Malus's law

6 minAS Level7.5.2

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can state and use Malus's law, $I = I_0\cos^2\theta$ , to find the intensity after one polarising filter or several in series.

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$I_2$	second (or new) intensity
$I_1$	first (or original) intensity
$\theta_1$	first angle
$\theta_2$	second angle

The second filter is called the analyser

Suppose light is already plane-polarised by a first filter. Its electric field oscillates along one line, and its intensity is $I_0$ .

Now put a second filter in the beam, with its transmission axis at an angle $\theta$ to the plane of polarisation. This second filter is called the analyser.

The analyser does not transmit all the light, and it does not block all of it (unless $\theta = 90^\circ$ ). It transmits the component of the electric field along its axis.

The component of a vector of length $E_0$ at angle $\theta$ is $E_0\cos\theta$ . So the transmitted electric field amplitude is

$E = E_0\cos\theta$

Intensity is proportional to amplitude squared, so the transmitted intensity is

$I \propto E^2 = E_0^2\cos^2\theta$

Dividing by the incident intensity gives Malus's law.

Definition

Malus's law

For plane-polarised light of intensity $I_0$ incident on an analyser whose transmission axis makes an angle $\theta$ with the plane of polarisation,

$I = I_0\cos^2\theta$

where $I$ is the transmitted intensity. $\theta$ is measured between the two filter axes, not between a filter and the vertical.

Why the square?

A filter passes the part of the oscillation that lies along its axis. That part is $E_0\cos\theta$ , so the amplitude falls by a factor $\cos\theta$ .

Intensity depends on amplitude squared, so it falls by $\cos^2\theta$ . At $\theta = 60^\circ$ , the amplitude is halved, but the intensity falls to a quarter, not a half.

This is why the graph of intensity against angle is a series of humps rather than a straight line.

Worked example: One analyser at 30°

Plane-polarised light of intensity $80\ \text{W m}^{-2}$ falls on an analyser whose axis is at $30^\circ$ to the plane of polarisation. Calculate the transmitted intensity.

Have a go first?

Worked example: Two analysers in series

Plane-polarised light of intensity $100\ \text{W m}^{-2}$ passes through a filter at $30^\circ$ to its plane of polarisation, then through a second filter at $60^\circ$ to the first filter's axis. Find the final intensity.

Have a go first?

Common mistakes

The intensity falls as $\cos\theta$ .

The amplitude falls as $\cos\theta$ ; the intensity falls as $\cos^2\theta$ . At $60^\circ$ the amplitude halves but the intensity quarters.

In the exam

Quote the law as $I = I_0\cos^2\theta$ and define $\theta$ as the angle between the analyser axis and the plane of polarisation (equivalently, between the two filter axes).
For a series, take the filters one at a time. A common error is measuring the second angle from the original plane instead of from the first filter's axis.
Keep intensities in $\text{W m}^{-2}$ and check that each step reduces the intensity. An answer larger than the incident intensity is impossible.
If asked for the angle giving a particular fraction, invert the law: $\cos^2\theta = I/I_0$ , so $\cos\theta = \sqrt{I/I_0}$ .
Do not calculate the effect of the first filter on unpolarised light unless the question gives you the polarised intensity. The guidance excludes that calculation.

Go deeper: What happens with unpolarised light?

If unpolarised light of intensity $I_{\text{unpolarised}}$ passes through a single ideal polarising filter, the transmitted intensity is about $\dfrac{I_{\text{unpolarised}}}{2}$ . This is because, on average, half the oscillation energy lies along the filter's axis.

This result is not required by the syllabus for calculation. It is included so that you understand why the first filter still transmits a bright beam even though it removes half the light. Once the beam is polarised, every later filter follows Malus's law with its own angle.

Key takeaways

Malus's law is $I = I_0\cos^2\theta$ for plane-polarised light through an analyser.
$\theta$ is the angle between the analyser axis and the plane of polarisation.
Amplitude falls as $\cos\theta$ ; intensity falls as $\cos^2\theta$ .

One filter

CalculateQuestion 1 · Recall

CalculateQuestion 2 · Apply

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