Q.Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?

The key idea is Malus Law: when plane-polarised light passes through an analyser, the transmitted intensity is $I = I_0 \cos^2\theta$, where $\theta$ is the angle between the polariser’s transmission axis and the incident polarisation direction.

Reasoning:

1. Two crossed polaroids (axes at $90^\circ$) transmit zero light: $I_{\text{final}} = 0$.
2. Insert a third polaroid between them, with its axis at an angle $\theta$ to the first polariser. After the first polaroid, intensity is $I_0/2$ (unpolarised light becomes polarised). After the middle polaroid: $I_1 = \frac{I_0}{2} \cos^2\theta$.
3. The light incident on the second crossed polaroid is polarised at angle $\theta$ relative to its axis, which is at $90^\circ$ to the first. So the angle between the middle polaroid’s output and the second polaroid’s axis is $90^\circ - \theta$. Applying Malus Law again: $I_2 = I_1 \cos^2(90^\circ - \theta) = I_1 \sin^2\theta$. …

The problem is a classic demonstration of Malus law in sequence. Two crossed polaroids (P₁ and P₂) have their transmission axes at 90° to each other, so normally no light passes through. But when you insert a third polaroid (P₃) between them and rotate it, something interesting happens: light can now get through, and the intensity changes in a specific way as you turn P₃.

Why does inserting an extra polaroid let light through? Because the intermediate polaroid acts as a "bridge" — it takes the linearly polarised light from P₁ and rotates its plane of polarisation by an angle, so that some component now aligns with P₂. Without P₃, the light from P₁ is perpendicular to P₂, giving zero transmission. With P₃, you get two successive projections, and the product is not zero unless P₃ is aligned exactly with either P₁ or P₂.

Let’s work through it step by step.

1. Set up the axes. Let the transmission axis of the first polaroid P₁ be along the vertical direction (say, the $y$-axis). The second polaroid P₂ is crossed to P₁, so its axis is horizontal (the $x$-axis). The intermediate polaroid P₃ has its transmission axis at an angle $\theta$ to the vertical. As you rotate P₃, $\theta$ changes from $0^\circ$ to $360^\circ$.

2. Light after P₁. Unpolarised light of intensity $I_0$ falls on P₁. After passing through P₁, the light becomes linearly polarised along the vertical, and its intensity is halved:

3. Light after P₃. This vertically polarised light now falls on P₃, whose axis is at angle $\theta$ to the vertical. By Malus law, the intensity transmitted through P₃ is:

The light emerging from P₃ is polarised along the direction of P₃’s axis — that is, at angle $\theta$ to the vertical.

4. Light after P₂. This light now encounters P₂, whose axis is horizontal (angle $90^\circ$ to the vertical). The angle between the polarisation direction of the light (angle $\theta$) and P₂’s axis is $90^\circ - \theta$. Applying Malus law again:

Substitute $I_2$:

5. Simplify the expression. Use the identity $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$:

Alternatively, using $\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}$, we can write:

Both forms are equivalent; the $\sin^2 2\theta$ form is often more intuitive for discussing maxima and minima.

6. Interpret the variation. As $\theta$ varies from $0^\circ$ to $360^\circ$: …

Method: Malus Law Analysis with Three Polaroids

We use Malus Law step-by-step to find the transmitted intensity when a middle polaroid rotates between two crossed polaroids.

Step 1 — Understand the Setup

• First polaroid (P₁) — fixed, vertically polarised.
• Second polaroid (P₂) — rotated by an angle $\theta$ relative to P₁.
• Third polaroid (P₃) — fixed, crossed with P₁ (i.e., its transmission axis is at $90^\circ$ to P₁).

So, the angle between P₁ and P₃ is $90^\circ$.

Step 2 — Apply Malus Law at Each Stage

Let the intensity of unpolarised light incident on P₁ be $I_0$.

After P₁:

Only half the intensity passes (for unpolarised light):

After P₂ (rotated by $\theta$ from P₁):

Malus Law: $I = I_1 \cos^2 \theta$

After P₃ (crossed with P₁, so at $90^\circ$ to P₁):

The angle between P₂ and P₃ is $(90^\circ - \theta)$.

Apply Malus Law again:

Recall: $\cos(90^\circ - \theta) = \sin \theta$

So:

Step 3 — Simplify the Expression

Use the identity: $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$

Thus:

$I_3 = \frac{I_0}{2} \left( \frac{1}{2} \sin 2\theta \right)^2$ …

Common Mistakes with Malus Law (Crossed Polaroids Problem)

Students often struggle with the three-polaroid problem — where one polaroid rotates between two crossed polaroids. Here are the most frequent errors and how to avoid each.

Mistake 1: Assuming zero intensity when the middle polaroid is at $45^\circ$

The error: Students think that because the outer two polaroids are crossed ($90^\circ$ apart), no light can pass through — regardless of what's in between.

Why it's wrong: The middle polaroid reorients the polarization direction. Even though the first and last are crossed, the middle one at $45^\circ$ allows some light through.

How to avoid: Always apply Malus Law step by step — treat each polaroid as a separate filter. Never skip the intermediate polaroid's effect.

Mistake 2: Applying Malus Law only once (instead of three times)

The error: Students write $I = I_0 \cos^2(90^\circ) = 0$ and stop.

Why it's wrong: Malus Law applies between successive polaroids, not just between the first and last. You must compute:

• Step 1: Intensity after first polaroid
• Step 2: Intensity after middle polaroid (at angle $\theta$)
• Step 3: Intensity after last polaroid (at $90^\circ$ from first)

How to avoid: Write three separate expressions:

Then combine:

Mistake 3: Forgetting that unpolarized light loses half its intensity

The error: Students take $I_1 = I_0$ after the first polaroid.

Why it's wrong: Unpolarized light passing through an ideal polarizer has its intensity halved — $I_1 = I_0/2$.

How to avoid: Memorise: First polaroid = half intensity. Always start with $I_0/2$ for unpolarized incident light.

Mistake 4: Using the wrong angle in the second Malus Law step

The error: Students use $90^\circ$ instead of $(90^\circ - \theta)$ for the angle between the middle and last polaroid.

Why it's wrong: The last polaroid is at $90^\circ$ to the first, not to the middle one. The angle between middle and last is $90^\circ - \theta$.

How to avoid: Draw a diagram. Label each polaroid's transmission axis. The angle between any two is the difference in their orientations.

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