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
In this section we recall and apply Malus’s law for a plane‑polarised electromagnetic wave passing through one or more ideal polarising filters. The law relates the transmitted intensity \$I\$ to the incident intensity \$I_0\$ and the angle \$\theta\$ between the transmission axes of the polariser and analyser.
\$\$
I = I_0 \cos^{2}\theta
\$\$
Key Concepts
Plane‑polarised wave: The electric field oscillates in a single plane.
Polariser (or analyser): An ideal filter that transmits only the component of the electric field parallel to its transmission axis.
Angle \$\theta\$: Measured between the transmission axis of the incident polarised wave and the axis of the filter.
Intensity: Proportional to the square of the amplitude of the electric field; therefore the transmitted intensity follows the \$\cos^{2}\theta\$ dependence.
Applying Malus’s Law – Single Filter
When a plane‑polarised wave of intensity \$I_0\$ meets a single polarising filter whose axis makes an angle \$\theta\$ with the wave’s polarisation direction, the transmitted intensity \$I\$ is given directly by Malus’s law.
Identify the incident intensity \$I_0\$ (the intensity just before the filter).
Measure or be given the angle \$\theta\$ between the wave’s polarisation direction and the filter’s axis.
Calculate \$I\$ using \$I = I_0 \cos^{2}\theta\$.
Series of Polarising Filters
When more than one filter is used, the intensity after each filter becomes the incident intensity for the next. For \$n\$ filters with successive angles \$\theta1, \theta2, \dots, \thetan\$ (each measured relative to the preceding filter), the final intensity \$In\$ is:
\$\$
In = I0 \prod{k=1}^{n} \cos^{2}\thetak
\$\$
where \$\thetak\$ is the angle between the transmission axes of filter \$k-1\$ and filter \$k\$ (with \$\theta1\$ being the angle between the initial polarisation direction and the first filter).
Worked Example – Two Filters
Suppose a plane‑polarised beam of intensity \$I0 = 100\ \text{W m}^{-2}\$ passes through two ideal polarisers. The first analyser is oriented at \$\theta1 = 30^{\circ}\$ to the incident polarisation, and the second analyser is at \$\theta_2 = 45^{\circ}\$ relative to the first.
Thus the final transmitted intensity is \$37.5\ \text{W m}^{-2}\$.
Common Pitfalls
Confusing the angle \$\theta\$ with the angle between the incident wave’s direction and the filter – the angle is always between the polarisation directions, not the propagation directions.
For a series of filters, using the absolute angle of each filter with respect to the original polarisation instead of the relative angle between successive filters.
Neglecting to square the cosine term; remember the law involves \$\cos^{2}\theta\$, not \$\cos\theta\$.
Summary Table – Steps for Calculating Transmitted Intensity
Step
Action
Formula Used
1
Identify initial intensity \$I_0\$.
—
2
Determine the angle \$\theta_1\$ between the incident polarisation and the first filter.
—
3
Calculate intensity after first filter.
\$I1 = I0 \cos^{2}\theta_1\$
4
For each subsequent filter \$k\$, find the relative angle \$\theta_k\$ to the previous filter.
—
5
Update intensity using \$Ik = I{k-1} \cos^{2}\theta_k\$.
\$Ik = I{k-1} \cos^{2}\theta_k\$
6
Repeat steps 4–5 for all filters; the final \$I_n\$ is the transmitted intensity.
Suggested diagram: A series of three polarising filters showing the initial polarisation direction and the successive angles \$\theta1\$, \$\theta2\$, \$\theta_3\$ between each pair of axes.
Practice Questions
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.
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.
Explain why inserting a third polariser at an intermediate angle can increase the transmitted intensity compared with using only two polarisers that are orthogonal (i.e., \$\theta = 90^{\circ}\$).
Further Reading (Textbook Sections)
Cambridge International AS & A Level Physics, Section 12.2 – Polarisation of Light.
Section on Malus’s Law – derivation and experimental verification.