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

Published by Patrick Mutisya · 8 days ago

Polarisation – Cambridge A‑Level Physics 9702

Polarisation

Learning objective

Recall and use Malus’s law

\$I = I_0 \cos^2\theta\$

to calculate the intensity of a plane‑polarised electromagnetic wave after transmission through a polarising filter or a series of polarising filters.

1. What is polarisation?

Electromagnetic waves consist of oscillating electric and magnetic fields. In an unpolarised beam the direction of the electric field vector changes randomly with time. In a plane‑polarised beam the electric field oscillates in a single, fixed direction.

2. Polarising filters

  • A polarising filter (or analyser) only allows the component of the electric field that is parallel to its transmission axis to pass.

  • Any component perpendicular to the transmission axis is absorbed or reflected.

  • When a plane‑polarised wave meets a filter, the transmitted intensity depends on the angle \$\theta\$ between the wave’s polarisation direction and the filter’s transmission axis.

Suggested diagram: A plane‑polarised wave incident on a polarising filter with transmission axis at angle \$\theta\$.

3. Malus’s law

For a plane‑polarised wave of initial intensity \$I_0\$ incident on a single polarising filter, the transmitted intensity \$I\$ is

\$I = I_0 \cos^2\theta\$

where \$\theta\$ is the angle between the wave’s polarisation direction and the filter’s transmission axis.

4. Series of polarising filters

If a wave passes through several filters, the law is applied successively. For two filters with a relative angle \$\theta_{12}\$, the intensity after the second filter is

\$I2 = I1 \cos^2\theta{12} = I0 \cos^2\theta{01}\cos^2\theta{12}\$

where \$\theta_{01}\$ is the angle between the incident polarisation and the first filter.

5. Worked examples

Example 1 – Single filter

A plane‑polarised beam of intensity \$I_0 = 120\ \text{W m}^{-2}\$ meets a polarising filter whose transmission axis makes an angle \$\theta = 30^\circ\$ with the polarisation direction. Find the transmitted intensity.

  1. Write Malus’s law: \$I = I_0\cos^2\theta\$.

  2. Calculate \$\cos30^\circ = \frac{\sqrt{3}}{2}\$.

  3. \$I = 120\left(\frac{\sqrt{3}}{2}\right)^2 = 120\left(\frac{3}{4}\right)=90\ \text{W m}^{-2}.\$

Example 2 – Two filters

A plane‑polarised beam of intensity \$I_0 = 200\ \text{W m}^{-2}\$ passes through two filters. The first filter’s axis is \$0^\circ\$ relative to the beam (i.e., it is aligned), and the second filter’s axis is \$45^\circ\$ relative to the first. Find the final intensity.

  1. After the first filter (aligned) \$I1 = I0\cos^2 0^\circ = I_0\$.

  2. Apply Malus’s law for the second filter: \$I2 = I1\cos^2 45^\circ\$.

  3. Since \$\cos45^\circ = \frac{1}{\sqrt{2}}\$, \$I_2 = 200\left(\frac{1}{\sqrt{2}}\right)^2 = 200\left(\frac{1}{2}\right)=100\ \text{W m}^{-2}.\$

Example 3 – Three filters (demonstrating “crossed” polarizers with an intermediate angle)

Three filters are placed in sequence. The first and third are crossed (\$90^\circ\$ apart). The middle filter is at \$45^\circ\$ to each. Initial intensity \$I_0 = 80\ \text{W m}^{-2}\$. Find the intensity after the third filter.

  1. After the first filter (aligned): \$I1 = I0\$.

  2. Second filter at \$45^\circ\$: \$I2 = I1\cos^2 45^\circ = I_0\left(\frac{1}{2}\right)=40\ \text{W m}^{-2}\$.

  3. Third filter \$45^\circ\$ relative to the second (i.e., \$90^\circ\$ relative to the first): \$I3 = I2\cos^2 45^\circ = 40\left(\frac{1}{2}\right)=20\ \text{W m}^{-2}\$.

6. Common pitfalls

  • Confusing the angle \$\theta\$ with the angle between successive filters; always use the angle between the current polarisation direction and the next filter’s axis.

  • For unpolarised light, the first polariser reduces intensity to \$I_0/2\$ before Malus’s law applies. (Not required for this objective.)

  • Remember that \$\cos^2\theta\$ is always positive; intensity never becomes negative.

7. Summary table

SituationExpression for transmitted intensityKey point
Single filter\$I = I_0\cos^2\theta\$Use angle between incident polarisation and filter axis.
Two filters\$I = I0\cos^2\theta{01}\cos^2\theta_{12}\$Apply Malus’s law sequentially.
Three or more filters\$I = I0\prod{k=1}^{n}\cos^2\theta_{k-1,k}\$Product of \$\cos^2\$ for each successive angle.

8. Practice questions

  1. A plane‑polarised beam of intensity \$150\ \text{W m}^{-2}\$ passes through a filter whose axis is \$60^\circ\$ to the polarisation direction. What is the transmitted intensity?

  2. Two polarising filters are placed in series. The first is aligned with the incident polarisation. The second is at \$30^\circ\$ to the first. If the initial intensity is \$100\ \text{W m}^{-2}\$, what intensity emerges from the second filter?

  3. Three filters are used. The first and third are crossed (\$90^\circ\$ apart). The middle filter is at \$30^\circ\$ to the first. Starting intensity \$I_0 = 120\ \text{W m}^{-2}\$. Calculate the final intensity.

9. Answers to practice questions

QuestionAnswer
1\$I = 150\cos^2 60^\circ = 150\left(\frac{1}{2}\right)^2 = 150\left(\frac{1}{4}\right)=37.5\ \text{W m}^{-2}\$
2\$I = 100\cos^2 30^\circ = 100\left(\frac{\sqrt{3}}{2}\right)^2 = 100\left(\frac{3}{4}\right)=75\ \text{W m}^{-2}\$
3

\$I_1 = 120\$ (first filter aligned)

\$I_2 = 120\cos^2 30^\circ = 120\left(\frac{\sqrt{3}}{2}\right)^2 = 120\left(\frac{3}{4}\right)=90\$

\$I_3 = 90\cos^2 60^\circ = 90\left(\frac{1}{2}\right)^2 = 90\left(\frac{1}{4}\right)=22.5\ \text{W m}^{-2}\$

10. Quick reference

  • Malus’s law: \$I = I_0\cos^2\theta\$.

  • For \$n\$ filters: \$I = I0\displaystyle\prod{k=1}^{n}\cos^2\theta_{k-1,k}\$.

  • Angles are measured between the electric‑field direction of the incoming wave and the transmission axis of the next filter.