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3. Polarization & Malus's Law

Module

Wave Optics · Module 3

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Aim

To verify Malus's Law by measuring the transmitted intensity through an analyzer as a function of the angle between the polarizer and analyzer transmission axes.

Theoretical Background

Ordinary light contains electric field oscillations in all directions transverse to the propagation direction. A polarizer transmits only the component of the field parallel to its transmission axis, producing linearly polarized light of amplitude E0. When this polarized beam is passed through a second polarizer (the analyzer) whose transmission axis makes an angle θ with the first, only the component E0 cosθ is transmitted.

Since intensity is proportional to the square of the field amplitude, the transmitted intensity follows Malus's Law, I = I0 cos²θ, where I0 is the intensity transmitted when θ = 0. At θ = 90° (crossed polarizers) the transmitted intensity theoretically falls to zero.

This principle underlies polarimeters, photoelastic stress analysis, LCD displays, and the design of quarter-wave and half-wave plates used to generate circularly and elliptically polarized light.

Governing Formulas

Malus's Law:

\[ I(\theta) = I_0\cos^2\theta \]

Labelled apparatus layout — original illustration prepared for this guide (not a photograph). See the drag-to-rotate 3-D schematic below for a spatial view of the same setup.

3-D apparatus schematic · drag to rotate, scroll to zoom · Source → polarizer → analyzer → detector

Simulator Controls

Control	Symbol	Range	Unit
Analyzer angle	θ	0 – 180	°
Incident intensity	I0	10 – 100	%

Procedure

1. Set the incident intensity I0 to a convenient fixed value (e.g. 100%).
2. Vary the analyzer angle θ in steps of 15° from 0° to 180° and record the transmitted intensity readout at each step.
3. Repeat the run for a second value of I0 to confirm that the cos²θ dependence is independent of the absolute intensity.
4. Plot I(θ) against θ and separately I(θ) against cos²θ.

Observation Table

Table: Transmitted intensity versus analyzer angle (θ = 0° to 180° in steps of 15°)

S. No.	θ (°)	cos²θ	I observed (%)	I0 cos²θ calculated (%)
1
2
3
4
5
6
7
8
9
10
11
12
13

Graph

Plot Transmitted intensity I (%) (y-axis) against cos²θ (x-axis).

Expected trend: a straight line through the origin of slope I0.

I vs cos²θ — a straight line confirms Malus's Law

📋 Click the grid, then paste (Ctrl+V) a copied photo of your camera/CRO/DSO trace to overlay it for tracing — nothing is uploaded, it stays on your screen only. Overlay opacity Clear pasted image

15×15 grid — choose your own scale per square; plot from your observation table

Calculation

1. Compute cos²θ for every recorded angle.
2. Plot I against cos²θ; the slope of the best-fit straight line gives the experimental I0.
3. Compare the graph-derived I0 with the value set on the simulator and compute the percentage deviation.

Result

Table: Verification of Malus's Law

I0 set (%)	I0 from graph slope (%)	% deviation
1
2
3

Maximum Permissible Error

I = I₀cos²θ, so d(ln I) = d(ln I₀) − 2 tanθ·dθ; the angle term grows rapidly as θ → 90°, so avoid taking your reading very close to the crossed-polarizer position.

Precautions

• Take enough angle readings near θ = 90° since the intensity changes most rapidly there.
• Always plot I against cos²θ (not θ) to obtain a straight line for slope analysis.

Self-Check Quiz

1. At θ = 45° between polarizer and analyzer, what fraction of I0 is transmitted?

• A) 100%
• B) 70.7%
• C) 50%
• D) 0%

Answer: C — I = I0 cos²(45°) = I0 × 0.5 = 50% of I0.

2. At what angle θ does Malus's Law predict zero transmitted intensity?

• A) 0°
• B) 45°
• C) 60°
• D) 90°

Answer: D — cos²(90°) = 0 — this is the 'crossed polarizers' condition.

3. Why is intensity proportional to cos²θ and not cosθ?

• A) Intensity is proportional to the square of the field amplitude, and only the E0 cosθ component of the field is transmitted
• B) It's an empirical fit with no physical reason
• C) Because θ is always small
• D) Because two polarizers are used instead of one

Answer: A — Only the field component E0 cosθ passes the analyzer; since intensity ∝ (field)², I ∝ (E0 cosθ)² = I0 cos²θ.

4. Ordinary (unpolarized) light passed through a single ideal polarizer emerges with intensity:

• A) Zero
• B) The full incident intensity
• C) Half the incident intensity (on average)
• D) I0 cos²θ, same as after the analyzer

Answer: C — Unpolarized light has field oscillations equally distributed in all transverse directions, so on average a single ideal polarizer transmits half the incident intensity — this I0 is then used as the reference for Malus's Law at the second polarizer.

Viva-Voce Questions

1. Why does natural (unpolarized) light not show a cos²θ dependence when passed through a single polarizer alone?
2. What is the difference between a polarizer and an analyzer in this experimental arrangement?
3. What happens to the transmitted intensity when the analyzer is rotated to 90° with respect to the polarizer?
4. Why does unpolarized light lose exactly half its intensity after passing through the first (ideal) polarizer?
5. Distinguish between polarization by selective absorption (as in a Polaroid) and polarization by reflection.
6. At what angle between polarizer and analyzer axes is the rate of change of transmitted intensity greatest?
7. How would you experimentally verify that a given beam of light is completely unpolarized?

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Author & Maintainer: Dr. S. K. Jain, Associate Professor in Physics, Department of Applied Sciences and Humanities, Invertis University, Bareilly, India 📧 sanjeev.j@invertis.org

© 2026 Author. All rights reserved.