| Date | May 2012 | Marks available | 2 | Reference code | 12M.3.HL.TZ2.12 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Show that | Question number | 12 | Adapted from | N/A |
Question
This question is about thin-film interference.
A piece of glass of refractive index 1.62 is covered with a thin film of magnesium fluoride of thickness t and refractive index 1.38. The diagram shows a ray of monochromatic light incident on the film at an angle θ to the normal.
X is a ray reflected from the surface of the film and Y is reflected from the surface of the glass.
Show that when θ=0 the condition for destructive interference between rays X and Y is
2t=(m+½)λ
where m is an integer and λ is the wavelength of light in the magnesium fluoride film.
Light of wavelength 640 nm in air is incident normally on the glass surface.
(i) Show that the wavelength of light in the magnesium fluoride film is 464 nm.
(ii) Calculate the minimum thickness of the film for which no light will be reflected back into the air.
Markscheme
the path difference (at normal incidence) is 2t;
since there is phase change of π at both surfaces there is destructive interference if the path difference is half integral multiple of the wavelength;
and so 2t=(m+½)λ
(i) \(\lambda = \frac{{{\lambda _{{\rm{air}}}}}}{n} = \frac{{640}}{{1.38}}\);
=464nm
(ii) \(m = 0 \Rightarrow t = \frac{\lambda }{4}\);
\(t = \left( {\frac{{464}}{4} = } \right)116{\rm{nm}} \approx 120{\rm{nm}}\);
Examiners report
