Physics 9702 · Paper 2

Malus's Law

Once light is plane-polarised, a second filter cuts its intensity by $\cos^2\theta$ , where $\theta$ is the angle between the light's polarisation and the filter's axis.
This rule is Malus's law.

Malus's law

I = I_0\cos^2\theta

Symbols

$I$ = intensity passed by the filter (W m⁻²)
$I_0$ = intensity of the polarised light hitting the filter (W m⁻²)
$\theta$ = angle between the polarisation direction and the filter axis (degrees)

$\theta = 0^\circ$ (axis lined up): $\cos^2 0 = 1$ , so all the light passes.
$\theta = 90^\circ$ (crossed): $\cos^2 90 = 0$ , so nothing passes.
Because $I \propto A^2$ , the amplitude form is $A = A_0\cos\theta$ .
Rotate the filter below to see the intensity change as $\cos^2\theta$ .

angle θ

θ = 30°·cos²θ = 0.750·I = I₀ cos²θ = 0.750·I₀

Worked example

One filter at an angle

Vertically polarised light of intensity $I_0$ meets a filter whose axis is 60° from the vertical. Find the transmitted intensity.

$I = I_0\cos^2 60^\circ = I_0\times(0.5)^2$ .
$I = 0.25\,I_0$ .

Answer

I = \tfrac{1}{4} I_0

Two filters, and unpolarised light

With two filters, apply Malus's law at each one in turn, using the angle between successive axes.
If the light starts unpolarised, the first filter always passes exactly half the intensity (the average of $\cos^2\theta$ over all angles), and the light is polarised from then on.

Worked example

Two filters in a row

Vertically polarised light of intensity $I_0$ passes a filter with its axis 50° from the vertical, then a second filter with its axis 20° from the vertical. Find the final intensity.

First filter (50° from the polarisation): $I_1 = I_0\cos^2 50^\circ$ .
The second axis is $50^\circ - 20^\circ = 30^\circ$ from the first: $I_2 = I_1\cos^2 30^\circ$ .
$I_2 = I_0\cos^2 50^\circ\cos^2 30^\circ \approx I_0(0.413)(0.75) \approx 0.31\,I_0$ .

Answer

I_2 \approx 0.31\,I_0

(9702/11/O/N/25 Q26)

Common mistake

The angle

\theta

is always measured between the light's polarisation direction and the filter's axis, not always from the vertical. With two filters, the second angle is the angle between the two axes.

Your turn— tap to reveal the worked answer (9702/22/O/N/23 Q5)

A polarising filter is slowly rotated through 360° in a beam of plane-polarised light. Sketch how the transmitted intensity varies with angle. (9702/22/O/N/23 Q5)

Answer: a $\cos^2\theta$ curve. It starts at maximum, drops to zero at 90°, back to maximum at 180°, zero at 270°, and maximum again at 360°, never going negative.

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Malus's Law

Once light is plane-polarised, a second filter cuts its intensity by $\cos^2\theta$ , where $\theta$ is the angle between the light's polarisation and the filter's axis.
This rule is Malus's law.

Malus's law

I = I_0\cos^2\theta

Symbols

$I$ = intensity passed by the filter (W m⁻²)
$I_0$ = intensity of the polarised light hitting the filter (W m⁻²)
$\theta$ = angle between the polarisation direction and the filter axis (degrees)

$\theta = 0^\circ$ (axis lined up): $\cos^2 0 = 1$ , so all the light passes.
$\theta = 90^\circ$ (crossed): $\cos^2 90 = 0$ , so nothing passes.
Because $I \propto A^2$ , the amplitude form is $A = A_0\cos\theta$ .
Rotate the filter below to see the intensity change as $\cos^2\theta$ .

angle θ

θ = 30°·cos²θ = 0.750·I = I₀ cos²θ = 0.750·I₀

Worked example

One filter at an angle

Vertically polarised light of intensity $I_0$ meets a filter whose axis is 60° from the vertical. Find the transmitted intensity.

$I = I_0\cos^2 60^\circ = I_0\times(0.5)^2$ .
$I = 0.25\,I_0$ .

Answer

I = \tfrac{1}{4} I_0

Two filters, and unpolarised light

With two filters, apply Malus's law at each one in turn, using the angle between successive axes.
If the light starts unpolarised, the first filter always passes exactly half the intensity (the average of $\cos^2\theta$ over all angles), and the light is polarised from then on.

Worked example

Two filters in a row

Vertically polarised light of intensity $I_0$ passes a filter with its axis 50° from the vertical, then a second filter with its axis 20° from the vertical. Find the final intensity.

First filter (50° from the polarisation): $I_1 = I_0\cos^2 50^\circ$ .
The second axis is $50^\circ - 20^\circ = 30^\circ$ from the first: $I_2 = I_1\cos^2 30^\circ$ .
$I_2 = I_0\cos^2 50^\circ\cos^2 30^\circ \approx I_0(0.413)(0.75) \approx 0.31\,I_0$ .

Answer

I_2 \approx 0.31\,I_0

(9702/11/O/N/25 Q26)

Common mistake

The angle

\theta

is always measured between the light's polarisation direction and the filter's axis, not always from the vertical. With two filters, the second angle is the angle between the two axes.

Your turn— tap to reveal the worked answer (9702/22/O/N/23 Q5)

A polarising filter is slowly rotated through 360° in a beam of plane-polarised light. Sketch how the transmitted intensity varies with angle. (9702/22/O/N/23 Q5)

Answer: a $\cos^2\theta$ curve. It starts at maximum, drops to zero at 90°, back to maximum at 180°, zero at 270°, and maximum again at 360°, never going negative.