A polariod (I) is placed in front of a monochromatic source. Another polatiod (II) is placed in front of this

NCERT Exemplar · Q14

Q. A polariod (I) is placed in front of a monochromatic source. Another polatiod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.

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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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When two crossed polaroids (I and II) block all light, inserting a third polaroid (III) at an intermediate angle allows some light to pass through all three. The final intensity is $I_0/8$ if III is at $45^\circ$ to both I and II.

The key insight is that polaroids work by transmitting only the component of light’s electric field aligned with their transmission axis. When two polaroids are crossed (axes at $90^\circ$ ), the first polaroid (I) produces linearly polarised light, and the second (II) blocks it completely because the electric field has zero component along its axis.

Now, inserting a third polaroid (III) between them changes the story. Even though I and II remain crossed, III can rotate the polarisation direction partially, allowing some light to reach II. This is a classic demonstration that polarisation is a vector phenomenon — you can’t just “block” light in one step if an intermediate axis exists.

Let’s work through it step by step.

Set up the axes.

Let the transmission axis of polaroid I be vertical ( $0^\circ$ ). Polaroid II is rotated to $90^\circ$ (horizontal) so that no light passes when only I and II are present. The source is monochromatic and unpolarised, but after I, the light is vertically polarised with intensity $I_0/2$ (since an ideal polaroid transmits half the intensity of unpolarised light).
Insert polaroid III at some angle $\theta$ .

Place III between I and II with its transmission axis at an angle $\theta$ to the vertical. The light emerging from I is vertically polarised. When it hits III, Malus’s law gives the intensity after III:

I_{III} = \frac{I_0}{2} \cos^2 \theta.

The light is now polarised along the axis of III (at angle $\theta$ ).

Light then passes through II. Polaroid II has its axis at $90^\circ$ (horizontal). The angle between the polarisation direction of light from III ( $\theta$ ) and the axis of II is $90^\circ - \theta$ . Applying Malus’s law again:

I_{final} = I_{III} \cos^2(90^\circ - \theta) = \frac{I_0}{2} \cos^2 \theta \cdot \sin^2 \theta.

Simplify the expression. Using $\cos^2 \theta \sin^2 \theta = \frac{1}{4} \sin^2 2\theta$ , we get: $I_{final} = \frac{I_0}{8} \sin^2 2\theta.$ …

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

TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
$4:1$
›Reveal solutionSolution
The key is to orient the middle polaroid at 45° to the first to maximize transmitted intensity through crossed polarizers; the ratio of intensity incident on the middle polaroid to the final emergent intensity is 2:1.

We have three polaroids: Q (first), P (middle), and R (last). Q and R are crossed (their transmission axes are perpendicular). Unpolarized light first hits Q. We place P between them so that the final intensity from R is maximum. We need the ratio: (intensity of polarized light incident on P) : (intensity of polarized light emerging from R).

Concept & Intuition

When unpolarized light passes through a polaroid, its intensity halves. Then, when polarized light passes through another polaroid at an angle θ, the transmitted intensity follows Malus’s law: $I = I_0 \cos^2\theta$ .

If Q and R are crossed, light from Q is polarized along Q’s axis; to get any light through R, we must rotate the polarization using P. The maximum final intensity occurs when P is at 45° to Q (and thus also 45° to R, since Q ⟂ R). This gives the largest possible product of two cos² factors.

Step-by-step reasoning

Light through Q Unpolarized light of initial intensity $I_0$ falls on Q. After Q, the light becomes polarized along Q’s axis, and its intensity is

$I_Q = \frac{I_0}{2}.$
This is the intensity of polarized light incident on P. So the numerator of our ratio is $\frac{I_0}{2}$ .

Light through P Let the angle between Q’s axis and P’s axis be θ. Then the intensity after P is

$I_P = I_Q \cos^2\theta = \frac{I_0}{2} \cos^2\theta.$
This light is now polarized along P’s axis.

Light through R R’s axis is perpendicular to Q’s axis. Since P makes angle θ with Q, it makes angle $90^\circ - \theta$ with R. The intensity after R is

$I_R = I_P \cos^2(90^\circ - \theta) = I_P \sin^2\theta.$
Substituting $I_P$ :
$I_R = \frac{I_0}{2} \cos^2\theta \sin^2\theta = \frac{I_0}{2} \cdot \frac{1}{4} \sin^2(2\theta) = \frac{I_0}{8} \sin^2(2\theta).$

Maximizing $I_R$ The maximum of $\sin^2(2\theta)$ is 1, achieved when $2\theta = 90^\circ$ i.e. $\theta = 45^\circ$ . So the maximum possible intensity from R is
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TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$45^\circ$
›Reveal solutionSolution
When a polaroid C is placed between two crossed polaroids A and B, the intensity of light transmitted through B is maximized when the pass axis of C is at $45^\circ$ to the pass axis of A. The maximum intensity is $\boxed{45^\circ}$ .

The problem describes a setup with three polaroids: A, C, and B. Polaroids A and B are "crossed," meaning their pass axes are perpendicular to each other. Polaroid C is rotated between them. We need to find the angle of C relative to A that maximizes the light transmitted through B.

The core concept here is Malus's Law, which describes how the intensity of plane-polarized light changes when it passes through a polaroid.

Understanding Polarization and Malus's Law

Polaroids and Pass Axis: A polaroid is a material that transmits light waves oscillating in a specific plane (or direction) and blocks waves oscillating in all other planes. This specific direction is called the pass axis (or transmission axis) of the polaroid.

Unpolarized Light: Natural light (like sunlight or light from a bulb) is unpolarized, meaning its electric field vectors oscillate randomly in all possible directions perpendicular to the direction of propagation.

Plane-Polarized Light: When unpolarized light passes through a polaroid, the transmitted light becomes plane-polarized. Its electric field vectors now oscillate only along the pass axis of that polaroid. The intensity of unpolarized light is halved upon passing through the first polaroid.

Malus's Law: If plane-polarized light of intensity $I_0$ is incident on a polaroid whose pass axis makes an angle $\theta$ with the plane of polarization of the incident light, the intensity $I$ of the transmitted light is given by: $I = I_0 \cos^2 \theta$

Applying Malus's Law to the Setup

Let's denote the intensity of light after each polaroid.

Light through Polaroid A:

Let the incident light be unpolarized. When it passes through polaroid A, it becomes plane-polarized along the pass axis of A. Let's assume the pass axis of A is along the $0^\circ$ direction (our reference).

Let the intensity of the plane-polarized light emerging from A be $I_A$ . (If the incident unpolarized light has intensity $I_{unpol}$ , then $I_A = I_{unpol}/2$ ). For simplicity, we will use $I_A$ as our starting intensity for the subsequent calculations.

Light through Polaroid C:

Polaroid C is rotated such that its pass axis makes an angle $\theta$ with the pass axis of polaroid A. The light incident on C is plane-polarized along A's pass axis (at $0^\circ$ ) with intensity $I_A$ .

According to Malus's Law, the intensity of light emerging from C, $I_C$ , will be:

$I_C = I_A \cos^2 \theta$
The light emerging from C is now plane-polarized along the pass axis of C, which is at an angle $\theta$ from A's pass axis.

3. Light through Polaroid B:

Polaroids A and B are "crossed." This means their pass axes are perpendicular. Since A's pass axis is at $0^\circ$ , B's pass axis must be at $90^\circ$ .

The light incident on B is plane-polarized along C's pass axis (at angle $\theta$ ) with intensity $I_C$ .

The angle between the pass axis of C (at $\theta$ ) and the pass axis of B (at $90^\circ$ ) is $(90^\circ - \theta)$ .

Applying Malus's Law again, the intensity of light emerging from B, $I_B$ , will be:
$I_B = I_C \cos^2 (90^\circ - \theta)$
Substitute the expression for $I_C$: …
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TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$60^\circ$
›Reveal solutionSolution
When unpolarized light passes through a polarizer, its intensity is halved. When this polarized light then passes through an analyzer set at $60^\circ$ to the polarizer's axis, its intensity is further reduced by a factor of $\cos^2(60^\circ)$ . The final intensity is $1/8$ th of the initial unpolarized intensity, making the ratio $8:1$ .

The problem asks for the ratio of the initial intensity of unpolarized light to the final intensity after passing through a polarizer and then an analyzer. This involves understanding how the intensity of light changes as it interacts with these optical components.

Concept and Intuition

Unpolarized Light and Polarizer:

Unpolarized light consists of electromagnetic waves where the electric field vectors oscillate randomly in all possible directions perpendicular to the direction of propagation. A polarizer is a device that allows only electric field oscillations parallel to a specific direction, called its transmission axis, to pass through.

When unpolarized light of intensity $I_0$ passes through an ideal polarizer, the light that emerges is plane-polarized. Since the electric field oscillations in unpolarized light are randomly oriented, on average, only half of the incident intensity is transmitted through the polarizer. The other half is absorbed or reflected.

Important

The intensity of unpolarized light after passing through an ideal polarizer is exactly half of its initial intensity.

Polarized Light and Analyzer (Malus's Law):

An analyzer is essentially another polarizer. When plane-polarized light (from the first polarizer) is incident on an analyzer, the intensity of the transmitted light depends on the angle between the plane of polarization of the incident light and the transmission axis of the analyzer.

Let $I_1$ be the intensity of the plane-polarized light incident on the analyzer, and let $\theta$ be the angle between the plane of polarization of this light (which is defined by the transmission axis of the first polarizer) and the transmission axis of the analyzer.

The electric field vector of the incident polarized light can be resolved into two components: one parallel to the analyzer's transmission axis ( $E_1 \cos\theta$ ) and one perpendicular to it ( $E_1 \sin\theta$ ). Only the parallel component is transmitted.

Since intensity is proportional to the square of the electric field amplitude ( $I \propto E^2$ ), the transmitted intensity $I_2$ will be proportional to $(E_1 \cos\theta)^2$ .
$I_2 = I_1 \cos^2 \theta$

Step-by-Step Solution

Intensity after the Polarizer: Let the intensity of the unpolarized light incident on the polarizer be $I_0$ . When this unpolarized light passes through the polarizer, its intensity is halved. So, the intensity of the plane-polarized light emerging from the polarizer, $I_1$ , is:

$I_1 = \frac{I_0}{2}$

Intensity after the Analyzer: …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$45^\circ$
›Reveal solutionSolution
Unpolarised light passes through two polaroids at $45^\circ$ ; Malus’s law gives the final intensity as $I_0/4$ , so the ratio is $1:4$ .

The key idea here is that unpolarised light, when passed through a polaroid, loses half its intensity regardless of the polaroid’s orientation. Then, the second polaroid further reduces the intensity according to Malus’s law, which depends on the cosine squared of the angle between the transmission axes.

Let’s walk through it step by step.

First polaroid (P) – unpolarised to polarised When unpolarised light of intensity $I_0$ falls on a polaroid, the transmitted light is polarised along the axis of the polaroid, and its intensity becomes exactly half:

$I_1 = \frac{I_0}{2}$
This is because only the component of the electric field parallel to the axis passes through, and averaging over all directions gives a factor of $1/2$ .

Second polaroid (Q) – Malus’s law The light emerging from P is now linearly polarised. When it meets Q, whose transmission axis is at an angle $\theta = 45^\circ$ to that of P, the transmitted intensity is given by Malus’s law:

$I_2 = I_1 \cos^2 \theta$
Substituting $I_1 = I_0/2$ and $\theta = 45^\circ$ :

$I_2 = \frac{I_0}{2} \cdot \cos^2 45^\circ = \frac{I_0}{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4}$ …
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Q.A polariod (I) is placed in front of a monochromatic source. Another polatiod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.

Concept understanding — Malus Law

Malus Law: How Light Gets Weaker Through a Polariser

The Intuition

The Precise Statement

A Quick Example

Why this formula?

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

1. What Does Polarized Light Look Like?

2. The Setup for Malus Law

3. The Core Reasoning: Electric Field Components

4. From Amplitude to Intensity

5. Why the Square? — Physical Meaning

6. Key Insight: Why Not $\cos \theta$ ?

Method: Chaining Malus's Law Through a Sequence of Polaroids

Steps

Concept and Intuition

Step-by-Step Solution

Q.A polariod (I) is placed in front of a monochromatic source. Another polatiod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.

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