Polarization & Malus's Law

Here is a fact that interference and diffraction, for all their power, could not settle. Both of those effects prove light is a wave — but a wave of what, oscillating in which direction? A sound wave shakes the air back and forth along its own line of travel; a wave on a rope can shake up-and-down or side-to-side, across the direction it runs. Longitudinal or transverse? Interference happens either way, so it stays silent on the question. Polarization is the experiment that breaks the tie, and it answers loudly: light is transverse. Its electric field oscillates strictly across the direction the wave travels, never along it. That single fact has a visible consequence you can hold in your hand. Take two lenses from a pair of polarizing sunglasses, or two squares of Polaroid film, and look through both at a lamp. Rotate one. The light dims, and at one orientation — the two axes at right angles — it goes almost completely black. Nothing has absorbed the light chemically; you have simply asked a transverse wave to squeeze through two gates whose slots are crossed, and a transverse oscillation cannot pass both. Do the same trick with sound and nothing happens at all, because there is no "sideways" to a longitudinal wave to block. This article turns that demonstration into physics. A polarizing filter transmits only the field component along its axis, so unpolarized light loses exactly half its intensity passing the first filter, and then a second filter obeys a clean law discovered by Étienne-Louis Malus in 1809: $I = I_0\cos^2\theta$ , where $\theta$ is the angle between the two axes. Full through when aligned, half at $45^\circ$ , dark when crossed. From that one $\cos^2$ come polarizing sunglasses, LCD screens, photographic filters, and the proof, two centuries old, that light shakes sideways.

Key takeaways

Light is a transverse electromagnetic wave: its electric field oscillates perpendicular to the travel direction, so it can be polarized — confined to a single plane.
A polarizing filter passes only the field component along its axis. Unpolarized light loses half its intensity through the first filter: $I \to I_0/2$ (the average of $\cos^2$ over all angles).
A second filter (analyser) at angle $\theta$ obeys Malus's law $I = I_0\cos^2\theta$ : full at $0^\circ$ , half at $45^\circ$ , zero when crossed at $90^\circ$ .
Two crossed filters pass no light — but slip a third filter at $45^\circ$ between them and light reappears, at $I_u/8$ . The order and angles matter.
Longitudinal waves (sound) cannot be polarized; polarization is the decisive proof that light is transverse.
Key idea: unpolarized light halves through one filter, then a second filter follows $I = I_0\cos^2\theta$ — one law that only a transverse wave could obey.

The core insight

Because light is a transverse wave whose electric field can be confined to one plane, a polarizing filter transmits only the field component along its axis: unpolarized light drops to $I_0/2$ through the first filter, and a second filter at angle $\theta$ then transmits $I = I_0\cos^2\theta$ (Malus's law) — full when aligned, zero when crossed.

Where this lives today

The two filters in your polarizing sunglasses are Malus's law you can wear. Glare from a wet road or a lake is light partially polarized by reflection, mostly horizontal; sunglasses set their transmission axis vertical, so $\cos^2\theta$ is near zero for that glare and it is cut, while the rest of the scene passes. Every LCD — phone, laptop, television — is a sandwich of two crossed polarizing filters with a liquid-crystal layer between that rotates polarization pixel by pixel, switching each dot between $I_u/8$ -style bright and crossed-dark exactly as in the three-filter demonstration. Photographers screw a rotatable polarizing filter onto a lens to deepen skies and kill reflections. Stress engineers shine polarized light through transparent plastic models to see stress patterns light up (photoelasticity), and 3-D cinema hands you glasses whose two lenses are polarized differently so each eye sees its own image. All of it runs on one transverse fact and one $\cos^2$ .

The physical picture

An electromagnetic wave, met earlier in this course, is an oscillating electric field $\vec{E}$ (with a partner magnetic field) travelling through space. The crucial detail for this article is the direction of that oscillation: $\vec{E}$ points across the direction of travel — it is a transverse wave. If all the light in a beam happens to oscillate its $\vec{E}$ in one fixed plane, the light is linearly polarized (or plane-polarized). Ordinary light from the Sun or a lamp is not: it is an incoherent jumble of countless wavetrains from independent atoms, each with its own transverse direction, so the net $\vec{E}$ points every which way across the beam, changing randomly millions of times a second. That is unpolarized light. A polarizing filter — a Polaroid sheet — has a built-in transmission axis. It behaves like a gate that passes the component of $\vec{E}$ along that axis and absorbs the component across it. Send polarized light in with its $\vec{E}$ at angle $\theta$ to the axis and only the projection $E\cos\theta$ gets through. Send unpolarized light in and, because its direction is random, on average exactly half the intensity survives — the filter keeps one of the two transverse directions and discards the other. After that first filter the light is polarized along the filter's axis, whatever it was before.

Figure

Unpolarized light through two filters — the first halves it to I₀ over 2, the second follows Malus's law with the transmitted intensity depending on the angle θ between the axes

Unpolarized light through the first filter

Why exactly half? Unpolarized light has its $\vec{E}$ pointing at every transverse angle $\phi$ with equal likelihood. Each such component, hitting a filter whose axis is fixed, is itself reduced by $\cos^2\phi$ (that is Malus's law, which we prove next). To get the transmitted intensity we average $\cos^2\phi$ over all directions $\phi$ from $0$ to $2\pi$ . The average value of $\cos^2$ over a full turn is exactly $\tfrac12$ , so

I_1 = \frac{I_0}{2} \qquad (1)

where: $I_0$ — the incident (unpolarized) intensity; $I_1$ — what survives the first filter, now polarized along its axis. This is not Malus's law with some particular angle — it is the average over all angles, and it is the single most common place students misapply $\cos^2\theta$ . Unpolarized light does not have "an angle" to the filter; it has all of them at once, and the average is one half.

Malus's law: the second filter

Now the light is polarized along the first filter's axis with intensity $I_1$ . Send it into a second filter — the analyser — whose axis makes angle $\theta$ with the first. The analyser passes only the component of the field along its own axis. If the incoming amplitude is $E_1$ , the transmitted amplitude is the projection $E_1\cos\theta$ . Because intensity is proportional to the square of the amplitude ( $I \propto E^2$ ), the transmitted intensity is

I = I_1\cos^2\theta \qquad (2)

where: $\theta$ — the angle between the analyser's axis and the polarization direction of the light reaching it; $I_1$ — the intensity arriving at the analyser. Equation (2) is Malus's law. Read it at the three angles you must know cold: at $\theta = 0$ the axes are aligned and $I = I_1$ (everything through); at $\theta = 90^\circ$ the axes are crossed and $I = 0$ (dark); at $\theta = 45^\circ$ , $\cos^2 45^\circ = \tfrac12$ , so exactly half gets through. The full chain from unpolarized light through two filters is therefore

I = \frac{I_0}{2}\cos^2\theta. \qquad (3)

The key values, tattooed into every exam: $\cos^2 30^\circ = 0.75$ , $\cos^2 45^\circ = 0.5$ , $\cos^2 60^\circ = 0.25$ , $\cos^2 90^\circ = 0$ . So a beam already polarized with intensity $I_1$ emerges at $0.75\,I_1$ , $0.5\,I_1$ , $0.25\,I_1$ , and $0$ respectively.

Figure

A transverse wave polarized in a single plane, its electric field arrows all vertical, contrasted with unpolarized light whose field points every way across the beam

Derivation AWhy unpolarized light halves and why the analyser follows cos-squared

The goal is to establish both master facts of this topic from one idea — that a filter passes the field component along its axis — giving the half-intensity rule $I_0/2$ for the first filter and Malus's law $I = I_1\cos^2\theta$ for the second. The strategy is to project the amplitude onto the axis, square it to get intensity, then average over all directions for the unpolarized case. Only algebra and one averaging fact are needed; no calculus is required beyond quoting that the mean of $\cos^2$ over a full turn is one half.

Step 1. Project the amplitude. Light of amplitude $E$ whose field makes angle $\theta$ with a filter's transmission axis has, along that axis, the component $E\cos\theta$ . The filter passes this and absorbs the perpendicular component $E\sin\theta$ .

Step 2. Square to intensity. Since $I \propto E^2$ , the transmitted intensity is $I = I_{\text{in}}\cos^2\theta$ , where $I_{\text{in}} \propto E^2$ is the incoming intensity. For polarized light entering an analyser at a definite angle $\theta$ , this is Malus's law, equation (2).

Step 3. Handle unpolarized input. Unpolarized light is an equal mixture of all directions $\theta$ . Each contributes $\cos^2\theta$ of its share, so the transmitted fraction is the average of $\cos^2\theta$ over $0$ to $2\pi$ . Using $\cos^2\theta = \tfrac12(1 + \cos 2\theta)$ , the $\cos 2\theta$ term averages to zero over a full turn, leaving the mean equal to $\tfrac12$ . Hence the first filter transmits $I_0/2$ , equation (1) — no matter which way its axis points.

Step 4. Chain them. Unpolarized $I_0$ → first filter → $I_0/2$ , polarized along axis 1 → analyser at $\theta$ → $(I_0/2)\cos^2\theta$ , equation (3). At $\theta = 45^\circ$ that is $(I_0/2)(0.5) = I_0/4$ ; at $\theta = 90^\circ$ it is zero. Verified: the average of $\cos^2$ over a full turn equals $1/2$ , and $\cos^2\theta$ at $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ equals $1, 0.75, 0.5, 0.25, 0$ — recomputed in verify_hs2-polarization.py.

The crossed-filters trick: three do what two cannot

Cross two filters ( $\theta = 90^\circ$ ) and nothing gets through — Malus's law gives $\cos^2 90^\circ = 0$ . Now insert a third filter between them, at $45^\circ$ to both. Astonishingly, light reappears. Track it with Malus's law twice. Unpolarized light $I_u$ enters and the first filter gives $I_u/2$ , polarized vertically. The middle filter at $45^\circ$ transmits $(I_u/2)\cos^2 45^\circ = I_u/4$ , and re-polarizes the light to $45^\circ$ . The last filter, at $90^\circ$ to the first but only $45^\circ$ from the middle one, transmits $(I_u/4)\cos^2 45^\circ = I_u/8$ :

I = \frac{I_u}{2}\cos^2 45^\circ \cos^2 45^\circ = \frac{I_u}{2}\cdot\frac12\cdot\frac12 = \frac{I_u}{8} = 0.125\,I_u. \qquad (4)

The middle filter rotates the polarization in two $45^\circ$ steps, so the crossed pair no longer sees a right angle across a single gap. Remove the middle filter and you are back to $\cos^2 90^\circ = 0$ . This is the famous demonstration that a polarizing filter does not merely "select" light already present — it re-projects the field, and inserting an element can increase what passes downstream.

Limiting cases

case	result	meaning
unpolarized into 1 filter	$I = I_0/2$	half survives, whatever the axis direction
axes aligned, $\theta = 0$	$I = I_1$	analyser fully open, Malus gives $\cos^2 0 = 1$
axes at $45^\circ$	$I = I_1/2$	half through; $\cos^2 45^\circ = 0.5$
axes crossed, $\theta = 90^\circ$	$I = 0$	dark; $\cos^2 90^\circ = 0$
third filter at $45^\circ$ between crossed pair	$I = I_u/8$	light reappears — the middle filter rotates the polarization
longitudinal wave (sound)	no effect	cannot be polarized — no transverse direction to block

Common trapThe first filter does not use Malus's law

The number-one error is to apply $\cos^2\theta$ to unpolarized light hitting the first filter. Unpolarized light has no single angle to the axis — it carries every angle at once — so the correct reduction is the average, exactly $\tfrac12$ , and the result $I_0/2$ is independent of how the filter is turned. Malus's law $I = I_1\cos^2\theta$ applies only after the light is polarized, to the second filter and beyond, where a definite angle $\theta$ exists. A useful discipline: for a stack of filters, write $I_0/2$ for the first, then multiply by $\cos^2(\text{angle between consecutive axes})$ for each filter after it. The angle in Malus's law is always measured between the light's current polarization direction and the next axis — never to some fixed reference.

Worked examples

Board level — two filters at an angle (CBSE and NEET pattern)

Unpolarized light of intensity $I_0$ passes through two polarizing filters whose transmission axes are at $30^\circ$ to each other. (a) What fraction emerges? (b) What if the angle is $60^\circ$ ?

Step 1. First filter halves the unpolarized light: $I_1 = I_0/2$ , polarized along axis 1.

Step 2. Second filter, Malus's law at $\theta = 30^\circ$ : $I = I_1\cos^2 30^\circ = (I_0/2)(0.75) = \boxed{0.375\,I_0}$ , i.e. three-eighths.

Step 3. For $\theta = 60^\circ$ : $I = (I_0/2)\cos^2 60^\circ = (I_0/2)(0.25) = \boxed{0.125\,I_0}$ , i.e. one-eighth.

Step 4. Sense-check: increasing the angle from $30^\circ$ to $60^\circ$ cut the output from $0.375\,I_0$ to $0.125\,I_0$ — dimmer, as it must, since the axes are further from aligned. At $90^\circ$ it would be zero.

Board level — reading off the angle (ICSE and JEE pattern)

Polarized light of intensity $I_1$ falls on an analyser. At what angle is the transmitted intensity (a) half of $I_1$ , and (b) one-quarter of $I_1$ ?

Step 1. Half: set $\cos^2\theta = 0.5$ , so $\cos\theta = 1/\sqrt2$ and $\theta = \boxed{45^\circ}$ .

Step 2. One-quarter: set $\cos^2\theta = 0.25$ , so $\cos\theta = 0.5$ and $\theta = \boxed{60^\circ}$ .

Step 3. Note the pattern for later recall: $30^\circ \to 0.75$ , $45^\circ \to 0.5$ , $60^\circ \to 0.25$ — the outputs step down as the angle climbs, symmetric about $45^\circ$ being the half-power point.

Step 4. Caution: these angles are between the analyser and the already-polarized beam. If the source were unpolarized, the first filter would separately contribute its own factor of $\tfrac12$ .

Competitive level — the three-filter surprise (JEE-Advanced and NEET pattern)

Two polarizing filters are crossed ( $90^\circ$ apart) so no light passes. Unpolarized light of intensity $I_u$ is incident. A third filter is now inserted between them at $45^\circ$ to each. (a) What intensity emerges? (b) What emerges if the third filter is instead placed at $30^\circ$ to the first?

Step 1. First filter: $I_u/2$ , polarized vertically. Middle filter at $45^\circ$ : $(I_u/2)\cos^2 45^\circ = I_u/4$ , now polarized at $45^\circ$ . Last filter (at $90^\circ$ from first, so $45^\circ$ from middle): $(I_u/4)\cos^2 45^\circ = \boxed{I_u/8 = 0.125\,I_u}$ .

Step 2. Part (b): middle filter at $30^\circ$ from the first, so $60^\circ$ from the last. After first: $I_u/2$ . After middle: $(I_u/2)\cos^2 30^\circ = (I_u/2)(0.75) = 0.375\,I_u$ . After last: $(0.375\,I_u)\cos^2 60^\circ = (0.375\,I_u)(0.25) = \boxed{0.09375\,I_u}$ .

Step 3. The maximum output over all middle-filter angles is at $45^\circ$ , giving $I_u/8$ — you can see this because the transmitted fraction $\tfrac12\cos^2\theta\cos^2(90^\circ-\theta) = \tfrac12\cos^2\theta\sin^2\theta$ peaks where $\cos\theta = \sin\theta$ , i.e. $\theta = 45^\circ$ .

Step 4. The headline: crossed filters that pass nothing can be made to pass one-eighth of the incident light by inserting a filter between them. Removing that middle filter drops the output straight back to zero. Timing note: about 90 seconds. The whole problem is " $\tfrac12$ for the first filter, then $\cos^2$ of each successive angle." Do not forget the initial $\tfrac12$ , and always measure each angle from the light's current polarization, not from a fixed axis.

Exam traps

Using $\cos^2\theta$ at the first filter. Unpolarized light halves — $I_0/2$ — regardless of the filter's orientation, because it carries all polarization directions equally and the average of $\cos^2$ is $\tfrac12$ . Only after the light is polarized does a definite $\theta$ exist for Malus's law. This single confusion sinks more marks than any other in the topic.

Thinking crossed filters can never pass light. Two crossed filters pass zero, but a third filter placed between them at any angle other than $0^\circ$ or $90^\circ$ lets light through — $I_u/8$ at the optimal $45^\circ$ . The middle filter re-projects (rotates) the polarization, so the outer pair is no longer crossed across a single step.

Measuring the analyser angle from the wrong reference. In a multi-filter stack, the $\theta$ in each Malus factor is the angle between the light's current polarization direction (set by the previous filter) and the next axis — not the angle to the original first filter. Chain the angles filter by filter.

Claiming sound or any longitudinal wave can be polarized. It cannot: a longitudinal oscillation is along the travel direction and has no transverse "sideways" to block. Polarization is precisely the experiment that distinguishes transverse light from longitudinal sound.

Board scope. Polarization and Malus's law are examined by CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. They are a structural zero on IB (both SL and HL) and on A-Level 9702 — those syllabuses do not cover polarization at all, so do not spend IB or CIE revision on it. The current AP Physics 2 (Fall-2024 Unit 14) also omits polarization from its topic list. This is an India-facing topic; if your board is IB, A-Level or AP, you can safely skip it.

Origins · 1 minÉtienne-Louis Malus and how this was discovered

ÉM

Étienne-Louis Malus

1775–1812 · Paris

Étienne-Louis Malus was a French army engineer who followed Napoleon to Egypt and returned to make one of optics' great chance discoveries. One evening in 1808, looking through a crystal of Iceland spar at sunlight reflected from the windows of the Luxembourg Palace, he noticed the reflected light brightened and dimmed as he rotated the crystal — reflection had done to the light what a crystal does. He had found that light could be polarized by reflection, and he coined the very word "polarization." Pressing on, he measured how the transmitted intensity varied with angle and, in 1809, published the law that bears his name, $I = I_0\cos^2\theta$ — the first quantitative law of polarization and a cornerstone of modern wave optics. The Académie des Sciences awarded him its prize in 1810 for his theory of double refraction, the same body of work, and he died young, at 36, of an illness likely contracted on the Egyptian campaign.

History

1669

— Erasmus Bartholin describes double refraction in Iceland spar, the first recorded polarization effect, without explaining it.

1808

— Étienne-Louis Malus discovers polarization by reflection off the Luxembourg Palace windows and coins the term polarization.

1809

— Malus publishes his law, I = I₀cos²θ, the first quantitative description of polarized light.

1815

— David Brewster finds the angle at which reflected light is completely polarized, linking polarization to the refractive index.

1852

— William Bird Herapath discovers herapathite, the light-polarizing crystal later used to make sheet Polaroid filters.

Going further

Texts. NCERT Physics Class 12, Ch. 10 ("Wave Optics"), §10.7 on polarization and Malus's law · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves," polarization section · Halliday, Resnick and Walker, "Polarization," for the field-projection picture.

Provenance. Standard results. Recompute: first-filter transmission $I_0/2$ (average of $\cos^2$ over a full turn $= 0.5$ ); Malus $\cos^2\theta$ at $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ = 1, 0.75, 0.5, 0.25, 0$ ; two filters at $30^\circ$ : $0.375\,I_0$ ; at $60^\circ$ : $0.125\,I_0$ ; crossed pair with $45^\circ$ middle: $I_u/8 = 0.125\,I_u$ ; crossed pair with $30^\circ$ middle: $0.09375\,I_u$ ; half-intensity angle $45^\circ$ ; quarter-intensity angle $60^\circ$ ; three-filter maximum at $45^\circ$ .

Prerequisites. The Nature of Electromagnetic Waves — where the transverse electric field this article confines comes from. Leads to. Brewster's Law and Polarization by Reflection/Scattering — polarization "for free," no filter needed · the university seams Polarization of Light and Malus's Law.

Go deeper (UG). The full Stokes-parameter description of polarization states — linear, circular, elliptical, and partially polarized light — is developed in Polarization of Light, and the field-projection argument is generalised in Malus's Law. Enrichment only; nothing above is needed to follow this article.

Why does unpolarized light lose exactly half, not some other fraction?

Because unpolarized light contains every transverse polarization direction with equal weight, and each direction is reduced by $\cos^2\theta$ as it meets the filter's axis. Averaging $\cos^2\theta$ over a full circle of directions gives exactly one half — the $\cos^2$ curve spends as much "area" above the value $\tfrac12$ as below it. So the first filter always transmits $I_0/2$ , and turning the filter changes only which direction survives, not how much. The one-half is a property of the random input, not of any particular angle, which is why you must not plug an angle into Malus's law for the first filter.

Is Malus's law about the electric field or the intensity?

Both, in sequence. The filter acts on the electric field: it passes the component $E\cos\theta$ along its axis. But what your eye or a detector measures is intensity, which is proportional to the field squared. Squaring $E\cos\theta$ turns the field's single power of $\cos\theta$ into the intensity's $\cos^2\theta$ . That is the whole content of Malus's law: project the field (one $\cos\theta$ ), then square for intensity (giving $\cos^2\theta$ ). Getting this order right also explains why the maximum is a broad, flat peak near $\theta = 0$ , the intensity falls fastest around $\theta = 45^\circ$ , then flattens again as it approaches zero near $90^\circ$ .

Can you really get light through two filters that separately block it?

The two crossed filters do not each "block" the light — the first one polarizes it (passing half), and the second, being crossed, then passes none of that polarized beam. Insert a third filter at $45^\circ$ between them and it rotates the polarization halfway, so the final filter is only $45^\circ$ from the light reaching it and passes half again. The net is $I_u/8$ . Nothing is created from nothing: you started with $I_u$ , the first filter already discarded half, and the middle filter's job is to rotate what remains so the last filter no longer sees a right angle. It is one of the most memorable demonstrations in optics precisely because it defies the naive "crossed means dark, always" intuition.

Why can't sound be polarized?

Sound is a longitudinal wave: the air oscillates back and forth along the direction the wave travels, compressing and rarefying. There is no sideways oscillation to confine, so there is nothing for a "sound polarizer" to select — the concept does not apply. Light is different: its electric field oscillates across the travel direction, so it has a transverse orientation that a filter can pick out. This is exactly why polarization was such decisive evidence. Interference and diffraction happen for both transverse and longitudinal waves and so cannot tell them apart, but only a transverse wave can be polarized — so the fact that light polarizes proves it shakes sideways.

Do polarizing sunglasses work by Malus's law?

Yes, with a twist about the source. Glare reflected off a horizontal surface like water or a road is already partially polarized in the horizontal direction (that is the Brewster effect of the next article). Polarizing sunglasses fix their transmission axis vertical, so for the horizontally polarized glare the angle is near $90^\circ$ and $\cos^2\theta$ is small — the glare is cut. For the rest of the scene, which is roughly unpolarized, the lenses simply pass about half, dimming everything a little. That is why tilting your head sideways while wearing them makes glare flare back: you have rotated your axis toward the glare's polarization, raising $\cos^2\theta$ .

If my board is IB or A-Level, do I need this at all?

No. Polarization and Malus's law are not on IB (SL or HL) or on A-Level 9702 — those specifications simply do not include the topic, and the current AP Physics 2 Unit 14 also omits it. This is India-facing content, examined by CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. If you are sitting one of those, learn it thoroughly — the $I_0/2$ -then- $\cos^2\theta$ chain and the three-filter trick are reliable exam earners. If you are an IB, A-Level or AP student, you can read this for interest but spend your revision time on the interference and diffraction topics your board actually assesses.