Questions: Malus's Law: Derivation and Applications

Question 1

Two polarizers are crossed (their transmission axes are 90° apart) and block all incident light. A third polarizer is inserted between them at 45° to the first. What fraction of the original intensity emerges after all three?

• A. Zero — crossed polarizers block all light regardless of what is inserted between them
• B. 1/√2 — since cos(45°) = 1/√2 and only one projection step occurs
• C. 1/2 — only the final polarizer reduces the intensity
• D. 1/4 — applying Malus's law twice: I₀ → I₀cos²(45°) = I₀/2, then I₀/2 → (I₀/2)cos²(45°) = I₀/4

Answer: (D)

Apply Malus's law at each polarizer. After the first polarizer, the light is polarized along 0°. The intermediate polarizer at 45° transmits I₁ = I₀cos²(45°) = I₀/2, and the polarization direction is now 45°. The final polarizer is at 90°, so the angle between incoming polarization (45°) and the final axis (90°) is 45°. It transmits I₂ = I₁cos²(45°) = (I₀/2)(1/2) = I₀/4. The key insight is that the intermediate polarizer rotates the polarization direction, creating a nonzero component along the final axis — something impossible if you think of polarizers as pure absorption filters.

Question 2

Malus's law gives I = I₀cos²θ. If intensity were directly proportional to electric field amplitude rather than amplitude squared, what would the transmitted intensity relationship look like?

• A. I = I₀cos²θ — no change, since the cosine projection already accounts for this
• B. I = I₀cosθ — intensity would follow the projected amplitude directly
• C. I = I₀sin²θ — the perpendicular component would be transmitted instead
• D. I = I₀/cos²θ — intensity would increase at larger angles to compensate

Answer: (B)

The derivation has two steps: (1) project the electric field amplitude onto the transmission axis, giving E_t = E₀cosθ; (2) square to get intensity, giving I ∝ E_t² = E₀²cos²θ. If intensity were proportional to amplitude instead of amplitude squared, only step 1 would contribute and the transmitted intensity would be I₀cosθ — a linear function of cosθ, not a quadratic one. The cos²θ (rather than cosθ) dependence is entirely due to the I ∝ E² relationship, which is fundamental to wave energy.

Question 3

According to Malus's law, at θ = 90° (crossed polarizers), the transmitted intensity is zero because the electric field is largely absorbed by the polarizer material.

• T. True
• F. False

Answer: False

The zero transmission at θ = 90° follows from cos²(90°) = 0, which means the component of the electric field parallel to the transmission axis is exactly zero — there is no projection along the allowed direction, so nothing passes through. It is not that the field is absorbed per se; it is that the field has no component in the direction that the polarizer transmits. The mechanism is vector projection (only the component along the transmission axis passes), not selective absorption of a particular amount.

Question 4

When polarized light at 45° strikes a polarizer, exactly half the original intensity is transmitted.

• T. True
• F. False

Answer: True

By Malus's law: I = I₀cos²(45°) = I₀ × (1/√2)² = I₀ × 1/2 = I₀/2. This result is also memorable as a check: at 0°, full transmission; at 90°, zero; at 45° (halfway between), exactly half. This is only exact because intensity goes as cos² rather than as cos (which would give I₀/√2 ≈ 0.707I₀ at 45°). The cos² dependence ensures the 45° case is a clean half.

Question 5

In the three-polarizer demonstration, why does inserting a polarizer at 45° between two crossed polarizers allow light through, when the crossed polarizers alone transmit nothing? What must you track to explain this correctly?

Model answer: You must track the electric field vector, not just the intensity. Two crossed polarizers transmit nothing because the first polarizer establishes a polarization direction (say 0°) and the second (at 90°) has zero component along 0°. When the intermediate polarizer at 45° is inserted, it projects the 0°-polarized light onto the 45° axis, transmitting half the intensity — but critically, the transmitted light is now polarized at 45°. The final polarizer (at 90°) now sees light polarized at 45°, which has a nonzero projection (cos²45° = 1/2) along 90°. The intermediate polarizer effectively rotates the polarization direction in a step, creating a path from 0° to 90° via 45°.

If you think of polarizers as filters that simply 'block a fraction of photons' without changing the polarization direction, the three-polarizer result is mysterious. The explanation requires understanding that each polarizer resets the polarization direction to its own transmission axis — the output is always polarized along the polarizer's axis, regardless of the input polarization direction. Malus's law gives the intensity cost of each step; the rotation of polarization direction is the key mechanism that allows a second step to occur at all.