Polarization | Physics - Wyatt's Notes

5.1 States of Polarization

For a plane wave propagating in the $z$ -direction:

$\mathbf{E} = E_{0x}\cos(kz - \omega t)\,\hat{\mathbf{x}} + E_{0y}\cos(kz - \omega t + \delta)\,\hat{\mathbf{y}}$

Linear polarization: $\delta = 0$ or $\delta = \pi$ . The E-field oscillates along a fixed line.
Circular polarization: $E_{0x} = E_{0y}$ and $\delta = \pm\pi/2$ . Right-handed ( $\delta = -\pi/2$ ) or left-handed ( $\delta = +\pi/2$ ).
Elliptical polarization: General case. The tip of $\mathbf{E}$ traces an ellipse.

5.2 Malus”s Law

When linearly polarised light of intensity $I_0$ passes through a polariser at angle $\theta$ to the Polarisation direction:

$I = I_0 \cos^2\theta$

Proof. The component of $\mathbf{E}$ along the polariser axis is $E\cos\theta$ . Since $I \propto E^2$ : $I = I_0 \cos^2\theta$ . $\blacksquare$

Worked Example: Three polarisers in series

Problem. Unpolarised light of intensity $I_0$ passes through three ideal linear polarisers. The First has its transmission axis vertical. The second is at $45°$ to the vertical. The third is Horizontal. What fraction of $I_0$ is transmitted?

Solution. After the first polariser: $I_1 = I_0/2$ (vertical polarisation).

After the second: $I_2 = I_1\cos^2 45° = (I_0/2)(1/2) = I_0/4$ (polarised at 45°).

After the third: $I_3 = I_2\cos^2 45° = (I_0/4)(1/2) = I_0/8$ (horizontal polarisation).

Answer: $I_0/8 = 12.5\%$ . Note that inserting the middle polariser increases the transmitted Intensity compared with just the crossed first and third polarisers (which would transmit zero).

5.3 Birefringence and Wave Plates

Birefringent crystals (e.g., calcite) have two refractive indices: $n_o$ (ordinary ray) and $n_e$ (extraordinary ray). The two rays have orthogonal polarisations and different phase velocities.

A wave plate of thickness $t$ introduces a relative phase shift between the two polarisation Components:

$\Delta\phi = \frac{2\pi}{\lambda}(n_o - n_e)\,t$

Quarter-wave plate (QWP): $\Delta\phi = \pi/2$ So $t_{\mathrm{QWP} = \lambda/(4|n_o - n_e|)}$ . Converts linear polarisation at $45°$ to the fast/slow axes into circular polarisation, and vice Versa.

Half-wave plate (HWP): $\Delta\phi = \pi$ So $t_{\mathrm{HWP} = \lambda/(2|n_o - n_e|)}$ . Rotates the plane of linear polarisation by $2\theta$ Where $\theta$ is the angle between the Input polarisation and the fast axis.

linear polarisation is at Exactly $45°$ to the fast and slow axes. For other input angles, the output is elliptically Polarised. A half-wave plate rotates linear polarisation by $2\theta$ Not $\theta$ .

Worked Example: Quarter-wave plate design and application

Problem. (a) Design a quarter-wave plate for $\lambda = 589$ nm using calcite ( $n_o = 1.658$ , $n_e = 1.486$ ). (b) If linearly polarised light at $30°$ to the fast axis Enters this QWP, describe the output polarisation.

Solution.

(a) Minimum thickness: $t = \lambda/(4|n_o - n_e|) = 589 \times 10^{-9}/(4 \times 0.172) = 8.56 \times 10^{-7}$ m $= 856$ nm.

(b) The components along the fast and slow axes are: $E_f = E_0\cos 30° = 0.866\,E_0$ and $E_s = E_0\sin 30° = 0.500\,E_0$ . After the QWP, these have a $\pi/2$ phase difference but unequal amplitudes ( $0.866 \neq 0.500$ ), So the output is elliptically polarised (not circular).

5.4 Optical Activity

Certain materials (sugars, quartz) rotate the plane of linearly polarised light. The specific Rotation is:

$[\alpha] = \frac{\theta}{c \cdot l}$

Where $\theta$ is the rotation angle, $c$ is the concentration, and $l$ is the path length.

Optical activity arises from the helical structure of molecules, which gives different refractive Indices for left- and right-circularly polarised light (circular birefringence). If $n_L$ and $n_R$ Are the refractive indices for left and right circular polarisation:

$\theta = \frac{\pi l}{\lambda}(n_L - n_R)$

Optical activity is reciprocal: if the beam is reflected back through the medium, the rotation Is cancelled.

5.5 Brewster’s Angle

At the Brewster angle $\theta_B$ The reflected beam for p-polarised light vanishes ( $r_p = 0$ ):

$\tan\theta_B = \frac{n_2}{n_1}$

Proof. Setting $r_p = 0$ requires $n_2\cos\theta_i = n_1\cos\theta_t$ . Using Snell’s law $n_1\sin\theta_i = n_2\sin\theta_t$ :

$\frac{\cos\theta_i}{\sin\theta_i} = \frac{\cos\theta_t}{\sin\theta_t} \implies \cot\theta_i = \cot\theta_t \implies \theta_i = \theta_t'$

Where $\theta_t' = 90° - \theta_t$ . This gives $\theta_i + \theta_t = 90°$ So:

$\tan\theta_i = \frac{n_2}{n_1} \quad \blacksquare$

At Brewster’s angle, the reflected beam is purely s-polarised, and the reflected and refracted beams Are perpendicular ( $\theta_B + \theta_t = 90°$ ). This principle is used in Brewster windows and Polarisation by reflection.

For an air-glass interface ( $n_1 = 1$ , $n_2 = 1.5$ ): $\theta_B = \arctan(1.5) = 56.3°$ .

Worked Example: Brewster's angle and reflected intensity

Problem. Unpolarised light is incident on a glass surface ( $n = 1.50$ ) at Brewster’s angle. What fraction of the incident intensity is reflected, and what is the polarisation state of the Reflected light?

Solution. $\theta_B = \arctan(1.50) = 56.3°$ .

The reflected light is purely s-polarised. The reflectance for s-polarisation at $\theta_B$ : $\theta_t = 90° - 56.3° = 33.7°$ . $r_s = \frac{n_1\cos\theta_B - n_2\cos\theta_t}{n_1\cos\theta_B + n_2\cos\theta_t} = \frac{\cos 56.3° - 1.50\cos 33.7°}{\cos 56.3° + 1.50\cos 33.7°} = \frac{0.555 - 1.50 \times 0.832}{0.555 + 1.50 \times 0.832} = \frac{0.555 - 1.248}{0.555 + 1.248} = \frac{-0.693}{1.803} = -0.384$

$R_s = (-0.384)^2 = 0.148$ .

The incident unpolarised light has equal s and p components ( $I_s = I_p = I_0/2$ ). Only the s Component is reflected: $I_{\mathrm{reflected} = R_s \times I_0/2 = 0.148 \times I_0/2 = 0.074\,I_0}$ .

The reflected light is 100% s-polarised with intensity $0.074\,I_0$ (about 7.4% of the incident).

5.6 Faraday Rotation

In a magneto-optical material with a magnetic field $\mathbf{B}$ applied along the propagation Direction, the plane of polarisation rotates by:

$\theta_F = V B l$

Where $V$ is the Verdet constant (rad/(T $\cdot$ M)), $B$ is the magnetic field strength, and $l$ is the path length through the material.

Mechanism. The magnetic field induces circular birefringence: left and right circular polarisations Experience different refractive indices ( $n_L \neq n_R$ ). Unlike natural optical activity, Faraday Rotation is non-reciprocal: if the beam is reflected back through the medium, the rotation Doubles rather than cancelling.

Applications. Optical isolators (one-way light valves), optical circulators, and magneto-optical Sensors. An optical isolator consists of a polariser, a Faraday rotator set to rotate by $45°$ And An analyser at $45°$ to the polariser. Forward-propagating light is transmitted; backward light Is rotated by another $45°$ (total $90°$ ) and blocked by the analyser.

Worked Example: Faraday rotation in flint glass

Problem. A Faraday rotator uses heavy flint glass with Verdet constant $V = 32$ rad/(T $\cdot$ M). (a) What magnetic field over a 10 cm length produces a $45°$ rotation? (b) If linearly polarised Light makes a round trip through the rotator, what is the total rotation?

Solution.

(a) $\theta_F = VBl \implies B = \theta_F/(Vl) = (\pi/4)/(32 \times 0.10) = 0.785/3.2 = 0.245$ T.

(b) Because Faraday rotation is non-reciprocal, the return trip adds another $45°$ : Total rotation $= 90°$ . The polarisation is rotated by $90°$ after the round trip, which is the Basis of optical isolation.

5.7 Polarization by Scattering

When light is scattered by particles much smaller than the wavelength (Rayleigh scattering), the Scattered light is partially polarised. Light scattered at $90°$ to the incident direction is completely linearly polarised in the plane perpendicular to the scattering plane.

Proof. Consider an incident unpolarised beam propagating along $\hat{\mathbf{z}}$ . The $\mathbf{E}$ -field oscillates in the $xy$ -plane. An observer along $\hat{\mathbf{x}}$ (scattering Angle $90°$ ) receives radiation from the accelerating electrons. The dipole radiation pattern of an Oscillator along $\hat{\mathbf{y}}$ has zero intensity along $\hat{\mathbf{y}}$ but maximum along $\hat{\mathbf{x}}$ . The oscillator along $\hat{\mathbf{x}}$ radiates zero along its own axis. Thus The observer along $\hat{\mathbf{x}}$ sees only the $y$ -component: the scattered light is Polarised along $\hat{\mathbf{y}}$ . $\blacksquare$

This explains why the sky is polarised at $90°$ from the sun and why polarising sunglasses reduce Glare from horizontal surfaces (Brewster’s angle reflection from road/water).

flowchart TD
    A[5_Polarization] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

Polarization describes the orientation of a light wave’s electric field oscillation. Unpolarised light has field vectors pointing randomly, while linearly polarised light oscillates in a single plane. Malus’s law reveals that intensity drops as the cosine squared of the angle because only the component along the axis gets through. Brewster’s angle is where reflected light becomes completely polarised because the reflected and refracted rays are perpendicular. Birefringent materials split light into two components that travel at different speeds, which is how wave plates convert between linear and circular polarisation. Faraday rotation is non-reciprocal, enabling optical isolators.

Cross-References

Fresnel Equations: Derives the reflection and transmission coefficients that depend on polarisation, including Brewster’s angle where p-polarised light has zero reflection.
Geometric Optics: Provides the ray-tracing framework for understanding polarisation by reflection and Brewster windows.
Coherence: Partially polarised light is described by the mutual coherence function, connecting polarisation to the coherence theory of partially polarised fields.
Vector Calculus

Common Mistakes

Mistake 1: Assuming a quarter-wave plate produces circular polarisation for any input angle A quarter-wave plate only converts linear polarisation to circular polarisation when the input polarisation is at exactly 45 degrees to the fast and slow axes. For other angles, the output is elliptically polarised because the two components have unequal amplitudes after passing through the plate. Students often overlook this angular dependence.

Mistake 2: Confusing the rotation direction of Faraday rotation with natural optical activity Natural optical activity is reciprocal: rotating the beam direction reverses the rotation, so a round trip produces no net rotation. Faraday rotation is non-reciprocal: reversing the propagation direction doubles the rotation. This distinction is the basis of optical isolators. Students frequently assume all polarisation rotation is reciprocal.

Mistake 3: Assuming unpolarised light becomes polarised after a single polariser at arbitrary intensity After passing through one ideal polariser, unpolarised light of intensity $I_0$ becomes linearly polarised with intensity $I_0/2$ , regardless of the polariser orientation. The factor of 1/2 comes from averaging $\cos^2\theta$ over all angles. Students sometimes expect the output intensity to depend on the polariser angle, which is only true for already-polarised input.

Polarization describes the orientation of a light wave’s electric field oscillation. Unpolarised light has field vectors pointing in random directions, while linearly polarised light oscillates in a single plane. Malus’s law states that when polarised light passes through a polariser, intensity drops as the cosine squared of the angle between them because only the component along the polariser’s axis gets through. Brewster’s angle is the angle at which reflected light becomes completely polarised because the reflected and refracted rays are perpendicular, eliminating one polarisation component. Scattering at 90 degrees also produces polarisation through the same dipole radiation pattern.

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