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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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

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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