Malus's Law & Crossed Polarisers

Set each filter's axis and see what is left of the beam. Adding a filter usually costs light. Sometimes it buys light back.

Worked example — Malus's Law & Crossed Polarisers

After filter 1 (0.0°)	50.000%
After filter 2 (45.0°)	25.000%
Through the stack	12.500%
Without filter 2	0%
Best angle for filter 2	45.0°
Reading at that angle	12.500%
Best with 3 filters over 90.0°	12.500%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Malus's Law & Crossed Polarisers

Malus's law says what a polarising filter leaves of light that is already polarised: I = I₀cos²θ, where θ is the angle between the light's plane and the filter's transmission axis. Unpolarised light is the other case and it has no θ at all. It arrives with every direction equally represented, so the first filter it meets takes exactly half of it and hands on a beam polarised along its own axis. Every filter after the first is therefore in the Malus case, because the one in front of it did the polarising.

What each symbol means

I₀: the intensity arriving at the filter, not the intensity that started out
θ: the angle between the light's plane and the filter's axis. Only this difference matters, which is why turning the whole stack together changes no reading on the page
φ: the plane of an already-polarised beam, where there is one. Natural light has no φ, which is why its first filter costs a flat half rather than a cosine
cos²θ: the surviving fraction: a half at 45°, a quarter at 60°, nothing at 90°

Assumes: Ideal filters, transmitting perfectly along the axis and blocking perfectly across it. A real sheet does neither, and Polaroid named its material for the gap: an HN-38 sheet passes 38% of natural light where the arithmetic here says 50%, the missing twelve points being absorbed rather than polarised.
Breaks when: On a tall stack of real sheets, which is where this page's most surprising claim lives. Each real filter multiplies in its own absorption alongside its cos², so the climb toward 50% turns around after a few sheets and then falls away. Ninety-one ideal filters read 48.648%; a stack of ninety-one real ones is opaque.

Two filters block everything. Three let an eighth through. 🖖

Set the first axis to 0° and the second to 90°. cos²(90°) is exactly zero, the meter reads 0%, and nothing you do to the lamp will change it.

Now slide a third filter between them, at 45°. The reading is 12.500%.

The new filter absorbs light like the other two. What it also does is re-aim it, so the light crosses two gaps of 45° instead of one of 90°, and cos²(45°) is a half. Half of a half of the half the first filter left is an eighth.

The Without filter 2 row sits directly under Through the stack, so both numbers stay on screen while you drag the middle axis around.

Ten filters are brighter than three, and 50% is the wall 🖖

Keep the first axis at 0° and the last at 90°, and fill the gap with evenly spaced filters. Two give 0%. Three give 12.500%. Four, 21.094%. Five, 26.540%. Ten, 37.957%. Ninety-one of them, one degree apart, give 48.648%.

Every filter in that row absorbs light, and every one added to it raises the reading, because one large turn costs more than the same turn taken in small steps.

The climb has a ceiling it never touches. Natural light loses exactly half at whichever filter it meets first, whatever that filter's angle, and nothing downstream can put that half back.

The middle third of the turn costs twice what either end costs 🖖

Switch the source to Polarised and leave one filter in the beam. At 0° the reading is 100.000%. Turn it to 30° and 75.000% is left. Turn it to 60° and 25.000% is left. Turn it to 90° and nothing is.

Three equal turns of 30°, and the middle one costs twice what either of its neighbours costs. cos²θ is flat where cos is near 1 and steepest at 45°.

Anyone who has turned a polarising filter on a camera knows the feel of it. The first part of the turn barely changes the sky, and the middle of it changes everything.

Problem solved in full

The filter that lets light through by getting in the way 6 steps
Natural light of intensity I₀ meets a filter with its axis at 0°, then a second at 90°. A third filter is slid in between them with its axis at 30°. What fraction of I₀ leaves the stack, where should that third filter be to do best, and what does the best setting buy?

Open this problem in the tool
1. The first filter is the only one that is not obeying Malus's law. Natural light has no single plane to measure an angle from, so there is no θ to put in cos²θ; it arrives as an even mixture of every direction and the filter passes the average, which is exactly a half. With this stack loaded the After filter 1 (0.0°) row reads 50.000%, and turning that first filter anywhere you like leaves it at 50.000%.
  
  $unpolarised light: \frac{I _{1}}{I _{0}} = \frac{1}{2} at every angle$
2. Everything downstream is in the Malus case, because the first filter polarised the beam along its own axis. The second filter in this stack is the one at 30°, so the gap is 30° and the factor is cos²(30°) = 0.75. Half of the light, three quarters of that: 37.500%, and the After filter 2 (30.0°) row says so.
  
  $\frac{I _{2}}{I _{0}} = \frac{1}{2} cos^{2} (3 0^{\circ}) = \frac{1}{2} \cdot 0.75 = 0.375$
3. The last filter sits at 90°, which is 60° on from 30°. cos²(60°) = 0.25, and a quarter of 37.500% is 9.375%. That is the answer to the first question, and the Through the stack row carries the whole chain that produced it: ½ · cos²(30°) · cos²(60°).
  
  $\frac{I _{3}}{I _{0}} = \frac{1}{2} cos^{2} (3 0^{\circ}) cos^{2} (6 0^{\circ}) = 0.375 \cdot 0.25 = 0.09375$
4. Now take the middle filter out, which is the step worth doing slowly. The beam then meets 0° and 90° with nothing in between, one gap of 90°, and cos²(90°) is zero. The Without filter 2 row reads 0%. A sheet of absorbing plastic, dropped into a beam that was already dead, took the reading from nothing to 9.375%.
  
  $filter 2 removed: \frac{1}{2} cos^{2} (9 0^{\circ}) = 0$
5. So where does it do most good? The two gaps have to add to 90°, and what is being maximised is cos²(δ)·cos²(90° − δ). Equal gaps win: ½ · ½ · ½ = 12.500% at 45°, against 9.375% at 30°. The Best angle for filter 2 row reads 45.0° and the Reading at that angle row reads 12.500%.
  
  $δ max \frac{1}{2} cos^{2} (δ) cos^{2} (9 0^{\circ} - δ) \Rightarrow δ = 4 5^{\circ}, \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = 0.125$
6. Equal gaps keep winning however many filters there are, and the reason is that ln cos is concave, so a fixed total turn is cheapest split evenly. Press Add filter to put a fourth in and Spread evenly to place them: 21.094%. Ten filters across the same quarter turn reach 37.957%.
  
  $\frac{I _{n}}{I _{0}}_{even} = \frac{1}{2} cos^{2 (n - 1)} (\frac{9 0 ^{\circ}}{n - 1}) : n = 4 \to 0.21094, n = 10 \to 0.37957$
Answer

9.375% at 30°, and 12.500% at 45°, which is the most three filters can do across a quarter turn. The figure worth carrying away is neither of those. It is the nothing that comes back the moment the middle filter is pulled out. That filter absorbs light like the other two, and the stack is brighter with it than without it, because what it does is turn the light, so that the last filter is asked an easier question.

$\frac{1}{2} cos^{2} (3 0^{\circ}) cos^{2} (6 0^{\circ}) = 9.375% against \frac{1}{2} cos^{4} (4 5^{\circ}) = 12.5%$

References (3)

The paper the law is named after, written by an army engineer who noticed it in the windows of the Luxembourg Palace: É. L. Malus, "Sur une propriété de la lumière réfléchie." Mémoires de physique et de chimie de la Société d'Arcueil 2, 1809, pp. 145–158.
Where the 38% comes from — the man who invented sheet polarisers, on how well they really work: E. H. Land, "Some Aspects of the Development of Sheet Polarizers." Journal of the Optical Society of America 41(12), 1951, pp. 957–963. DOI 10.1364/JOSA.41.000957
The textbook treatment of ideal polarisers and the stacked-filter result: E. Hecht, Optics, 5th edition, Pearson, 2017, chapter 8 (Polarization).

By the 42Math team · Reviewed against published sources. Corrections welcome.