Q.What do you understand by polarized light? When a third polaroid is rotated between two crossed polaroids, then discuss the change in the intensity of the transmitted light. OR A lamp “50 watt and 100 volt” is to be connected to AC mains of 200 volt 50 Hz. Calculate the capacity of condenser required in series of lamp.
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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. …
Polarized light has its electric-field vibrations confined to a single plane, and a polaroid transmits only the component along its own axis, so a third polaroid inserted between two crossed ones passes an intensity that depends on its orientation. …
Light with E vibrations in one plane is polarized. Between crossed polaroids, a middle polaroid at angle θ transmits I=4I0sin22θ, peaking at θ=45∘. (An OR numerical on a series capacitor is also offered.)
Polarized light. Ordinary light has electric-field vibrations in all directions perpendicular to propagation (unpolarised). Polarized light has its E-vibrations restricted to one plane. A polaroid passes only the component along its axis (Malus's law: I=I0cos2θ).
Three-polaroid experiment. Let P1 (polariser) and P3 (analyser) be crossed (axes at 90∘). Alone, they transmit zero light. Now insert a third polaroid P2 between them, with its axis at angle θ to P1.
- After P1: intensity I0 (say), polarized along P1.
- After P2 (at angle θ): I2=I0cos2θ, now polarized along P2.
- P3 is at 90∘ to P1, hence at (90∘−θ) to P2. After P3: I=I2cos2(90∘−θ)=I0cos2θsin2θ=4I0sin22θ.
Discussion as P2 is rotated.
- θ=0∘ or 90∘: I=0 (no light — as with just the crossed pair).
- θ=45∘: I=4I0 (maximum). …
- CBSE 2025Set ANNUAL1 markQ.When light from a sodium lamp passes through a polaroid sheet then its intensity becomes ____________.
›Reveal solutionSolution
Ordinary (unpolarised) light passing through a single polaroid always emerges at exactly half the incident intensity, regardless of the polaroid's orientation.
Sodium light is unpolarised — its electric field vibrates randomly in all directions perpendicular to propagation. When such light of intensity I0 passes through a single polaroid, the polaroid transmits only the component of vibration along its transmission axis. Averaging over all random orientations of the incident field gives a transmitted intensity of exact …
- CBSE 2025Set ANNUAL1 markMCQQ.In Sunglasses and 3D movie cameras, polaroids are used to control:(a) frequency of light(b) intensity of light(c) wavelength of light(d) both(a) and (b).
›Reveal solutionSolution
Polaroids work by selectively absorbing one plane of polarisation, which controls the intensity of transmitted light.
A polaroid consists of long chain molecules aligned in a particular direction; it transmits light whose electric vector is parallel to this pass-axis and absorbs the perpendicular component. When two polaroids are used together, rotating one relative to the other varies the transmitted intensity according to Malus's law, I=I0cos2θ. In sunglasses, polaroids cut down glare (reduce the intensity of reflected/scattered light); in 3D movie syst …
- CBSE 2023Set ANNUAL1 markMCQQ.Two polaroids are kept with their transmission axes inclined at 30°. Unpolarised light of intensity I falls on the first polaroid. Intensity of light emerging from the second polaroid :(a) 81I(b) 41I(c) 83I(d) 43I
›Reveal solutionSolution
Unpolarised light is halved by the first polaroid, then reduced further by cos2θ (Malus' law) through the second polaroid at 30°, giving 3I/8.
Working
Step 1 — first polaroid: Unpolarised light of intensity I falling on a polaroid always transmits half its intensity as plane-polarised light:
I1=2I
Step 2 — second polaroid: This linearly polarised light of intensity I1 now falls on the second polaroid, whose transmission axis is inclined at θ=30° to the first. By Malus' law: …
- CBSE 2022Set ANNUAL1 markQ.How should a polarizer and an analyzer be placed so that intensity is maximum?
›Reveal solutionSolution
By Malus's law, I=I0cos2θ, transmitted intensity is maximum when the analyzer's pass axis is parallel to the polarizer's.
When unpolarized light passes through a polarizer, it becomes linearly polarized along the polarizer's transmission axis, with intensity I0 (half the incident intensity). If this polarized light then falls on an analyzer whose pass axis makes an angle θ with the polarizer's axis, Malus's law gives the transmitted intensity as
I=I0cos2θ …
- CBSE 2019Set ANNUAL1 markQ.The figure shows four pairs of polaroids, seen from front. Each pair is mounted in the path of initially unpolarized light. The axis of polaroids is shown with dotted lines. Through which pair the intensity of light seen is :(i) maximum ?(ii) minimum ?
›Reveal solutionSolution
Intensity transmitted through a pair of polaroids follows Malus's law, I=I0cos2θ, where θ is the angle between the two polaroid axes — maximum when the axes are parallel, minimum (zero) when they are perpendicular.
Unpolarized light first becomes polarized (and halved in intensity, I0/2) after passing through the first (upper) polaroid. Passing through the second (lower) polaroid then attenuates it further according to Malus's law:
I=2I0cos2θ
where θ is the angle between the transmission axes of the upper and lower polaroids.
In each of the four pairs (a)–(d), the lower polaroid's axis is marked at a fixed angle and the upper polaroid's axis is rotated to a different absolute orientation, which changes θ (and hence the relative angle between the two axes) from pair to pair. To identify the answer:
- The pair where the two marked axis lines are CLOSEST to being parallel to each other (smallest θ, ideally θ→0°) transmits the MAXIMUM intensity (I→I0/2, its largest possible value).
- The pair where the two marked axis lines are CLOSEST to being perpendicular (θ→90°) transmits the MINIMUM intensity (I→0). …
- CBSE 2018Set ANNUAL1 markQ.Write Malus law.
›Reveal solutionSolution
The intensity of polarised light passing through an analyser is I = I₀ cos²θ.
When plane-polarised light of intensity I0 falls on an analyser (a polaroid), only the component of its electric field along the analyser's transmission axis is passed.
If θ is the angle between the plane of polarisation of the incident light and the analyser's axis, the transmitted amplitude is E0cosθ, and since intensity is proportional to the square of amplitude: …
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