Part III: Optics | Chapter 7

Polarization

The vector nature of light and how to manipulate it

7.1 Polarization States

Derivation 1: General Polarization from Superposition

An electromagnetic wave propagating in the $\hat{z}$ direction has$\vec{E}$ in the $xy$-plane. The most general monochromatic wave can be written:

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

The polarization state is determined by the amplitudes $E_{0x}$,$E_{0y}$ and the relative phase $\delta$:

Linear

$\delta = 0$ or $\pi$

E traces a line

Circular

$E_{0x} = E_{0y}$, $\delta = \pm\pi/2$

E traces a circle

Elliptical

General $E_{0x}$, $E_{0y}$, $\delta$

E traces an ellipse

The tip of the electric field vector traces an ellipse in the $xy$-plane. Eliminating $t$ from the components gives the polarization ellipse equation:

Polarization Ellipse

$$\frac{E_x^2}{E_{0x}^2} + \frac{E_y^2}{E_{0y}^2} - \frac{2E_x E_y}{E_{0x}E_{0y}}\cos\delta = \sin^2\delta$$

7.2 Jones Vectors and Matrices

Derivation 2: Matrix Formalism for Polarization

For fully polarized light, we represent the polarization state as a two-component complex vector (Jones vector):

$$\vec{J} = \begin{pmatrix} E_{0x} \\ E_{0y} e^{i\delta} \end{pmatrix}$$

Common polarization states:

Horizontal

$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$

Vertical

$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$

Right Circular

$\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}$

Left Circular

$\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}$

Optical elements that transform polarization are represented by 2x2 Jones matrices. The output is $\vec{J}_{\text{out}} = \mathbf{M}\vec{J}_{\text{in}}$:

Common Jones Matrices

Horizontal polarizer:$\quad \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$

Quarter-wave plate (fast axis horizontal):$\quad e^{-i\pi/4}\begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$

Half-wave plate:$\quad \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Rotation by angle $\theta$:$\quad \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

7.3 Malus's Law

Derivation 3: Intensity Through Polarizers

When linearly polarized light passes through a polarizer whose transmission axis makes an angle $\theta$ with the polarization direction, only the component along the transmission axis is transmitted. The transmitted electric field is$E_t = E_0\cos\theta$, so the intensity is:

Malus's Law

$$I = I_0 \cos^2\theta$$

At $\theta = 0$, all light passes; at $\theta = 90°$, no light passes (crossed polarizers). This is the basis of LCD displays.

For unpolarized light through a single polarizer, the average value of$\cos^2\theta$ is $1/2$, so the transmitted intensity is$I_0/2$. The transmitted light is now linearly polarized.

Three-Polarizer Paradox

Two crossed polarizers block all light. But inserting a third polarizer at 45 degrees between them allows some light to pass! This follows directly from Malus's law:

$$I = I_0 \cdot \frac{1}{2} \cdot \cos^2(45°) \cdot \cos^2(45°) = \frac{I_0}{8}$$

7.4 Stokes Parameters

Derivation 4: Describing Partially Polarized Light

Jones vectors can only describe fully polarized light. For partially polarized or unpolarized light, we need the Stokes parameters $(S_0, S_1, S_2, S_3)$:

Stokes Parameters

$$S_0 = \langle E_{0x}^2 \rangle + \langle E_{0y}^2 \rangle \quad \text{(total intensity)}$$

$$S_1 = \langle E_{0x}^2 \rangle - \langle E_{0y}^2 \rangle \quad \text{(horizontal vs vertical)}$$

$$S_2 = 2\langle E_{0x} E_{0y}\cos\delta \rangle \quad \text{(+45° vs -45°)}$$

$$S_3 = 2\langle E_{0x} E_{0y}\sin\delta \rangle \quad \text{(right vs left circular)}$$

The degree of polarization is:

$$p = \frac{\sqrt{S_1^2 + S_2^2 + S_3^2}}{S_0} \quad (0 \leq p \leq 1)$$

The Stokes parameters can be measured with a set of intensity measurements through different polarizer/wave-plate combinations. They are the observable quantities of polarization, and their manipulation uses Mueller matrices (4x4 real matrices).

7.5 Birefringence and Wave Plates

Derivation 5: Phase Retardation

In an anisotropic crystal (e.g., calcite, quartz), the refractive index depends on the polarization direction. The two principal refractive indices $n_o$ (ordinary) and $n_e$ (extraordinary) cause the two polarization components to travel at different speeds, accumulating a relative phase:

Phase Retardation

$$\Gamma = \frac{2\pi}{\lambda}(n_e - n_o)d$$

where $d$ is the crystal thickness. A quarter-wave platehas $\Gamma = \pi/2$ (converts linear to circular polarization). A half-wave plate has $\Gamma = \pi$ (rotates the polarization plane).

Quarter-Wave Plate ($\lambda/4$)

Converts linear polarization at 45 degrees to the fast axis into circular polarization. Thickness: $d = \lambda/[4(n_e - n_o)]$

Half-Wave Plate ($\lambda/2$)

Rotates linear polarization by twice the angle between the polarization and the fast axis. Used in optical isolators and polarization rotators.

7.5b Optical Activity and the Faraday Effect

Certain materials rotate the plane of polarization of linearly polarized light. This optical activity occurs in chiral molecules (like sugar solutions) and in crystals lacking a center of symmetry (like quartz):

$$\theta_{\text{rot}} = [\alpha] \cdot c \cdot l$$

where $[\alpha]$ is the specific rotation, $c$ is the concentration, and $l$ is the path length. This rotation isreciprocal: it undoes itself on the return path.

The Faraday Effect

An applied magnetic field along the propagation direction causes polarization rotation in any transparent medium:

Faraday Rotation

$$\theta_F = VBl$$

where $V$ is the Verdet constant, $B$ is the magnetic field, and $l$ is the path length. Unlike optical activity, the Faraday effect is non-reciprocal: the rotation doubles on the return path. This is what makes optical isolators possible — light can only pass in one direction.

Saccharimetry: The concentration of sugar solutions has been measured using optical rotation since the 19th century. A polarimeter passes linearly polarized light through the solution and measures the rotation angle. This is still used in the food industry for quality control.

7.5c Worked Example: Designing a Circular Polarizer

Problem: Design a device that converts unpolarized light into right-circularly polarized light using a linear polarizer and a quarter-wave plate.

Solution using Jones Calculus

Step 1: Pass unpolarized light through a linear polarizer oriented at 45 degrees. Output (normalized): $\vec{J}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

Step 2: Pass through a QWP with fast axis horizontal. The Jones matrix is$\begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/2} \end{pmatrix}$:

$$\vec{J}_{\text{out}} = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}$$

This is left-circular polarization. For right-circular, rotate the polarizer to -45 degrees, or rotate the QWP by 90 degrees. The intensity is $I_0/2$ (half the input is lost at the linear polarizer).

7.5d Mueller Matrices for Partially Polarized Light

The Jones calculus handles only fully polarized light. For partially polarized light (including unpolarized), we use the Mueller matrix formalism, which operates on Stokes vectors $\vec{S} = (S_0, S_1, S_2, S_3)^T$:

$$\vec{S}_{\text{out}} = \mathbf{M}\vec{S}_{\text{in}}$$

Key Mueller matrices (4x4 real matrices):

Horizontal Polarizer

$\frac{1}{2}\begin{pmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

Depolarizer

$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

Scrambles polarization; p goes to 0

Mueller matrices can describe depolarizing optical elements (like scattering surfaces) that have no Jones matrix representation. They are essential in remote sensing, ellipsometry, and biomedical optics where depolarization is significant.

7.6 Applications

LCD Displays

Liquid crystal displays use crossed polarizers with a twisted nematic liquid crystal between them. Applying a voltage changes the crystal alignment, controlling how much light passes through each pixel. This is the basis of most flat screens.

3D Cinema

3D movies project left and right eye images with different circular polarizations. Polarized glasses filter the correct image to each eye. Circular polarization is preferred because it works regardless of head tilt.

Optical Isolators

Faraday rotators combined with polarizers create optical isolators that allow light to pass in one direction only. Essential for protecting lasers from back-reflections, using the non-reciprocal Faraday effect.

Stress Analysis

Photoelasticity uses birefringence induced by mechanical stress in transparent materials. Viewing stressed parts between crossed polarizers reveals colorful fringe patterns that map the stress distribution.

7.7 Historical Context

Christiaan Huygens (1690): Discovered double refraction (birefringence) in calcite crystals, observing that each incident ray produces two refracted rays. He could explain this with his wave theory but could not explain polarization, which required the later recognition that light is a transverse wave.

Etienne-Louis Malus (1809): Discovered polarization by reflection while looking through a calcite crystal at sunlight reflected from a window of the Luxembourg Palace in Paris. He formulated the law of intensity through a polarizer that bears his name.

Augustin-Jean Fresnel (1821): Showed that light is a transverse wave, resolving the puzzle of polarization. This was initially controversial because longitudinal vibrations were expected for waves in a fluid medium (the "luminiferous ether").

R. Clark Jones (1941): Developed the Jones calculus at Harvard, providing the elegant matrix formalism for polarization optics that is now standard. The Stokes parameters were introduced earlier by George Gabriel Stokes in 1852.

7.7b Polarization in Nature and Technology

Polarization phenomena are ubiquitous in the natural world and underpin many technologies:

Sky Polarization

Rayleigh scattering from atmospheric molecules partially polarizes sunlight. The degree of polarization is maximum at 90 degrees from the Sun. Many insects (bees, ants) use this polarization pattern for navigation, detecting it with specialized photoreceptors. Viking navigators may have used polarizing calcite crystals ("sunstones") to find the Sun's position on overcast days.

Cosmic Microwave Background Polarization

The CMB is weakly polarized by Thomson scattering at the epoch of recombination. E-mode polarization (curl-free) has been detected and confirms cosmological models. B-mode polarization (divergence-free) from primordial gravitational waves is the target of current experiments (BICEP, LiteBIRD) as it would confirm cosmic inflation.

Quantum Key Distribution

The BB84 protocol for quantum cryptography encodes information in the polarization states of individual photons (horizontal/vertical or $\pm 45°$). The no-cloning theorem guarantees that any eavesdropping attempt disturbs the quantum states, alerting the legitimate users.

7.8 Python Simulation

This simulation visualizes polarization ellipses, Malus's law, the effect of wave plates, and the Poincare sphere representation.

Polarization: States, Malus's Law, Wave Plates, and Poincare Sphere

Python

script.py172 lines

import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt

# ============================================================
# Polarization: States, Malus's Law, Wave Plates
# ============================================================

fig, axes = plt.subplots(2, 2, figsize=(12, 10))
fig.suptitle('Polarization of Light', fontsize=16, fontweight='bold', color='white')
fig.patch.set_facecolor('#0a0a0a')

for ax in axes.flat:
    ax.set_facecolor('#0a0a0a')
    ax.tick_params(colors='white')
    ax.xaxis.label.set_color('white')
    ax.yaxis.label.set_color('white')
    ax.title.set_color('white')
    for spine in ax.spines.values():
        spine.set_color('#333')

# --- Panel 1: Various polarization ellipses ---
ax1 = axes[0, 0]
t = np.linspace(0, 2 * np.pi, 500)

states = [
    ('Linear (H)', 1.0, 0.0, 0, '#22d3ee'),
    ('Linear (45)', 1.0, 1.0, 0, '#f97316'),
    ('Circular (R)', 1.0, 1.0, -np.pi/2, '#a855f7'),
    ('Elliptical', 1.0, 0.6, np.pi/4, '#4ade80'),
]

for name, Ex, Ey, delta, color in states:
    x = Ex * np.cos(t)
    y = Ey * np.cos(t + delta)
    ax1.plot(x, y, color=color, linewidth=2, label=name)
    # Arrow showing direction
    idx = len(t) // 8
    ax1.annotate('', xy=(x[idx+1], y[idx+1]), xytext=(x[idx], y[idx]),
                arrowprops=dict(arrowstyle='->', color=color, lw=1.5))

ax1.set_xlabel('E_x')
ax1.set_ylabel('E_y')
ax1.set_title('Polarization States')
ax1.set_aspect('equal')
ax1.legend(fontsize=7, facecolor='#0a0a0a', edgecolor='#333', labelcolor='white')
ax1.set_xlim(-1.3, 1.3)
ax1.set_ylim(-1.3, 1.3)
ax1.axhline(y=0, color='white', linewidth=0.3, alpha=0.3)
ax1.axvline(x=0, color='white', linewidth=0.3, alpha=0.3)

# --- Panel 2: Malus's law ---
ax2 = axes[0, 1]
theta = np.linspace(0, 360, 500)
theta_rad = np.radians(theta)
I_malus = np.cos(theta_rad)**2

ax2.plot(theta, I_malus, color='#22d3ee', linewidth=2.5, label="Malus's law: I = I_0 cos^2(theta)")
ax2.fill_between(theta, 0, I_malus, color='#22d3ee', alpha=0.15)

# Three-polarizer case
theta_3p = np.linspace(0, 90, 100)
theta_3p_rad = np.radians(theta_3p)
I_3pol = 0.5 * np.cos(theta_3p_rad)**2 * np.cos(np.pi/2 - theta_3p_rad)**2
ax2.plot(theta_3p, I_3pol * 4, color='#f97316', linewidth=2, linestyle='--',
         label='Three-polarizer (0, theta, 90)')

ax2.set_xlabel('Angle theta (degrees)')
ax2.set_ylabel('I / I_0')
ax2.set_title("Malus's Law")
ax2.legend(fontsize=7, facecolor='#0a0a0a', edgecolor='#333', labelcolor='white')
ax2.set_xlim(0, 360)
ax2.set_ylim(0, 1.1)

# --- Panel 3: Effect of wave plates on linear input ---
ax3 = axes[1, 0]

# Input: linear at 45 degrees
Ex_in = 1.0 / np.sqrt(2)
Ey_in = 1.0 / np.sqrt(2)

retardations = [
    ('No plate (linear)', 0, '#22d3ee'),
    ('lambda/8 plate', np.pi/4, '#f97316'),
    ('lambda/4 (circular)', np.pi/2, '#a855f7'),
    ('3lambda/8 plate', 3*np.pi/4, '#4ade80'),
    ('lambda/2 (linear rotated)', np.pi, '#ef4444'),
]

for name, Gamma, color in retardations:
    x = Ex_in * np.cos(t)
    y = Ey_in * np.cos(t + Gamma)
    ax3.plot(x, y, color=color, linewidth=2, label=name, alpha=0.8)

ax3.set_xlabel('E_x')
ax3.set_ylabel('E_y')
ax3.set_title('Wave Plate Effects (45-deg Linear Input)')
ax3.set_aspect('equal')
ax3.legend(fontsize=6, facecolor='#0a0a0a', edgecolor='#333', labelcolor='white', loc='upper left')
ax3.set_xlim(-0.9, 0.9)
ax3.set_ylim(-0.9, 0.9)
ax3.axhline(y=0, color='white', linewidth=0.3, alpha=0.3)
ax3.axvline(x=0, color='white', linewidth=0.3, alpha=0.3)

# --- Panel 4: Stokes parameters on Poincare sphere projection ---
ax4 = axes[1, 1]

# Draw the Poincare sphere equator and meridians
phi_arr = np.linspace(0, 2 * np.pi, 200)
ax4.plot(np.cos(phi_arr), np.sin(phi_arr), color='white', linewidth=0.5, alpha=0.3)
ax4.plot([-1, 1], [0, 0], color='white', linewidth=0.3, alpha=0.3)
ax4.plot([0, 0], [-1, 1], color='white', linewidth=0.3, alpha=0.3)

# Mark common polarization states on S1-S2 plane
pol_states = {
    'H': (1, 0), 'V': (-1, 0),
    '+45': (0, 1), '-45': (0, -1),
}

for name, (s1, s2) in pol_states.items():
    ax4.plot(s1, s2, 'o', color='#22d3ee', markersize=10)
    ax4.annotate(name, (s1, s2), textcoords='offset points', xytext=(8, 8),
                color='#22d3ee', fontsize=10, fontweight='bold')

# Trace various polarization angles on equator
theta_poincare = np.linspace(0, np.pi, 50)
for th in theta_poincare:
    s1 = np.cos(2 * th)
    s2 = np.sin(2 * th)
    ax4.plot(s1, s2, '.', color='#f97316', markersize=3, alpha=0.5)

# Add RCP and LCP above/below
ax4.annotate('RCP (S3=+1)', (0, 0), textcoords='offset points', xytext=(-35, 40),
            color='#a855f7', fontsize=9, fontweight='bold')
ax4.annotate('LCP (S3=-1)', (0, 0), textcoords='offset points', xytext=(-35, -50),
            color='#4ade80', fontsize=9, fontweight='bold')

ax4.set_xlabel('S1 / S0')
ax4.set_ylabel('S2 / S0')
ax4.set_title('Poincare Sphere (S1-S2 Equatorial Plane)')
ax4.set_aspect('equal')
ax4.set_xlim(-1.4, 1.4)
ax4.set_ylim(-1.4, 1.4)

plt.tight_layout()
plt.savefig('output.png', dpi=150, bbox_inches='tight', facecolor='#0a0a0a')
plt.close()
print("Plot saved to output.png")

print("\n" + "=" * 60)
print("POLARIZATION: RESULTS")
print("=" * 60)
print("\n--- Polarization States ---")
print("  Linear: delta = 0 or pi")
print("  Circular: E_0x = E_0y, delta = +/- pi/2")
print("  Elliptical: general case")
print("\n--- Malus's Law ---")
print("  I(theta) = I_0 cos^2(theta)")
print("  Crossed polarizers: I = 0")
print("  Three polarizers (0, 45, 90): I = I_0 / 8")
print("\n--- Jones Matrices ---")
print("  Horizontal polarizer: diag(1, 0)")
print("  QWP (fast H): diag(1, i) * exp(-i pi/4)")
print("  HWP: diag(1, -1)")
print("\n--- Birefringence ---")
print("  Calcite: n_o = 1.658, n_e = 1.486, Delta n = 0.172")
print("  Quartz: n_o = 1.544, n_e = 1.553, Delta n = 0.009")
print("\n--- Stokes Parameters ---")
print("  Degree of polarization: p = sqrt(S1^2 + S2^2 + S3^2) / S0")
print("=" * 60)

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Summary

Core Equations

Malus's law: $I = I_0\cos^2\theta$
Jones: $\vec{J}_{\text{out}} = \mathbf{M}\vec{J}_{\text{in}}$
Retardation: $\Gamma = 2\pi(n_e - n_o)d/\lambda$
Degree of polarization: $p = \sqrt{S_1^2+S_2^2+S_3^2}/S_0$

Key Insights

Light is a transverse wave with two polarizations
Jones calculus: 2x2 matrices for fully polarized light
Stokes/Mueller: handles partial polarization
Wave plates convert between polarization states
Brewster angle produces pure polarization

Practice Problems

Problem 1: Malus's Law Intensity CalculationUnpolarized light of intensity $I_0 = 100$ W/m$^2$ passes through three ideal linear polarizers. The first is vertical, the second is at 30° to the vertical, and the third is at 75° to the vertical. Find the transmitted intensity.

Solution:

1. After the first polarizer, unpolarized light becomes linearly polarized with half the intensity:

$$I_1 = \frac{I_0}{2} = \frac{100}{2} = 50\;\text{W/m}^2$$

2. Apply Malus's law at the second polarizer. The angle between polarizers 1 and 2 is 30°:

$$I_2 = I_1\cos^2(30°) = 50 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 50 \times 0.75 = 37.5\;\text{W/m}^2$$

3. Apply Malus's law at the third polarizer. The angle between polarizers 2 and 3 is $75° - 30° = 45°$:

$$I_3 = I_2\cos^2(45°) = 37.5 \times \left(\frac{1}{\sqrt{2}}\right)^2 = 37.5 \times 0.5$$

4. The final transmitted intensity is:

$$\boxed{I_3 = 18.75\;\text{W/m}^2}$$

5. Note that without the middle polarizer, the angle between polarizers 1 and 3 would be 75°, giving $I = 50\cos^2(75°) = 3.35$ W/m$^2$. The intermediate polarizer actually increases transmission by rotating the polarization in steps — a result that can seem counterintuitive.

Problem 2: Jones Matrix for a Quarter-Wave PlateLinearly polarized light at 45° to the fast axis enters a quarter-wave plate. Using the Jones calculus, find the output polarization state and verify it is circularly polarized.

Solution:

1. The input Jones vector for light polarized at 45° is:

$$\vec{J}_{\text{in}} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

2. The Jones matrix for a quarter-wave plate (QWP) with fast axis horizontal introduces a $\pi/2$ phase retardation on the slow (vertical) component:

$$M_{\text{QWP}} = e^{-i\pi/4}\begin{pmatrix} 1 & 0 \\ 0 & e^{-i\pi/2} \end{pmatrix} = e^{-i\pi/4}\begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix}$$

3. The output Jones vector is:

$$\vec{J}_{\text{out}} = M_{\text{QWP}}\,\vec{J}_{\text{in}} = \frac{e^{-i\pi/4}}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix}$$

4. Verify circular polarization: the x and y components have equal magnitudes ($|1| = |-i| = 1$) and a phase difference of $-\pi/2$:

$$\boxed{\vec{J}_{\text{out}} \propto \begin{pmatrix} 1 \\ -i \end{pmatrix} = \text{Right Circular Polarization (RCP)}}$$

5. The electric field traces a circle: $E_x = \cos\omega t$, $E_y = \sin\omega t$ (rotating clockwise when viewed facing the source). If the input were at $-45°$, we would get LCP. The global phase $e^{-i\pi/4}$ is physically unobservable.

Problem 3: Brewster Angle for GlassLight in air ($n_1 = 1.00$) strikes a glass surface ($n_2 = 1.52$). Calculate Brewster's angle, the corresponding refraction angle, and the reflectance of s-polarized light at this angle.

Solution:

1. Brewster's angle is defined by the condition that reflected and refracted rays are perpendicular:

$$\tan\theta_B = \frac{n_2}{n_1} = \frac{1.52}{1.00} = 1.52$$

2. Therefore:

$$\boxed{\theta_B = \arctan(1.52) = 56.7°}$$

3. The refraction angle at Brewster's incidence satisfies $\theta_r = 90° - \theta_B$:

$$\theta_r = 90° - 56.7° = 33.3°$$

4. Verify with Snell's law: $n_1\sin\theta_B = n_2\sin\theta_r$: $1.00 \times \sin(56.7°) = 1.52 \times \sin(33.3°) = 0.836$. Confirmed.

5. The s-polarized reflectance at Brewster's angle using the Fresnel equation:

$$r_s = \frac{n_1\cos\theta_B - n_2\cos\theta_r}{n_1\cos\theta_B + n_2\cos\theta_r} = \frac{1.00(0.549) - 1.52(0.836)}{1.00(0.549) + 1.52(0.836)} = \frac{0.549 - 1.271}{0.549 + 1.271} = \frac{-0.722}{1.820}$$

$$\boxed{R_s = r_s^2 = (-0.397)^2 = 0.157 \approx 15.7\%}$$

At Brewster's angle, $R_p = 0$ exactly (p-polarized light is fully transmitted), while s-polarized light has significant reflection. This is the basis for polarization by reflection, used in polarizing beam splitters.

Problem 4: Degree of Polarization from Stokes ParametersA light beam has Stokes parameters $S_0 = 10$, $S_1 = 6$, $S_2 = -4$, $S_3 = 2$. Determine the degree of polarization, the orientation angle of the polarization ellipse, and the ellipticity angle.

Solution:

1. The degree of polarization is:

$$p = \frac{\sqrt{S_1^2 + S_2^2 + S_3^2}}{S_0} = \frac{\sqrt{36 + 16 + 4}}{10} = \frac{\sqrt{56}}{10} = \frac{7.483}{10}$$

$$\boxed{p = 0.748 \approx 74.8\%}$$

2. The light is partially polarized. To find the polarization ellipse parameters, use the polarized part of the Stokes vector: $(S_1, S_2, S_3) = (6, -4, 2)$.

3. The orientation angle $\psi$ of the major axis of the polarization ellipse:

$$\tan(2\psi) = \frac{S_2}{S_1} = \frac{-4}{6} = -0.667 \implies 2\psi = -33.7° \implies \boxed{\psi = -16.8°}$$

4. The ellipticity angle $\chi$:

$$\sin(2\chi) = \frac{S_3}{\sqrt{S_1^2+S_2^2+S_3^2}} = \frac{2}{7.483} = 0.267 \implies \boxed{\chi = 7.75°}$$

5. Since $S_3 > 0$, the polarized component is right-handed elliptically polarized with the major axis tilted 16.8° clockwise from horizontal. The unpolarized fraction is $1 - p = 25.2\%$ of the total intensity.

Problem 5: Mueller Matrix for Cascaded Optical ElementsUnpolarized light passes through a horizontal linear polarizer followed by a half-wave plate with fast axis at 22.5° to horizontal. Find the output Stokes vector using Mueller matrices.

Solution:

1. The input Stokes vector for unpolarized light of intensity $I_0$ is:

$$\vec{S}_{\text{in}} = I_0\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

2. After the horizontal linear polarizer (Mueller matrix $M_P$):

$$\vec{S}_1 = M_P \vec{S}_{\text{in}} = \frac{I_0}{2}\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$$

3. A half-wave plate at angle $\theta = 22.5°$ rotates the polarization direction by $2\theta = 45°$. Its Mueller matrix transforms $S_1 \to S_1\cos(4\theta) + S_2\sin(4\theta)$ and $S_2 \to S_1\sin(4\theta) - S_2\cos(4\theta)$:

$$4\theta = 90°, \quad \cos 90° = 0, \quad \sin 90° = 1$$

4. Applying the HWP Mueller matrix:

$$\vec{S}_{\text{out}} = \frac{I_0}{2}\begin{pmatrix} 1 \\ S_1\cos 90° + S_2\sin 90° \\ S_1\sin 90° - S_2\cos 90° \\ -S_3 \end{pmatrix} = \frac{I_0}{2}\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$

5. The output is:

$$\boxed{\vec{S}_{\text{out}} = \frac{I_0}{2}\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} \implies \text{linearly polarized at } 45°, \quad I = I_0/2}$$

The horizontal polarizer creates horizontal linear polarization, and the HWP rotates it by $2 \times 22.5° = 45°$. The degree of polarization is $p = 1$ (fully polarized), with half the original intensity lost at the polarizer.

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Polarization

7.1 Polarization States

Derivation 1: General Polarization from Superposition

Linear

Circular

Elliptical

Polarization Ellipse

7.2 Jones Vectors and Matrices

Derivation 2: Matrix Formalism for Polarization

Horizontal

Vertical

Right Circular

Left Circular

Common Jones Matrices

7.3 Malus's Law

Derivation 3: Intensity Through Polarizers

Malus's Law

Three-Polarizer Paradox

7.4 Stokes Parameters

Derivation 4: Describing Partially Polarized Light

Stokes Parameters

7.5 Birefringence and Wave Plates

Derivation 5: Phase Retardation

Phase Retardation

Quarter-Wave Plate (\(\lambda/4\))

Half-Wave Plate (\(\lambda/2\))

7.5b Optical Activity and the Faraday Effect

The Faraday Effect

Faraday Rotation

7.5c Worked Example: Designing a Circular Polarizer

Solution using Jones Calculus

7.5d Mueller Matrices for Partially Polarized Light

Horizontal Polarizer

Depolarizer

7.6 Applications

LCD Displays

3D Cinema

Optical Isolators

Stress Analysis

7.7 Historical Context

7.7b Polarization in Nature and Technology

Sky Polarization

Cosmic Microwave Background Polarization

Quantum Key Distribution

7.8 Python Simulation

Polarization: States, Malus's Law, Wave Plates, and Poincare Sphere

Summary

Core Equations

Key Insights

Practice Problems

Solution:

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