Polarisation

1. Overview

Polarisation is a phenomenon exclusive to transverse waves. It occurs when the oscillations of a wave are restricted to a single plane that is perpendicular to the direction of energy transfer. Because longitudinal waves (such as sound) oscillate only along the axis of propagation, they possess no directional variety in their oscillations and therefore cannot be polarised. Consequently, demonstrating that a wave can be polarised is the definitive experimental proof that the wave is transverse. In the context of electromagnetic (EM) waves, polarisation refers to the orientation of the oscillating electric field vector.

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Key Definitions

• Polarisation: The process by which the oscillations of a transverse wave are restricted to a single plane which includes the direction of energy transfer.
• Plane-polarised wave: A transverse wave in which the oscillations are limited to a single plane perpendicular to the direction of propagation.
• Unpolarised wave: A transverse wave where oscillations occur in all possible planes perpendicular to the direction of propagation.
• Transmission Axis: The specific direction or orientation of a polarising filter that allows the component of the wave's oscillations parallel to that axis to pass through with minimal absorption.
• Intensity ($I$): The power per unit area incident on a surface, measured in Watts per square metre ($W , m^{-2}$). For any wave, intensity is directly proportional to the square of the amplitude ($I \propto A^2$).
• Analyser: A second polarising filter placed in the path of a plane-polarised wave used to vary the transmitted intensity or determine the plane of polarisation of the incident wave.

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Content

3.1 The Nature of Polarisation

In an unpolarised transverse wave, such as light from the Sun or an incandescent bulb, the oscillations occur in every possible plane perpendicular to the direction of travel. A polarising filter (often called a "Polaroid") acts as a molecular "slit." It consists of long-chain polymer molecules aligned in a specific direction.

• Mechanism: The filter absorbs the energy of the oscillations that are parallel to the alignment of the molecules and allows the oscillations perpendicular to the molecules to pass through.
• Transverse vs. Longitudinal:
  • Transverse waves (e.g., light, microwaves, waves on a string) have oscillations at $90^\circ$ to the velocity. These oscillations can be filtered into a single plane.
  • Longitudinal waves (e.g., sound waves, ultrasound, P-waves) have oscillations parallel to the velocity. There is no "side-to-side" component to filter out; thus, they cannot be polarised.

3.2 Malus’s Law

Malus’s Law quantifies how the intensity of plane-polarised light changes when it passes through a second polarising filter (the analyser).

The Derivation of Malus’s Law:

1. Consider a plane-polarised wave with electric field amplitude $A_0$ incident on a filter.
2. The filter has a transmission axis at an angle $\theta$ to the plane of the incident oscillations.
3. Only the component of the amplitude that is parallel to the transmission axis can pass through.
4. By resolving the amplitude vector into two perpendicular components, the transmitted amplitude $A$ is: $A = A_0 \cos \theta$
5. Since the intensity $I$ of a wave is proportional to the square of its amplitude ($I \propto A^2$):
   • Incident intensity: $I_0 = k A_0^2$
   • Transmitted intensity: $I = k A^2 = k (A_0 \cos \theta)^2$
6. Substituting $I_0$ into the equation yields Malus’s Law: $I = I_0 \cos^2 \theta$

3.3 Intensity and Angle Relationships

The relationship between the angle of the analyser and the transmitted intensity follows a $\cos^2$ distribution:

• Parallel ($\theta = 0^\circ$ or $180^\circ$ or $360^\circ$): The transmission axis aligns with the incident polarisation. $\cos^2(0) = 1$. Therefore, $I = I_0$ (Maximum intensity).
• Perpendicular / Crossed ($\theta = 90^\circ$ or $270^\circ$): The transmission axis is at right angles to the incident polarisation. $\cos^2(90) = 0$. Therefore, $I = 0$ (Minimum/Zero intensity).
• Intermediate ($\theta = 45^\circ$): $\cos^2(45) = 0.5$. Therefore, $I = 0.5 I_0$ (Intensity is halved).

3.4 Conceptual Note: Unpolarised Light

While the syllabus focuses on the effect of filters on already polarised light, it is important to understand that when unpolarised light of intensity $I_{un}$ passes through a single ideal polariser, the transmitted intensity is exactly $\frac{1}{2} I_{un}$. This is because the average value of $\cos^2 \theta$ over all possible angles ($0$ to $2\pi$) is $0.5$.

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4. Worked Examples

Worked example 1 — Calculating Transmitted Intensity

A plane-polarised laser beam has an initial intensity of $4.50 \times 10^3 , W , m^{-2}$. It passes through a polarising filter whose transmission axis is oriented at $25.0^\circ$ to the plane of the laser's polarisation. Calculate the intensity of the beam after it emerges from the filter.

Step 1: Identify the known variables

• Incident intensity $I_0 = 4.50 \times 10^3 , W , m^{-2}$
• Angle $\theta = 25.0^\circ$

Step 2: State the relevant equation

• $I = I_0 \cos^2 \theta$

Step 3: Substitute and calculate

• $I = (4.50 \times 10^3) \times (\cos 25.0^\circ)^2$
• $I = (4500) \times (0.9063...)^2$
• $I = 4500 \times 0.82139...$
• $I = 3696.2... , W , m^{-2}$

Step 4: Final answer with units and significant figures

• $I = 3.70 \times 10^3 , W , m^{-2}$ (to 3 s.f.)

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Worked example 2 — Finding the Required Angle

A student wishes to reduce the intensity of a plane-polarised light source to exactly one-eighth ($12.5%$) of its original value using a single polarising filter. Determine the angle at which the filter's transmission axis must be set relative to the light's plane of polarisation.

Step 1: Identify the ratio of intensities

• $I = \frac{1}{8} I_0$ or $\frac{I}{I_0} = 0.125$

Step 2: Set up Malus’s Law

• $\frac{I}{I_0} = \cos^2 \theta$
• $0.125 = \cos^2 \theta$

Step 3: Solve for $\theta$

• $\cos \theta = \sqrt{0.125}$
• $\cos \theta = 0.35355...$
• $\theta = \cos^{-1}(0.35355...)$
• $\theta = 69.295...^\circ$

Step 4: Final answer

• $\theta = 69.3^\circ$

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Worked example 3 — Multiple Filters (The Three-Filter Problem)

Unpolarised light is incident on a series of three polarising filters.

1. The first filter polarises the light vertically, resulting in an intensity $I_1$.
2. The second filter has its transmission axis at $30^\circ$ to the vertical.
3. The third filter has its transmission axis at $90^\circ$ to the vertical (horizontal). Calculate the final intensity $I_3$ as a fraction of $I_1$.

Step 1: Calculate the intensity after the second filter ($I_2$)

• The light entering the second filter is already polarised (from the first filter).
• $I_2 = I_1 \cos^2(30^\circ)$
• $I_2 = I_1 \times (0.866...)^2 = 0.75 I_1$

Step 2: Calculate the intensity after the third filter ($I_3$)

• The light entering the third filter is polarised at $30^\circ$ to the vertical.
• The third filter is at $90^\circ$ to the vertical.
• The angle $\theta$ between the light's polarisation ($30^\circ$) and the filter's axis ($90^\circ$) is: $\Delta \theta = 90^\circ - 30^\circ = 60^\circ$.
• $I_3 = I_2 \cos^2(60^\circ)$
• $I_3 = (0.75 I_1) \times (0.5)^2$
• $I_3 = 0.75 I_1 \times 0.25$

Step 3: Final Ratio

• $I_3 = 0.1875 I_1$
• $I_3 = \frac{3}{16} I_1$ (or $18.8%$ of $I_1$)

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Key Equations

Equation	Description	Status
$I = I_0 \cos^2 \theta$	Malus's Law: Relates transmitted intensity ($I$) to incident polarised intensity ($I_0$) and the angle ($\theta$) between the plane of polarisation and the transmission axis.	On Formula Sheet
$I \propto A^2$	Intensity-Amplitude Relationship: Essential for deriving Malus's Law or solving problems involving amplitude changes.	Must Memorise
$A = A_0 \cos \theta$	Amplitude Resolution: The component of amplitude transmitted through a filter.	Must Memorise

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Common Mistakes to Avoid

• ❌ The "Square" Error: Forgetting to square the cosine. Students often write $I = I_0 \cos \theta$.
  • ✅ Right: Always use $I = I_0 \cos^2 \theta$. Remember that the $\cos \theta$ comes from resolving the amplitude, and intensity is proportional to amplitude squared.
• ❌ Longitudinal Polarisation: Attempting to apply polarisation concepts to sound waves.
  • ✅ Right: If a question asks why sound cannot be polarised, the answer is: "Sound is a longitudinal wave; oscillations are parallel to the direction of energy transfer."
• ❌ Misidentifying the Angle $\theta$: Using the angle relative to the "horizontal" or "vertical" instead of the angle between the light's plane of polarisation and the filter's axis.
  • ✅ Right: Always draw a quick sketch of the two axes and find the difference between them to get $\theta$.
• ❌ Unpolarised vs. Polarised $I_0$: Using Malus's Law to calculate the effect of the first filter on unpolarised light.
  • ✅ Right: Malus's Law only applies when the incident light is already plane-polarised. For unpolarised light hitting the first filter, the intensity is simply halved (though this specific calculation is technically "extended" beyond the core 9702 syllabus requirements, it is vital for multi-filter problems).
• ❌ Calculator Units: Having the calculator in Radians mode when the angle is given in Degrees.
  • ✅ Right: 9702 exams almost exclusively use degrees for polarisation. Double-check your "D" indicator.

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Exam Tips

1. The "Definitive Test": If an exam question asks "State how it can be shown that light waves are transverse," the answer is always: "By showing they can be polarised."
2. Graph Sketching: When sketching the $I$ vs $\theta$ graph, ensure the curve never goes below the x-axis (intensity cannot be negative) and that the peaks are at $0, 180, 360$ and the zeros are exactly at $90$ and $270$.
3. Vector Resolution: If a question asks you to "Explain Malus's Law," you must mention that the filter only transmits the component of the amplitude parallel to its transmission axis ($A \cos \theta$) and that intensity is proportional to amplitude squared.
4. Microwave Experiments: Polarisation is often tested using microwaves and a metal grille. Remember: Microwaves are absorbed when the metal rods are parallel to the electric field oscillations (because the electrons in the metal can flow along the rods and absorb the energy). This is counter-intuitive compared to a "slit" model, so read the question carefully.
5. Ratio Problems: Many 9702 questions ask for the ratio $I/I_0$. In these cases, you don't need the actual values of intensity; just calculate $\cos^2 \theta$.