Thin Film Interference

Thin film interference occurs when light reflects from the different surfaces of a thin transparent layer and the reflected waves overlap.

Examples include:

• soap bubbles
• oil films on water
• thin coatings on lenses
• protective optical coatings
• interference filters

Part of the incoming light reflects from the top surface of the film. Another part enters the film, reflects from the bottom surface, and emerges again.

Because these reflected waves have travelled different optical paths—and may also experience different phase changes upon reflection—they can interfere constructively or destructively.

The result can be bright reflections, dark reflections, or striking patterns of colour.

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How Thin Film Interference Occurs

Consider a thin transparent film surrounded by other materials.

When a light wave reaches the upper surface:

• some light reflects immediately
• some light enters the film

The transmitted light travels through the film and reaches the lower surface.

Again:

• some light reflects
• some light continues into the material below

The wave reflected from the lower surface travels back through the film and emerges from the top.

The two main reflected rays can then overlap.

Their superposition produces interference.

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The Two Main Reflected Rays

We can identify two important rays.

Ray 1 reflects from the upper surface of the film.

Ray 2:

• enters the film
• travels through the film
• reflects from the lower surface
• travels back through the film
• emerges into the original medium

Ray 2 therefore travels an additional distance through the film.

This creates a phase difference between Ray 1 and Ray 2.

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Why the Film Must Be Thin

Interference can occur whenever coherent waves overlap, but visible thin-film effects are especially important when the film thickness is comparable to the wavelength of light.

Visible-light wavelengths are only a few hundred nanometres.

Films with thicknesses on the micrometre or sub-micrometre scale can therefore produce strong wavelength-dependent interference.

Different film thicknesses can reinforce different wavelengths.

This is why thin films often appear colourful.

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Two Sources of Phase Difference

There are two major sources of relative phase difference between the reflected rays:

• different optical path lengths
• phase changes upon reflection

Both must be considered.

Ignoring either one can lead to the wrong prediction about whether the reflected light is bright or dark.

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Optical Path Difference

Suppose the film has:

• physical thickness t
• refractive index n

At normal incidence, Ray 2 travels approximately an additional physical distance:

2t

inside the film.

Because light travels more slowly in the film, the corresponding optical path difference is:

2nt

This quantity is central to thin-film interference calculations.

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Why Refractive Index Appears

The wavelength of light inside a material is:

λ_medium = λ₀/n

where:

• λ_medium = wavelength inside the material
• λ₀ = wavelength in vacuum
• n = refractive index

Therefore, a given physical distance inside a material corresponds to more wavelengths than the same distance in vacuum.

Using the optical path:

2nt

allows calculations to be made conveniently using the vacuum wavelength λ₀.

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Phase Changes Upon Reflection

Reflection itself can change the phase of a wave.

For the standard dielectric situations considered here:

reflection toward a higher refractive index → π phase change

reflection toward a lower refractive index → no π phase change

A π phase change corresponds to:

180° = half a cycle

It has the same phase effect as an additional path difference of:

λ/2

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Examining the Top Reflection

Suppose light travels from air into a film.

If:

n_air < n_film

Ray 1 reflects from a boundary leading toward a higher refractive index.

Therefore, Ray 1 receives a:

π phase shift

This is very common in thin-film problems.

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Examining the Bottom Reflection

Now consider Ray 2 reflecting from the lower surface.

If:

n_below > n_film

Ray 2 also undergoes a π phase shift.

If:

n_below < n_film

Ray 2 undergoes no π phase shift.

Therefore, the refractive indices of all the materials must be checked before choosing an interference condition.

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Relative Phase Changes

There are three possibilities.

Neither ray undergoes a π shift

Relative reflection phase difference = 0.

Both rays undergo a π shift

Relative reflection phase difference = 0 because both waves have been shifted equally.

Only one ray undergoes a π shift

Relative reflection phase difference = π.

For interference, the relative phase difference is what matters.

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The One-Phase-Reversal Case

A very common arrangement has:

n₁ < n_film > n₃

For example, a film may have a greater refractive index than the materials on either side.

The upper reflection is:

lower n → higher n

so it receives a π shift.

The lower reflection is:

higher n → lower n

so it receives no π shift.

There is therefore one relative phase reversal.

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Constructive Reflection with One Phase Reversal

If exactly one of the reflected rays undergoes a π phase reversal, constructive interference in the reflected light occurs when:

2nt = (m + ½)λ₀

where:

• n = refractive index of film
• t = film thickness
• m = 0, 1, 2, ...
• λ₀ = wavelength in vacuum

At these wavelengths, the two reflected rays reinforce one another.

The reflected light is therefore relatively bright.

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Destructive Reflection with One Phase Reversal

For the same one-phase-reversal situation, destructive reflected interference occurs when:

2nt = mλ₀

where:

m = 0, 1, 2, ...

The reflected waves arrive out of phase and reduce the reflected intensity.

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Zero or Two Phase Reversals

If neither reflected ray undergoes a π phase shift, or both undergo one, there is no additional relative phase difference from reflection.

The interference conditions are then reversed.

For constructive reflected interference:

2nt = mλ₀

For destructive reflected interference:

2nt = (m + ½)λ₀

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Thin-Film Conditions

This table applies to the normal-incidence situations considered here.

The first step in almost every thin-film problem should therefore be:

determine the reflection phase changes.

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Constructive Interference

Constructive interference occurs when the two reflected waves arrive in phase.

Their electric-field amplitudes reinforce.

The reflected intensity increases.

For visible light, this means a particular wavelength may appear especially bright.

If several wavelengths are present, some may be reinforced while others are suppressed.

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Destructive Interference

Destructive interference occurs when the reflected waves arrive out of phase.

Their electric-field amplitudes partially or completely cancel.

The reflected intensity decreases.

Ideally, equal-amplitude waves that are exactly 180° out of phase could cancel completely.

In real films, the amplitudes are often unequal, so the minimum may not be perfectly dark.

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Worked Example: Identifying Phase Reversals

Consider:

n_air = 1.00

n_film = 1.40

n_below = 1.30

At the top surface:

1.00 → 1.40

Reflection is toward higher n.

Therefore:

π phase change

At the bottom surface:

1.40 → 1.30

Reflection is toward lower n.

Therefore:

no π phase change

There is exactly one relative phase reversal.

Therefore, bright reflected light satisfies:

2nt = (m + ½)λ₀

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Worked Example: Two Phase Reversals

Suppose:

n_air = 1.00

n_film = 1.30

n_glass = 1.50

At the top:

1.00 → 1.30

π phase shift.

At the bottom:

1.30 → 1.50

π phase shift.

Both rays experience the same reflection phase change.

Therefore:

relative reflection phase difference = 0

Bright reflected light therefore satisfies:

2nt = mλ₀

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Finding the Minimum Thickness for Bright Reflection

Suppose a film has:

n = 1.40

and exactly one reflected ray experiences a π phase shift.

We want constructive reflection for:

λ₀ = 560 nm

The condition is:

2nt = (m + ½)λ₀

For the smallest positive thickness:

m = 0

Therefore:

2nt = λ₀/2

and:

t = λ₀/(4n)

Substitute:

t = 560 nm/(4 × 1.40)

t = 100 nm

Therefore, the minimum film thickness is:

100 nm

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Quarter-Wave Thickness

The previous example gives an important result:

t = λ₀/(4n)

or:

nt = λ₀/4

This is called a quarter-wave optical thickness.

After travelling down and back through the film, the second ray gains an optical path difference:

2nt = λ₀/2

This corresponds to a propagation phase difference of π.

Quarter-wave layers are extremely important in optical coatings.

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Worked Example: Dark Reflected Light

A film has:

n = 1.50

and exactly one reflection undergoes a π phase reversal.

Find a non-zero film thickness that produces destructive reflection for:

λ₀ = 600 nm

For destructive interference:

2nt = mλ₀

Choose the first non-zero case:

m = 1

Then:

t = λ₀/(2n)

t = 600 nm/(2 × 1.50)

t = 200 nm

Therefore, a thickness of:

200 nm

produces destructive reflected interference under these assumptions.

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Why Soap Bubbles Show Colours

White light contains many wavelengths.

A soap film has a thickness that can vary from one location to another.

At one location, the film may satisfy the constructive condition for red light.

At another location, it may reinforce green.

Elsewhere, it may reinforce blue.

The result is a changing pattern of colours.

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Film Thickness and Colour

For one relative phase reversal:

2nt = (m + ½)λ₀

for constructive reflected interference.

Therefore, changing t changes the wavelength that is strongly reflected.

A thicker part of the film may reinforce a different wavelength from a thinner part.

This creates coloured bands and patches.

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Why Soap Films Change with Time

A soap film is not static.

Gravity can cause liquid to drain downward.

Evaporation can also reduce its thickness.

Therefore, t changes with time.

Since the interference condition depends on t, the reflected colours also change.

Eventually, extremely thin regions can appear dark because visible wavelengths may undergo strong destructive reflection.

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Oil on Water

A thin layer of oil on water can also produce interference colours.

Light reflects from:

• the air-oil surface
• the oil-water surface

The phase changes depend on the refractive indices of air, oil, and water.

The thickness of the oil layer may vary across the surface.

Different locations therefore reinforce different wavelengths.

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Why Oil Films Can Look Rainbow-Coloured

The colours are not necessarily pigments in the oil.

Instead, the film selectively reinforces and suppresses wavelengths through interference.

The observed colour therefore depends on:

• wavelength
• film thickness
• refractive index
• viewing angle
• illumination angle
• phase changes at the boundaries

The rainbow appearance is a wave-interference effect.

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Effect of Viewing Angle

At non-normal incidence, the distance travelled through the film depends on the angle inside the film.

The optical path difference becomes approximately:

2nt cos θ_film

where θ_film is the angle of the light inside the film.

Therefore, changing the viewing angle changes the interference condition.

This is why the colour of a thin film can change as the observer moves.

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Refraction Inside the Film

When light enters the film at an angle, its direction changes according to Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

The angle inside the film must therefore be determined from refraction.

For more advanced thin-film calculations, both:

• refraction
• interference

may need to be considered.

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Thin Films and White Light

White light contains a continuous range of visible wavelengths.

For a particular film thickness:

• some wavelengths undergo constructive interference
• some undergo destructive interference
• others undergo partial interference

Therefore, the reflected light can have a colour even when the incident light is white.

The transmitted light can also have a complementary change in its spectrum.

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Reflected and Transmitted Light

If a wavelength is strongly reflected, less of that wavelength is generally transmitted, assuming absorption is small.

Similarly, if reflection is strongly reduced through destructive interference, transmission can be increased.

This principle is central to optical coating design.

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Anti-Reflection Coatings

An anti-reflection coating is a thin layer designed to reduce reflected light.

The coating is chosen so that reflected waves interfere destructively at a desired wavelength.

Applications include:

• eyeglasses
• camera lenses
• microscopes
• telescopes
• binoculars
• solar cells
• optical sensors

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How an Anti-Reflection Coating Works

Consider light reflecting from:

• the top of the coating
• the coating-glass boundary

The coating thickness can be selected so that the two reflected waves arrive approximately 180° out of phase.

They then interfere destructively.

A common design uses:

t = λ₀/(4n)

although the complete design must also consider the refractive indices and reflection amplitudes at both boundaries.

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Why Refractive Index Also Matters

Perfect cancellation requires more than the correct phase difference.

The two reflected waves should also have similar amplitudes.

If one reflected wave is much stronger than the other, complete cancellation cannot occur.

Therefore, high-quality anti-reflection coatings are designed using both:

• film thickness
• refractive index

This allows the amplitudes and phases of the reflected waves to be controlled.

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Reflective Coatings

Thin films can also be designed to do the opposite: increase reflection.

Multiple layers can be arranged so that reflected waves from many boundaries interfere constructively.

Such coatings are used in:

• high-reflectivity mirrors
• laser systems
• optical instruments
• wavelength-selective devices

Thin-film interference can therefore be used to either reduce or enhance reflection.

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Interference Filters

An interference filter uses thin-film structures to transmit or reflect selected wavelength ranges.

By controlling:

• film thickness
• refractive index
• number of layers

engineers can design filters with specific optical properties.

Applications include:

• cameras
• astronomy
• spectroscopy
• scientific imaging
• optical communications

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Newton's Rings

Thin-film interference can also occur in the thin layer between curved and flat surfaces.

A famous example is Newton's rings.

When a curved lens rests near a flat glass surface, the thickness of the thin air layer varies with distance from the center.

Different thicknesses satisfy different interference conditions.

The result is a series of concentric bright and dark rings.

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Why Newton's Rings Form

The air-film thickness increases gradually away from the point of closest contact.

Because the thickness changes continuously, the path difference also changes.

At some radii:

constructive interference occurs

At others:

destructive interference occurs

The circular geometry produces circular interference fringes.

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Wedge-Shaped Films

A thin film can also have a gradually increasing thickness, forming a wedge.

As the thickness changes, the interference condition changes.

This can produce a series of approximately parallel bright and dark fringes.

Such patterns can be used to investigate:

• very small thicknesses
• surface flatness
• small angles
• spacing between surfaces

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Analyzing a Thin-Film Pattern

When a thin film displays alternating bright and dark regions, each region corresponds to a particular phase relationship.

A bright region indicates:

constructive interference of reflected light

A dark region indicates:

destructive interference of reflected light

If the film thickness changes gradually, successive bright and dark regions correspond to successive interference orders.

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Interference Order

The integer m identifies different interference conditions.

For one phase reversal, constructive reflection occurs when:

2nt = (m + ½)λ₀

For:

m = 0

the condition is:

2nt = λ₀/2

For:

m = 1

2nt = 3λ₀/2

For:

m = 2

2nt = 5λ₀/2

Each value represents another possible film thickness producing constructive reflection for the same wavelength.

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Adjacent Interference Orders

For one phase reversal, two successive constructive conditions are:

2nt₁ = (m + ½)λ₀

and:

2nt₂ = (m + 3/2)λ₀

Subtracting gives:

2n(t₂ − t₁) = λ₀

Therefore:

Δt = λ₀/(2n)

So an increase in thickness of λ₀/(2n) moves the reflected light from one constructive order to the next.

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Worked Example: Successive Bright Regions

Suppose:

λ₀ = 600 nm

n = 1.50

The thickness difference between successive constructive orders is:

Δt = λ₀/(2n)

Δt = 600 nm/(3.00)

Δt = 200 nm

Therefore, increasing the film thickness by 200 nm produces the next constructive order for that wavelength.

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Thin-Film Interference and Energy

Destructive interference does not destroy light energy.

If reflected intensity is reduced, the energy is redistributed.

In a transparent, low-absorption system, reduced reflection generally corresponds to increased transmission.

Similarly, strong constructive reflection can reduce the transmitted intensity.

This is an important application of conservation of energy.

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A Complete Thin-Film Problem

A thin film has:

n = 1.25

Light of wavelength:

λ₀ = 500 nm

is incident normally.

Exactly one of the two main reflected rays experiences a π phase reversal.

Find the minimum positive thickness that produces bright reflected light.

For one phase reversal:

2nt = (m + ½)λ₀

For minimum thickness:

m = 0

Therefore:

t = λ₀/(4n)

t = 500 nm/(4 × 1.25)

t = 100 nm

The minimum positive thickness is:

100 nm

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Another Complete Problem

A film has:

n = 1.50

and exactly one relative phase reversal.

Its thickness is:

t = 300 nm

Find wavelengths that could undergo destructive reflection.

For destructive reflection:

2nt = mλ₀

Calculate:

2nt = 2(1.50)(300 nm)

2nt = 900 nm

Therefore:

λ₀ = 900 nm/m

Possible integer values include:

m = 1 → λ₀ = 900 nm

m = 2 → λ₀ = 450 nm

m = 3 → λ₀ = 300 nm

Of these, 450 nm lies in the visible region.

Therefore, blue light around 450 nm could be strongly suppressed in reflection.

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A Reliable Problem-Solving Strategy

When solving a thin-film problem, use the following sequence:

Identify the refractive indices.

Write them in order:

n₁ → n_film → n₃

Determine reflection phase changes.

For each reflected ray:

higher n boundary → π shift

lower n boundary → no π shift

Determine the relative reflection phase.

Ask whether there are:

• zero relative reversals
• one relative reversal
• two reversals

Calculate the optical path difference.

At normal incidence:

2nt

Choose the correct interference condition.

With one relative π reversal:

bright:

2nt = (m + ½)λ₀

dark:

2nt = mλ₀

With zero or two:

bright:

2nt = mλ₀

dark:

2nt = (m + ½)λ₀

Solve and check units.

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Common Mistakes

Considering only one reflected ray

Thin-film interference requires at least two overlapping reflected waves.

Ignoring reflection phase changes

The path difference alone does not determine the interference result.

Assuming every reflection causes a π shift

A π shift occurs for reflection toward a higher refractive index in the standard cases considered here.

Thinking two π shifts produce a relative π shift

If both rays undergo π shifts, their relative reflection phase remains unchanged.

Using physical path 2t instead of optical path 2nt

The refractive index must be included when using the vacuum wavelength.

Using the wrong bright/dark condition

Determine the number of relative phase reversals before selecting an equation.

Assuming destructive interference destroys energy

Energy is redistributed between reflection, transmission, and possible absorption.

Thinking film colour must come from pigment

Thin transparent films can produce colour through wavelength-dependent interference.

Ignoring viewing angle

At non-normal incidence, changing the viewing angle can change the optical path difference and therefore the observed colour.

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Key Terms

Thin film — transparent layer whose thickness is sufficiently small for reflections from its surfaces to interfere noticeably.

Thin-film interference — interference between waves reflected or transmitted at different boundaries of a thin layer.

Reflected ray — portion of light returned from a boundary.

Transmitted ray — portion of light passing through a boundary.

Optical path length — refractive index multiplied by physical distance travelled.

Optical path difference — difference between the optical path lengths of interfering waves.

Phase difference — difference in phase between waves.

Phase reversal — π-radian phase change that can occur upon reflection.

Constructive interference — reinforcement of waves producing increased intensity.

Destructive interference — cancellation of waves producing reduced intensity.

Interference order — integer m identifying successive interference conditions.

Quarter-wave thickness — film optical thickness of approximately λ₀/4.

Anti-reflection coating — thin film designed to reduce reflected light through destructive interference.

Interference filter — optical device using thin-film interference to select wavelengths.

Newton's rings — circular interference fringes produced by a thin layer of varying thickness between curved and flat surfaces.

Wedge film — film whose thickness gradually changes across its surface.

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Key Takeaways

• Thin-film interference occurs when reflected waves from different film boundaries overlap.
• One important ray reflects from the upper surface while another travels through the film and reflects from the lower surface.
• The waves acquire a phase difference from both propagation and reflection.
• At normal incidence, the additional optical path through a film is approximately 2nt.
• Reflection toward a higher refractive index produces a π phase shift.
• Reflection toward a lower refractive index produces no standard π phase shift.
• Interference depends on the relative phase between the reflected waves.
• Exactly one relative π phase reversal changes the conditions for constructive and destructive reflection.
• With one relative phase reversal, bright reflected light satisfies 2nt = (m + ½)λ₀.
• With one relative phase reversal, dark reflected light satisfies 2nt = mλ₀.
• With zero or two phase reversals, these bright and dark conditions are reversed.
• Different wavelengths can experience different interference conditions in the same film.
• Varying film thickness therefore produces colour patterns under white light.
• Viewing angle can change the interference condition and observed colour.
• Soap bubbles and oil films provide common examples of thin-film interference.
• Anti-reflection coatings use destructive interference to reduce reflected light.
• Highly reflective coatings can use constructive interference to increase reflection.
• Newton's rings and wedge films produce patterns because film thickness varies across the surface.
• Destructive interference does not destroy energy; it redistributes it.
• Thin-film interference allows extremely small thicknesses and optical properties to be measured and controlled.

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Check Your Understanding

1. Explain how two reflected rays are produced in a thin transparent film.
2. Why does the ray reflected from the lower surface usually travel a greater optical path than the ray reflected from the upper surface?
3. What two factors determine the relative phase of the reflected rays?
4. State the rule for a π phase change when light reflects from a boundary between two dielectric media.
5. A film has n = 1.40 and lies between air, n = 1.00, and a material with n = 1.20. Determine the phase changes experienced by the two main reflected rays.
6. A film has n = 1.30 and lies between air, n = 1.00, and glass, n = 1.50. Determine whether the two reflected rays have a relative reflection phase difference of 0 or π.
7. State the constructive and destructive reflected-light conditions when exactly one reflected ray undergoes a π phase reversal.
8. State the constructive and destructive conditions when both reflected rays undergo the same phase reversal.
9. A film has refractive index 1.50. Exactly one reflected ray undergoes a π phase reversal. Calculate the minimum positive film thickness for constructive reflection of 600 nm light.
10. A film with n = 1.40 is 200 nm thick. Exactly one relative π phase reversal occurs. Determine whether 560 nm light undergoes constructive or destructive reflection at normal incidence.
11. Explain why a soap bubble illuminated with white light can display many different colours.
12. Explain why the colours of a soap bubble can change as the film becomes thinner.
13. Explain how changing the viewing angle can change the colour observed from a thin film.
14. Describe how a thin-film anti-reflection coating reduces the amount of light reflected from a lens.
15. A student calculates the path difference between two reflected rays and immediately concludes that they interfere constructively. Explain why the student must check the refractive indices at both reflecting boundaries before deciding.