Explain polarisation of light as evidence for the transverse-wave nature of light, and apply Malus's law $I = I_0 \cos^2(\theta)$ to determine the intensity of light transmitted by an ideal polariser

What this dot point is asking

VCAA wants you to explain polarisation as evidence for the transverse-wave nature of light, apply Malus's law to compute the intensity of polarised light transmitted through a polariser, and handle both the unpolarised-then-polariser and polariser-then-polariser cases. Cross-link: see the Malus law calculator.

What is polarised light

Light is a transverse electromagnetic wave. The electric field $\vec{E}$ oscillates perpendicular to the direction of propagation. The plane in which the electric field oscillates is the plane of polarisation.

Unpolarised light has electric field oscillations randomly distributed over all directions perpendicular to the propagation direction. A typical light source (incandescent bulb, sun) emits unpolarised light because the many radiating atoms are oriented randomly.

Polarised light has electric field oscillations confined to one specific direction. Polarised light can be produced by:

• Passing unpolarised light through a polarising filter. The filter transmits the component of $\vec{E}$ along its transmission axis and absorbs the perpendicular component.
• Reflection. Light reflecting off non-metallic surfaces (water, glass) becomes partially or fully polarised, especially at Brewster's angle.
• Scattering. Light scattered through 90 degrees by atmospheric molecules is largely polarised (the blue sky has measurable polarisation).
• Specific sources. Lasers (depending on type), some LCDs.

Polarisation as evidence for transverse-wave nature

Only transverse waves can be polarised. A longitudinal wave (compression-rarefaction along the direction of travel) has no direction-of-oscillation choice perpendicular to the propagation direction; rotating a "longitudinal polariser" would have no effect.

The observation that light's intensity through a rotating polariser changes with angle is therefore direct evidence that light is a transverse wave. The first polariser-based experiments (early 1800s) established the transverse nature of light, supporting the wave model.

Malus's law

When polarised light of intensity $I_0$ passes through an ideal polariser whose transmission axis makes angle $\theta$ with the polarisation direction of the incoming light, the transmitted intensity is:

This is Malus's law.

Why $\cos^2$?

The electric field amplitude after the polariser is the projection of the incoming field onto the transmission axis: $E = E_0 \cos(\theta)$.

Intensity is proportional to the square of the field amplitude: $I \propto E^2$.

So $I / I_0 = (E / E_0)^2 = \cos^2(\theta)$.

Limit cases

• $\theta = 0^\circ$: $\cos^2 = 1$, full transmission. The polariser's axis aligns with the light's polarisation.
• $\theta = 90^\circ$: $\cos^2 = 0$, no transmission. The axes are crossed.
• $\theta = 45^\circ$: $\cos^2 = 0.5$, half transmission.

Common values

The unpolarised-then-polariser rule

When unpolarised light passes through an ideal polariser, the transmitted intensity is half the incoming intensity:

This is because unpolarised light has equal contributions from all possible polarisation directions. The polariser transmits only the component along its axis, which averages to half over all orientations.

After the first polariser, the light is polarised along the polariser's axis. Subsequent polarisers behave according to Malus's law with $\theta$ measured from the previous polariser's axis.

Two-polariser problems

The standard VCAA exam polarisation problem (Section B) involves an unpolarised source followed by two (or more) polarisers.

Procedure

1. First polariser, unpolarised input. $I_1 = I_0 / 2$.
2. Second polariser, polarised input. $I_2 = I_1 \cos^2(\theta_{12})$ where $\theta_{12}$ is the angle between the first and second polarisers' axes.
3. Third polariser, if present. $I_3 = I_2 \cos^2(\theta_{23})$ where $\theta_{23}$ is between the second and third axes.

The angle in each Malus application is the angle between adjacent polarisers, not the cumulative angle from the source.

Worked example. Three polarisers

Unpolarised light of intensity $I_0$ passes through three polarisers oriented at $0^\circ$, $45^\circ$ and $90^\circ$.

After first polariser: $I_1 = I_0 / 2$.

After second polariser: $\theta_{12} = 45^\circ$, so $I_2 = (I_0 / 2) \cos^2(45^\circ) = (I_0 / 2) (1/2) = I_0 / 4$.

After third polariser: $\theta_{23} = 45^\circ$ (between 45 and 90), so $I_3 = (I_0 / 4) \cos^2(45^\circ) = (I_0 / 4)(1/2) = I_0 / 8$.

Final intensity is $I_0 / 8$.

Note: removing the middle polariser would give $I_3 = (I_0 / 2) \cos^2(90^\circ) = 0$. A counter-intuitive result: inserting an extra polariser between two crossed polarisers actually increases the transmitted intensity. This is the classic three-polariser demonstration of the quantum nature of polarisation (each polariser projects, not filters; the second polariser "rotates" the polarisation toward its axis).

Applications

• Polarising sunglasses reduce glare from horizontal surfaces (water, roads). The glare is largely horizontally polarised by reflection at grazing angles, so vertical-axis polarising lenses block it.
• LCD displays use crossed polarisers with liquid-crystal rotators in between. Applying voltage to a pixel rotates the polarisation, changing transmitted intensity.
• Stress analysis. Stressed transparent plastics become birefringent and show coloured patterns under polarised light, revealing stress concentrations.
• 3D cinema. Some 3D systems use orthogonal circular polarisations for the two eyes.

Examples in context

Example 1. Polarised sunglasses on a Bass Strait fishing boat. Light reflected from the sea off Bass Strait is partially polarised horizontally (Brewster's effect at the air-water interface). Anglers wear polarised sunglasses oriented with a vertical transmission axis, blocking the horizontally polarised reflected glare. If the reflected light is $80\%$ horizontally polarised with intensity $1000$ W m$^{-2}$, the polarised lens transmits $I_0 \cos^2 \theta$ from the polarised component and half the unpolarised. So total transmitted $= 800 \times 0 + 200 \times 0.5 = 100$ W m$^{-2}$, a tenfold reduction. The same sunglasses rotated $90^\circ$ would transmit $800 \times 1 + 100 = 900$ W m$^{-2}$, almost no reduction.

Example 2. LCD pixel modulation in a Federation Square display. Federation Square's outdoor LCD displays use crossed polarisers with a liquid-crystal layer in between. Backlight passes through a vertical polariser ($I_0$), then enters the liquid crystal. With no voltage applied, the crystal rotates the polarisation by $90^\circ$, so it passes the horizontal output polariser. With voltage applied, the crystal axis aligns and the polarisation is not rotated, so the output polariser blocks it ($\cos 90^\circ = 0$). Partial-voltage states produce intermediate intensities $I = I_0 \cos^2(\theta)$ where $\theta$ varies from $0$ to $90^\circ$, giving each pixel a continuous greyscale.

Try this

Q1. State Malus's law and explain why it provides evidence for the transverse-wave nature of light. [2 marks]

• Cue. $I = I_0 \cos^2\theta$. Light intensity transmitted by an ideal polariser depends on angle between the wave's E-field direction and the polariser axis; only transverse waves can be polarised.

Q2. Unpolarised light of intensity $200$ W m$^{-2}$ passes through two polarisers with axes at $30^\circ$ to each other. Calculate the intensity after each polariser. [4 marks]

• Cue. After first: $100$ W m$^{-2}$ (half). After second: $100 \cos^2 30^\circ = 100 \times 0.75 = 75$ W m$^{-2}$.

Q3. Refer to LCD pixel modulation. (a) Outline the function of the input and output polarisers. (b) Calculate the transmitted intensity if the liquid-crystal rotates polarisation by $30^\circ$, given input intensity $100$ W m$^{-2}$ after first polariser. (c) Explain how greyscale modulation works. [2+2+2 marks]

• Cue. (a) Input polariser selects single direction; output polariser is crossed and blocks unrotated light. (b) Crystal axes give crossed transmission of $\cos^2(90 - 30) = \cos^2 60 = 0.25$ so $25$ W m$^{-2}$. (c) Varying voltage varies rotation angle continuously between $0$ and $90^\circ$.