§ 21. Light Wavefront Polarization & Malus's Law (Grade 11 · Uni)
Transmission of a linearly polarized light wavefront through a polarizer and rotatable analyzer. Because light is a transverse wave, a crystalline analyzer transmits only the projection of E parallel to its optical axis.
Theory Summary & Scientific Mechanism
Because light is a transverse electromagnetic wave (E ⊥ k), the electric field vector E oscillates in the plane perpendicular to propagation. Unpolarized natural light consists of rapidly varying random polarization orientations, whereas a linear polarizer transmits only the component of E parallel to its transmission axis.
When linearly polarized light of amplitude E₀ and intensity I₀ ∝ E₀² encounters a second polarizer (analyzer) rotated by angle θ, only the projected amplitude E = E₀cos θ passes through. Since optical intensity scales as the square of the electric field amplitude, the transmitted intensity obeys Malus’s Law: I(θ) = I₀cos ²θ, vanishing completely when the axes are crossed (θ = 90^°).
Polarization also occurs by reflection off a dielectric of refractive index n. At Brewster’s angle tan θ_B = n (where reflected and refracted rays are orthogonal, θ_B + γ = 90^°), the p-polarized component cannot radiate along the reflection direction, rendering the reflected beam 100% linearly s-polarized, which is the principle behind anti-glare polarized sunglasses.
Everyday Analogy
Why do polarized sunglasses magically erase blinding glare off water and wet roads? Because light reflecting at Brewster’s angle becomes 100% horizontally polarized, allowing a vertical polarizing filter to block the glare completely!
Interactive Question & Derivation
A linearly polarized light beam of intensity I₀ strikes a second polarizer (analyzer) whose transmission axis is rotated by θ = 60^° relative to the beam's polarization plane. What intensity I emerges from the analyzer?
- A: The intensity is reduced by half: I = 0.5 I₀
- B: The electric field amplitude drops as E = E₀cos 60^° = 0.5 E₀, so by Malus's law the intensity drops to one-fourth: I = I₀cos ² 60^° = 0.25 I₀
- C: The light is completely extinguished already at 60^°
Study Hint: At Brewster’s angle (tan θ_B = n₂/n₁), what angle do the reflected and refracted rays make with each other (θ_B + θ₂)?
Proof & Explanation: The analyzer passes only the electric field component E = E₀ cos θ along its axis. Since optical intensity is proportional to the square of field amplitude (I ∝ E²), we obtain Malus's Law: I(θ) = I₀ cos ²θ. At θ = 60^°, cos 60^° = 0.5 so I = 0.25 I₀, and at θ = 90^° (crossed polarizers) the wavefront is completely blocked.
Key Equations & 3D Simulation Parameters
I(θ) = I₀ cos ²θ, E(θ) = E₀ |cos θ|, θ_B = arctan (n)
- Analyzer axis angle θ: 0 … 90 ° (default: 35 °)
- Polarized intensity I₀: 40 … 160 W/m² (default: 100 W/m²)
- Light wavelength λ: 420 … 680 nm (default: 530 nm)
- Glass refractive index n (for θ_B): 1.3 … 1.9 (default: 1.52)