How to Sketch Intensity Against Analyser Angle in HSC Physics

A light sensor sits behind a fixed polariser and a rotatable analyser. As you turn the analyser, the measured intensity falls, reaches zero, and then increases again. What should the graph of intensity against analyser angle actually look like?

The answer is a smooth cosine-squared curve. Intensity is maximum when the analyser and incoming polarisation are parallel, falls to zero when they are perpendicular, and returns to maximum after a further \(90^\circ\) of rotation.

The important detail is how you label the horizontal axis. Use the angle between the incoming polarisation direction and the analyser’s transmission axis, not automatically the analyser’s absolute angle from the horizontal.

There is another detail that catches students out: the maximum intensity on your graph is the intensity reaching the analyser, not necessarily the original intensity from the light source.

01Understand the shape of the graph

Start with plane-polarised light entering an ideal analyser.

The analyser only transmits the component of the electric field parallel to its transmission axis. As the analyser rotates, this component changes.

Malus’ law describes the resulting intensity:

\[ I=I_{\mathrm{pre}}\cos^2\theta \]

Where:

• \(I\) is the intensity transmitted through the analyser, measured in \(\mathrm{W\,m^{-2}}\).
• \(I_{\mathrm{pre}}\) is the intensity immediately before the analyser.
• \(\theta\) is the angle between the incoming polarisation direction and the analyser’s transmission axis.

For this graph, \(I_{\mathrm{pre}}\) stays constant while you rotate the analyser. It is the angle that changes.

What happens at important angles?

Before drawing a smooth curve, identify the positions where you know the intensity exactly.

Three positions are particularly important.

At \(0^\circ\): maximum intensity.

The transmission axes are parallel. The analyser allows the entire incoming electric-field component through.

At \(90^\circ\): zero intensity.

The transmission axes are perpendicular. There is no component of the incoming electric field parallel to the analyser’s axis.

At \(180^\circ\): maximum intensity again.

The axes are parallel again. Remember that a polariser’s transmission axis is a line, not an arrow. An axis at \(180^\circ\) is physically identical to one at \(0^\circ\).

Notice that the graph repeats every \(180^\circ\), not every \(360^\circ\).

This is a common sketching mistake. Students sometimes draw one complete cycle over \(360^\circ\), forgetting that the transmission axis has no preferred arrow direction.

Also, the graph is not a straight line between its maximum and minimum. It follows a smooth cosine-squared relationship.

02Use relative angles, not automatically absolute angles

This is probably the most important decision when interpreting an HSC question.

Suppose a fixed polariser has its transmission axis at \(30^\circ\) above the horizontal. The analyser rotates, and its angle is measured from the horizontal.

Should your graph have its maximum at \(0^\circ\)?

No. The maximum occurs when the analyser aligns with the incoming polarisation direction, which is \(30^\circ\).

There are two ways to describe the horizontal axis.

Option A: Relative angle

Let \(\theta\) represent the angle between the incoming polarisation direction and the analyser.

Your equation is simply:

\[ I(\theta)=I_{\mathrm{pre}}\cos^2\theta \]

The maximum occurs at \(0^\circ\), and the minimum occurs at \(90^\circ\).

This is usually the easiest way to sketch the relationship.

Option B: Absolute analyser angle

Let \(\alpha\) represent the analyser’s angle measured anticlockwise from the horizontal.

Let \(\phi\) represent the fixed incoming polarisation direction, measured from the same reference.

Then:

\[ I(\alpha)=I_{\mathrm{pre}}\cos^2(\alpha-\phi) \]

The graph has the same shape, but it is shifted horizontally.

For example, if \(\phi=30^\circ\):

• Maximum intensity occurs at \(\alpha=30^\circ\).
• Zero intensity occurs at \(\alpha=120^\circ\).
• Maximum intensity returns at \(\alpha=210^\circ\).

The key is to check what the question actually asks you to plot.

If it asks for intensity against the angle between the polarisers, use relative angle. If it asks for intensity against the analyser’s absolute orientation, shift the curve to match the incoming polarisation direction.

03Track the intensity before each filter

Now for the other major source of mistakes: choosing the correct maximum intensity.

Imagine a beam travelling through several fixed polarising filters before reaching a rotatable analyser.

You cannot automatically use the original source intensity as the maximum of your graph.

You need to calculate how much intensity reaches the analyser first.

Consider this sequence:

\[ \text{Source}\rightarrow P_1\rightarrow P_2\rightarrow A \]

Here:

• \(P_1\) is the first fixed polariser.
• \(P_2\) is the second fixed polariser.
• \(A\) is the rotatable analyser.

Track the intensity at each stage.

The important idea is that each polariser changes the polarisation direction of the surviving light.

After passing through \(P_1\), the light is polarised along \(P_1\). After passing through \(P_2\), it is polarised along \(P_2\).

Therefore, the rotatable analyser compares its transmission axis with \(P_2\), not with the original source or \(P_1\).

The maximum of your graph is the intensity immediately before \(A\).

A useful rule for multiple filters

For each fixed filter:

1. Identify the intensity immediately before it.
2. Find the angle between the incoming polarisation direction and its transmission axis.
3. Calculate the transmitted intensity.
4. Update the polarisation direction to the new filter’s axis.
5. Repeat until you reach the rotatable analyser.

If the original light is already plane-polarised, apply Malus’ law at the first filter. Do not halve its intensity.

The half-intensity rule applies when unpolarised light first encounters an ideal polariser.

For a more detailed treatment of this sequence, see Malus’ Law for HSC Physics: Intensity and Polariser Angles.

04A contrasting case: what if the incoming light is unpolarised?

Here is a situation where drawing a cosine-squared curve would actually be incorrect.

Suppose unpolarised light travels directly through a single rotatable analyser.

Would rotating the analyser produce a maximum at \(0^\circ\) and zero intensity at \(90^\circ\)?

No. The transmitted intensity stays constant at half the original intensity.

Why?

Unpolarised light does not have one fixed polarisation direction. An ideal analyser transmits half of its intensity, regardless of its orientation.

Therefore:

\[ I=\frac{I_0}{2} \]

The graph is a horizontal line.

Now add a fixed polariser before the analyser. The first filter produces plane-polarised light, giving the analyser a definite incoming polarisation direction.

The graph now follows:

\[ I= \frac{I_0}{2}\cos^2\theta \]

This is why a fixed polariser is needed to demonstrate the angle-dependent Malus’ law relationship using initially unpolarised light.

The difference is worth remembering:

This is a useful diagnostic when an exam question gives you an unfamiliar arrangement of filters.

05Worked examples

06A reliable method for any HSC sketching question

When you see a question asking you to sketch intensity against analyser angle, work through these decisions in order.

First: Identify the light entering the analyser.

Is it unpolarised or already plane-polarised? If it is unpolarised and meets its first ideal polariser, halve its intensity.

Second: Track the intensity through every fixed filter.

Apply Malus’ law using the angle between consecutive transmission axes. Update the polarisation direction after each filter.

Third: Identify the horizontal axis.

• Relative angle: maximum at \(0^\circ\), minimum at \(90^\circ\).
• Absolute analyser angle: maximum when the analyser aligns with the last incoming polarisation direction.

Fourth: Identify the vertical scale.

The maximum intensity is the intensity immediately before the rotatable analyser.

Fifth: Mark the important points.

At minimum, identify the maximum, the zero-intensity point, and the next maximum. Add useful intermediate points such as \(45^\circ\) and \(135^\circ\).

Finally: Draw a smooth cosine-squared curve.

Remember the \(180^\circ\) period.

07Questions and solutions

Question 1

An unpolarised light source produces an intensity of \(80\ \mathrm{W\,m^{-2}}\). The light passes through a fixed ideal polariser and then a rotatable ideal analyser.

The relative angle between the two transmission axes is \(\theta\).

Calculate:

a) The intensity immediately before the analyser.

b) The intensity transmitted when \(\theta=60^\circ\).

c) The maximum and minimum intensities, including the angles at which they occur.

d) The shape of the intensity-versus-angle graph between \(0^\circ\) and \(180^\circ\).

Solution 1

The maximum intensity is \(40\ \mathrm{W\,m^{-2}}\), the minimum is zero, and the intensity at \(60^\circ\) is \(10\ \mathrm{W\,m^{-2}}\).

a) Intensity before the analyser

The first polariser receives unpolarised light, so it transmits half:

\[ \begin{aligned} I_{\mathrm{pre}}&=\frac{I_0}{2}\\ &=\frac{80}{2}\\ &=40\ \mathrm{W\,m^{-2}} \end{aligned} \]

b) Intensity at \(60^\circ\)

Apply Malus’ law:

\[ \begin{aligned} I&=I_{\mathrm{pre}}\cos^2\theta\\ &=40\cos^2(60^\circ)\\ &=40(0.25)\\ &=10\ \mathrm{W\,m^{-2}} \end{aligned} \]

c) Maximum and minimum

The maximum occurs when the axes are parallel:

\[ I_{\max}=40\ \mathrm{W\,m^{-2}} \]

This happens at \(0^\circ\) and \(180^\circ\).

The minimum occurs when the axes are perpendicular:

\[ I_{\min}=0\ \mathrm{W\,m^{-2}} \]

This happens at \(90^\circ\).

d) Graph

The graph starts at 40, decreases smoothly to zero at \(90^\circ\), and rises smoothly back to 40 at \(180^\circ\).

The curve is symmetric about \(90^\circ\).

The important point is that the maximum is 40, not 80. The first polariser has already reduced the incoming intensity.

Question 2

A beam of plane-polarised light has an intensity of \(48\ \mathrm{W\,m^{-2}}\), with its polarisation direction horizontal.

Two experiments are performed.

• Run A: The beam passes directly through a rotatable analyser.
• Run B: An additional fixed ideal polariser \(Q\) is inserted before the analyser.

The analyser’s absolute angle \(\alpha\) is measured anticlockwise from the horizontal.

In Run B, the intensity graph has a maximum of \(12\ \mathrm{W\,m^{-2}}\) at \(\alpha=60^\circ\), and zero intensity at \(\alpha=150^\circ\).

Determine:

a) The transmission axis of \(Q\).

b) The equation for the intensity in Run B.

c) The intensity when \(\alpha=105^\circ\).

d) Why the height and horizontal position of the maximum provide different information.

Solution 2

The transmission axis of \(Q\) is at \(60^\circ\), the Run B equation is \(I=12\cos^2(\alpha-60^\circ)\), and the intensity at \(105^\circ\) is \(6\ \mathrm{W\,m^{-2}}\).

a) Find the transmission axis

The incoming light is horizontally polarised, so its direction is \(0^\circ\).

The maximum intensity occurs when the analyser is parallel to the polarisation direction of the light reaching it.

Since the maximum occurs at \(\alpha=60^\circ\), the light leaving \(Q\) must be polarised at \(60^\circ\).

Therefore, \(Q\) has its transmission axis at \(60^\circ\).

There is another useful way to check this using the maximum intensity.

b) Write the intensity equation

The intensity entering the analyser in Run B is the maximum transmitted intensity:

\[ I_{\mathrm{pre}}=12\ \mathrm{W\,m^{-2}} \]

The incoming polarisation direction is \(60^\circ\), so:

\[ I(\alpha)=12\cos^2(\alpha-60^\circ) \]

This graph has a maximum at \(60^\circ\) and a minimum at \(150^\circ\), matching the supplied information.

We can also check the transmission through \(Q\):

\[ \begin{aligned} 12&=48\cos^2\Delta\theta\\ \frac{12}{48}&=\cos^2\Delta\theta\\ 0.25&=\cos^2\Delta\theta \end{aligned} \]

Therefore:

\[ \Delta\theta=60^\circ \]

The intensity ratio confirms that the filter is \(60^\circ\) from the original polarisation direction.

Importantly, the ratio alone would also allow a transmission axis at \(120^\circ\). The position of the graph’s maximum resolves this ambiguity.

c) Calculate the intensity at \(105^\circ\)

The relative angle is:

\[ \theta=105^\circ-60^\circ=45^\circ \]

Therefore:

\[ \begin{aligned} I&=12\cos^2(45^\circ)\\ &=12(0.50)\\ &=6\ \mathrm{W\,m^{-2}} \end{aligned} \]

d) Interpret the graph

The height of the maximum tells us the intensity reaching the analyser.

The horizontal position of the maximum tells us the direction of the incoming polarisation.

These are two separate pieces of information. A graph can shift horizontally without changing its maximum height, or change its maximum height without changing the position of its peak.

Question 3

An unpolarised light source produces an intensity of \(120\ \mathrm{W\,m^{-2}}\).

Two arrangements are tested:

• Arrangement A: The light passes directly through a rotatable ideal analyser.
• Arrangement B: The light first passes through a fixed ideal polariser, then through the same type of rotatable analyser.

Both arrangements use the same source.

Sketch both intensity-versus-angle graphs on the same axes, between \(0^\circ\) and \(180^\circ\).

Calculate the transmitted intensity for each arrangement at:

a) \(0^\circ\).

b) \(45^\circ\).

c) \(90^\circ\).

Explain why the graphs have different shapes.

Solution 3

Arrangement A produces a constant intensity of \(60\ \mathrm{W\,m^{-2}}\), while Arrangement B produces a cosine-squared curve with a maximum of \(60\ \mathrm{W\,m^{-2}}\) and a minimum of zero.

Arrangement A: no fixed polariser

The analyser receives unpolarised light.

An ideal analyser transmits half of the incoming intensity, regardless of its orientation.

\[ I_A=\frac{120}{2} \]

\[ I_A=60\ \mathrm{W\,m^{-2}} \]

Therefore, its graph is a horizontal line at 60.

The readings are:

Arrangement B: fixed polariser before the analyser

The first polariser transmits half of the unpolarised light:

\[ I_{\mathrm{pre}}=\frac{120}{2}=60\ \mathrm{W\,m^{-2}} \]

The light is now plane-polarised, so Malus’ law applies:

\[ I_B=60\cos^2\theta \]

At \(0^\circ\):

\[ I_B=60(1)=60\ \mathrm{W\,m^{-2}} \]

At \(45^\circ\):

\[ I_B=60(0.5)=30\ \mathrm{W\,m^{-2}} \]

At \(90^\circ\):

\[ I_B=60(0)=0\ \mathrm{W\,m^{-2}} \]

The two graphs therefore share the same starting intensity but behave differently as the analyser rotates.

Why?

Arrangement A has no fixed incoming polarisation direction. Rotating the analyser does not change the fraction transmitted.

Arrangement B creates a fixed polarisation direction before the analyser. The relative angle now determines how much intensity passes through.

This is why you must identify the state of the light entering the analyser before deciding what graph to draw.

Question 4

A beam of plane-polarised light has an intensity of \(50\ \mathrm{W\,m^{-2}}\) immediately before a rotatable analyser.

The fixed polariser has its transmission axis at \(20^\circ\) above the horizontal.

The analyser initially has its axis at \(150^\circ\). It is then rotated clockwise.

Let \(\delta\) represent the clockwise rotation from its starting position, with \(\delta=0^\circ\) initially.

Sketch the intensity against \(\delta\) from \(0^\circ\) to \(180^\circ\).

Determine:

a) The analyser’s absolute angle after a clockwise rotation of \(30^\circ\).

b) The intensity after this rotation.

c) The angles of maximum and minimum intensity on your graph.

Solution 4

The intensity after a \(30^\circ\) clockwise rotation is approximately \(1.51\ \mathrm{W\,m^{-2}}\). The graph reaches its minimum at \(\delta=40^\circ\) and its maximum at \(\delta=130^\circ\).

a) Find the new absolute analyser angle

Clockwise rotation reduces the absolute angle.

Therefore:

\[ \begin{aligned} \alpha&=150^\circ-30^\circ\\ &=120^\circ \end{aligned} \]

The analyser is now at \(120^\circ\).

b) Calculate the transmitted intensity

The incoming polarisation direction is still \(20^\circ\).

The absolute difference between the two axes is:

\[ \theta=120^\circ-20^\circ=100^\circ \]

However, transmission axes are lines, so the smaller angle between them is:

\[ \theta=180^\circ-100^\circ=80^\circ \]

Apply Malus’ law:

\[ \begin{aligned} I&=50\cos^2(80^\circ)\\ &=50(0.03015)\\ &=1.5075\ \mathrm{W\,m^{-2}} \end{aligned} \]

Therefore:

\[ I\approx1.51\ \mathrm{W\,m^{-2}} \]

Notice that using the analyser’s rotation of \(30^\circ\) directly in Malus’ law would give the wrong answer.

The angle through which the analyser rotates is not necessarily the angle between the two transmission axes.

c) Find the maximum and minimum

At any point:

\[ \alpha=150^\circ-\delta \]

The incoming polarisation direction remains \(20^\circ\), so:

\[ I(\delta)=50\cos^2(130^\circ-\delta) \]

The maximum occurs when the analyser aligns with the fixed polariser.

This happens when:

\[ 130^\circ-\delta=0^\circ \]

Therefore:

\[ \delta=130^\circ \]

The maximum intensity is:

\[ I_{\max}=50\ \mathrm{W\,m^{-2}} \]

The minimum occurs when the analyser is perpendicular to the fixed polariser:

\[ 130^\circ-\delta=90^\circ \]

Therefore:

\[ \delta=40^\circ \]

And:

\[ I_{\min}=0 \]

Useful graph points are:

The graph falls to zero at \(40^\circ\), rises to its maximum at \(130^\circ\), and then begins falling again.

The important distinction is between the analyser’s absolute orientation, its rotation from the starting position, and its relative angle to the incoming polarisation.

Question 5

An unpolarised light source produces an intensity of \(160\ \mathrm{W\,m^{-2}}\).

The light passes through two fixed ideal polarisers, \(P_1\) and \(P_2\), followed by a rotatable ideal analyser \(A\).

The transmission axis of \(P_1\) is fixed at \(20^\circ\) above the horizontal. The transmission axis of \(P_2\) is initially unknown.

The detector produces an intensity-versus-analyser-angle graph with the following features:

• Maximum intensity of \(20\ \mathrm{W\,m^{-2}}\) at an absolute analyser angle of \(80^\circ\).
• Zero intensity at an absolute analyser angle of \(170^\circ\).
• The pattern repeats every \(180^\circ\).

Assume the source is steady, all filters are ideal, and there is no background light.

Part A

Determine the transmission-axis angle of \(P_2\).

Calculate the intensity immediately after \(P_1\) and immediately after \(P_2\). Check whether your result agrees with the graph’s maximum.

Part B

Now rotate \(P_2\) until its transmission axis is at \(110^\circ\), while keeping \(P_1\) fixed at \(20^\circ\).

Predict the new intensity-versus-analyser-angle graph.

Explain why rotating the analyser cannot recover the original transmitted intensity, even when the analyser is aligned with one of the fixed polarisers.

Solution 5

The original transmission axis of \(P_2\) is at \(80^\circ\), and the intensity after the two fixed polarisers is \(20\ \mathrm{W\,m^{-2}}\). When \(P_2\) is rotated to \(110^\circ\), the new graph is a horizontal line at zero intensity.

Part A: Determine the original orientation of \(P_2\)

The maximum of the graph occurs when the analyser is aligned with the polarisation direction of the incoming light.

The last fixed polariser before the analyser is \(P_2\).

Therefore, the maximum at an analyser angle of \(80^\circ\) tells us that the transmission axis of \(P_2\) is also at \(80^\circ\).

This gives:

\[ \phi_2=80^\circ \]

Calculate the intensity after \(P_1\)

The source is unpolarised, so the first ideal polariser transmits half:

\[ \begin{aligned} I_1&=\frac{I_0}{2}\\ &=\frac{160}{2}\\ &=80\ \mathrm{W\,m^{-2}} \end{aligned} \]

The surviving light is now polarised at \(20^\circ\).

Calculate the intensity after \(P_2\)

The relative angle between the fixed polarisers is:

\[ \Delta\theta=80^\circ-20^\circ=60^\circ \]

Apply Malus’ law:

\[ \begin{aligned} I_2&=80\cos^2(60^\circ)\\ &=80(0.25)\\ &=20\ \mathrm{W\,m^{-2}} \end{aligned} \]

This agrees exactly with the maximum shown on the detector graph.

The intensity immediately before the analyser is therefore:

\[ I_{\mathrm{pre}}=20\ \mathrm{W\,m^{-2}} \]

The original graph can be described by:

\[ I(\alpha)=20\cos^2(\alpha-80^\circ) \]

The maximum occurs at \(80^\circ\), while the minimum occurs at \(170^\circ\).

The graph’s horizontal position identifies the final incoming polarisation direction. Its maximum height identifies the intensity reaching the analyser.

Part B: Rotate \(P_2\) to \(110^\circ\)

This is where the problem changes.

The first polariser remains at \(20^\circ\), while the second polariser is now at \(110^\circ\).

Their relative angle is:

\[ \Delta\theta=110^\circ-20^\circ=90^\circ \]

The two fixed polarisers are now crossed.

Apply Malus’ law to \(P_2\):

\[ \begin{aligned} I_2&=80\cos^2(90^\circ)\\ &=80(0)\\ &=0\ \mathrm{W\,m^{-2}} \end{aligned} \]

No light reaches the rotatable analyser.

Therefore:

\[ I(\alpha)=0 \]

for every analyser angle.

The new graph is a horizontal line along the zero-intensity axis.

Why can’t rotating the analyser restore the light?

A tempting prediction is that the analyser should transmit some light when it aligns with either \(P_1\) or \(P_2\).

But this ignores what happens before the analyser.

The crossed fixed polarisers have already reduced the transmitted intensity to zero. The analyser cannot recover light that never reaches it.

This is an important distinction between changing the orientation of the final analyser and changing the arrangement of the filters before it.

In the original experiment, rotating \(P_2\) changed both the intensity reaching the analyser and the direction of its incoming polarisation. In the modified experiment, the intensity reaching the analyser becomes zero, so the entire angle-dependent curve disappears.

That is why a good sketch starts by tracking the light through every fixed filter, before considering the analyser’s rotation.

The next useful step is to apply the same reasoning to experimental intensity-angle data: compare the measured curve with the predicted cosine-squared relationship, and consider whether any non-zero reading at crossed polarisers comes from imperfect filters or background light.