recall and use Malus’s law ( I = I0 cos2θ ) to calculate the intensity of a plane-polarised electromagnetic wave after transmission through a polarising filter or a series of polarising filters (calculation of the effect of a polarising filter on the

Polarisation – Using Malus’s Law (Cambridge International AS & A Level Physics 9702)

Learning Outcome (Syllabus 7.5)

• Recall that all electromagnetic (EM) waves are transverse.
• State and apply Malus’s law \(I = I_{0}\cos^{2}\theta\) to determine the intensity of a plane‑polarised wave after one or more ideal polarising filters.
• Explain how an un‑polarised beam becomes plane‑polarised after passing through a single ideal polariser.
• Identify practical applications and recognise the limitations of the ideal‑filter model.

1. Fundamental Concepts

1.1 Transverse nature of EM waves

An EM wave consists of mutually perpendicular electric (\(\mathbf{E}\)) and magnetic (\(\mathbf{B}\)) fields that both oscillate perpendicular to the direction of propagation. Because the electric‑field vector defines the direction of the wave’s polarisation, the wave is transverse.

1.2 Plane‑polarised and un‑polarised light

• Plane‑polarised wave: the electric‑field vector oscillates in a single fixed plane (the plane of polarisation).
• Un‑polarised light: the electric‑field direction varies randomly over time; the intensity is the average of many random polarisations.

1.3 How a polariser produces plane‑polarised light

An ideal linear polariser transmits only the component of \(\mathbf{E}\) that is parallel to its transmission axis. For an un‑polarised incident beam each random electric‑field vector can be resolved into a component parallel to the axis and a component perpendicular to it; the perpendicular component is completely blocked. The average transmitted intensity is therefore half of the incident intensity, and the emerging beam is plane‑polarised with its electric field aligned with the transmission axis:

\[

I{\text{after 1st polariser}} = \tfrac{1}{2}\,I{0}

\]

2. Derivation of Malus’s Law

Consider a plane‑polarised wave of electric‑field amplitude \(E_{0}\) incident on an ideal analyser whose transmission axis makes an angle \(\theta\) with the wave’s polarisation direction.

1. The transmitted electric‑field component is the projection:

   \[

   E = E_{0}\cos\theta

   \]

2. Intensity of an EM wave is proportional to the square of the field amplitude:

   \[

   I \propto E^{2}

   \]

   Hence,

   \[

   I = I_{0}\cos^{2}\theta,

   \]

   where \(I_{0}\) is the intensity just before the analyser.

3. Applying Malus’s Law

3.1 Single polarising filter

1. Write down the incident intensity \(I_{0}\) (W m\(^{-2}\)).
2. Determine the angle \(\theta\) between the incident polarisation direction and the filter’s transmission axis.
3. Calculate the transmitted intensity:

   \[

   I = I_{0}\cos^{2}\theta.

   \]

3.2 Series of polarising filters

When several ideal filters are used, the intensity after each filter becomes the incident intensity for the next. If the relative angles between successive filters are \(\theta{1},\theta{2},\dots,\theta{n}\) (with \(\theta{1}\) measured from the original polarisation), the final intensity is

\[

I{n}=I{0}\prod{k=1}^{n}\cos^{2}\theta{k}.

\]

3.3 Practical tip – relative vs. absolute angles

Use the relative angle between each pair of successive transmission axes. Using the absolute angle of every filter with respect to the original polarisation will give an incorrect result.

4. Limitations of the Ideal‑Filter Model

When required, these factors can be introduced by multiplying each \(\cos^{2}\theta\) term by the appropriate transmission coefficient.

5. Applications (Illustrative)

• Polarised sunglasses – reduce glare by blocking horizontally polarised reflected light.
• Liquid‑crystal displays (LCDs) – use a pair of crossed polarisers with a liquid‑crystal layer that rotates the plane of polarisation.
• Stress analysis in transparent plastics (photo‑elasticity) – changes in polarisation reveal internal stresses.
• Optical isolators and lasers – control the direction of light propagation.
• Polarising microscope: combines crossed polarisers with a sample stage to examine birefringent materials, a technique explicitly mentioned in the Cambridge textbook.

6. Common Pitfalls

• Angle definition: \(\theta\) is the angle between polarisation directions, not between the beam’s propagation direction and the filter.
• Squaring the cosine: The law involves \(\cos^{2}\theta\); forgetting the square over‑estimates the transmitted intensity.
• Series calculations: Use the relative angle between each successive pair of axes, not the absolute angle from the original polarisation.
• Significant figures: Carry the appropriate number of significant figures through each calculation and round only at the final step.

7. Summary – Step‑by‑Step Procedure

8. Worked Examples

Example 1 – Single filter

A plane‑polarised beam of intensity \(I_{0}=80\ \text{W m}^{-2}\) meets an analyser set at \(\theta=60^{\circ}\) to the beam’s polarisation.

\[

I = 80\cos^{2}60^{\circ}=80\left(\frac{1}{2}\right)^{2}=20\ \text{W m}^{-2}

\]

Result (2 sf): \(I = 20\ \text{W m}^{-2}\).

Example 2 – Three filters (non‑orthogonal)

Incident intensity \(I_{0}=120\ \text{W m}^{-2}\).

Angles: \(\theta{1}=0^{\circ}\) (first filter aligned), \(\theta{2}=20^{\circ}\), \(\theta_{3}=30^{\circ}\) (relative to the preceding filter).

\[

\begin{aligned}

I_{1}&=120\cos^{2}0^{\circ}=120\\[4pt]

I_{2}&=120\cos^{2}20^{\circ}=120(0.9397)^{2}=120\times0.883=106\ \text{W m}^{-2}\\[4pt]

I_{3}&=106\cos^{2}30^{\circ}=106\left(\frac{\sqrt{3}}{2}\right)^{2}=106\times\frac{3}{4}=79.5\ \text{W m}^{-2}

\end{aligned}

\]

Final transmitted intensity (3 sf): \(I_{3}=79.5\ \text{W m}^{-2}\).

Example 3 – “Unlocking” light with a third polariser

Two ideal polarisers are crossed (\(\theta_{12}=90^{\circ}\)); without a third filter, \(I=0\). Insert a third polariser between them, each successive pair being at \(\theta=45^{\circ}\).

\[

I = I{0}\cos^{2}45^{\circ}\cos^{2}45^{\circ}=I{0}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{I_{0}}{4}.

\]

Thus 25 % of the original intensity emerges – a classic demonstration of Malus’s law.

9. Practice Questions

1. A plane‑polarised beam of intensity \(80\ \text{W m}^{-2}\) passes through a single analyser set at \(60^{\circ}\) to the beam’s polarisation. Calculate the transmitted intensity (2 sf).
2. Three ideal polarisers are placed in sequence. The first is aligned with the incident polarisation, the second is at \(20^{\circ}\) to the first, and the third is at \(30^{\circ}\) to the second. If the incident intensity is \(120\ \text{W m}^{-2}\), find the intensity after the third polariser (3 sf).
3. Two ideal polarisers are crossed (\(90^{\circ}\) apart) so that no light is transmitted. Explain, using Malus’s law, how inserting a third polariser at an intermediate angle (e.g. \(45^{\circ}\) to each of the others) can increase the transmitted intensity.
4. Un‑polarised light of intensity \(200\ \text{W m}^{-2}\) passes through a single ideal polariser. What is the intensity and the polarisation state of the emerging beam?
5. Real polarisers have a transmission efficiency of 95 % for the parallel component and an extinction ratio of 1 : 10⁴. For a single filter set at \(30^{\circ}\) to the incident polarisation, calculate the transmitted intensity, taking the efficiency into account (2 sf).

10. Further Reading (Cambridge Textbook Sections)

• Cambridge International AS & A Level Physics, 9702 – Section 12.2: “Polarisation of Light”.
• Section 12.3: “Malus’s Law – Derivation and Experimental Verification”.
• Appendix B: “Properties of Polarising Materials”.