Malus's Law: What Actually Gets Through Two Polarisers

Q: Why is there a square? Where does cos²θ come from rather than cos θ?

The filter acts on the electric field, which is a vector. It passes the component along its axis, and that component is E₀cosθ by ordinary vector resolution. But what we measure — brightness, power per square metre — goes as the square of the field amplitude. Square E₀cosθ and you get I₀cos2θ. The rows above print cosθ and cos2θ separately for exactly this reason: at 60° half the field survives and only a quarter of the intensity does, and knowing which one a question is asking about is half the battle.

Q: Do polarising sunglasses work this way?

Partly. Their axis is vertical, so they block horizontally polarised light — and the glare reflected off a horizontal surface such as water or a wet road is strongly horizontally polarised. The Cornell lecture cited below says just this: sunglasses “transmit only vertically polarized light to cut glare due to reflections from water”. So they do far more to glare than to the rest of the scene, which is why they work so well and why tilting your head sideways brings the glare back. That polarisation on reflection is not Malus's law — it is Brewster's, and it has its own page.

I = I₀cos²θ is the easy half. The half that costs marks is what I₀ is: if the light arriving was unpolarised, the first filter has already thrown away half of it before the cosine squared ever applies. This page gives you the fraction of the original beam that survives both filters, keeps the halving visible, and draws the two polarisers with the second one turned to your angle.

Last updated 11 Nov 2026

Transmission through two polarisers

The cos² law with the unpolarised halving kept in view

Intensity of the light arriving at the first filterAny units you like — the headline is a percentage of this number, so it cancels out and only the absolute row below depends on it. 100 is set as the default so the percentage and the W/m² row read the same at a glance. If a question gives you an amplitude rather than an intensity, square it first: intensity goes as the square of the electric field, which is where the square in cos² comes from.

What the light arriving is likeThis is the switch that catches people out. Unpolarised light has its electric field in every direction across the beam, equally, so a polaroid passes exactly half of it whatever angle the polaroid is set to — NCERT states this directly: “the intensity coming out of a single polaroid is half of the incident intensity”. Light that is already polarised and lined up with the first filter loses nothing there. Either way, Malus’s law applies at the SECOND filter, to whatever the first one let through — which is what NCERT means when it defines its I₀ as the intensity of the polarised light after P₁.

Angle between the two filters’ transmission axesDegrees, measured between the two transmission axes. 0 means parallel and everything the first filter passed gets through; 90 means crossed and nothing does. Beyond 90 the pattern simply repeats: cos² has a period of 180°, so 120° behaves exactly like 60° and 170° like 10°. The field stops at 90° to keep the diagram readable. Note that the half-way point is not 45° of the way to dark — cos²45° = 0.5 exactly, so 45° is where half goes through, and the curve is flat near 0 and steepest at 45°.

Two polarisers seen from the side, with the beam running left to right. The thickness of each blue band is the intensity at that point as a fraction of what arrived, so you can watch the first filter halve an unpolarised beam before the second filter has done anything at all. The double-headed line on each plate is its transmission axis, drawn as you would see it looking down the beam: the first filter's is dark and fixed upright, the second filter's is red and turns to your angle, and the red arc measures that angle from the first one. At 90 degrees the outgoing band has no thickness, which is the correct picture of a crossed pair.

37.50%Example

Unpolarised light of 100 W/m² into a pair of polaroids with their axes 30° apart

Malus’s law, and the two things that happen before it

I = I₀cos²θ · unpolarised in: I₀ = I_in/2 · so I = ½I_incos²θ · three filters at 0°, θ, 90°: I = I₀cos²θ sin²θ

I₀: the intensity of the polarised light leaving the first filter, which is NCERT’s own wording for Equation 10.18 — not the intensity of whatever arrived at it. If the incoming light was unpolarised, I₀ is half of it; if it was polarised and aligned, I₀ is all of it. Getting this wrong is a factor-of-two error and it is the commonest mistake on the topic
θ: the angle between the two transmission axes, not between the light and anything. Because only the difference matters, rotating both filters together changes nothing — which is how you can tell polarised light from unpolarised with a single filter: turn it and watch whether the brightness changes
cos²: the square is there because intensity goes as the square of the electric field. The filter passes the component of E along its axis, which is E cosθ — an amplitude — and the intensity is that squared. Both numbers are given in the rows above, and at 60° they are 0.5 and 0.25, which is a useful pair to keep in mind
the factor of ½: unpolarised light has E in every transverse direction equally, and the average of cos² over all angles is exactly ½. So a single polaroid passes half of unpolarised light whatever angle it is set to, which is why the first filter’s orientation never appears in the arithmetic
cos²θ sin²θ: the three-filter case: apply Malus twice, once from the first filter to the middle one and once from the middle to the last. It equals ¼sin²2θ, so it is zero at 0° and 90° and peaks at 45° with a value of ¼ — one eighth of the original unpolarised beam

Worked example

Unpolarised light of 100 W/m² into a pair of polaroids with their axes 30° apart
THE FIRST FILTER, AND HALF THE MARKS. The light arriving is unpolarised, so the first polaroid passes exactly half of it whatever angle it is set to: I₁ = 100/2 = 50 W/m². NCERT states this on its own: “the intensity coming out of a single polaroid is half of the incident intensity”. Malus's law has not been used yet.
NOW MALUS, AND ON THE RIGHT NUMBER. I = I₀cos²θ with I₀ = 50 W/m², the light leaving P₁. cos 30° = √3/2 = 0.8660, so cos²30° = 0.75.
MULTIPLY. I = 50 × 0.75 = 37.5 W/m², which is 37.5% of the 100 W/m² you started with. Write the fraction of the original down as well as the absolute figure — questions ask for either and they differ by the factor of two.
SEPARATE THE TWO LOSSES. The first filter cost you 50% and the second cost 25% of what was left. In decibels that is −3.01 dB then −1.25 dB, −4.26 dB in all. The decibel form is worth seeing once, because losses in dB simply add while fractions have to be multiplied.
THE AMPLITUDE, WHICH IS THE PART PEOPLE SKIP. The electric field that gets through the second filter is E₀cos30° = 0.866E₀ — the field is down by 13%, while the intensity is down by 25%. Every time you see a square in optics, that is why.
NOW THE TRICK QUESTION. Cross the two filters at 90° and nothing comes out. Slide a third polaroid between them at 45°, and I = 50 × cos²45° × sin²45° = 50 × 0.5 × 0.5 = 12.5 W/m² — an eighth of the original. Adding a filter made it brighter. The middle filter rotates the plane of the light that passes it; it does not unblock anything.

What gets through, angle by angle, for both kinds of input

Three things to take from this. **The fourth column is exactly half the fifth, everywhere**, and that factor of two is the first filter, not Malus. **cosθ goes negative past 90° and cos²θ does not**, which is why the pattern repeats every 180° rather than every 360° — a polariser cannot tell “up” from “down”. And the intensity column falls far faster than the amplitude column: at 60° half the field is still there but only a quarter of the energy. The input field on this page stops at 90°, so read the last three rows as 60°, 30° and 0°.
θ	cos θ (amplitude)	cos²θ (intensity)	% of original, unpolarised in	% of original, polarised in
0°	1.000	1.000	50.0	100.0
30°	0.866	0.750	37.5	75.0
45°	0.707	0.500	25.0	50.0
60°	0.500	0.250	12.5	25.0
90°	0.000	0.000	0.0	0.0
120°	−0.500	0.250	12.5	25.0
150°	−0.866	0.750	37.5	75.0
180°	−1.000	1.000	50.0	100.0

Why half the light vanishes before the formula even starts

Light is a transverse wave: its electric field points across the direction of travel, not along it. Polarisation is the question of which way across. Light from the sun, a filament or a flame is unpolarised — the field direction scrambles randomly millions of times a second, with no direction preferred. A polaroid sheet passes the component of that field along one line, its transmission axis, and absorbs the rest.

What a polariser does to unpolarised light. Average cos² over every possible field direction and you get exactly one half. So the first polaroid passes 50% — whatever angle you set it to. Turn it and nothing changes, which is itself the test for whether light is polarised. NCERT states the result plainly: the intensity coming out of a single polaroid is half of the incident intensity.

And what it does to light that is already polarised. Now the angle matters, and this is Malus’s law. The filter passes only the component of E along its axis, E cosθ. Intensity goes as the square of the field, so I = I₀cos²θ — NCERT’s Equation 10.18, with I₀ defined in that chapter as the intensity of the polarised light after passing through P₁. That definition is the whole difficulty. Students read I₀ as “the light I started with”, apply cos²θ to it, and come out exactly twice too high.

Two filters, in order. Unpolarised light in: the first filter halves it and polarises it, then the second applies cos²θ to what is left. Overall ½cos²θ, with a ceiling of 50% that no angle can beat. Already-polarised light in, aligned with the first filter: it loses nothing at the first and only cos²θ at the second, so a parallel pair passes everything. The headline on this page is always a percentage of what arrived at the first filter, so both cases are on the same scale and you can see the factor of two directly.

The three-polariser result, which is the one worth understanding. Cross two polaroids at 90° and no light comes out. Now slide a third between them at 45°. Light reappears — up to an eighth of the original unpolarised beam. It looks like adding an absorber made things brighter, and students reasonably think something has gone wrong. Nothing has. A polariser is not a sieve that removes photons of the wrong kind; it projects the field onto its own axis, and the light that emerges is polarised along that axis, not the old one. The middle filter rotates the plane by 45°, so what reaches the last filter is 45° from it rather than 90°, and half of it gets through. Remove the middle filter and the plane is never rotated.

What the syllabus documents actually say. NCERT’s Chapter 10 has polarisation as section 10.7 and gives Malus’s law as Equation 10.18. The CBSE Class XII curriculum for 2025-26 does not: the whole document was searched, and “polarisation”, “Malus” and “Brewster” appear nowhere in the Optics unit. That is a statement about that one document for that one year and not a prediction about any paper; check your own board’s current syllabus. It is recorded here because it would be dishonest to call this page examinable content on a syllabus this batch read and did not find it in. For the companion pages, Brewster’s angle is how reflection polarises light without a filter at all, and Young’s double slit is the other piece of evidence that light is a wave.

Frequently asked questions

Why does a single polaroid always pass half of unpolarised light?

Because unpolarised light has its electric field in every transverse direction with equal probability, and the average of cos² over all directions is exactly one half. The filter passes a fraction cos²θ of the light whose field happens to be at θ to its axis, and averaging that over a uniform spread of θ gives ½. The result does not depend on the filter’s orientation, which is the useful part: rotate a single polaroid in front of a light source and if the brightness does not change, the light was unpolarised. If it does change, it was at least partly polarised.

Is I₀ in Malus’s law the light I started with or the light after the first filter?

After the first filter. NCERT’s own words for Equation 10.18 are “where I₀ is the intensity of the polarized light after passing through P₁”. Malus’s law describes what one polariser does to light that is already polarised, so it simply does not apply to an unpolarised beam — for that you use the factor of ½, which comes from averaging, not from the cosine. If a question starts with unpolarised light and two filters, you need both steps. If it starts with polarised light, you need only Malus. An answer out by exactly two is nearly always this.

How can adding a third polariser make the light brighter?

Because a polariser changes the light rather than just reducing it. Two filters crossed at 90° pass nothing: the field arriving at the second is exactly perpendicular to its axis, so there is no component to transmit. Put a third at 45° in between and the light leaving it is polarised at 45° — the middle filter rotated the plane — so it is now only 45° from the last filter and half of it gets through. The total is I₀cos²45 × sin²45 = I₀/4, or an eighth of the original unpolarised beam. No energy is created; less is absorbed in the middle than you gain at the end.

Why is there a square? Where does cos²θ come from rather than cos θ?

The filter acts on the electric field, which is a vector. It passes the component along its axis, and that component is E₀cosθ by ordinary vector resolution. But what we measure — brightness, power per square metre — goes as the square of the field amplitude. Square E₀cosθ and you get I₀cos²θ. The rows above print cosθ and cos²θ separately for exactly this reason: at 60° half the field survives and only a quarter of the intensity does, and knowing which one a question is asking about is half the battle.

Do polarising sunglasses work this way?

Partly. Their axis is vertical, so they block horizontally polarised light — and the glare reflected off a horizontal surface such as water or a wet road is strongly horizontally polarised. The Cornell lecture cited below says just this: sunglasses “transmit only vertically polarized light to cut glare due to reflections from water”. So they do far more to glare than to the rest of the scene, which is why they work so well and why tilting your head sideways brings the glare back. That polarisation on reflection is not Malus’s law — it is Brewster’s, and it has its own page.

Related calculators

References

NCERT, Physics, Class XII, Part II, Chapter 10, Wave Optics (same caveat about the hosted copy). The chapter’s sections are 10.1 Introduction, 10.2 Huygens Principle, 10.3 Refraction and Reflection of Plane Waves using Huygens Principle, 10.4 Coherent and Incoherent Addition of Waves, 10.5 Interference of Light Waves and Young’s Experiment, 10.6 Diffraction and 10.7 Polarisation. Section 10.7 gives Equation (10.18), I = I₀cos²θ, and names it: “This is known as Malus’ law.” It is explicit about the thing that catches people out — “where I₀ is the intensity of the polarized light after passing through P₁”, not the intensity of the unpolarised beam that arrived — and states separately that “the intensity coming out of a single polaroid is half of the incident intensity”. Brewster’s law does not appear anywhere in the chapter. Section 10.7 runs from transverse waves through polaroids to Malus’ law and its applications and stops; there is no tan i_B = n. The chapter was searched for the word. Figures are absent from the extracted copy and several equations are garbled, so only the statements quoted above are relied on.
Central Board of Secondary Education, Physics (Code No. 042), Classes XI–XII, Senior Secondary Curriculum 2025-26, read from cbseacademic.nic.in. Unit VI Optics, Chapter 9 Ray Optics lists the lens maker’s formula by name, alongside the thin lens formula, magnification, power of a lens, thin lenses in contact and refraction through a prism. Chapter 10 Wave Optics lists Huygens’ principle, the proof of the laws of reflection and refraction, coherent sources, interference, “Young’s double slit experiment and expression for fringe width (No derivation final expression only)” and “diffraction due to a single slit, width of central maxima (qualitative treatment only)” — and then stops. “Polarisation”, “Malus” and “Brewster” appear nowhere in the Optics unit. The whole document was searched: the only occurrence of the word at all is “Dielectrics and electric polarization” in Unit I, which is electrostatics and a different thing. This is reported because it is load-bearing: on that document the lens maker’s formula is named examinable content and the two polarisation laws are not.
Cornell University, Physics Department / CLASSE, P3330 Experimental Optics, Lecture 3 (I. Bazarov). Gives the Fresnel amplitude reflection coefficients in the compact form r_⊥ = (1 − nm)/(1 + nm) and r_∥ = (m − n)/(m + n), with n = n_t/n_i and m = cosθ_t/cosθ_i. Those are algebraically identical to the long forms this page implements: divide the usual numerator and denominator through by n_icosθ_i and you get them, which is the check that this page’s Fresnel arithmetic is that source’s Fresnel arithmetic. For the air-to-glass case, with n_air ≈ 1 and n_glass ≈ 1.5, it states “Zero reflection for parallel polarization at Brewster’s angle (56.3° for these values of n_i and n_t)” — and arctan(1.5) = 56.3099°, which is the independent check that the arctangent on this page is the right arctangent. For polarisation in use it says sunglasses “transmit only vertically polarized light to cut glare due to reflections from water”. What this source does NOT contain, and is therefore not attributed to it: the formula tanθ_B = n₂/n₁, and any statement that the reflected and refracted rays are 90° apart. Both are derived from the geometry in the worked example on this page instead. Several of the slides’ fractions were damaged in extraction and are read from the layout, so they are reported as a reading rather than a verbatim quotation.