Polarization

AP Physics 2· 12 min read

1. Polarized vs Unpolarized Light★★☆☆☆⏱ 3 min

Visible light is a transverse electromagnetic wave, where electric and magnetic fields oscillate perpendicular to the direction the wave travels. Polarization describes the orientation of the electric field oscillation for a given light beam.

📘 Definition

Linearly Polarized Light

Light where the electric field oscillates along exactly one fixed axis for every wavefront in the beam.

Example:

Light that has passed through a single ideal polarizing filter

✓ Quick check

Test your baseline understanding before proceeding:

Unpolarized light has electric fields oscillating in how many distinct axes?
- 1
- 2
- All possible perpendicular axes
- Zero
Reveal answer
All possible perpendicular axes —
Unpolarized light has random, evenly distributed E-field orientations across every axis perpendicular to travel.

2. Malus's Law for Ideal Polarizers★★★☆☆⏱ 4 min

An ideal linear polarizer only transmits the component of the incident electric field that is aligned to its transmission axis. The transmitted intensity is proportional to the square of the electric field amplitude, leading directly to Malus's Law for incident polarized light.

I = I_0 \cos^2(\theta)

📐 Worked Example

Unpolarized light of intensity 800 W/m² passes first through a vertical polarizer, then through a second polarizer oriented 60° from vertical. Calculate the final transmitted intensity.

1
Step 1: Process the first polarizer. Since incident light is unpolarized, transmitted intensity is halved:
2
$I_1 = \frac{800}{2} = 400 \text{ W/m}^2$
3
Step 2: The light after the first polarizer is vertically polarized. The angle between its polarization axis and the second polarizer is 60°, so apply Malus's Law:
4
$I_2 = 400 \times \cos^2(60^\circ) = 400 \times (0.5)^2 = 100 \text{ W/m}^2$
5
Final transmitted intensity is 100 W/m².

Exam tip:

AP graders will deduct partial credit if you forget to halve the intensity of unpolarized light before applying Malus's Law for subsequent filters.

3. Multi-Polarizer Stack Problems★★★★☆⏱ 4 min

A common exam trick question involves three or more sequential polarizers, often with the first and last oriented 90° apart (called crossed polarizers). Many students incorrectly assume the final intensity is zero, but an intermediate polarizer reorients the polarization to allow non-zero transmission.

📐 Worked Example

Unpolarized 1000 W/m² light passes through 3 polarizers: first vertical, second at 45° from vertical, third horizontal. Find the final transmitted intensity.

1
Step 1: First polarizer halves the unpolarized intensity:
2
$I_1 = 1000 / 2 = 500 \text{ W/m}^2$
3
Step 2: Angle between first and second polarizer is 45°, apply Malus's Law:
4
$I_2 = 500 \times \cos^2(45^\circ) = 500 \times 0.5 = 250 \text{ W/m}^2$
5
Step 3: Angle between second and third polarizer is 45°, apply Malus's Law again:
6
$I_3 = 250 \times \cos^2(45^\circ) = 125 \text{ W/m}^2$
7
Final intensity is 125 W/m², not zero.

✓ Quick check

Quick check: What would the final intensity be if you removed the middle 45° polarizer?

What is the transmitted intensity for two crossed polarizers with unpolarized incident light?
- 500 W/m²
- 250 W/m²
- 0 W/m²
- 125 W/m²
Reveal answer
0 W/m² —
With no intermediate polarizer, the second horizontal polarizer is 90° offset from the first vertical polarizer, so cos²(90°) = 0.

4. Polarization by Reflection and Brewster's Angle★★★☆☆⏱ 3 min

When light reflects off a smooth dielectric surface like water or glass, the reflected light is partially polarized. At one specific incident angle called Brewster's angle, the reflected light is 100% linearly polarized parallel to the surface (horizontally polarized for a flat lake surface).

\tan(\theta_B) = \frac{n_2}{n_1}

📘 Definition

Brewster's Angle

Incident angle where reflected and refracted rays are exactly perpendicular to each other, eliminating all p-polarized light from the reflected beam.

5. Common Pitfalls

Wrong move:

Applying Malus's Law directly to unpolarized incident light instead of halving intensity first

Why:

Unpolarized light has evenly distributed E-field orientations, so averaging cos²θ over all angles gives 0.5, not cos² of any single angle

Correct move:

Always reduce unpolarized light intensity by 50% after the first polarizer before applying Malus's Law for subsequent filters

Wrong move:

Claiming two crossed polarizers will always produce zero transmitted intensity even if there is a third polarizer in between

Why:

The intermediate polarizer reorients the E-field to a non-90° angle relative to the final polarizer, allowing non-zero transmission

Correct move:

Process each polarizer sequentially, calculating transmitted intensity and new polarization orientation at every step

Wrong move:

Using Snell's Law sine ratio to solve for Brewster's angle instead of the tangent relation

Why:

Brewster's angle is derived from the condition that reflected and refracted rays are perpendicular, leading to the unique tangent form of the law

Correct move:

Use \tan(\theta_B) = n2/n1 for Brewster's angle calculations, confirm θ_B + refracted angle = 90°

Wrong move:

Assuming all reflected light off any surface is fully polarized at all incident angles

Why:

Full polarization by reflection only occurs exactly at Brewster's angle, light is only partially polarized at all other incident angles

Correct move:

Explicitly state that 100% polarized reflected light only occurs at the specific Brewster's angle for the media pair

Wrong move:

Claiming scattered blue skylight is randomly unpolarized

Why:

Rayleigh scattering of sunlight in the atmosphere produces strongly polarized light oriented perpendicular to the line between observer and sun

Correct move:

Note that skylight polarization is the core design principle behind polarizing sunglasses

6. Quick Reference Cheatsheet

Scenario	Formula / Rule
Unpolarized light after first polarizer	I = I₀ / 2
Polarized light through polarizer at angle θ	I = I₀ cos²θ (Malus's Law)
Brewster's Angle (n₁ → n₂)	tan(θ_B) = n₂ / n₁
Two crossed polarizers (no intermediate filter)	Final intensity = 0
3 polarizers at 0°, 45°, 90° (unpolarized incident)	Final I = I₀ / 8

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Set 1 MCQ
3-polarizer stack intensity calculation
2022 · FRQ Part B
Polarization by reflection explanation
2019 · MCQ
Sky light polarization reasoning

What's Next

Mastering polarization is a critical stepping stone for the remaining physical optics topics on your AP Physics 2 exam. You will regularly combine Malus's Law calculations with wave interference and diffraction problems in multi-part FRQs that test cross-topic mastery. Polarization concepts also appear frequently in lab-based questions, where you may be asked to design an experiment to verify Malus's Law using a light sensor and rotatable polarizer. To reinforce your understanding, move to the linked topics below to practice applying polarization rules alongside related optics content, and work through our dedicated problem set for this sub-topic to lock in your score on this high-frequency exam concept.