En anglais

Colours from a thin skin of light

Science · Waves, Light & Sound · ages 19-21

Name ______________________ Date ____________

A soap film in air has optical thickness nt much smaller than every visible wavelength. Under white light at normal incidence, what should you expect in reflection?
- Strong white reflection across the visible band
- Very weak reflection across the visible band
- Only red reflection
Light travels in glass of index 1.50 and reflects from air of index 1.00 at normal incidence. What reflection phase change occurs?
- A half-cycle flip at every reflection
- A quarter-cycle shift on leaving the glass
- No phase flip
At normal incidence, a transparent coating has n_air < n_film < n_glass, so both reflections flip. What smallest positive thickness gives a reflection phase minimum for the target wavelength?
- One quarter of the wavelength inside the film
- One whole wavelength in air
- One half of the wavelength inside the glass substrate
A soap film in air drains under steady white light. Its reflected colours change at a fixed viewing angle. Why can changing thickness cause this?
- Thickness changes the frequency of each reflected wave
- Thickness changes the reflected waves' phase gap
- Thickness changes white illumination into pure red light
At normal incidence, 500 nm vacuum-wavelength light meets a coating of index 1.38 between air of index 1.00 and glass of index 1.52. Enter the smallest quarter-wave thickness in nm, rounded to one decimal place.
Answer: ______________
One oil film of index 1.46 lies between air and water of index 1.33. Under white light at a fixed viewing angle, it shows colour bands. What can explain them without different oil types?
- Different types of oil form bands that reflect different colours
- Each patch converts the incident waves to a new light frequency
- Thickness varies, changing which wavelengths reflect strongly
At normal incidence, Sam uses t = 650/(2 × 1.33) ≈ 244 nm for a soap film in air. Sam calls this the smallest bright thickness for 650 nm vacuum-wavelength red light. Which correction fits?
- 244 nm is near a red minimum; the first red maximum is about 122 nm
- 244 nm cannot interfere; the first red maximum is 650 nm
- 244 nm is the first red maximum because reflection flips cannot change phase
A single uniform quarter-wave coating designed for one wavelength has zero reflection at every visible wavelength at normal incidence.
Circle one: True False

Answer key

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Very weak reflection across the visible band · The propagation phase is small and only the front reflection flips. Reflection is weak across the visible band, rather than exactly zero for every finite thickness.
No phase flip · At normal incidence, reflection toward the lower-index medium has no electric-field phase flip.
One quarter of the wavelength inside the film · The reflection flips give no relative phase gap. Propagation must add half a cycle on the round trip, so the layer is a quarter wavelength thick.
Thickness changes the reflected waves' phase gap · Changing thickness changes the extra optical path and phase gap. The wavelengths reflected strongly therefore change.
90.6 · t = 500/(4 × 1.38) = 90.5797... nm, which rounds to 90.6 nm. This is a reflection minimum, not zero reflection.
Thickness varies, changing which wavelengths reflect strongly · The same oil can have different thicknesses across the film. The resulting phase differences favour different reflected wavelength mixtures.
244 nm is near a red minimum; the first red maximum is about 122 nm · Sam used t = lambda0/(2n), the first nonzero minimum rule. The smallest maximum uses t = lambda0/(4n), giving about 122.2 nm.
False · The phase condition depends on wavelength. Even at the design wavelength, zero reflection also requires the appropriate index match.