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

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

1. 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

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

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

This light is now polarized along P’s axis.

3. 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

Substituting $I_P$:

4. 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 $$$$

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

1. 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.
2. 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.
3. 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.
4. 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:

   $I = I_0 \cos^2 \theta$

   This law is crucial for solving problems involving multiple polaroids.

Applying Malus's Law to the Setup

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

1. 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.

2. 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:

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:

Substitute the expression for $I_C$: …

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

1. 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.

2. 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 $(E_1 \cos\theta)^2$.

   The intensity of polarized light after passing through an analyzer is given by Malus's Law:

   $I_2 = I_1 \cos^2 \theta$

   where $I_1$ is the intensity of the incident polarized light, and $\theta$ is the angle between the plane of polarization of the incident light and the transmission axis of the analyzer.

Step-by-Step Solution

1. 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:

2. Intensity after the Analyzer: …

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.

1. 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:

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 $1/2$.

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:

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

$I_2 = \frac{I_0}{2} \cdot \cos^2 45^\circ = \frac{I_0}{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4}$ …