Audit thin-film phase shifts before predicting cancellation

Establish the model and evidence

Two reflected contributions can return from the front and back of a thin film. Their relative phase includes both the extra propagation through the film and any reversal upon reflection. A reflection toward a higher-index medium reverses phase by 180°; reflection toward a lower-index medium does not. Transmission alone does not add that reflection reversal. Count each interface separately before classifying the result.

At normal incidence the extra optical path is 2nt, giving propagation phase 4πnt/λvacuum. Equivalently, use the wavelength inside the film, λfilm=λvacuum/n. Counting round-trip distance with vacuum wavelength while omitting n would give a wrong thickness. The current AP Physics 2 quantitative boundary is normal incidence; oblique-angle film calculations are not requested here.

For air, an index-1.5 film and an index-2 substrate, both reflected contributions reverse. Their reflection reversals cancel in the relative phase, so a quarter-wave film adds the half-cycle needed for destructive interference. If the substrate instead has index 1, only the front reflection reverses, so the same quarter-wave round trip gives constructive phase. The substrate is therefore part of the evidence, not an irrelevant background.

The plotted two-ray intensity deliberately assumes equal amplitudes to display phase contrast. It is not the exact reflectance of the stated multilayer stack. In actual antireflection design, destructive phase does not guarantee zero reflected intensity unless amplitudes also match, and repeated internal reflections or absorption can matter. No oil, chemical film, coating fabrication or physical optical exposure is requested; the evidence is virtual. Different thicknesses select different wavelengths for stronger reflection, so a film under white light can show changing reflected colors; a monochromatic phase plot is not a full perceived-color prediction.

Work a reference case

For λvacuum=600 nm and nfilm=1.5, λfilm=400 nm and a quarter-wave film is 100 nm thick. With index-2 substrate, both reflections reverse and the round-trip phase is 180°, giving destructive phase. With index-1 substrate, the reflection-phase difference changes by 180° and the same thickness gives constructive phase.

Predict before investigating

Can the same film thickness switch between constructive and destructive reflection when only the substrate index changes?

Compare 0, 100 and 200 nm for each substrate choice. Retain top and bottom reflection reversals, optical path and net phase separately. Explain why the normalized equal-amplitude zero is not a measured zero-reflectance certificate.

Transfer to changed conditions

At normal incidence, use vacuum wavelength 480 nm, film index 1.2 and substrate index 1.8 with air above. Find the quarter-wave thickness and classify the reflected relative phase.

Compare your reasoning

Inside-film wavelength is 400 nm, so thickness is 100 nm. Both reflections reverse; the round-trip half-cycle gives destructive relative phase. Full cancellation still requires the amplitude condition, not phase alone.

Reflect and qualify the evidence

What independent amplitude condition is missing if someone claims quarter-wave thickness always eliminates all reflected light?

Sources and boundaries

Original text, calculations and diagrams; checked 30 October 2026 against the current AP Physics 2 course framework and primary textbook verification. Controls generate virtual evidence, not actual measurements or independent practical competence. No lasers, sunlight focusing, loud audio, traffic observations or chemical-film preparation are requested. Public feedback is practice; retain your derivation and critique in the course evidence workspace.