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 Malusâs Law
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. Malusâs 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, Brewsterâs Law and Double Refraction in Calcite
1. Brewsterâs 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 **Brewsterâs 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:- 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.
- 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.
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.
§5.3 Wave Retardation Plates: Quarter-Wave and Half-Wave Plates
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, Biotâs Laws and Laurentâs Half-Shade Polarimeter
1. Fresnelâs 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. Biotâs 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).
3. Laurentâs 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 $2\phi$.
đ Chapter Worked Examples & Exercises
Complete derivations & analytical proofsA $20.0\text{ cm}$ 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.
The sugar concentration is exactly $0.100\text{ g/cm}^3$ (or $100\text{ g/L}$).
There are $50.0\text{ grams}$ of sugar in the $500\text{ mL}$ solution.