Introductory Physics Volume Two

7.9 Thin Film Interference 153
passes into the oil, were it proceeds until it reaches the water, at which
point part of the light is reflected back toward the detector. The light
that is not reflected passes into the water and eventually strikes the
pavement, where it is mostly absorbed, since the pavement is black. In
the end then, we have light from the sun reflected into the detector,
by two paths, one path is reflected from the air-oil interface and the
other path is reflected from the oil-water interface. If the light is nearly
normal to the surface the path difference will be twice the thickness of
the oil: ∆r = 2t. The light will be strongly reflected when the two
paths are in-phase, that is when ∆r = m λ 2
with m even. Thus in order
for the light to be strongly reflected we need,
2t = m λ 4t
−→ λ =
2
m
So only wavelengths that “match” the thickness of the oil will be reflected.
This is why you see swirling colors, what you are seeing is the
different thicknesses of the oil film, and for each thickness there is a
particular color that gets reflected.
There are a two complications to thin films that we need to consider.
The first complication is that the wavelength of light changes when
it passes into the oil. This is because the light slows down in the oil.
Definition: Index of Refraction
The index of refraction of an optical medium is the ratio of the
speed of light in a vacuum and the speed of light in the medium.
n = c v
Let λ be the wavelength in a vacuum, then λf = c. Let λ ′ be the
wavelength in the medium, then λ ′ f = v. Taking the ratio of these two
equations we find
λf
λ ′ f = c v = n
Solving for λ ′ we find the following result.
Theorem: Wavelength in a Medium
λ ′ = λ n