Polarisation, analyser chains and phase
| English | 中文 | Pinyin |
|---|---|---|
| Malus’s law | 马吕斯定律 | mǎ lǚ sī dìng lǜ |
| Brewster angle | 布儒斯特角 | bù rú sī tè jiǎo |
A decision before an answer
- Inserting a third polariser between crossed filters can increase the transmitted intensity, even though every filter removes energy.
- Your goal: Calculate successive ideal analyser transmissions from the immediate input state.
Project the electric field
- For propagation along z, a transverse electric field can have x and y components. Linear polarisation means the field oscillates along one fixed line; an ideal analyser transmits the projection along its axis. Field amplitude becomes E cosθ, so intensity becomes I cos²θ. Here θ is between the incoming polarisation and that analyser, and I is the intensity immediately before it.
- An unpolarised beam is a statistical mixture of transverse orientations; an ideal first polariser passes half its mean intensity. The outgoing beam is then linearly polarised along the filter axis. Do not apply a new factor of one half at every later filter, or use one angle to the original source for the whole chain.
Unpolarised intensity I₀ passes ideal axes 0°, 30°, 90°. Final intensity is:
The first output is I₀/2; later angles are 30° and 60°. Multiply (1/2)(3/4)(1/4)=3/32.
Update each analyser input
- After each analyser, update both the intensity and the polarisation direction. For an initially unpolarised beam and ideal axes α₁, α₂, α₃, the final intensity is (I₀/2)cos²(α₂−α₁)cos²(α₃−α₂). Two crossed filters transmit zero in this model; a middle oblique axis changes the direction before the final projection.
- This increase relative to the crossed pair does not create energy: every step still has transmission between zero and one. Real filters have absorption, imperfect extinction and wavelength dependence. Use the ideal law only when those losses are excluded or separately specified.
At one point E_x=2 cosωt and E_y=2 sinωt. The polarisation is:
Equal nonzero amplitudes with quarter-cycle phase give constant magnitude and a rotating transverse direction.
Track transverse phase
- Write E_x=A cosωt and E_y=B cos(ωt+δ) at a fixed point. Equal or opposite phases give a line, including a line at an oblique angle; equal nonzero amplitudes and a quarter-cycle phase difference give a circle. Unequal amplitudes at quarter-cycle phase give an ellipse.
- A quarter-wave plate adds a relative phase of π/2 between its principal axes under its design conditions. A linear input at 45° supplies equal components, so the outgoing field can be circular. A linear input along a principal axis has only one component and remains linear. For circular input, every ideal linear analyser passes half the intensity; handedness requires a declared viewing direction and phase convention.
Unpolarised I₀=80 W/m² passes axes 0°, 45°, 90°. Intensities are 40, 20 and 10 W/m²; without the middle filter the crossed pair would give zero. Equal E_x=3 cosωt and E_y=3 sinωt trace a circle with E_x²+E_y²=9. From n₁=1 to n₂=1.5, θ_B=atan(1.5)=56.31° and refraction is 33.69°; the reflected beam is s polarised.
Polarised input intensity 24 W/m² meets an ideal analyser at 60°. Output is ____ W/m².
24 cos²60°=24/4=6.
Apply Brewster geometry
- For incidence from transparent nonmagnetic medium n₁ into n₂, Brewster’s angle measured from the normal satisfies tanθ_B=n₂/n₁. At that angle, the reflected p component, whose electric field lies in the plane of incidence, vanishes in the ideal dielectric model. Reflected light from unpolarised input is then s polarised, perpendicular to that plane.
- Snell’s law gives a refracted angle complementary to θ_B. This is distinct from the critical-angle condition sinθ_c=n₂/n₁, which needs n₁>n₂ and concerns total internal reflection. Brewster reflection does not imply that the entire incident intensity is reflected or that the transmitted beam is completely polarised.
Malus intensity uses cos² of the angle to the immediately preceding polarisation; a phase plate changes relative phase, while a polariser removes a field component.
Which answer fits this case?
Calculate successive ideal analyser transmissions from the immediate input state
At Brewster incidence for the ideal dielectric model, reflected light from unpolarised input is polarised perpendicular to the plane of incidence.
The reflected p component is zero; the surviving s electric field is perpendicular to the incidence plane.
Keep the distinctions
- Malus’s law 马吕斯定律 — Ideal linear-analyser intensity law I_out=I_in cos²θ for linearly polarised input.
- Brewster angle 布儒斯特角 — Incidence angle at which the reflected p component vanishes for the ideal dielectric interface.
- Calculate successive ideal analyser transmissions from the immediate input state.
- Distinguish linear, circular and elliptical field motion using relative phase.
- Use Brewster incidence with refractive-index and plane-of-incidence conventions.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.