Q.(a) When an unpolarized light of intensity Io is passed through a polaroid, what is the intensity of the linearly polarized light? Does it depend on the orientation of the polaroid? Explain your answer.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Malus Law
Malus Law: How Light Gets Weaker Through a Polariser
Imagine you're trying to push a rope through a narrow fence. If the rope is aligned with the gap, it passes through easily. If you twist the rope sideways, it gets blocked. Light behaves similarly — it's a transverse wave, meaning its electric field oscillates in a direction perpendicular to its travel. A polariser is like that fence: it only lets through light whose electric field oscillates in one specific direction (its "pass axis").
Now, what happens when you take already-polarised light and send it through a second polariser? That's exactly what Malus Law describes.
The Intuition
Suppose you have two polarisers. The first one takes ordinary (unpolarised) light and makes it polarised along some direction. The second polariser is rotated by an angle θ relative to the first.
- When θ=0∘ (both aligned), all the polarised light passes through — maximum intensity.
- When θ=90∘ (crossed), no light passes through — zero intensity.
- For any angle in between, only the component of the electric field that lies along the second polariser's axis gets through.
That component is E0cosθ, where E0 is the amplitude of the incident polarised light. Since intensity I is proportional to the square of the amplitude (I∝E2), the transmitted intensity becomes:
I=I0cos2θ
where I0 is the intensity of the light incident on the second polariser (i.e., after the first polariser).
I=I0cos2θ
The Precise Statement
Malus Law states: When completely plane-polarised light of intensity I0 is incident on an analyser (a polariser), the intensity I of the transmitted light is proportional to the square of the cosine of the angle θ between the transmission axes of the polariser and the analyser.
Key points to remember for exams:
- The law applies only when the incident light is already fully polarised. If the light is unpolarised, the first polariser reduces its intensity by half (I0/2), and then Malus Law applies to that reduced intensity.
- θ is the angle between the two transmission axes, not the angle of incidence or any other angle.
- The result is always I≤I0, with equality only at θ=0∘ or 180∘.
A common mistake: applying Malus Law directly to unpolarised light. Unpolarised light has no fixed θ, so you cannot use cos2θ on it. First, pass it through a polariser to get I0/2, then apply Malus Law.
A Quick Example
Unpolarised light of intensity 100W/m2 passes through two polarisers whose axes are at 60∘ to each other. What is the final intensity? …
Why this formula?
Malus Law: Why Intensity Varies as cos2θ
Malus Law describes how the intensity of polarized light changes when it passes through a second polarizer (called an analyzer). Let's build the understanding step-by-step.
1. What Does Polarized Light Look Like?
- Unpolarized light has electric field vectors vibrating in all directions perpendicular to propagation.
- After passing through a polarizer, only the component of the electric field parallel to the polarizer's transmission axis survives.
- The result: linearly polarized light — the electric field oscillates in a single plane.
2. The Setup for Malus Law
Imagine:
- A polarizer (first filter) produces vertically polarized light.
- An analyzer (second filter) has its transmission axis at an angle θ to the vertical.
The key question: How much light gets through the analyzer?
3. The Core Reasoning: Electric Field Components
The incident polarized light has an electric field amplitude E0 (along the polarizer's axis).
When this field reaches the analyzer at angle θ:
- Only the component of E0 parallel to the analyzer's axis passes through.
- That component is:
Etransmitted=E0cosθ
Why cosθ?
Because the electric field is a vector. The projection of E0 onto the analyzer's axis is E0cosθ — just like resolving a force into components.
4. From Amplitude to Intensity
Intensity I is proportional to the square of the amplitude of the electric field:
I∝E2
So:
- Incident intensity: I0∝E02
- Transmitted intensity: I∝(E0cosθ)2=E02cos2θ
Therefore:
I=I0cos2θ
This is Malus Law.
5. Why the Square? — Physical Meaning
- If θ=0∘: cos20=1 → maximum intensity (all light passes).
- If θ=90∘: cos290∘=0 → zero intensity (crossed polarizers, no light).
- If θ=45∘: cos245∘=21 → half intensity.
The cos2 dependence arises because intensity is energy per unit time, and energy is proportional to the square of the field amplitude — not the amplitude itself.
6. Key Insight: Why Not cosθ?
A common mistake is to think intensity varies as cosθ. But:
- Amplitude varies as cosθ (field component).
- Intensity (energy) varies as (cosθ)2 because energy ∝ (amplitude)2. …
An ideal polaroid transmits only the component of the field along its axis, so unpolarised light (all vibration directions equally likely) always emerges at half intensity, independent of orientation; a plane-polarised beam then follows Malus's law. …
Unpolarised light → I=I0/2 (orientation-independent); polarised beam obeys Malus's law I=I0cos2θ, a curve peaking every 180∘.
(a) Unpolarised light has electric-field vibrations in all directions perpendicular to propagation with equal probability. A polaroid passes only the component along its pass-axis. Averaging over all angles, ⟨cos2θ⟩=21, so the transmitted (now linearly polarised) intensity is
I=2I0.
Rotating the polaroid does not change this value — every orientation still faces an isotropic incident beam. (The emerging light is, however, polarised along whichever direction the axis points.)
…
- AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQQ.The percentage of decrease in the intensity of polarized light when it is passed through an analyser at an angle of 60° is (A) 25% (B) 50% (C) 75% (D) 60%
›Reveal solutionSolution
This tests Malus's Law for intensity of polarized light through an analyser. Answer: 75% decrease in intensity at 60°.
Concept and Intuition
When already-polarized light passes through an analyser (a second polarizer) whose transmission axis is at angle θ to the light's plane of polarization, only the component of the electric field along the analyser's axis is transmitted. Since intensity depends on the square of the amplitude, the transmitted intensity follows I=I0cos2θ (Malus's Law).
Step-by-Step Solution
- Malus's Law: I=I0cos2θ, where I0 is incident (already polarized) intensity and θ is the angle between polarization direction and analyser axis.
- Given θ=60°: cos60°=0.5, so cos260°=0.25.
- Transmitted intensity: I=0.25I0. …
- AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQQ.An unpolarised beam of light incidents on a group of three polarising sheets arranged such that the angle between the axes of any two adjacent sheets is 30°. The ratio of the intensities of polarised light emerging from the second and third sheets is (A) 1:1 (B) 2:1 (C) 4:3 (D) 3:2
›Reveal solutionSolution
Malus's law applied twice (30° between each successive pair of sheets) gives the intensity ratio after the 2nd and 3rd sheets as 4:3.
Concept and Intuition
An unpolarised beam loses half its intensity at the very first polariser it meets (this converts it into fully polarised light). Every subsequent polariser then obeys Malus's law, I=I0cos2θ, where θ is the angle between the incoming polarisation direction and the new polariser's axis. Since each adjacent pair here differs by the same 30°, going from sheet 2 to sheet 3 multiplies intensity by the same factor cos230°, so the ratio of intensities emerging from sheet 2 and sheet 3 is simply 1:cos230°.
Step-by-Step Solution
- After sheet 1 (unpolarised → polarised): I1=2I0.
- After sheet 2 (30° from sheet 1's axis): I2=I1cos2(30°)=I1×43.
- After sheet 3 (30° from sheet 2's axis): I3=I2cos2(30°)=I2×43. …
- AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQQ.In an experiment, two polariods are arranged such that the intensity of the polarised light emerged from the second polaroid is 37.5% of the intensity of the unpolarised light incident on the first polaroid. Then the angle between the axes of the two polaroids is (A) 60° (B) 90° (C) 45° (D) 30°
›Reveal solutionSolution
This tests Malus's law applied after an initial unpolarised-to-polarised step; the answer is (D) 30°.
Concept and Intuition
Unpolarised light passing through the first polaroid always loses exactly half its intensity (by symmetry over all polarisation directions), regardless of the polaroid's orientation. The resulting linearly polarised light then obeys Malus's law, I=I1cos2θ, when passing through a second polaroid at angle θ to the first.
Step-by-Step Solution
- After the first polaroid: I1=2I0 (standard result for unpolarised light through one polaroid).
- After the second polaroid, by Malus's law: I2=I1cos2θ=2I0cos2θ.
- We are given I2=0.375I0, so 2I0cos2θ=0.375I0⇒cos2θ=0.75. …
- AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQQ.The angle between the axes of a polariser and an analyser is 45∘. If the intensity of the unpolarized light incident on the polariser is I, then the intensity of the light emerged from the analyser is (A) 2I (B) 2I (C) I (D) 4I
›Reveal solutionSolution
Unpolarized light is first halved in intensity by the polarizer, then reduced further by cos2θ (Malus's law) passing through the analyzer at 45∘.
Concept and Intuition
Unpolarized light contains all possible directions of oscillation with equal probability; a polarizer always transmits exactly half the incident intensity regardless of its orientation, producing linearly polarized light along its own transmission axis. That polarized light then encounters the analyzer, whose transmitted intensity follows Malus's law, I=I0cos2θ, where θ is the angle between the polarizer's and analyzer's axes.
Step-by-Step Solution
- Intensity after the polarizer (unpolarized input I): I1=2I.
- Angle between polarizer and analyzer axes: θ=45∘. …
- AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQQ.The relation I=I0cos2θ is (I0 – Intensity of incident light on the analyser, I – intensity of emergent light from the analyser, θ – angle between plane of polarization and the axis of analyser) (A) Newton's law (B) Snell's law (C) Brewster's law (D) Malus's law
›Reveal solutionSolution
This is a direct factual identification: I=I0cos2θ for light through a polarizer/analyser is Malus's law.
Concept and Intuition
When plane-polarized light of intensity I0 passes through an analyser whose transmission axis makes angle θ with the plane of polarization, only the component of the electric field along the analyser's axis is transmitted. Since intensity is proportional to the square of the field amplitude, and the transmitted amplitude scales as cosθ, the transmitted intensity scales as cos2θ — this named relationship is Malus's law.
Step-by-Step Solution
- The transmitted electric field amplitude through the analyser is E=E0cosθ (only the component along the analyser axis passes through).
- Since intensity ∝ amplitude2: I=I0cos2θ.
- This specific relation is named after Étienne-Louis Malus, who discovered polarization by reflection and this intensity law — hence "Malus's law". …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.