load would change this result; combine the load in parallel with the LDR first.
若负载电流不可忽略，应先把负载与光敏电阻并联等效。
First state the environmental change, then explain carriers or scattering. Next state the resistance change. Finally follow the divider fraction to the named output.
先说明环境变化，再解释载流子或散射变化，然后说明电阻变化，最后利用分压比例判断指定输出。
An LDR below the fixed resistor gives a falling output when illuminated; above the fixed resistor it gives a rising output across the fixed lower resistor.
光敏电阻在固定电阻下方时，照明增强会使其两端输出下降；若光敏电阻在上方，下方固定电阻两端的输出则上升。
Electromotive force is energy supplied per unit charge, measured in volts. It is not a mechanical force.
电动势是电源每单位电荷提供的能量，单位是伏，并不是机械力。
When current flows, internal resistance causes an internal voltage drop, leaving a smaller terminal voltage.
有电流时，内阻产生内部电压降，使端电压小于电动势。
Source power is e.m.f. times current; terminal power plus current squared times internal resistance accounts for it.
电源功率等于电动势乘电流，它等于外部获得的功率加上电流平方乘内阻的内部发热功率。
Pause and calculate. The terminal-voltage line passes through zero point two amps, one point four volts, and zero point six amps, one point two volts. Find internal resistance and electromotive force.
端电压与电流的直线经过零点二安、一点四伏和零点六安、一点二伏。
Check both points.
求内阻和电动势。
Use a variable external load and ammeter in series with the cell, and a high-resistance voltmeter across its terminals.
将可变外部负载和电流表与电池串联，高阻电压表接在电池两端。
Keep currents limited and open the switch between readings.
限制电流，并在读数之间断开开关。
From a line through zero point two amperes at one point four volts and zero point six amperes at one point two volts, internal resistance is half an ohm and e.m.f. is one point five volts.
如果直线经过零点二安、一点四伏和零点六安、一点二伏两点，则内阻为零点五欧，电动势为一点五伏。
Close the reference and answer these three prompts.
合上讲义，回答这三个问题。
Use a physical explanation, not only a formula.
给出物理解释，不要只写公式。
Then check your answer against the reference and return to the matching skill sheet where needed.
然后对照讲义检查，必要时回到对应技能练习。
The authentic sets deliberately sample demands; original practice remains essential for syllabus requirements absent from those papers.
精选真题只抽样考查部分要求，因此，大纲中未被这些试卷覆盖的内容仍需要原创练习。