Group: ____________ Names: ______________________________________________________
Group Worksheet on Thin Films
1. A disabled tanker leaks kerosene n=1.20 into the Persian Gulf, creating a large slick on top of the
water, n=1.30. (a) If you are looking straight down from an airplane, while the Sun is overhead,
at a region of the slick where its thickness is 460 nm, for which wavelength(s) of visible light is
the reflection brightest because of constructive interference? (b) If you are scuba diving directly
under this same region of the slick, for which wavelength(s) of visible light is the transmitted
intensity strongest?
a) Let the indices of refraction of the air, oil and water be n1,n2, and n3, respectively. Since n1<n2
and n2<n3, there is a phase change of λ from each interfaces. The second wave also travels an
additional distance of 2L (for a shift of 2n2L), so the overall phase difference is λ +2n2L. If this
constructively interferes (bright reflections), we have m λ = λ +2n2L for m=0,1,2… Which leaves
us with λ=2n2L/(m-1) or
λ=2n2L/m where m=1,2…
Plugging in numbers gives λ=2(1.2)(460nm)/m=1104 nm, 552 nm, 368 nm… but only the 552 nm
wavelength falls within the visible light range.
b) Maximum transmission into the water occurs for wavelengths for which reflection is a
minimum. This changes our conditions to look at destructive interference. So now we have
(m+ ½ ) λ= λ+2n2L or λ=4n2L/(2m+1).
Plugging in numbers here yields λ=2208nm, 736nm, 442 nm…but only the 442 nm wavelength is
in the visible range.
2. A broad beam of light of wavelength 683 nm is sent directly downward through the top plate of
a pair of glass plates. The plates are 120 mm long, touch at the left end, and are separated by 48
microns at the right end. The air between the plates acts as a thin film. How many bright fringes
will be seen by an observer looking down through the top plate?
The wave reflected from the upper surface does not change phase on reflection but the wave
reflected from the bottom surface changes phase by λ. At a place where the thickness of the air
film is L, the condition for fully constructive interference (bright fringes), remembering that
nair=1, is 2L=(m+ ½ )λ. Solving for m=2L/λ – ½ = 140. So there are 140 bright fringes.
3. What is the minimum, non-zero thickness of a soap film (n=1.34) that would appear black if
illuminated normally with red light of wavelength 630 nm. Assume there is air on both sides of
the soap film.
Appear black means destructively interfere. Off the top surface of the soap bubble there is a phase shift
of λ/2 but not from the bottom surface (soap to air). For the rays to destructively interfere we have
λ/2=2nL+(m+ ½ )λ. Solving for d yields: L=mλ/2n for m=1,2… The minimum thickness is with m=1.
L=630nm/(2*1.34)=235.1 nm film.
4.
Antireflective coatings are used on lenses to eliminate various colors on reflection. What is the
minimum, non-zero thickness for such a coating (n=1.4) applied to a glass lens (n=1.55) such that
red light of wavelength 630 nm will be eliminated for light at normal incidence?
2nL=(m+ ½ )λ where n=1.4 and λ=630 nm . Solving for L (with m=0), L=λ/4n=112.5 nm