Example 23-6 A Soapy Film
Monochromatic light that has wavelength 560 nm in air strikes a layer of soapy water, which has an index of refraction of
1.40 and rests on a bathroom tile. (a) If the layer of soapy water is 700 nm thick, does constructive interference, destructive
interference, or neither occur when the light strikes the surface close to the normal? (b) What is the minimum thickness of
soapy water that would result in no (or minimum) reflection from the surface?
Set Up
The situation is the same as in Figure 23-25.
Interference occurs between (i) light that is
reflected from the top surface of the soapy layer
and (ii) light that enters the soapy layer and is
eventually reflected back out. In part (a) we’ll
compare the given values of wavelength and
film thickness to Equations 23-17 and 23-19 to
decide whether the interference is constructive,
destructive, or something in between. In part
(b) we’ll use Equation 23-20 to find the minimum thickness for destructive interference. In
both parts, Equation 23-16 will help us relate
the wavelength of the light in air to its wavelength in the soapy water.
Constructive interference:
2D = ml2 = m
(a) In this situation medium 1 is air (n1 = 1.00)
and medium 2, of which the film is made, is
soapy water (n2 = 1.40). The wavelength in the
soapy water is therefore shorter than in air.
Find how many wavelengths of the wavelength
l2 in the soapy water fit into the path length
difference  pl = 2D, where D = 700 nm is the
film thickness.
(23-17)
Destructive interference:
l2
n1l1
= 12m - 12
, m = 1, 2, 3, c
2
2n2
(23-19)
Minimum thickness for destructive interference:
2D = 12m - 12
D min =
n1l1
4n2
(23-20)
Wavelength in two different media:
l2 =
Solve
n1l1
, m = 1, 2, 3, c n2
n1l1
n2
(23-16)
Wavelength in air: l1 = 560 nm
Wavelength in soapy water:
l2 =
11.002 1560 nm2
n1l1
=
= 400 nm
n2
1.40
The path length difference between the waves that reflect off the top
and bottom surfaces of the film is
 pl = 2D = 2(700 nm) = 1400 nm
The number of wavelengths that fit into this path length difference
equals  pl divided by l2:
 pl
l2
=
1400 nm
7
= 3.5 =
400 nm
2
The path length difference is an odd number of half wavelengths.
We conclude that there is destructive interference between the light
that reflects from the top surface of the soapy water and the light that
reflects from the bottom surface (where the soapy water touches
the tile).
(b) The minimum thickness required for
destructive interference is such that the path
length difference 2D is one half-wavelength,
so 2D = l2 >2 and D = l2 >4.
From Equation 23-20, the minimum thickness for destructive
interference is
Dmin =
Alternatively, since l2 = n1l1 >n2 from Equation 23-16,
Dmin =
Reflect
11.002 1560 nm2
n1l1
=
= 100 nm
4n2
411.402
l2
400 nm
=
= 100 nm
4
4
When light of wavelength 560 nm in air strikes the soapy layer close to the normal, destructive interference occurs both
when the layer is 100 nm thick (so twice the thickness is 200 nm, or 1>2 the wavelength in the soapy water) and when
the layer is 700 nm thick (so twice the thickness is 1400 nm, or 7>2 the wavelength in the soapy water). For these thicknesses, reflections from the surface would be minimized, and the surface would look dark.
Destructive interference always occurs when twice the thickness of the layer is an odd multiple of one-half of the
wavelength. Can you see that destructive interference would also occur if the film of soapy water were either 300 nm
thick or 500 nm thick?