5cPolarization and Intensity
Apply Malus's Law to calculate transmitted intensity through polarizers, use the inverse square law for point sources, and explain applications of polarization in technology.
Polarization is the basis of LCD screens, polarized sunglasses, 3D cinema, stress analysis in engineering, and optical communications. Malus's Law and the inverse square law are fundamental to understanding how light intensity is controlled and measured.
Why do polarized sunglasses reduce glare from water but not from a white wall — and how do LCD screens use polarization to display images?
Lesson Overview
Polarization is a property of transverse waves (including light) describing the orientation of the oscillation. Ordinary (unpolarized) light has electric field oscillations in all directions perpendicular to propagation. A polarizer transmits only the component of light oscillating along its transmission axis, reducing intensity by half: I = I₀/2. When polarized light passes through a second polarizer (the analyzer) at angle θ to the first, Malus's Law gives the transmitted intensity: I = I₀ cos² θ. At θ = 0° (parallel), all light passes; at θ = 90° (crossed), no light passes. Light can also be polarized by reflection (Brewster's angle), scattering, and birefringence. The inverse square law for intensity states that I = P/(4πr²) for a point source — intensity decreases with the square of distance.
Key Equations
Malus's Law — Transmitted Intensity vs Angle
| Angle θ | cos θ | cos² θ | I / I₀ |
|---|---|---|---|
| 0° | 1.000 | 1.000 | 100% |
| 30° | 0.866 | 0.750 | 75% |
| 45° | 0.707 | 0.500 | 50% |
| 60° | 0.500 | 0.250 | 25% |
| 90° | 0.000 | 0.000 | 0% |
Worked Examples
Unpolarized light of intensity I₀ = 800 W/m² passes through a polarizer. What is the transmitted intensity?
Polarized light of intensity I₀ = 400 W/m² passes through an analyzer at θ = 30° to the polarization axis. Find the transmitted intensity.
Unpolarized light of intensity 1000 W/m² passes through two polarizers. The first is vertical; the second is at 60° to the first. Find the final intensity.
A point source emits 50 W of light. Find the intensity at (a) r = 1.0 m and (b) r = 5.0 m.
Find Brewster's angle for light traveling from air (n₁ = 1.00) into glass (n₂ = 1.52).
Guided Problems
Polarized light of intensity 600 W/m² passes through an analyzer at 45° to the polarization axis. Find the transmitted intensity.
Hint: Use Malus's Law: I = I₀ cos² θ. cos 45° = 0.707, so cos² 45° = 0.5.
Unpolarized light passes through two crossed polarizers (θ = 90°). What is the transmitted intensity?
Hint: After the first polarizer: I₁ = I₀/2. After the second (at 90°): I₂ = I₁ cos² 90° = I₁ × 0 = 0.
A light source emits 100 W. At what distance is the intensity equal to 1.0 W/m²?
Hint: Use I = P/(4πr²). Rearrange: r = √(P/(4πI)) = √(100/(4π×1.0)).
The intensity of light at 2.0 m from a source is 50 W/m². What is the intensity at 6.0 m?
Hint: Use the inverse square law: I₂ = I₁ × (r₁/r₂)² = 50 × (2/6)² = 50 × (1/9).
Polarized light passes through an analyzer. The transmitted intensity is 25% of the incident intensity. Find the angle between the polarizer and analyzer.
Hint: I/I₀ = cos² θ = 0.25. So cos θ = 0.5. Find θ = cos⁻¹(0.5).
Key Vocabulary
Polarization
A property of transverse waves describing the direction of oscillation. In polarized light, the electric field oscillates in a single plane. Unpolarized light has oscillations in all planes perpendicular to propagation.
Example: Sunlight is unpolarized; light reflected from a horizontal surface is partially polarized horizontally.
Polarizer
An optical filter that transmits light oscillating in only one direction (its transmission axis). When unpolarized light passes through a polarizer, the transmitted intensity is I₀/2.
Example: Polaroid film is a common polarizer used in sunglasses and camera filters.
Malus's Law
The intensity of polarized light transmitted through an analyzer at angle θ to the polarization axis: I = I₀ cos² θ. Maximum transmission at θ = 0°; zero transmission at θ = 90°.
Example: Rotating a polarized lens 60° from alignment reduces intensity to 25% of the incident value.
Analyzer
A second polarizer used to analyze the polarization state of light. The transmitted intensity depends on the angle between the analyzer's transmission axis and the polarization direction of the incident light.
Example: In an LCD screen, the analyzer (second polarizer) controls which pixels appear bright or dark.
Brewster's Angle
The angle of incidence at which reflected light is completely polarized parallel to the surface: tan θ_B = n₂/n₁. At this angle, the reflected and refracted rays are perpendicular.
Example: Glare from water or roads is partially polarized at Brewster's angle — polarized sunglasses block this horizontal polarization.
Inverse Square Law
The intensity of radiation from a point source decreases with the square of the distance: I = P/(4πr²). Doubling the distance reduces intensity to one-quarter.
Example: A lamp that produces 1 W/m² at 1 m produces only 0.25 W/m² at 2 m.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
Unpolarized light of intensity 200 W/m² passes through a polarizer. The transmitted intensity is:
Polarized light of intensity 80 W/m² passes through an analyzer at 60°. The transmitted intensity is:
Two polarizers are crossed (θ = 90°). What fraction of the original unpolarized light is transmitted?
A point source emits 40 W. The intensity at 2 m is:
Polarized light passes through an analyzer. The transmitted intensity is 50% of the incident intensity. The angle between them is:
Independent Practice
Unpolarized light of intensity 500 W/m² passes through a polarizer. Find the transmitted intensity.
Polarized light of intensity 300 W/m² passes through an analyzer at 30°. Find the transmitted intensity.
Unpolarized light passes through two polarizers. The second is at 45° to the first. Find the final intensity as a fraction of the original.
A light source emits 200 W. Find the intensity at (a) 2.0 m and (b) 10.0 m from the source.
★ Unpolarized light of intensity I₀ passes through three polarizers. The first is vertical, the second is at 45°, and the third is horizontal (90° from the first). (a) Find the intensity after each polarizer. (b) Compare this to the case where the middle polarizer is removed. Explain why adding a polarizer can increase the transmitted intensity.
ChallengeCommon Mistakes
Applying Malus's Law to unpolarized light hitting the first polarizer
Malus's Law (I = I₀ cos² θ) applies only to already-polarized light. When unpolarized light hits the first polarizer, use I = I₀/2. Then apply Malus's Law for any subsequent polarizers.
Thinking crossed polarizers (90°) transmit half the light
Crossed polarizers transmit zero light: I = I₀ cos² 90° = 0. This is because no component of the polarized light aligns with the second polarizer's axis.
Confusing intensity with amplitude
Intensity is proportional to amplitude squared: I ∝ A². Malus's Law gives I = I₀ cos² θ, which means the amplitude is A = A₀ cos θ. Intensity and amplitude are not the same quantity.
Forgetting that the inverse square law applies to point sources only
I = P/(4πr²) assumes the source radiates uniformly in all directions (a point source). For a laser beam or directed source, the intensity does not follow the inverse square law.
Math Tips
Malus's Law: I = I₀ cos² θ. Key values: θ=0° → I=I₀ (100%); θ=45° → I=I₀/2 (50%); θ=60° → I=I₀/4 (25%); θ=90° → I=0 (0%).
Two-polarizer sequence: (1) Unpolarized → first polarizer: I₁ = I₀/2. (2) Polarized → second polarizer: I₂ = I₁ cos² θ. Combined: I₂ = (I₀/2) cos² θ.
Inverse square law: I = P/(4πr²). Ratio form: I₂/I₁ = (r₁/r₂)². If r doubles, I drops by factor 4. If r triples, I drops by factor 9.
Brewster's angle: tan θ_B = n₂/n₁. At this angle, reflected light is 100% polarized parallel to the surface. Polarized sunglasses block this horizontal polarization to reduce glare.