Cực hóa, chuỗi bộ phân tích và pha
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| polariser/ˈpəʊlɑːraɪzə/ | A device that selectively transmits light in particular polarisation states. |
| Brewster's angle/ˈbruːstəz ˈæŋɡl/ | The incidence angle at which reflected light is fully s-polarised. |
| quarter-wave plate/ˈkwɔːtə weɪv pleɪt/ | A crystal plate that adds a quarter-cycle phase between two polarisation components. |
A third filter lets more light through
- Crossed polarisers 偏振片 block everything. Slip a 45° polariser between them and a quarter of the light from the first filter gets through.
- GRE polarisation questions test Malus's law in chains, polarisation states, wave plates and Brewster's angle 布儒斯特角.
Malus's law, step by step
- An ideal analyser passes $E\cos\theta$, so $I = I_{\text{in}}\cos^2\theta$, with $\theta$ measured from the current polarisation.
- Unpolarised light through the first polariser: $I_0/2$, now polarised along that axis.
$I_0 = 80$ W/m² through axes 0°, 45°, 90°: $40 \to 40\cos^2 45° = 20 \to 20\cos^2 45° = 10$ W/m². Remove the middle filter and the answer is 0.
Only unpolarised light loses half at a polariser. Do not apply another factor of ½ at each later filter.
Unpolarised 160 W/m² passes ideal axes 0°, 30°, 90°. Final intensity in W/m²?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
80 × cos²30° × cos²60° = 80 × 0.75 × 0.25.
Polarisation states
- $E_x = A\cos\omega t$, $E_y = B\cos(\omega t + \delta)$. $\delta = 0$ or $\pi$: linear. $A = B$, $\delta = \pm\pi/2$: circular. Otherwise elliptical.
- Circular light through any ideal linear analyser: half the intensity.
Linear, circular or elliptical?
Read the polarisation state from amplitudes and phase.
Circular light of 50 W/m² meets an ideal linear analyser. Transmitted intensity in W/m²?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Half, at any analyser angle.
Wave plates
- A quarter-wave plate 四分之一波片 adds a phase of $\pi/2$ between its axes. Linear light at 45° to the axes becomes circular; light along an axis stays linear.
- A polariser removes a component; a wave plate only shifts phase.
Linear light at 45° to the axes of a quarter-wave plate emerges
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Equal components, π/2 apart.
An ideal wave plate removes one field component, as a polariser does.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
It only changes relative phase.
Brewster's angle
- $\tan\theta_B = n_2/n_1$. The reflected $p$-component vanishes, so reflected light is $s$-polarised; the refracted ray is at $90°$ to the reflected one.
- Air to glass ($n = 1.5$): $\theta_B = 56.3°$. Not the critical angle, which needs $n_1 > n_2$.
Brewster's angle from air into water ($n = 1.33$), in degrees (one decimal place)?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Brewster's angle satisfies $\tan\theta_B = n_2/n_1 = 1.33$, so $\theta_B = 53.1°$.
Recap
- First polariser: $I_0/2$; then $\cos^2$ of the angle to the previous axis.
- Equal amplitudes and quarter-cycle phase: circular.
- $\tan\theta_B = n_2/n_1$; reflection is $s$-polarised.
- Brewster's angle: The incidence angle at which reflected light is fully s-polarised.
- quarter-wave plate: A crystal plate that adds a quarter-cycle phase between two polarisation components.