Basic Concepts of Polarization in Optics

Brewster's Law Derivation

Conditions: Unpolarized light incident on a dielectric boundary at polarizing angle p

• Let light be incident at angle i_p such that the reflected ray is completely polarized.
• According to Brewster's law, at angle of incidence $i_p = i_p$, the reflected ray and refracted ray (angle r) are perpendicular: i_p + $r = 90\degree$, hence $r = 90\degree - i_p$.
• Applying Snell's Law at the interface: \frac{\sin i_p}{\sin r} = \mu_2 / \m$u_1 = \mu$.
• Substitute $r = 90\degree - i_p into Snell's law: \frac{\sin i_p}{\sin(90\degree - i_p)} = \frac{\sin i_p}{\cos i_p} = \tan i_p = \mu$.

Result: $\mu = \tan i_p$

Intensity of Transmitted Light via Malus's Law

Conditions: Plane-polarized electric field vector E_0 incident on an analyzer at angle θ

• Let the electric field vector of the incident polarized light be E_0 with amplitude \vec{E}_0.
• When incident on an analyzer with a transmission axis at angle θ to E_0, resolve E_0 into two rectangular components: E_0 \cos θ parallel to the axis and E_0 \sin θ perpendicular to the axis.
• The component that passes through the analyzer is the parallel component: $E = E_0 \cos \theta$.
• Since light intensity is proportional to the square of the electric field amplitude (I \propto E^2), the transmitted intensity $I = I_0 \cos^2 \theta$.

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

Malus's Law Intensity Variation Plot

Draw: A polar or Cartesian plot of transmitted intensity I versus angle θ showing symmetric lobes with maxima at 0\degree, 180\degree, 360\degree and zero intensity at 90\degree, 270\degree.

Interpret: Demonstrates the cosine-squared dependence of light intensity passing through crossed and parallel polarizers.

Exam use: Tested in conceptual questions regarding rotating analyzers and identifying extinction points.

Polarization by Reflection Setup

Draw: Ray diagram showing unpolarized light striking a glass slab at angle i_p, with a reflected ray perpendicular to the refracted ray, indicating complete polarization of the reflected beam.

Interpret: Visualizes Brewster's condition where the reflected beam contains only s-polarized light (electric field perpendicular to the plane of incidence).

Exam use: Used in numerical problems to calculate refractive index given the angle of incidence.

Double Refraction in Calcite Crystal

Draw: Unsplit ray entering a calcite crystal surface and splitting into ordinary (O) and extraordinary (E) rays bending at different angles.

Interpret: Illustrates optical anisotropy and the existence of two distinct refractive indices in birefringent media.

Exam use: Core concept for PGT-level questions regarding wavefront propagation in crystals.

At what angle of incidence should a ray of light be incident on the surface of water (refractive index = 1.33) so that the reflected light is completely polarized?

• Identify the given refractive index: \mu = 1.33 = \frac{4}{3}.
• Apply Brewster's law: \tan $i_p = \mu$.
• Calculate the polarizing angle: $i_p = \arctan(1.33) \approx 53.1\degree$.

Answer: 53.1\degree (or \arctan(4/3))

Unpolarized light of intensity 32 W/m^2 is passed through a polarizer. The transmitted light is then passed through an analyzer whose transmission axis makes an angle of 60\degree with the polarizer axis. Find the final transmitted intensity.

• Find the intensity after the first polarizer: $I_1 = \frac{1}{2} I_0 = \frac{1}{2} \times 32 = 16 \text{ W/m}^2$.
• Apply Malus's law for the analyzer with $\theta = 60\degree: I = I_1 \cos^2(60\degree)$.
• Substitute \cos(60\degree) = \frac{1}{2}: $I = 16 \times \left(\frac{1}{2}\right)^2 = 16 \times \frac{1}{4} = 4 \text{ W/m}^2$.

Answer: 4 W/m^2

A quarter-wave plate is designed for sodium light of wavelength \lambda = 589 \text{ nm}. If the principal refractive indices are $n_e = 1.553$ and $n_o = 1.544$, calculate the minimum thickness of the plate.

• Identify the given parameters: $\lambda = 589 \times 10^{-9} \text{ m}$, $n_e = 1.553$, $n_o = 1.544$.
• Calculate the birefringence: |n_e - $n_o| = |1.553 - 1.544| = 0.009$.
• Use the formula for the minimum thickness of a quarter-wave plate: $t = \frac{\lambda}{4|n_e - n_o|}$.
• Substitute the values: $t = \frac{589 \times 10^{-9}}{4 \times 0.009} = \frac{589 \times 10^{-9}}{0.036} \approx 1.636 \times 10^{-5} \text{ m} = 16.36 \mu\text{m}$.

Answer: 16.36 \mu\text{m}