(a) Good quality sunglasses made of polaroids are preferred over ordinary coloured glasses. Explain why. (b) H

Question

Q. (a) Good quality sunglasses made of polaroids are preferred over ordinary coloured glasses. Explain why.

(b) How is plane polarized light defined?

(c) A beam of plane polarised light is passed through a polaroid. Show graphically, variation of the intensity of the transmitted light with angle of rotation of the polaroid.

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★

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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 $\theta$ relative to the first.

When $\theta = 0^\circ$ (both aligned), all the polarised light passes through — maximum intensity.
When $\theta = 90^\circ$ (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 $E_0 \cos\theta$ , where $E_0$ is the amplitude of the incident polarised light. Since intensity $I$ is proportional to the square of the amplitude ( $I \propto E^2$ ), the transmitted intensity becomes:

I = I_0 \cos^2 \theta

where $I_0$ is the intensity of the light incident on the second polariser (i.e., after the first polariser).

I = I_0 \cos^2 \theta

The Precise Statement

Malus Law states: When completely plane-polarised light of intensity $I_0$ 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 $\theta$ 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 ( $I_0/2$ ), and then Malus Law applies to that reduced intensity.
$\theta$ is the angle between the two transmission axes, not the angle of incidence or any other angle.
The result is always $I \leq I_0$ , with equality only at $\theta = 0^\circ$ or $180^\circ$ .

Watch out

A common mistake: applying Malus Law directly to unpolarised light. Unpolarised light has no fixed $\theta$ , so you cannot use $\cos^2\theta$ on it. First, pass it through a polariser to get $I_0/2$ , then apply Malus Law.

A Quick Example

Unpolarised light of intensity $100 \, \text{W/m}^2$ passes through two polarisers whose axes are at $60^\circ$ to each other. What is the final intensity? …

Why this formula?

Malus Law: Why Intensity Varies as $\cos^2 \theta$

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 $\theta$ 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 $E_0$ (along the polarizer's axis).

When this field reaches the analyzer at angle $\theta$ :

Only the component of $E_0$ parallel to the analyzer's axis passes through.
That component is:

E_{\text{transmitted}} = E_0 \cos \theta

Why $\cos \theta$ ?

Because the electric field is a vector. The projection of $E_0$ onto the analyzer's axis is $E_0 \cos \theta$ — 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 \propto E^2

So:

Incident intensity: $I_0 \propto E_0^2$
Transmitted intensity: $I \propto (E_0 \cos \theta)^2 = E_0^2 \cos^2 \theta$

Therefore:

\boxed{I = I_0 \cos^2 \theta}

This is Malus Law.

5. Why the Square? — Physical Meaning

If $\theta = 0^\circ$ : $\cos^2 0 = 1$ → maximum intensity (all light passes).
If $\theta = 90^\circ$ : $\cos^2 90^\circ = 0$ → zero intensity (crossed polarizers, no light).
If $\theta = 45^\circ$ : $\cos^2 45^\circ = \frac{1}{2}$ → half intensity.

The $\cos^2$ 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 $\cos \theta$ ?

A common mistake is to think intensity varies as $\cos \theta$ . But:

Amplitude varies as $\cos \theta$ (field component).
Intensity (energy) varies as $(\cos \theta)^2$ because energy ∝ (amplitude) $^2$ . …

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Polaroid sunglasses block glare by selectively absorbing light polarised parallel to the surface (horizontal), while ordinary coloured glasses only dim all light equally. Plane polarised light has its electric field confined to one plane. The transmitted intensity through a polaroid follows Malus’ law: $I = I_0 \cos^2\theta$ , giving a $\cos^2$ curve.

Why polaroid sunglasses are better than ordinary coloured glasses The key difference lies in what they block. Ordinary coloured glasses simply absorb a fixed fraction of light across all directions — they reduce overall brightness but do nothing to remove glare. Glare, especially from flat surfaces like water, roads, or snow, is largely horizontally polarised light. Sunlight reflecting off a horizontal surface becomes polarised with its electric field oscillating mostly parallel to the surface (horizontal). Polaroid sunglasses are made with their transmission axis oriented vertically. This means they strongly absorb the horizontally polarised glare while still transmitting vertically polarised light (which carries useful visual information). The result: glare is cut dramatically, contrast improves, and colours appear richer — something no ordinary tinted lens can achieve.
Watch out

A common mistake is thinking polaroids block all polarised light. They only block light polarised perpendicular to their transmission axis. Rotate the sunglasses by 90°, and the glare comes right through.
Definition of plane polarised light Plane polarised light (also called linearly polarised light) is a transverse electromagnetic wave in which the electric field vector oscillates in a single, fixed plane containing the direction of propagation. All other orientations of the electric field are absent. $z$
Variation of transmitted intensity with angle of rotation When plane polarised light of intensity $I_0$ passes through a polaroid (analyser), the transmitted intensity depends on the angle $\theta$ between the polarisation direction of the incident light and the transmission axis of the polaroid. This is Malus’ law: $I = I_0 \cos^2 \theta$ Here’s the step-by-step reasoning:

At $\theta = 0^\circ$ : The polarisation direction is perfectly aligned with the transmission axis. All the light passes through: $I = I_0$ .
As $\theta$ increases: Only the component of the electric field parallel to the transmission axis is transmitted. That component is $E_0 \cos\theta$ , so intensity (proportional to $E^2$ ) becomes $I_0 \cos^2\theta$ .
At $\theta = 90^\circ$ : The polarisation is perpendicular to the transmission axis. No light passes: $I = 0$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 mark
$Q. When light from a sodium lamp passes through a polaroid sheet then its intensity becomes ____________.$
›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 $I_0$ 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 …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In Sunglasses and 3D movie cameras, polaroids are used to control:$
(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, $I = I_0\cos^2\theta$ . In sunglasses, polaroids cut down glare (reduce the intensity of reflected/scattered light); in 3D movie syst …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Two polaroids are kept with their transmission axes inclined at 30°. Unpolarised light of intensity I falls on the first polaroid. Intensity of light emerging from the second polaroid :$
(a) $\dfrac{1}{8}I$
(b) $\dfrac{1}{4}I$
(c) $\dfrac{3}{8}I$
(d) $\dfrac{3}{4}I$
›Reveal solutionSolution

Unpolarised light is halved by the first polaroid, then reduced further by $\cos^2\theta$ (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:

$I_1 = \dfrac{I}{2}$

Step 2 — second polaroid: This linearly polarised light of intensity $I_1$ now falls on the second polaroid, whose transmission axis is inclined at $\theta = 30°$ to the first. By Malus' law: …

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CBSE 2022Set ANNUAL1 mark
$Q. How should a polarizer and an analyzer be placed so that intensity is maximum?$
›Reveal solutionSolution

By Malus's law, $I=I_0\cos^2\theta$ , 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 $I_0$ (half the incident intensity). If this polarized light then falls on an analyzer whose pass axis makes an angle $\theta$ with the polarizer's axis, Malus's law gives the transmitted intensity as

$I = I_0 \cos^2\theta$ …

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CBSE 2019Set ANNUAL1 mark
$Q. The figure shows four pairs of polaroids, seen from front. Each pair is mounted in the path of initially unpolarized light. The axis of polaroids is shown with dotted lines. Through which pair the intensity of light seen is :$
(i) maximum ?
(ii) minimum ?
›Reveal solutionSolution
Intensity transmitted through a pair of polaroids follows Malus's law, $I=I_0\cos^2\theta$ , where $\theta$ 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, $I_0/2$ ) after passing through the first (upper) polaroid. Passing through the second (lower) polaroid then attenuates it further according to Malus's law:

$I = \frac{I_0}{2}\cos^2\theta$

where $\theta$ 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 $\theta$ (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 $\theta$ , ideally $\theta \to 0°$ ) transmits the MAXIMUM intensity ( $I \to I_0/2$ , its largest possible value).

The pair where the two marked axis lines are CLOSEST to being perpendicular ( $\theta \to 90°$ ) transmits the MINIMUM intensity ( $I \to 0$ ). …
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CBSE 2018Set ANNUAL1 mark
$Q. Write Malus law.$
›Reveal solutionSolution

The intensity of polarised light passing through an analyser is I = I₀ cos²θ.

When plane-polarised light of intensity $I_0$ falls on an analyser (a polaroid), only the component of its electric field along the analyser's transmission axis is passed.

If $\theta$ is the angle between the plane of polarisation of the incident light and the analyser's axis, the transmitted amplitude is $E_0\cos\theta$ , and since intensity is proportional to the square of amplitude: …

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Q.(a) Good quality sunglasses made of polaroids are preferred over ordinary coloured glasses. Explain why.

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