Thin-Film Interference
Combine optical path and reflection phase, then state the amplitude condition needed for complete cancellation.
AP Physics 2 Topic 14.9 · College Board CED printed p. 126, 127 / PDF p. 133, 134 · See this topic in the course outline
A film creates two returning waves. Compare both their travel phase and their reflection phase before calling them bright or dark.
Before you start
- Index of refraction
- Interference
- Wave phase
Reflection adds zero or π phase
At an interface, some light is reflected, some transmitted, and some may be absorbed. In a lossless model absorption is negligible. Reflected electric-field phase reverses by 180° (π radians) when the wave, traveling in index n₁, reflects from a higher index n₂. Reflection from a lower index produces no phase reversal. Apply this rule at the medium the ray approaches from, not from the overall stack.
Transmission/refraction does not add the abrupt reflection phase reversal in this model. The frequency stays fixed and the wavelength inside index n is λ₀/n; traveling through the film still accumulates phase because of the extra optical path. “No phase change on refraction” does not mean “no phase accumulated during travel.”
Combine optical path and reflection phase
When a film thickness is comparable to wavelength, a wave reflected at the top overlaps a wave that travels into the film, reflects at the bottom, and exits. Their electric fields add. At normal incidence the second wave travels an extra geometric distance 2t in index n, giving optical path 2nt. Its propagation phase difference is . Add the bottom reflection phase minus the top reflection phase.
| Relative reflection reversals | Constructive reflected phase | Destructive reflected phase |
|---|---|---|
| Exactly one reversal | 2nt = (m + 1/2)λ₀ | 2nt = mλ₀ |
| Zero or two reversals | 2nt = mλ₀ | 2nt = (m + 1/2)λ₀ |
Here m is a nonnegative integer and λ₀ is the wavelength in vacuum/approximately air; equivalently replace 2nt/λ₀ by 2t/λfilm. Do not mix vacuum wavelength with geometric path. These conditions identify phase reinforcement or opposition; complete cancellation also requires equal amplitudes. Changing incidence angle changes optical path and shifts the favored colors, but AP quantitative thin-film analysis is limited to normal incidence.
Thickness chooses colors; coatings suppress reflections
A soap film in air has one reflection reversal (air to film at the top, film to air at the bottom). Different thicknesses reinforce different visible wavelengths, producing changing colors. An oil film can have one or two reversals depending on the underlying material; inspect its indices rather than assuming the soap-film rule.
For an antireflection layer with , both reflections reverse phase, so there is no relative reflection reversal. The smallest nonzero destructive thickness is . The two reflected waves then differ by half a cycle from travel. For exact elimination at the design wavelength, their amplitudes also need matching; an ideal lossless normal-incidence design uses . A layer with another intermediate index still reduces reflection but may not cancel it completely. A single layer is best at its design wavelength and angle, so it need not suppress all white light equally.
Quarter-wave reflection suppression
A coating of index 1.30 lies between air (1.00) and glass (1.69). For vacuum wavelength 520 nm at normal incidence, find the minimum thickness that cancels reflected waves in the ideal lossless model.
Air→coating and coating→glass are both increases in index, so both reflected waves acquire π. Their relative reflection phase is zero.
Why this step? Count both interfaces before selecting constructive or destructive travel phase.
A half-cycle travel phase requires 2nt = λ₀/2, so .
Why this step? The extra optical path is 2nt, using vacuum wavelength.
1.30 = √(1.00 × 1.69), so the ideal normal-incidence reflected amplitudes match.
Why this step? Opposite phase eliminates reflection only when the returning amplitudes match.
Answer and check
100 nm for the design wavelength and normal incidence.
Try it from memory
Write your answer and reason before opening the worked answer. Your self-assessment is a study decision; it is not an automatically graded score.
Question 1
For a soap film in air, how many relative reflection reversals occur? Is a very thin region bright or dark in reflected light?
Worked answer and reason
One reversal. As thickness approaches zero the travel phase approaches zero, so reflected waves oppose and the region is dark in the ideal model.
Top reflection is low-to-high index; bottom reflection is high-to-low.
Question 2
Why can one soap bubble show different reflected colors at different positions?
Worked answer and reason
Thickness varies, so the optical path 2nt selects different wavelengths for reinforcement at different positions.
Colors track wavelength-dependent interference, not a change in light frequency during refraction.
JavaScript enables saved drafts and self-assessment. You can always work on paper.
Catch the tempting mistake
- A quarter vacuum wavelength is always the best film thickness.
- Use the wavelength inside the coating and inspect both reflection phases. The quarter-wave rule here is λ₀/(4n) for an intermediate-index antireflection layer at normal incidence.
Put it to work
Give the physical reason as well as the calculation. The worked solutions are on a separate page.
- Problem 17 · mixed
- Problem 18 · mixed
- Problem 21 · mixed
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For the teacher
- Make a two-row interface table first. Complete cancellation requires suitable amplitudes as well as opposite phase; a quarter-wave thickness alone can only minimize reflection.
Source objectives and scope
- 14.9.A — Describe the behavior of light that interacts with a thin film.
Quantitative analysis of thin-film interference is limited to waves that are normal to the incident surface.
Suggested skills are teaching guidance. The course applies all science practices across its objectives.
Science references for this teaching
- AP Physics 2 Course and Exam Description, official Unit 14 — Official topic/LO/EK scope and boundaries; local pinned source pages 117–134, equations visually checked on 118,120,130,132,134.
- OpenStax University Physics 1, Energy and Power of a Wave — Verification of fixed-amplitude mechanical-wave frequency scaling and distinction from electromagnetic intensity.
- OpenStax University Physics 1, Interference of Waves — Linear superposition and pointwise addition model.
- OpenStax University Physics 3, Single-Slit Diffraction — Verification of aperture minima and small-angle conditions.
- OpenStax University Physics 3, Interference in Thin Films — Verification of reflection phase and normal-incidence thin-film paths.
Glossary and deeper reading · 11 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Oscillation phase
Phase specifies the position within a repeating oscillation cycle. A phase difference tells how much one oscillation leads or lags another at a chosen place and time. For example, two sinusoidal waves half a cycle out of phase can cancel when their amplitudes are equal.
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Reflection
Reflection occurs when a wave returns from a boundary into the region from which it arrived. For a light ray at a surface, the reflection angle equals the incidence angle when both are measured from the local normal. For example, a smooth mirror produces an orderly image while a rough surface scatters reflected rays in many directions.
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Refractive index
Refractive index is the ratio of light’s speed in vacuum to its speed in the material at the relevant frequency. It is dimensionless and can depend on wavelength. For example, glass usually has a larger refractive index than air, so visible light travels more slowly through the glass.
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Radian
A radian is an angle measure defined by arc length divided by radius. One complete turn is two pi radians, and angle formulas connecting linear and rotational motion use this measure. For example, an arc whose length equals the circle’s radius subtends one radian at the center.
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Wave
A wave is a propagating disturbance that can transfer energy without transporting the medium as a whole. Mechanical waves require a material medium, while electromagnetic waves can propagate through vacuum. For example, a pulse travels along a rope while individual rope segments move locally around their equilibrium positions.
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Refraction
Refraction is a change in wave propagation at a boundary where wave speed changes. For light entering a new medium, frequency stays fixed while speed and wavelength can change, and the direction generally bends unless incidence is normal. For example, a light ray entering water obliquely from air bends toward the normal.
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Frequency
Frequency is the number of complete repeating cycles per unit time. It is measured in hertz, meaning cycles per second, and is the reciprocal of the period. For example, a vibrating string with a higher frequency produces more cycles each second.
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Wavelength
Wavelength is the spatial distance between neighboring points at the same phase of a periodic wave. It connects propagation speed and frequency through speed equal to frequency times wavelength. For example, neighboring crests of a sinusoidal water wave are separated by one wavelength.
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Electric field
An electric field assigns the electric force per unit positive test charge to each location. The field is produced by the sources and is a vector, while the force on a negative charge points opposite the field. For example, the electric field between ideal parallel capacitor plates can accelerate an electron opposite the field arrows.
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Amplitude
Amplitude is the maximum magnitude of a repeating disturbance measured from its equilibrium value. It describes the size of the oscillation rather than its frequency or propagation speed. For example, pulling a spring-mounted mass farther from equilibrium increases its oscillation amplitude when it is released from rest.
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Optical path length
Optical path length accounts for wave phase accumulated along a path through a medium. For a uniform medium it equals refractive index times geometric path length at the relevant frequency. For example, equal geometric distances through air and glass generally correspond to different optical path lengths.