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.
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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. …
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.)
…
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.If unpolarized light is incident on polaroid, then intensity of emergent light is ______ of the intensity of incident light.(a) Double(b) Half(c) Four times(d) One fourth
›Reveal solutionSolution
Unpolarized light can be thought of as a mix of all polarization directions; a polaroid transmits only the component along its axis.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.The intensity of incident unpolarized light on a polaroid is I_1 and the intensity of emergent polarized light from this polaroid is I_2. The relation between I_1 & I_2 is ___.(a) I_1 < I_2(b) I_1 > I_2(c) I_1 = I_2(d) I_1 = 2 I_2
›Reveal solutionSolution
A polaroid passes on average half of unpolarized light, so I_2 = I_1/2, meaning I_1 > I_2.
Unpolarized light has all vibration directions equally. A polaroid transmits only the component along its axis, which averages to half the incident intensity: …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.The amount of Rayleigh scattering is ___.(a) inversely proportional to fourth power of wavelength(b) inversely proportional to wavelength(c) directly proportional to wavelength(d) directly proportional to fourth power of wavelength
›Reveal solutionSolution
Rayleigh scattering by particles much smaller than the wavelength varies as 1/lambda^4, so short (blue) wavelengths scatter far more than long (red) ones.
For scattering by particles small compared with the wavelength (Rayleigh scattering), the scattered intensity is proportional to 1/lambda^4.
…
- GUJCET 2022Set 171 markMCQQ.Unpolarised light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other? (A) 56° (B) 57° (C) 58° (D) 59°
›Reveal solutionSolution
[!TLDR] Brewster angle θB=tan−1(1.5)≈56∘.
Concept
At Brewster's angle the reflected and refracted rays are mutually perpendicular and the reflected light is completely plane-polarised; tanθB=n (NCERT Wave Optics).
Solution …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.Stokes and antistokes lines observed in Raman scattering is due to ___ of light.(a) reflection(b) elastic scattering(c) inelastic scattering(d) dispersion
›Reveal solutionSolution
Raman Stokes/anti-Stokes lines appear because photons scatter inelastically, gaining or losing energy to the molecule's vibrational/rotational states.
In Raman scattering, most light scatters elastically (Rayleigh line, no energy change). A small fraction scatters inelastically, exchanging energy with the molecule:
- Stokes lines: the photon loses energy to the molecule (lower frequency). …
- GUJCET 2015Set C1 markMCQQ.A plane polarized light is incident normally on a tourmaline plate. Its E vectors make an angle of 60∘ with the optic axis of the plate. Find the percentage difference between initial and final intensities. (A) 50% (B) 25% (C) 75% (D) 90%
›Reveal solutionSolution
[!TLDR] I=I0cos260∘=0.25I0, so the drop from the initial intensity is 75%.
Concept
Malus's law: when plane-polarised light of intensity I0 passes through a polariser (here the tourmaline plate) whose transmission axis makes angle θ with the light's polarisation (E-vector), the transmitted intensity is I=I0cos2θ.
Solution …
- GUJCET 2014Set A1 markMCQQ.Unpolarized light falls first on polarizer (P) and then on analyzer (A). If the intensity of the transmitted light from the analyzer is 81th of the incident unpolarized light. What will be the angle between optic axes of P & A? (A) 30° (B) 45° (C) 0° (D) 60°
›Reveal solutionSolution
[!TLDR] 21cos2θ=81⇒cosθ=21⇒θ=60∘.
Concept
Unpolarised light passing a polariser becomes half-intensity; a following analyser obeys Malus's law I=IPcos2θ (NCERT Wave Optics).
Solution
After the polariser: IP=2I0.
After the analyser at angle θ: I=2I0cos2θ. …
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