Paper 2

Physics9702

Paper 2

Physics9702

Malus's Law

Plane-polarised light already vibrates in one direction. A second filter passes most light when its axis points in that direction.
As the second filter turns away, less light passes. When the two directions are perpendicular, no light passes.

First watch the filter turn

0° — aligned

The filter axis matches the light's polarisation. All the intensity passes.

45° — halfway

The transmitted intensity is half the incident intensity.

90° — crossed

The filter axis is perpendicular to the polarisation. No intensity passes.

angle θ

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

The angle is measured between the incoming light's polarisation direction and the new filter's axis.

Malus's law

Malus's law turns the pattern above into an exact intensity rule.

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)

Because $I \propto A^2$ , the amplitude form is $A = A_0\cos\theta$ .

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

Successive filters

With two filters, apply Malus's law at each one in turn, using the angle between successive axes.

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 turn9702/23/M/J/25 Q5(b)

A filter starts with its axis perpendicular to the incident polarisation. It is rotated through 20°. Calculate the ratio of transmitted to incident amplitude. (9702/23/M/J/25 Q5(b))

Show worked answer

The angle between the final axis and the incident polarisation is $90^\circ-20^\circ=70^\circ$ .
$I/I_0=\cos^270^\circ=0.117$ .
Since $I\propto A^2$ , $A/A_0=\sqrt{0.117}=0.34$ .

Answer

A/A_0=0.34

PreviousPolarisation NextThe Principle of Superposition

Malus's Law

Plane-polarised light already vibrates in one direction. A second filter passes most light when its axis points in that direction.
As the second filter turns away, less light passes. When the two directions are perpendicular, no light passes.

First watch the filter turn

0° — aligned

The filter axis matches the light's polarisation. All the intensity passes.

45° — halfway

The transmitted intensity is half the incident intensity.

90° — crossed

The filter axis is perpendicular to the polarisation. No intensity passes.

angle θ

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

The angle is measured between the incoming light's polarisation direction and the new filter's axis.

Malus's law

Malus's law turns the pattern above into an exact intensity rule.

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)

Because $I \propto A^2$ , the amplitude form is $A = A_0\cos\theta$ .

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

Successive filters

With two filters, apply Malus's law at each one in turn, using the angle between successive axes.

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 turn9702/23/M/J/25 Q5(b)

A filter starts with its axis perpendicular to the incident polarisation. It is rotated through 20°. Calculate the ratio of transmitted to incident amplitude. (9702/23/M/J/25 Q5(b))

Show worked answer

The angle between the final axis and the incident polarisation is $90^\circ-20^\circ=70^\circ$ .
$I/I_0=\cos^270^\circ=0.117$ .
Since $I\propto A^2$ , $A/A_0=\sqrt{0.117}=0.34$ .

Answer

A/A_0=0.34

PreviousPolarisation NextThe Principle of Superposition