Polarisation · 偏振
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| polarisation/ˌpəʊləraɪˈzeɪʃn/ | 偏振 | piān zhèn |
| transverse/trænsˈvɜːs/ | 横波 | héng bō |
| polarising filter/ˈpəʊləraɪzɪŋ ˈfɪltə/ | 偏振片 | piān zhèn piàn |
| transmission axis/trænˈsmɪʃn ˈæksɪs/ | 透光轴 | tòu guāng zhóu |
| longitudinal/ˌlɒŋɡɪˈtjuːdɪnl/ | 纵波 | zòng bō |
| unpolarised/ʌnˈpəʊləraɪzd/ | 非偏振 | fēi piān zhèn |
| Malus's law/ˈmæləsɪz lɔː/ | 马吕斯定律 | mǎ lǚ sī dìng lǜ |
| intensity/ɪnˈtensɪti/ | 强度 | qiáng dù |
Sunglasses that cut glare
- Polarising sunglasses block the glare off water and roads.
- Rotate them and the brightness changes — even though the scene hasn't.
- They work by letting light through in just one plane.
减少眩光的太阳镜
- 偏振太阳镜能挡住水面和路面的眩光。
- 转动它,亮度会变化——尽管景物并没有变。
- 它的原理是只让 一个平面 内的光通过。
Polarisation intensity lab · 偏振强度实验
intensity changes with polariser angle · 强度随偏振片角度变化
Rotate a polariser and see why only transverse waves can be polarised. · 旋转偏振片，观察为何只有横波可以被偏振。
What polarisation 偏振 is
- Polarisation means making a transverse 横波 wave vibrate in one plane only.
- A polarising filter 偏振片 passes the part lined up with its transmission axis 透光轴 and blocks the rest.
什么是偏振
- 偏振(polarisation) 是让横波只在 一个平面 内振动。
- 偏振片让与它的 透光轴(transmission axis) 对齐的部分通过,挡住其余部分。
Polarising a transverse wave means making it vibrate: · 使横波偏振意味着使其振动：
A polarising filter passes only the vibration lined up with its axis, leaving the wave in a single plane. · 偏振滤光片只允许与其轴线对齐的振动通过，使波保持在单一平面内。
Only transverse waves
- A transverse wave has many planes to choose from, so it can be polarised.
- A longitudinal 纵波 wave (sound) vibrates along the travel — there is no other plane, so it cannot.
- So: if a wave can be polarised, it must be transverse.
Unpolarised waves vibrate in many planes; a polarised wave vibrates in one plane
只有横波
- 横波有很多平面可选,所以它 可以 被偏振。
- 纵波(声音)沿传播方向振动——没有别的平面,所以它 不能。
- 所以:如果一列波能被偏振,它一定是 横波。
非偏振波在许多平面内振动;偏振波在一个平面内振动
Sound waves can be polarised. · 声波可以被偏振。
No — sound is longitudinal (it vibrates along the travel), so there is no other plane to pick out. · 不能——声波是纵波（沿传播方向振动），因此没有其他平面可供选择。
If a wave can be polarised, it must be . · 如果波可以被偏振，它必须是。
Only transverse waves have vibration directions perpendicular to travel, so only they can be polarised. · 只有横波的振动方向垂直于传播方向，因此只有它们可以被偏振。
Sound cannot be polarised because it is a ____ wave. · 声波不能被偏振，因为它是____波。
Polarisation restricts oscillations to one plane, and a longitudinal wave already oscillates along only one line, the direction of travel. There is nothing left to restrict. · 偏振将振荡限制在一个平面内，而纵波已经仅沿一条线振荡，即传播方向。已无剩余内容可限制。
The exam sentence for "why not sound?"
- Two marks, two ideas: sound is longitudinal, so its vibrations are parallel to the direction of energy transfer.
- There is only one possible direction of vibration, so there is no plane for a filter to select.
- Light is transverse, so its vibrations can be in any plane perpendicular to the travel — a filter can pick one.
"为什么声音不行?"的考试答句
- 两分,两个要点:声音是 纵波,所以它的振动 平行于 能量传递的方向。
- 只有 一个 可能的振动方向,所以没有平面可供偏振片选择。
- 光是横波,它的振动可以在垂直于传播方向的 任何 平面内——偏振片可以选出一个。
Match each wave to whether it can be polarised. · 将每种波与其能否被偏振相匹配。
Polarisation restricts the oscillation to one plane, and a longitudinal wave already oscillates along a single line. Only transverse waves can be polarised. · 偏振将振荡限制在一个平面内，而纵波已经沿单一线条振荡。只有横波可以被偏振。
Malus's law 马吕斯定律
- Through a filter at angle $\theta$ to the polarisation: $I = I_0\cos^{2}\theta$.
- $\theta = 0^{\circ}$: all passes. $\theta = 90^{\circ}$: all blocked.
Crossed filters (a) block the light; parallel filters (b) let it pass
马吕斯定律
- 通过与偏振方向成 $\theta$ 角的偏振片:$I = I_0\cos^{2}\theta$。
- $\theta = 0^{\circ}$:全部通过。$\theta = 90^{\circ}$:全部挡住。
正交的偏振片 (a) 挡住光;平行的偏振片 (b) 让光通过
Polarised light of intensity $100\ \dfrac{\text{W}}{\text{m}^2}$ meets a filter at $60^{\circ}$ to its plane. What intensity gets through? · 强度为 $100\ \dfrac{\text{W}}{\text{m}^2}$ 的偏振光遇到与偏振面成 $60^{\circ}$ 角的滤光片。透射光的强度是多少？
$I = I_0\cos^{2}\theta = 100 \times \cos^{2}60^{\circ} = 100 \times 0.25 = 25\ \dfrac{\text{W}}{\text{m}^2}$.
Worked example: one filter
- Polarised light of intensity 强度 $I_0$ meets a filter at $\theta = 60^{\circ}$.
- $I = I_0\cos^{2}60^{\circ} = I_0 \times (0.5)^{2} = \dfrac{I_0}{4}$.
- Amplitude: intensity goes with amplitude squared, so the amplitude ratio is the square root: $\dfrac{A}{A_0} = \cos 60^{\circ} = 0.50$.
- Two crossed filters ($90^{\circ}$) let through nothing at all.
例题:一片偏振片
- 强度为 $I_0$ 的偏振光遇到一片 $\theta = 60^{\circ}$ 的偏振片。
- $I = I_0\cos^{2}60^{\circ} = I_0 \times (0.5)^{2} = \dfrac{I_0}{4}$。
- 振幅: 强度与振幅的平方成正比,所以振幅之比是平方根:$\dfrac{A}{A_0} = \cos 60^{\circ} = 0.50$。
- 两片 正交 的偏振片($90^{\circ}$)什么都不让通过。
Two polarising filters are crossed at $90^{\circ}$. How much light gets through? · 两个偏振滤光片正交放置（夹角 $90^{\circ}$）。有多少光能透过？
$I = I_0\cos^{2}90^{\circ} = 0$ — crossed filters block the light completely. · $I = I_0\cos^{2}90^{\circ} = 0$ ——正交滤光片完全阻挡光线。
Polarised light passes through a filter whose axis is at $60^{\circ}$ to its plane of polarisation. What is the ratio of the transmitted amplitude to the incident amplitude? · 偏振光通过一个偏振轴与其偏振面成 $60^{\circ}$ 角的滤光片。透射振幅与入射振幅的比值是多少？
$\dfrac{I}{I_0} = \cos^{2}60^{\circ} = 0.25$, and $I \propto A^{2}$, so $\dfrac{A}{A_0} = \sqrt{0.25} = \cos 60^{\circ} = 0.50$. · $\dfrac{I}{I_0} = \cos^{2}60^{\circ} = 0.25$，且 $I \propto A^{2}$，所以 $\dfrac{A}{A_0} = \sqrt{0.25} = \cos 60^{\circ} = 0.50$。
Polarised light of intensity 80 W/m^2 meets a filter at 60 degrees to its plane. What intensity passes, in W/m^2? · 强度为 80 W/m^2 的偏振光与滤光片相遇，滤光片与其平面成 60 度角。透过的强度是多少 W/m^2？
Malus: I = I0 cos^2(theta) = 80 x 0.25 = 20 W/m^2. Squaring the cosine is the step people drop, and dropping it gives 40. · 马吕斯定律：I = I0 cos^2(theta) = 80 x 0.25 = 20 W/m^2。人们通常会遗漏余弦的平方，如果遗漏了，结果是 40。
A third filter in the middle
- Two crossed filters block everything. Slide a third filter between them at $45^{\circ}$ and light gets through again.
- The middle filter passes $\cos^{2}45^{\circ} = \dfrac{1}{2}$ of the light and turns its plane to $45^{\circ}$.
- The last filter is now only $45^{\circ}$ from that new plane, so it passes another half.
- Each filter works on the light as it arrives, not on the original beam.
中间插入第三片
- 两片正交的偏振片挡住一切。在它们之间插入 第三片、与之成 $45^{\circ}$,光又能通过了。
- 中间那片让 $\cos^{2}45^{\circ} = \dfrac{1}{2}$ 的光通过,并把偏振面转到 $45^{\circ}$。
- 最后一片现在离这个新平面只有 $45^{\circ}$,所以又让一半通过。
- 每一片作用的是 到达它时 的光,而不是原来的光束。
{/callout}
Unpolarised light meets two crossed filters. Which are true? Select all · 所有 that apply. · 非偏振光遇到两个正交滤光片。哪些陈述是正确的？选择所有适用项。
Each filter re-polarises along its own axis, so every following angle is measured from the previous filter, which is exactly why a third filter can add light rather than remove it. · 每个滤光片都会沿其自身轴线重新偏振，因此后续每个角度都是从上一个滤光片测量的，这正是为什么第三个滤光片可以增加光强而非减少光强的原因。
Worked example: three filters
Unpolarised 非偏振 light of intensity $I_0$ passes through three filters: the first with a vertical axis, the second at $45^{\circ}$, the third horizontal. Find the final intensity.
- First filter: unpolarised light contains every plane equally, so a single filter passes half: $\dfrac{I_0}{2}$, now vertically polarised.
- Second filter at $45^{\circ}$ to vertical: $\dfrac{I_0}{2}\cos^{2}45^{\circ} = \dfrac{I_0}{4}$, now polarised at $45^{\circ}$.
- Third filter at $45^{\circ}$ to that: $\dfrac{I_0}{4}\cos^{2}45^{\circ} = \dfrac{I_0}{8}$.
- Check: remove the middle filter and the first and third are crossed — $\cos^{2}90^{\circ} = 0$, nothing passes. Adding a filter increased the light; that is the surprise the examiner is testing.
例题:三片偏振片
强度为 $I_0$ 的非偏振(unpolarised)光依次通过三片偏振片:第一片透光轴竖直,第二片成 $45^{\circ}$,第三片水平。求最终强度。
- 第一片: 非偏振光均等地包含每个平面,所以一片偏振片让 一半 通过:$\dfrac{I_0}{2}$,此时为竖直偏振。
- 第二片 与竖直成 $45^{\circ}$:$\dfrac{I_0}{2}\cos^{2}45^{\circ} = \dfrac{I_0}{4}$,此时偏振方向为 $45^{\circ}$。
- 第三片 与之成 $45^{\circ}$:$\dfrac{I_0}{4}\cos^{2}45^{\circ} = \dfrac{I_0}{8}$。
- 检查: 拿掉中间那片,第一片和第三片就是正交的——$\cos^{2}90^{\circ} = 0$,什么都不通过。多加一片反而 增加 了光;这正是阅卷人要考的意外之处。
Vertically polarised light of intensity $I_0$ passes through a filter at $45^{\circ}$ and then a horizontal filter. What fraction of $I_0$ comes out? · 强度为 $I_0$ 的垂直偏振光通过 $45^{\circ}$ 角的滤光片，然后是一个水平滤光片。$I_0$ 中有几分之几射出？
After the $45^{\circ}$ filter: $I_0\cos^{2}45^{\circ} = \dfrac{I_0}{2}$, now polarised at $45^{\circ}$. The horizontal filter is $45^{\circ}$ from that: $\dfrac{I_0}{2} \times \dfrac{1}{2} = \dfrac{I_0}{4}$. · 经过 $45^{\circ}$ 滤光片后：$I_0\cos^{2}45^{\circ} = \dfrac{I_0}{2}$，现在偏振方向为 $45^{\circ}$。水平滤光片与该方向成 $45^{\circ}$ 角：$\dfrac{I_0}{2} \times \dfrac{1}{2} = \dfrac{I_0}{4}$。
When two crossed filters block all light, inserting a third filter at 45 degrees between them lets some light through. · 当两个正交滤光片阻挡所有光时，在它们之间插入第三个 45 度角的滤光片会让部分光透过。
Each filter re-polarises along its own axis, so the angle for the next one is measured from THERE, not from the original plane. Adding a filter can increase the transmitted light. · 每个滤光片都会沿其自身轴线重新偏振，因此下一个滤光片的角度应从此处测量，而不是从原始平面测量。增加滤光片可以增加透射光。
$\theta$ is the angle between the light's plane of polarisation and the filter's transmission axis — not the angle you rotated the filter through, unless it started aligned. Rotating a filter through a full $360^{\circ}$ gives two maxima and two minima. And unpolarised light through one filter always comes out at half the intensity, whatever the filter's angle.
$\theta$ 是光的 偏振面 与偏振片 透光轴 之间的夹角——不是你把偏振片转过的角度,除非它一开始就是对齐的。把偏振片转过整整 $360^{\circ}$ 会出现 两个 最大值和 两个 最小值。而且 非偏振 光通过一片偏振片后总是变成一半的强度,与偏振片的角度无关。
Unpolarised light passes through a single polarising filter. Rotating the filter changes the intensity that comes out. · 非偏振光通过单个偏振滤光片。旋转滤光片会改变出射光的强度。
Unpolarised light has every plane equally, so one filter always passes half of it — the angle makes no difference. Malus's law only applies to light that is already polarised. · 非偏振光包含所有平面，因此一个滤光片总是透过其中一半——角度没有影响。马吕斯定律仅适用于已经偏振的光。
You've got it
- polarisation = vibrating in one plane; only transverse waves can do it (sound cannot: its vibrations are parallel to the energy transfer)
- Malus's law: $I = I_0\cos^{2}\theta$; amplitude ratio $= \cos\theta$; crossed filters block all light
- filters act one after another on the light as it arrives — a middle filter lets light through crossed ones
你掌握了
- 偏振 = 在一个平面内振动;只有 横波 能做到(声音不能:它的振动平行于能量传递方向)
- 马吕斯定律:$I = I_0\cos^{2}\theta$;振幅之比 $= \cos\theta$;正交的偏振片挡住全部的光
- 偏振片依次作用于 到达时 的光——中间一片能让光穿过正交的两片