| Date | November 2015 | Marks available | 2 | Reference code | 15N.3.HL.TZ0.11 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Explain | Question number | 11 | Adapted from | N/A |
Question
This question is about thin-film interference.
Monochromatic light with wavelength 572 nm is incident from air on a thin soap film.
The soap solution has a refractive index of 1.3.
Calculate the wavelength of the light within the soap solution.
Calculate the minimum thickness of the soap film that results in constructive interference for the reflected light.
Without a calculation, explain why a soap film that is twice as thick as that calculated in (b) results in destructive interference.
Markscheme
\(\lambda ' = \frac{\lambda }{{1.33}} = \frac{{572}}{{1.3}} = 440{\text{ nm}}\);
110 nm;
there would be a full wavelength within the film;
but the phase change at the first surface means that there is always destructive interference;
Examiners report
Vast majority of candidates calculated the values well and also explained destructive interference properly.
Vast majority of candidates calculated the values well and also explained destructive interference properly.
Vast majority of candidates calculated the values well and also explained destructive interference properly.
