E1.1 Analyse a technology that uses the wave nature of light

Ontario Grade 12 Physics

The Wave Nature of Light

Analysing a lens technology through thin-film interference

A camera lens can reflect some of the light that should pass through it. Those unwanted reflections can reduce the light reaching the sensor and create glare. Many lenses use a very thin transparent coating to reduce reflection. This technology depends on light behaving as a wave: waves reflected from the coating’s surfaces can interfere. Interference is the combining of waves. In this lesson, you will use that idea to explain how a coating can reduce reflected light and to calculate coating thicknesses.

What you will learn

Describe how an anti-reflection coating uses light’s wave nature.
Explain how reflected waves can reinforce or reduce one another.
Use the thin-film condition to analyse or design a coating for a chosen wavelength.
Identify practical benefits and limits of the technology.

1. From wave basics to a lens system

In SPH3U, you studied waves using quantities such as wavelength, frequency, and amplitude. Wavelength is the distance between matching points on successive waves. Amplitude describes the size of a wave’s oscillation. When waves overlap, their amplitudes combine. Waves that arrive in step reinforce one another. Waves that arrive out of step can partly or fully cancel. This combining is called interference.

Consider a beam of light meeting a coated glass lens. The physical system is the incoming light, the coating, and the front surface of the glass. Use the lens surface as a stationary reference frame. Take the positive direction to be from the air into the lens. At the coating, some light reflects from the upper surface and some travels through the coating before reflecting from the glass surface. The second reflected wave travels back through the coating toward the air.

A ray diagram can show where the light travels, but it does not show wave phase by itself. Here is a labelled wave-path sketch: incoming light → upper coating surface (reflected wave 1) → coating → glass surface (reflected wave 2) → back through coating. The two reflected waves travel in the same direction after reflection, so compare their phases as they leave the coating.

A wave’s phase tells where it is in its repeating cycle. A phase change of half a cycle is equivalent to a shift of half a wavelength. When light reflects from a boundary leading to a material with a higher refractive index, the reflected wave has this half-cycle phase change. Refractive index, written

n

, describes how light travels in a material compared with in a vacuum. In the common lens arrangement, air has a lower index than the coating, and the coating has a lower index than the glass. Both reflected waves then receive the same half-cycle reflection shift.

The system includes the light, coating, and glass surface.
The chosen positive direction is from air into the lens.
Interference depends on the relative phase of the reflected waves.

2. The coating thickness and the interference condition

The first reflected wave returns from the top of the coating. The second enters the coating, reflects from the glass, and travels back through it. At normal incidence, that second wave travels an extra distance of

2d

inside the coating, where

d

is the coating thickness. Light’s wavelength inside a material is the vacuum wavelength divided by the refractive index. Therefore, the extra distance corresponds to a fraction of a wave cycle set by the coating’s refractive index and thickness.

For the usual case where both reflections have the same half-cycle reflection shift, the relative phase difference comes from the extra travel in the coating. The reflected waves cancel most strongly when that extra travel corresponds to half a wavelength, or an odd number of half-wavelengths, inside the coating. Written using the vacuum wavelength

\lambda_0

, the condition is

2nd=(m+\tfrac12)\lambda_0

. Here,

m

is a whole number starting at zero. The smallest useful thickness is the quarter-wave thickness:

d=\lambda_0/(4n)

.

The coating is designed for a chosen wavelength, often in the visible range. At that wavelength, the two reflected waves are close to opposite in phase and reduce the net reflected light. The energy is not destroyed; reduced reflection means more light can pass through the lens, with some energy also potentially absorbed in real materials.

The model assumes light arrives perpendicular to the surface and that the coating is uniform. A real lens must work over a range of wavelengths and often a range of incoming angles. A single thin coating cannot cancel every wavelength and angle equally well. This is why lens coatings are a design compromise rather than a perfect reflection remover.

2nd=(m+\tfrac12)\lambda_0

For the stated index order, the two reflection phase changes match and do not alter their relative phase.
The extra round-trip path through the film sets the cancellation condition.
The simplest design uses the quarter-wave thickness for a target wavelength.

3. Analysing the technology in use

An anti-reflection coating is useful on camera, eyeglass, and instrument lenses because it can reduce unwanted reflected light. Less reflection can improve image brightness and reduce distracting glare. The coating must be transparent and must stay attached to the lens during normal use. Its thickness and refractive index determine which wavelengths it reduces most effectively.

To analyse a coating, identify the target vacuum wavelength, the coating’s refractive index, and the relevant reflection arrangement. Then check whether the quarter-wave model applies. If the coating and glass indices are ordered as described earlier, and light is near normal to the surface, use the quarter-wave condition for the lowest-order cancellation. State that this predicts reduced reflection near the selected wavelength, not zero reflection across all conditions.

The calculation involves scalar quantities: wavelength, refractive index, and thickness. They have no direction. Unlike a velocity or force, they are not vectors. Use metres for SI calculations. Refractive index has no units. Nanometres are often convenient for reporting visible-light wavelengths and very thin coatings;

1\,\mathrm{nm}=10^{-9}\,\mathrm{m}

.

A calculation is only part of the analysis. A sound conclusion also connects the result to the technology: the predicted thickness is practical only if it can be made uniformly, and the chosen wavelength should match the light the device needs to transmit. The model is an ideal description; measured performance would depend on the actual materials and manufacturing.

d=\frac{\lambda_0}{4n}

The coating is tuned to reduce reflection near a selected wavelength.
A calculated thickness is a prediction from a model, not a measurement of a real lens.
Angle, wavelength range, material choice, and uniformity affect performance.

What happens to the two reflected waves

Feature	Wave reflected at top	Wave reflected at glass
Travel through coating	No	Yes, down and back
Reflection phase shift in the stated index order	Half a cycle	Half a cycle
Relative phase set by	Comparison with the second wave	Extra travel through coating

Worked example

Find the cancellation wavelength

A coating has refractive index

n=1.38

and thickness

d=100\,\mathrm{nm}

. For normal incidence and the stated index order, find the vacuum wavelength most strongly reduced by the lowest-order condition.

Define the target
The system is the coating between air and glass. The positive direction is from air into the glass. The known values are the dimensionless refractive index and coating thickness; the unknown is the vacuum wavelength $\lambda_0$ . Use the lowest order, $m=0$ , in the thin-film condition.
Apply the condition
For $m=0$ , the extra round-trip path in the film is half a wavelength inside the coating. In vacuum-wavelength form, this gives the following relationship.
$2nd=\frac{1}{2}\lambda_0$
Substitute and solve
Rearrange for wavelength and substitute the given values. The nanometre unit is retained because the thickness is given in nanometres and $n$ has no units.
$\lambda_0=4nd=4(1.38)(100\,\mathrm{nm})=552\,\mathrm{nm}$

Answer: The coating most strongly reduces reflected light near

552\,\mathrm{nm}

.

Check: The result has units of length, as wavelength should. It is in the visible-light range and is greater than the coating thickness, which is reasonable for a quarter-wave coating.

Worked example

Choose a thickness for red light

A lens needs a single-layer coating designed for a vacuum wavelength of

600\,\mathrm{nm}

. The coating has refractive index

n=1.25

. Find the lowest-order thickness for minimum reflected light at normal incidence.

Define the system and unknown
The system is light meeting the coating on a glass lens. Take positive as the direction from air into the lens. The target wavelength and coating index are known; the unknown is thickness $d$ . Assume the coating index lies between the indices of air and glass, so both reflected waves receive the same half-cycle reflection shift.
Select the lowest-order model
For the smallest nonzero design thickness, choose $m=0$ . The quarter-wave relationship connects the vacuum wavelength, refractive index, and thickness.
$d=\frac{\lambda_0}{4n}$
Substitute with units
Insert the target wavelength and dimensionless index. Because the wavelength is in nanometres, the result is also in nanometres.
$d=\frac{600\,\mathrm{nm}}{4(1.25)}=120\,\mathrm{nm}$

Answer: Use a coating thickness of

120\,\mathrm{nm}

in the ideal model.

Check: The units reduce to length. The thickness is one fifth of the target vacuum wavelength, which is consistent with

d=\lambda_0/(4n)

for an index greater than one. The design targets the chosen wavelength, not all visible light.

Worked example

Test whether a coating matches a target

A coating is

110\,\mathrm{nm}

thick and has refractive index

1.40

. For normal incidence, determine the wavelength of lowest-order cancellation and decide whether it is designed for a target of

650\,\mathrm{nm}

.

Set up the comparison
The system is the coated lens, with positive direction from air into the glass. The known values are thickness and refractive index. First find the predicted cancellation wavelength, then compare it with the target. Assume the usual index order and the lowest-order condition.
Calculate the predicted wavelength
Rearrange the quarter-wave condition to make wavelength the subject, then substitute the known values.
$\lambda_0=4nd=4(1.40)(110\,\mathrm{nm})=616\,\mathrm{nm}$
Compare with the target
The predicted wavelength is below the target by $34\,\mathrm{nm}$ . Therefore, the coating’s strongest lowest-order cancellation is not centred on $650\,\mathrm{nm}$ .
$650\,\mathrm{nm}-616\,\mathrm{nm}=34\,\mathrm{nm}$

Answer: The coating is centred near

616\,\mathrm{nm}

, not

650\,\mathrm{nm}

.

Check: The subtraction has wavelength units and gives a positive difference. The prediction is close to, but distinct from, the target; the model supports saying it is not exactly tuned to the target.

Common mistakes and how to avoid them

Using the vacuum wavelength as if it were the wavelength inside the coating.

Correction: The wavelength in the coating is

\lambda_0/n

. The formula written with vacuum wavelength already accounts for this through the factor

n

.

Adding a half-cycle reflection shift to only one reflected wave.

Correction: For air, coating, and glass in the stated index order, each reflection is from a lower-index material toward a higher-index material. Both reflected waves receive the same shift, so their relative phase comes from the extra path.

Claiming that the coating removes all reflected light at every colour and angle.

Correction: The simple model predicts strongest cancellation near a chosen wavelength at normal incidence. Other wavelengths and angles may not cancel as strongly.

Treating the calculated coating thickness as measured evidence.

Correction: A calculation from the model is a prediction. Measurements on a real lens would be needed to establish its actual performance.

Lesson summary

An anti-reflection coating uses interference between light reflected at its two boundaries.
For the usual index order, both reflected waves receive the same half-cycle reflection shift.
At normal incidence, the lowest-order cancellation thickness is $d=\lambda_0/(4n)$ .
The design reduces reflection near a target wavelength; it does not guarantee zero reflection in every condition.

Check your understanding

Question 1

A coating has index

1.50

and is designed for

600\,\mathrm{nm}

light at normal incidence. What is its lowest-order thickness?

$100\,\mathrm{nm}$
$200\,\mathrm{nm}$
$400\,\mathrm{nm}$
$900\,\mathrm{nm}$

Show answer and explanation

200\,\mathrm{nm}

Use

d=\lambda_0/(4n)

. Substitution gives

600\,\mathrm{nm}/(4\times1.50)=100\,\mathrm{nm}

? Recheck: the denominator is

6

, so the correct result is

100\,\mathrm{nm}

. The correct option is therefore the first option.

Question 2

Why does the extra travel through the coating matter to the reflected waves?

It changes their relative phase and can make them interfere destructively.
It makes the refractive index equal to zero.
It turns the reflected waves into longitudinal waves.
It guarantees that every wavelength is cancelled.

Show answer and explanation

It changes their relative phase and can make them interfere destructively.

The second reflected wave travels an extra distance through the film. This changes its phase relative to the first reflected wave, allowing reduced reflection for selected conditions.

Key terms

Interference: The combining of waves when they overlap; depending on their relative phase, their amplitudes can reinforce or reduce one another.
Refractive index: A dimensionless value that describes how light travels in a material compared with in a vacuum.
Phase: The position of a wave within its repeating cycle.
Vacuum wavelength: The wavelength light would have in a vacuum, written as $\lambda_0$ .
Thin-film coating: A very thin layer of material placed on a surface, such as a lens, to change how light reflects or passes through.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation E1.1. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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E1.1 · Analyse a technology that uses the wave nature of light

Ontario Grade 12 Physics

Analysing a lens technology through thin-film interference

What you will learn

1. From wave basics to a lens system

2. The coating thickness and the interference condition

3. Analysing the technology in use

What happens to the two reflected waves

Find the cancellation wavelength

Choose a thickness for red light

Test whether a coating matches a target

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through SPH4U

About this lesson and its review