1. What Malus’s law describes
Malus’s law explains how the intensity of a beam of plane-polarised light changes when it passes through another polarising device called an analyser. The idea is simple: if the analyser is aligned with the polariser, maximum light passes through. If it is turned, the transmitted intensity decreases.
2. Understanding the setup
The experiment involves two polaroid sheets:
- Polariser → creates plane-polarised light
- Analyser → tests how the intensity changes
When the analyser is rotated, the amount of light it allows through depends on the angle between the transmission axes of the two sheets.
3. Mathematical form of Malus’s law
If the initial intensity after the polariser is \(I_0\), and the angle between the polariser and analyser is \(\theta\), then the transmitted intensity \(I\) is:
\( I = I_0 \cos^2 \theta \)
This shows that intensity depends on the square of the cosine of the angle.
4. Meaning of the cosine-squared dependence
The electric field of the polarised light has a direction. Only the component of this field that aligns with the analyser’s axis passes through. This component is proportional to \(\cos \theta\), and since intensity is proportional to the square of the electric field, the final expression becomes \(I = I_0 \cos^2 \theta\).
5. Special cases of Malus’s law
5.1. When \(\theta = 0^\circ\)
The polariser and analyser are aligned.
\( I = I_0 \cos^2 0 = I_0 \)
Maximum light passes through.
5.2. When \(\theta = 90^\circ\)
The analyser is perpendicular to the polariser.
\( I = I_0 \cos^2 90^\circ = 0 \)
No light passes — the field component along the analyser axis is zero.
5.3. When \(\theta = 45^\circ\)
\( I = I_0 \cos^2 45^\circ = I_0/2 \)
Half the initial intensity is transmitted.
6. Graph of intensity vs angle
If I plot \(I\) against \(\theta\), I get a smooth curve that drops from \(I_0\) at 0°, falls to zero at 90°, and rises again at 180° because the axes align once more. This matches the periodic behaviour of the cosine function.
7. Example calculation
If the intensity after the polariser is \(I_0 = 20 \, \text{mW/m}^2\), and the analyser is turned by 30°:
\( I = 20 \cos^2 30^\circ \)
\( I = 20 \left(\dfrac{\sqrt{3}}{2}\right)^2 = 20 \times \dfrac{3}{4} = 15 \, \text{mW/m}^2 \)
The transmitted intensity becomes 15 mW/m².
8. Why Malus’s law is important
Malus’s law helps explain and design systems involving polarised light:
- Light intensity control in optical instruments
- LCD screen functioning
- Glare reduction using polarised sunglasses
- Laser alignment tools
It is one of the key tools to understand how polarised light behaves when passed through different orientations.