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Chapter 5 • Theory & Derivations

Polarization of Light and Crystal Optics

States of polarization, Malus's law, Brewster's angle, double refraction in calcite, Nicol prism, quarter/half-wave plates, and Laurent half-shade polarimeter.

§5.1 States of Polarization, Mathematical Representation and Malus’s Law

Light is a transverse electromagnetic wave where the electric field vector $\mathbf{E}$ oscillates strictly perpendicular to the direction of propagation $\hat{\mathbf{k}}$. The polarization state describes the geometric locus traced by the tip of the $\mathbf{E}$-vector in a transverse plane over one temporal cycle.

1. Mathematical Superposition of Orthogonal Harmonic Field Components

Consider a monochromatic wave propagating along the $+z$ axis: $$\mathbf{E}(z, t) = E_x(z, t) \hat{\mathbf{x}} + E_y(z, t) \hat{\mathbf{y}}$$ $$E_x(z, t) = E_{0x} \cos(k z - \omega t), \quad E_y(z, t) = E_{0y} \cos(k z - \omega t + \delta)$$ where $\delta = \phi_y - \phi_x$ is the relative phase difference between the two orthogonal components. Eliminating the temporal parameter $(kz - \omega t)$ algebraically: $$\left(\frac{E_x}{E_{0x}}\right)^2 + \left(\frac{E_y}{E_{0y}}\right)^2 - 2 \left(\frac{E_x}{E_{0x}}\right)\left(\frac{E_y}{E_{0y}}\right) \cos \delta = \sin^2 \delta$$ This is the general equation of an **ellipse** inscribed in a rectangle of dimensions $2E_{0x} \times 2E_{0y}$.

2. Special Polarization States

  • Linearly Polarized Light: When $\delta = m \pi$ ($m \in \mathbb{Z}$): $\sin \delta = 0$, giving $\frac{E_y}{E_{0y}} = \pm \frac{E_x}{E_{0x}} \implies E_y = \pm \left(\frac{E_{0y}}{E_{0x}}\right) E_x$. The electric field oscillates along a fixed straight line at angle $\theta = \arctan(E_{0y}/E_{0x})$.
  • Circularly Polarized Light: When $E_{0x} = E_{0y} = E_0$ and $\delta = \pm \frac{\pi}{2}$: $$\cos \delta = 0, \quad \sin^2 \delta = 1 \implies E_x^2 + E_y^2 = E_0^2$$ The tip of $\mathbf{E}$ traces a circle of constant radius $E_0$.
    • $\delta = +\pi/2$: Right Circularly Polarized (RCP) (clockwise looking toward source).
    • $\delta = -\pi/2$: Left Circularly Polarized (LCP) (counterclockwise).
  • Elliptically Polarized Light: The general state where $E_{0x} \neq E_{0y}$ and $\delta \neq 0, \pi$.

3. Malus’s Law

When linearly polarized light of intensity $I_0$ is incident upon an ideal linear analyzer whose transmission axis forms an angle $\theta$ with the electric field vector: $$E_{\text{transmitted}} = E_0 \cos \theta \implies I(\theta) = I_0 \cos^2 \theta$$

§5.2 Polarization by Reflection, Brewster’s Law and Double Refraction in Calcite

Light can be polarized through four fundamental physical mechanisms: reflection, refraction, selective absorption (dichroism), and scattering.

1. Brewster’s Law and Polarization by Reflection

From Fresnel's reflection formulas for parallel polarization ($p$-polarization): $$r_{\parallel} = \frac{\tan(\theta_i - \theta_r)}{\tan(\theta_i + \theta_r)}$$ When the reflected and refracted rays are strictly mutually perpendicular: $$\theta_i + \theta_r = 90^\circ \implies \tan(\theta_i + \theta_r) \to \infty \implies r_{\parallel} = 0$$ At this critical angle, known as **Brewster’s Angle** $\theta_B$, all $p$-polarized light is completely transmitted into the medium. The reflected beam contains purely $s$-polarized light (electric field perpendicular to plane of incidence). Using Snell's law: $$\sin \theta_r = \sin(90^\circ - \theta_B) = \cos \theta_B$$ $$n_1 \sin \theta_B = n_2 \sin \theta_r = n_2 \cos \theta_B \implies \tan \theta_B = \frac{n_2}{n_1}$$

2. Double Refraction (Birefringence) in Uniaxial Anisotropic Crystals

In optically isotropic materials (glass, water), the dielectric permittivity is a scalar $\epsilon$. In anisotropic crystals (calcite $\text{CaCO}_3$, quartz $\text{SiO}_2$), the crystal lattice exhibits direction-dependent dielectric properties, described by a tensor. When an unpolarized beam enters a uniaxial crystal, it splits into two orthogonally polarized rays traveling with different phase velocities:
  1. Ordinary Ray ($o$-ray): Obeys Snell's law of refraction in all directions. It sees a constant refractive index $n_o$ regardless of propagation angle. Its wavefront is spherical.
  2. Extraordinary Ray ($e$-ray): Violates Snell's law. Its refractive index $n_e(\theta)$ varies continuously between $n_o$ and the principal extraordinary index $n_e$: $$\frac{1}{n_e^2(\theta)} = \frac{\cos^2 \theta}{n_o^2} + \frac{\sin^2 \theta}{n_e^2}$$ Its wavefront is an ellipsoid of revolution.
In negative uniaxial crystals like Calcite: $n_e < n_o$ ($v_e > v_o$). In positive uniaxial crystals like Quartz: $n_e > n_o$ ($v_e < v_o$).

3. The Nicol Prism Polarizer

Invented by William Nicol in 1828, the Nicol prism consists of a calcite rhomb cut diagonally at $68^\circ$ to its face and cemented together with a thin layer of Canada balsam ($n_{\text{cb}} = 1.550$). For sodium light, calcite has: $$n_o = 1.658, \quad n_e = 1.486$$ Since $n_o > n_{\text{cb}} > n_e$:
  • The $o$-ray strikes the Canada balsam interface traveling from denser ($1.658$) to rarer ($1.550$). The angle of incidence exceeds the critical angle $\theta_c = \arcsin(1.550/1.658) \approx 69^\circ$, undergoing **Total Internal Reflection (TIR)** and being absorbed in the black coating of the prism wall.
  • The $e$-ray passes from rarer ($1.486$) to denser ($1.550$) and is transmitted cleanly through the prism.
The transmitted beam emerges as **$100\%$ linearly polarized light**.

§5.3 Wave Retardation Plates: Quarter-Wave and Half-Wave Plates

Retardation plates (wave plates) are precision optical elements cut from birefringent crystals parallel to the optic axis, used to alter the polarization state of light by introducing a controlled phase delay between the $o$-ray and $e$-ray.

1. Phase Retardation Formulation

Let a birefringent plate of thickness $d$ have refractive indices $n_o$ and $n_e$. For a wave propagating normally through the plate, the optical path difference between the extraordinary and ordinary components is: $$\Delta = |n_e - n_o| d$$ The resulting optical phase retardation $\delta$ is: $$\delta = \frac{2\pi}{\lambda} |n_e - n_o| d$$

2. Half-Wave Plate (HWP)

A half-wave plate introduces a phase retardation of $\delta = \pi$ (path difference $\lambda/2$): $$|n_e - n_o| d_{\text{HWP}} = \frac{\lambda}{2} \implies d_{\text{HWP}} = \frac{\lambda}{2|n_e - n_o|}$$ Action on Linearly Polarized Light: If incident linearly polarized light makes an angle $\theta$ with the fast axis, the half-wave plate rotates the plane of polarization by exactly $2\theta$.

3. Quarter-Wave Plate (QWP)

A quarter-wave plate introduces a phase retardation of $\delta = \frac{\pi}{2}$ (path difference $\lambda/4$): $$|n_e - n_o| d_{\text{QWP}} = \frac{\lambda}{4} \implies d_{\text{QWP}} = \frac{\lambda}{4|n_e - n_o|}$$ Action:
  • Converts linearly polarized light oriented at $45^\circ$ to the axes into **circularly polarized light** ($E_{0x} = E_{0y}, \delta = \pi/2$).
  • Converts circularly polarized light into **linearly polarized light**.
  • Converts linearly polarized light at any arbitrary angle $\theta \neq 0, 45^\circ, 90^\circ$ into **elliptically polarized light**.

§5.4 Optical Activity, Biot’s Laws and Laurent’s Half-Shade Polarimeter

Certain chiral substances (such as quartz, cane sugar solutions, and tartaric acid) possess optical activity—the ability to rotate the plane of polarization of linearly polarized light.

1. Fresnel’s Theory of Optical Activity

Fresnel explained optical activity by demonstrating that a linearly polarized wave can be mathematically decomposed into two coherent circularly polarized waves of opposite handedness: Right Circularly Polarized ($E_R$) and Left Circularly Polarized ($E_L$): $$\mathbf{E} = \mathbf{E}_R + \mathbf{E}_L$$ In an optically active medium, the chiral molecular geometry causes $E_R$ and $E_L$ to propagate with slightly different phase velocities ($n_R \neq n_L$). After traversing path length $l$, the phase difference between the two circular components is: $$\delta = \frac{2\pi}{\lambda} (n_L - n_R) l$$ Recombining $E_R$ and $E_L$ produces a linearly polarized wave whose plane of polarization is rotated by angle $\theta$: $$\theta = \frac{\delta}{2} = \frac{\pi l}{\lambda} (n_L - n_R)$$

2. Biot’s Laws and Specific Rotation

For optically active solutions, the observed rotation angle $\theta$ satisfies Biot's empirical relations:
  • $\theta \propto l$ (proportional to path length in decimeters).
  • $\theta \propto c$ (proportional to solute concentration in $\text{g/cm}^3$).
  • $\theta \propto 1/\lambda^2$ (rotatory dispersion).
The **Specific Rotation** $[\alpha]_\lambda^T$ of a compound at temperature $T$ and wavelength $\lambda$ is defined as: $$[\alpha]_\lambda^T = \frac{\theta}{l \cdot c}$$ where $\theta$ is in degrees, $l$ is the tube length in decimeters ($1\text{ dm} = 10\text{ cm}$), and $c$ is the concentration in $\text{g/cm}^3$.

3. Laurent’s Half-Shade Polarimeter

To eliminate observer subjective error when determining the zero-rotation extinction angle, Laurent inserted a semicircular half-wave plate covering half the field of view:
  • Light traversing the plain glass half is transmitted unchanged.
  • Light traversing the quartz half-wave plate has its polarization rotated by $2\phi$.
The observer matches the brightness of the two halves until both are equally illuminated, achieving measurement precision better than $0.01^\circ$.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Medium Example 5.1: Sugar Concentration Determination via Polarimetry

A $20.0\text{ cm}$ long polarimeter tube containing an aqueous solution of cane sugar produces an optical rotation of $\theta = +13.2^\circ$ for the sodium D-line ($\lambda = 589.3\text{ nm}$). Given that the specific rotation of cane sugar is $[\alpha]_D^{20} = +66.0^\circ \text{ dm}^{-1} \cdot (\text{g/cm}^3)^{-1}$, calculate: (a) the concentration of the sugar solution in $\text{g/cm}^3$, and (b) the mass of sugar dissolved in $500\text{ mL}$ of solution.

Step 1: Convert path length to decimeters and apply Biot's specific rotation formula
$$l = 20.0 \text{ cm} = 2.00 \text{ dm}$$ $$[\alpha] = \frac{\theta}{l \cdot c} \implies c = \frac{\theta}{l \cdot [\alpha]}$$ $$c = \frac{13.2^\circ}{(2.00 \text{ dm}) \times (66.0^\circ \text{ dm}^{-1} \cdot \text{cm}^3/\text{g})} = \frac{13.2}{132.0} = 0.100 \text{ g/cm}^3$$

The sugar concentration is exactly $0.100\text{ g/cm}^3$ (or $100\text{ g/L}$).

Step 2: Calculate the total mass in 500 mL
$$m = c \times V = (0.100 \text{ g/cm}^3) \times 500 \text{ cm}^3 = 50.0 \text{ g}$$

There are $50.0\text{ grams}$ of sugar in the $500\text{ mL}$ solution.