Polarisation
Introduced| English | Meaning |
|---|---|
| polarisation/ˌpəʊləraɪˈzeɪʃn/ | Here, linear polarisation: transverse oscillations are restricted to one direction perpendicular to propagation; for light, this describes the electric field. |
| transverse/trænsˈvɜːs/ | Describing a wave whose oscillations are perpendicular to the direction in which the disturbance travels. |
| polarising filter/ˈpəʊləraɪzɪŋ ˈfɪltə/ | A filter that transmits a selected direction of transverse light oscillation. |
| transmission axis/trænˈsmɪʃn ˈæksɪs/ | The direction of light's electric-field oscillation that a linear polarising filter transmits. |
| longitudinal/ˌlɒŋɡɪˈtjuːdɪnl/ | Describing a wave in which particle oscillations are parallel to the direction of energy transfer. |
| unpolarised/ʌnˈpəʊləraɪzd/ | unpolarised |
| Malus's law/ˈmæləsɪz lɔː/ | The transmitted intensity of linearly polarised light through an ideal analyser equals its incident intensity times the square of the cosine of the angle between their polarisation axes. |
| intensity/ɪnˈtensɪti/ | The power carried by a wave per unit area perpendicular to its direction of propagation. |
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.
Polarisation intensity lab
intensity changes with polariser angle
Rotate a polariser and see why only transverse waves can be polarised.
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.
Polarising a transverse wave means making it vibrate:
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A polarising filter passes only the vibration lined up with its axis, leaving the wave in a single plane.
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
Sound waves can be polarised.
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No — sound is longitudinal (it vibrates along the travel), so there is no other plane to pick out.
If a wave can be polarised, it must be ____.
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Only transverse waves have vibration directions perpendicular to travel, so only they can be polarised.
Sound cannot be polarised because it is a ____ wave.
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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.
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.
Match each wave to whether it can be polarised.
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Polarisation restricts the oscillation to one plane, and a longitudinal wave already oscillates along a single line. Only transverse waves can be polarised.
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
Polarised light of intensity $100\ \dfrac{\text{W}}{\text{m}^2}$ meets a filter at $60^{\circ}$ to its plane. What intensity gets through?
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$I = I_0\cos^{2}\theta = 100 \times \cos^{2}60^{\circ} = 100 \times 0.25 = 25\ \dfrac{\text{W}}{\text{m}^2}$.
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.
Two polarising filters are crossed at $90^{\circ}$. How much light gets through?
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$I = I_0\cos^{2}90^{\circ} = 0$ — crossed filters block the light completely.
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?
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$\dfrac{I}{I_0} = \cos^{2}60^{\circ} = 0.25$, and $I \propto A^{2}$, so $\dfrac{A}{A_0} = \sqrt{0.25} = \cos 60^{\circ} = 0.50$.
Polarised light of intensity 80 W/m^2 meets a filter at 60 degrees to its plane. What intensity passes, in W/m^2?
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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.
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.
Unpolarised light meets two crossed filters. Which are true? Select all that apply.
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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.
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.
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?
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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: $\dfrac{I_0}{2} \times \dfrac{1}{2} = \dfrac{I_0}{4}$.
When two crossed filters block all light, inserting a third filter at 45 degrees between them lets some light through.
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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.
$\theta$ 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.
Unpolarised light passes through a single polarising filter. Rotating the filter changes the intensity that comes out.
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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.
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