Polarizing Filters

A polarizing filter, or polarizer, controls light according to the orientation of its electric field. It has a preferred direction called its transmission axis.

When unpolarized light passes through an ideal polarizing filter, the emerging light becomes linearly polarized along the filter's transmission axis.

Polarizing filters allow us to investigate one of the most important properties of light: light is a transverse electromagnetic wave.

They also have many practical uses, including:

• polarized sunglasses
• photography
• LCD screens
• scientific instruments
• stress analysis
• optical experiments

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How a Polarizing Filter Works

Light is an electromagnetic wave.

For unpolarized light, the electric field oscillates in many different directions perpendicular to the direction of travel.

A linear polarizer preferentially transmits the component of the electric field that lies along its transmission axis.

The perpendicular component is strongly reduced.

Therefore:

unpolarized light → polarizer → linearly polarized light

If the transmission axis is vertical, the emerging light is vertically polarized.

If the filter is rotated, the polarization direction of the transmitted light rotates with it.

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The Transmission Axis

The transmission axis specifies the electric-field direction that an ideal linear polarizer transmits.

For example, consider a vertical polarizer.

The incoming electric field can be resolved into:

• vertical component
• horizontal component

The vertical component is transmitted.

The horizontal component is blocked or strongly absorbed.

The outgoing light is therefore vertically polarized.

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Unpolarized Light Through One Polarizer

For ideal unpolarized light passing through an ideal polarizer, the average transmitted intensity is:

I = I₀/2

where:

• I₀ = incident intensity
• I = transmitted intensity

Therefore, approximately half of the incident intensity is transmitted.

The transmitted light is now linearly polarized.

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Why Half the Intensity Is Transmitted

The electric-field directions in unpolarized light are randomly distributed.

For a particular electric-field orientation θ relative to the transmission axis, the transmitted electric-field component is proportional to:

E cos θ

Intensity is proportional to the square of electric-field amplitude:

I ∝ E²

Therefore, the transmitted fraction involves:

cos² θ

Averaging cos² θ over all possible polarization orientations gives:

1/2

Hence:

I = I₀/2

for ideal unpolarized light.

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Worked Example: One Polarizer

Unpolarized light has intensity:

I₀ = 300 W/m²

It passes through an ideal polarizer.

Therefore:

I = I₀/2

I = 300/2

I = 150 W/m²

The emerging light has an intensity of 150 W/m² and is linearly polarized along the filter's transmission axis.

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Two Polarizers

Now suppose polarized light passes through a second polarizing filter.

The second filter is often called an analyzer.

The amount of light transmitted depends on the angle θ between:

• the polarization direction of the incoming light
• the transmission axis of the analyzer

This relationship is described by Malus's law.

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Malus's Law

For linearly polarized light passing through an ideal analyzer:

I = I₀ cos² θ

where:

• I = transmitted intensity
• I₀ = intensity entering the analyzer
• θ = angle between the incoming polarization direction and analyzer axis

This equation allows the transmitted intensity to be predicted for any analyzer angle.

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Parallel Polarizers

Suppose the first and second polarizers have parallel transmission axes.

Then:

θ = 0°

Using Malus's law:

I = I₀ cos² 0°

Since:

cos 0° = 1

then:

I = I₀

Ideally, the second polarizer does not reduce the intensity further.

Real polarizers do absorb or reflect some light, so actual transmission is lower.

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Polarizers at 30°

If:

θ = 30°

then:

I = I₀ cos² 30°

Since:

cos² 30° = 0.75

then:

I = 0.75I₀

The analyzer transmits 75% of the polarized light reaching it.

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Polarizers at 45°

If:

θ = 45°

then:

I = I₀ cos² 45°

Since:

cos² 45° = 0.50

then:

I = 0.50I₀

Half of the polarized light entering the second filter is transmitted.

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Polarizers at 60°

If:

θ = 60°

then:

I = I₀ cos² 60°

Since:

cos² 60° = 0.25

then:

I = 0.25I₀

Only one quarter of the polarized light entering the analyzer is transmitted.

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Crossed Polarizers

If the two transmission axes are perpendicular:

θ = 90°

Using Malus's law:

I = I₀ cos² 90°

Since:

cos 90° = 0

then:

I = 0

Ideally, no light is transmitted.

Two polarizers arranged at 90° are called crossed polarizers.

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Rotation and Intensity

As one polarizer is rotated relative to another, the transmitted intensity changes continuously.

The change is not linear.

For example, rotating from 0° to 45° does not reduce the intensity by 45/90 = 50% because of the angle alone.

The result follows:

cos² θ

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Intensity During a Full Rotation

If the analyzer is rotated through 360°, the intensity follows:

I = I₀ cos² θ

Maximum intensity occurs at:

0°, 180°, 360°

Minimum intensity occurs at:

90°, 270°

The pattern therefore repeats every:

180°

This periodic variation is strong experimental evidence of polarization.

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Worked Example: Two Polarizers

Unpolarized light has intensity:

I₀ = 400 W/m²

It passes through a vertical ideal polarizer.

After the first polarizer:

I₁ = I₀/2

I₁ = 200 W/m²

The light then passes through another polarizer at 60°.

Using Malus's law:

I₂ = I₁ cos²60°

I₂ = 200(0.25)

I₂ = 50 W/m²

Therefore, the final intensity is:

50 W/m²

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A Common Calculation Error

Suppose unpolarized light passes through two polarizers.

It is incorrect to immediately write:

I = I₀ cos² θ

using the original unpolarized intensity.

The first polarizer must be considered first:

I₁ = I₀/2

Then apply Malus's law to the second polarizer:

I₂ = I₁ cos² θ

Therefore:

I₂ = (I₀/2)cos² θ

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Finding the Angle

Malus's law can also be rearranged to determine the angle.

Suppose:

I = 0.25I₀

Then:

0.25 = cos² θ

Take the square root:

cos θ = 0.5

Therefore:

θ = 60°

So an analyzer at 60° transmits 25% of the incident linearly polarized intensity.

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Worked Example: Finding an Unknown Angle

Polarized light enters an analyzer with intensity:

I₀ = 120 W/m²

The transmitted intensity is:

I = 90 W/m²

Then:

I/I₀ = cos² θ

90/120 = cos² θ

0.75 = cos² θ

cos θ = √0.75

cos θ ≈ 0.866

Therefore:

θ ≈ 30°

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Three Polarizing Filters

A surprising result occurs when a third polarizer is placed between two crossed polarizers.

Suppose:

• Polarizer A = 0°
• Polarizer C = 90°

Without another filter between them:

I = 0

Now insert Polarizer B at:

45°

Some light can pass through all three filters.

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Why the Middle Polarizer Allows Light Through

After the first polarizer, the light is polarized at 0°.

The 45° polarizer transmits the component along 45°.

The emerging light is now polarized at 45°.

The final 90° polarizer can transmit the component of this 45°-polarized light along its own axis.

Therefore, the middle polarizer establishes a new polarization direction.

The light is not simply travelling unchanged through each filter.

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Worked Example: Three Polarizers

Unpolarized light has intensity:

I₀ = 160 W/m²

After the first polarizer:

I₁ = 160/2

I₁ = 80 W/m²

The second polarizer is at 45°:

I₂ = 80 cos²45°

I₂ = 80(0.5)

I₂ = 40 W/m²

The third polarizer is another 45° away:

I₃ = 40 cos²45°

I₃ = 40(0.5)

I₃ = 20 W/m²

Therefore:

I_final = 20 W/m²

Without the middle polarizer, the crossed first and third filters would ideally transmit zero.

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Polarizing Filters and Glare

Polarizing filters have an important practical application: reducing glare.

Light reflected from surfaces such as:

• water
• wet roads
• glass
• snow

can become strongly polarized.

For a horizontal reflecting surface, much of the glare is polarized approximately horizontally under suitable viewing conditions.

A polarizing filter with a vertical transmission axis can strongly reduce this horizontally polarized light.

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Polarized Sunglasses

Polarized sunglasses typically contain filters oriented to reduce horizontally polarized glare.

They are particularly useful for:

• driving
• boating
• fishing
• skiing
• outdoor activities near water

Reducing glare can improve:

• visibility
• contrast
• visual comfort

The filters do not simply darken all light equally. They selectively reduce certain polarization components.

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Why Reflected Light Becomes Polarized

When unpolarized light strikes a dielectric surface at an angle, the two polarization components do not reflect equally.

One component can be reflected more strongly than the other.

As a result, the reflected light becomes partially polarized.

At a particular angle called Brewster's angle, one component has zero ideal reflection and the reflected light is completely linearly polarized.

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Brewster's Angle

For light travelling from a medium with refractive index n₁ into a medium with refractive index n₂:

tan θ_B = n₂/n₁

For air and glass:

n₁ = 1.00

n₂ = 1.50

tan θ_B = 1.50

Therefore:

θ_B ≈ 56°

Around this angle, reflected glare can be strongly polarized.

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Polarizing Filters in Photography

Photographers often use polarizing filters in front of camera lenses.

Rotating the filter changes which polarization direction is transmitted.

This can:

• reduce reflections from water
• reduce some reflections from glass
• reduce glare from vegetation
• increase apparent colour saturation
• alter the appearance of the sky

The photographer can rotate the filter while observing the scene to select the desired effect.

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Why Rotating a Camera Polarizer Matters

Suppose reflected glare is predominantly horizontally polarized.

If the filter's transmission axis is horizontal, much of the glare passes.

Rotate the filter toward vertical and the horizontal component is increasingly suppressed.

Therefore, the amount of glare changes as the filter rotates.

This is a direct real-world application of polarization.

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Seeing Through Water

Surface reflections can make it difficult to see beneath water.

A correctly oriented polarizing filter can reduce much of the reflected glare.

More of the light coming from objects beneath the surface then contributes to the observed image.

This is useful in:

• fishing
• aquatic photography
• environmental observations
• outdoor recreation

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Looking Through Glass

A polarizing filter can sometimes reduce reflections from glass.

For example, it may help a photographer see through a shop window by reducing reflected light from the surroundings.

However, the effectiveness depends on:

• viewing angle
• glass properties
• polarization of the reflected light

Polarization does not remove every possible reflection.

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LCD Screens and Polarizers

Liquid crystal displays use polarizing filters as essential components.

A simplified LCD contains:

• a polarizer
• liquid-crystal material
• another polarizing layer

The liquid crystals affect the polarization of light.

Electrical signals change the behaviour of the liquid crystals.

This controls how much light can pass through the final polarizer.

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Testing an LCD with a Polarizer

A simple polarization experiment can be performed by viewing an LCD through a polarizing filter.

As the filter rotates:

• the screen may become brighter
• then darker
• sometimes nearly black
• then brighter again

This demonstrates that the light leaving the screen has a strong polarization state.

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Investigating Polarizers Experimentally

A basic classroom experiment requires:

• light source
• two polarizing filters
• light sensor or lux meter
• rotating mount or angle scale

The first filter produces linearly polarized light.

The second filter acts as the analyzer.

Students can rotate the analyzer and measure transmitted intensity.

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Experimental Procedure

Place the polarizers in the light path.

Set their transmission axes parallel:

θ = 0°

Measure the transmitted intensity.

Rotate the analyzer in regular steps, such as:

• 0°
• 15°
• 30°
• 45°
• 60°
• 75°
• 90°

Record the intensity at each angle.

Then compare the results with:

I = I₀ cos² θ

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Expected Experimental Pattern

The intensity should be largest when:

θ = 0°

It should decrease as θ increases.

At:

θ = 45°

the ideal intensity is:

I = 0.5I₀

At:

θ = 90°

the ideal intensity is:

I = 0

A graph of intensity against angle therefore produces a curved, periodic pattern rather than a straight line.

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Testing Malus's Law

A stronger analysis is to calculate:

cos² θ

for every measured angle.

Then compare I with cos² θ.

According to Malus's law:

I ∝ cos² θ

Therefore, a graph of:

I against cos² θ

should be approximately linear for an ideal experimental setup.

This provides a useful way to test the mathematical model.

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Example Experimental Results

Suppose the intensity entering an analyzer is:

100 units

Ideal predictions would be:

Real measurements may differ because real filters and detectors are not ideal.

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Why Real Crossed Polarizers May Not Be Completely Dark

In an ideal model:

I = 0

at 90°.

Real experiments may still show a small amount of light because of:

• imperfect polarizers
• stray light
• reflections
• detector background
• imperfect alignment
• scattering
• incomplete polarization

Therefore, a non-zero experimental value does not automatically disprove Malus's law.

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Extinction Ratio

The quality of a polarizer can be described partly by its extinction ratio.

This compares how strongly the filter transmits the desired polarization with how much unwanted perpendicular polarization leaks through.

A high-quality polarizer has:

• high transmission along its preferred direction
• very low transmission in the perpendicular direction

Therefore, high-quality crossed polarizers can produce a very dark field.

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Experimental Variables

In a polarizer experiment, possible variables include:

Independent variable

Angle θ between the polarizers.

Dependent variable

Transmitted light intensity.

Control variables

These might include:

• light-source intensity
• distance from source
• detector position
• polarizer type
• ambient lighting
• wavelength or colour of light

Controlling these variables improves the reliability of the experiment.

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Sources of Experimental Error

Possible errors include:

• inaccurate angle measurements
• changing source intensity
• ambient light reaching the detector
• polarizers not centered correctly
• detector movement
• imperfect polarizers
• reflections from surrounding surfaces

Repeating measurements and averaging results can improve reliability.

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Analyzing Unexpected Results

Suppose the intensity at 90° is much larger than expected.

Possible explanations include:

• the filters are not actually at 90°
• one filter is a poor polarizer
• ambient light is reaching the detector
• light is scattering around the filters

A good experimental analysis should compare observations with theory and then explain reasonable causes of differences.

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A Complete Analysis Example

Unpolarized light has intensity:

I₀ = 240 W/m²

It passes through an ideal horizontal polarizer.

After the first polarizer:

I₁ = I₀/2

I₁ = 120 W/m²

A second polarizer is positioned at 30°.

Using Malus's law:

I₂ = I₁ cos²30°

I₂ = 120(0.75)

I₂ = 90 W/m²

Now rotate the second polarizer to 60°:

I₂ = 120 cos²60°

I₂ = 120(0.25)

I₂ = 30 W/m²

Increasing the relative angle from 30° to 60° reduces the transmitted intensity from 90 W/m² to 30 W/m².

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Predicting Changes Without Calculating

Malus's law can also be used qualitatively.

If θ increases from:

0° → 90°

intensity generally decreases.

If θ decreases toward:

0°

intensity increases.

At:

0° → maximum

At:

90° → minimum

The pattern is symmetric because:

cos² θ

repeats every 180°.

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Polarizers and Energy

A polarizer does not simply make the unwanted polarization component disappear without an energy transfer.

Depending on the type of polarizer, unwanted light may be:

• absorbed
• reflected
• redirected

In an absorbing polarizing sheet, much of the rejected component is converted into internal energy.

Energy conservation still applies.

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Common Mistakes

Thinking a polarizer changes the direction the light travels

A linear polarizer mainly selects electric-field orientation; it does not normally redirect the transmitted beam.

Confusing transmission axis with propagation direction

The transmission axis describes the electric-field direction that passes through the filter.

Using I = I₀ cos θ

Malus's law is:

I = I₀ cos² θ

Applying Malus's law directly to unpolarized light without considering the first polarizer

For ideal unpolarized light:

I₁ = I₀/2

first.

Thinking two polarizers always halve the intensity twice

The second polarizer depends on its angle relative to the first.

Thinking 45° means 25% transmission

cos²45° = 0.5, so the transmitted fraction is 50% of the polarized light entering that analyzer.

Thinking crossed polarizers must transmit half the light

Ideal crossed polarizers transmit zero.

Assuming real experiments must give exactly zero at 90°

Real polarizers, detectors, and experimental environments are imperfect.

Thinking polarized sunglasses simply reduce all light equally

Their main advantage is preferentially reducing polarized glare.

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Key Terms

Polarizing filter — optical device that preferentially transmits a particular polarization direction.

Polarizer — device used to produce or select polarized light.

Analyzer — polarizer used to determine or control the polarization of incoming light.

Transmission axis — electric-field direction preferentially transmitted by a linear polarizer.

Unpolarized light — light whose electric-field orientation varies rapidly and randomly.

Linearly polarized light — light whose electric field oscillates along a fixed direction.

Malus's law — relationship I = I₀ cos² θ for linearly polarized light passing through an ideal analyzer.

Crossed polarizers — polarizers whose transmission axes are perpendicular.

Parallel polarizers — polarizers whose transmission axes point in the same direction.

Glare — intense reflected light that can reduce visibility and visual comfort.

Brewster's angle — incidence angle at which reflected light from an ideal dielectric boundary is completely linearly polarized.

Extinction ratio — measure of how effectively a polarizer suppresses the unwanted polarization direction.

Light intensity — optical power passing through a unit area.

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Key Takeaways

• A polarizing filter selectively transmits a component of the electric field.
• Light emerging from an ideal linear polarizer is linearly polarized along its transmission axis.
• Ideal unpolarized light loses half its intensity through the first ideal polarizer.
• A second polarizer is commonly called an analyzer.
• The intensity through an analyzer follows I = I₀ cos² θ.
• θ is the angle between the incoming polarization direction and the analyzer's transmission axis.
• Parallel ideal polarizers provide maximum transmission.
• Polarizers at 45° transmit half of the polarized intensity entering the analyzer.
• Polarizers at 60° transmit one quarter of the polarized intensity.
• Crossed ideal polarizers transmit zero light.
• Rotating one polarizer causes transmitted intensity to vary periodically.
• Inserting another polarizer between crossed polarizers can allow light to pass.
• Reflected glare from horizontal surfaces is often strongly horizontally polarized.
• Polarized sunglasses use a suitable transmission axis to reduce this glare.
• Camera polarizers can reduce reflections and alter the appearance of outdoor scenes.
• LCD screens use polarization to control transmitted light.
• Malus's law can be tested experimentally by measuring intensity as a polarizer rotates.
• A graph of I against cos² θ should be approximately linear under ideal conditions.
• Real polarizers may transmit a small amount of light even when crossed.
• Polarizer experiments provide a direct way to investigate the transverse nature and intensity behaviour of light.

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Check Your Understanding

1. Explain how a linear polarizing filter changes unpolarized light.
2. What is meant by the transmission axis of a polarizer?
3. Unpolarized light has an intensity of 500 W/m². Calculate its intensity after passing through one ideal polarizer.
4. State Malus's law and define each variable.
5. Polarized light with intensity 80 W/m² reaches an analyzer at 30°. Calculate the transmitted intensity.
6. Repeat Question 5 for an analyzer angle of 60°.
7. Explain why two parallel ideal polarizers transmit much more light than two crossed polarizers.
8. Unpolarized light with intensity 200 W/m² passes through two ideal polarizers whose axes differ by 45°. Calculate the final intensity.
9. Two ideal polarizers are crossed. Predict the transmitted intensity and explain your reasoning using Malus's law.
10. Explain why inserting a third polarizer at 45° between two crossed polarizers allows some light to pass.
11. Unpolarized light of intensity 320 W/m² passes through polarizers oriented at 0°, 30°, and 90°. Calculate the final intensity.
12. Explain how polarized sunglasses reduce glare from a wet road or the surface of a lake.
13. A student rotates an analyzer from 0° to 180° while recording light intensity. Describe the expected pattern and identify the angles of maximum and minimum intensity.
14. In an experiment, the detector records a small intensity when the polarizers are at 90°. Give three experimental reasons why the measured intensity may not be exactly zero.
15. Design an experiment to test Malus's law. Identify the independent variable, dependent variable, important control variables, measurements you would take, and the graph you could use to determine whether the results support I ∝ cos² θ.