### Thin Film Interference

```Thin Film Interference
AP Physics B
Montwood High School
R. Casao
• You often see bright bands of color when light
reflects from a thin layer of oil floating on water or
from a soap bubble as a result of interference.
• Light waves are reflected from the front and back
surfaces of the thin film, and constructive
interference between the two reflected waves with
different wavelengths occurs in different places for
different wavelengths.
• Light shining on the upper surface of a thin film with
thickness t is partly reflected at the upper surface
(path abc).
• Light transmitted
through the upper
surface is partly
reflected at the lower
surface (path abdef).
• The two reflected
waves come together
at point P on the
retina of the eye.
• Depending on the phase relationship, they may
interfere constructively or destructively.
• Different colors have
different wavelengths,
so the interference
may be constructive
for some colors and
destructive for others,
which is why we see
colored rings or
fringes.
• The complex shapes of the colored rings in the photo
result from differences
in the thickness of the
film.
• The bottom figure shows
a thin transparent film of
uniform thickness L and
index of refraction n2
illuminated by bright light
of wavelength λ from a distant
point source.
• Assume that air lies on both sides of the film; so
n1 = n3.
• Assume that the incident light ray is almost
perpendicular to the film (  ).
• We are interested in whether the film is bright or
dark to an observer viewing it almost
perpendicularly.
• Incident light ray i
strikes the left front
surface of the film at point a
and undergoes both reflection
and refraction there.
• The reflected ray r1 is intercepted by the observer’s
eye.
• The refracted light crosses the film to point b on the
back surface, where it undergoes both reflection and
refraction.
• The light reflected at b crosses back through the film
to point c, where it
undergoes both
reflection and refraction.
• The light refracted at point c,
represented by ray r2 is
intercepted by the observer’s eye.
• If the light waves of rays r1 and r2 are exactly in
phase at the eye, they produce an interference
maximum and region ac on the film is bright to the
observer.
• It rays r1 and r2 are out of phase at the eye, they
produce in interference minimum and region ac is
dark to the observer,
even though it is
illuminated.
• If there is some intermediate
phase difference, there are
intermediate interference and
brightness.
• The key to what the observer sees is the phase
difference between the waves of rays r1 and r2.
• Both rays are derived from the same ray i, but the
path involved in producing r2 involves light traveling
twice across the film (a to b, and then b to c),
whereas the path involved in producing r1 involves
no travel through the
film.
• Because  is about zero, we
approximate the path length
difference between the waves
of r1 and r2 as 2·L.
• To find the phase difference between the waves, we
cannot just find the number of wavelengths λ that is
equivalent to a path difference of 2·L.
• This approach is not possible for two reasons:
▫ the path length difference occurs in a medium other
than air, and
▫ reflections are
involved, which can
change the phase.
• The phase difference between
two waves can change if one or
both are reflected.
Reflection Phase Shifts
• Refraction at an interface never causes a phase
change – but reflection can, depending on the
indexes of refraction on the two sides of the
interface.
• Consider what happens when reflection causes a
phase change using the example of pulses on a
denser string (along which pulse travel is slow) and
a lighter string (along which pulse travel is fast).
• When a pulse traveling slowly along the denser
string reaches the interface with the lighter string,
the pulse is partially transmitted and partially
reflected, with no change in orientation.
• For light, this corresponds to the incident wave
traveling in the medium of greater index of
refraction (recall that greater n means slower
speed).
• The wave that is reflected at the interface does not
undergo a change in phase; its reflection phase shift
is zero.
• Light traveling from a
more dense medium to a
less dense medium is
reflected from the
interface with no phase
change.
• A crest is reflected as a
crest and a trough is
reflected as a trough.
• When a pulse traveling fast along a lighter string
reaches the interface with a denser string, the pulse
is again partially transmitted and partially reflected.
• The transmitted pulse has the same orientation as
the incident pulse, but now the reflected pulse in
inverted.
• For a sinusoidal wave, the inversion involves a phase
change of π rad, or half a wavelength (½·λ).
• For light, the situation corresponds to the incident
wave traveling in the medium of
lesser index of
refraction (with
greater speed).
• The wave that is reflected at the interface undergoes
a phase shift of π rad, or half a wavelength (½·λ).
•Light traveling from a
less dense medium to a
more dense medium is
reflected from the
interface with a phase
change of π rad or ½·λ.
•A crest is reflected as a
trough and a trough is
reflected as a crest.
• If the index of refraction
for both media is the same,
then the incident and
transmitted waves have
the same speed and there
is no reflection from the
interface.
• Summary:
Reflection
Reflection Phase Shift
Off lower index of refraction

Off higher index of refraction
½ wavelength
Remember as “higher means half”
Equations for Thin-Film Interference
• Three ways in which the phase difference between
two waves can change:
▫ By reflection
▫ By the waves traveling along paths of different lengths
▫ By the waves traveling through media of different
indexes of refraction
• When light reflects from a
thin film, producing the
waves of rays r1 and r2
shown, all three ways are
involved.
• Examine the two reflections in the figure.
▫ At point a on the front interface, the incident wave in
air reflects from the medium having the higher of the
two indexes of refraction; so the reflected ray r1 has its
phase shifted by ½·λ.
▫ At point b on the back interface, the incident wave
reflects from the medium (air) having
the lower of the two indexes of
refraction; so the reflected wave
is not shifted in phase by the
reflection, and neither is the
portion of it that exits the film as
ray r2.
• As a result of the reflection phase shifts, the waves r1
and r2 have a phase difference of ½·λ and are exactly
out of phase.
• Now consider the path length difference 2·L that
occurs because the waves of ray r2 crosses the film
twice.
• If the waves of r1 and r2 are to be
exactly in phase so that they produce
fully constructive
interference, the path length
2·L must cause an additional
phase difference of 0.5, 1.5, 2.5, …
wavelengths.
• Only then will the net phase difference be an integer
number of wavelengths.
• For a bright film, we must have:
odd number
2L 

2
(for in-phase waves)
• The wavelength we need here is the
wavelength λn2 of the light in the
medium containing path
length 2·L (in the medium
with index of refraction n2)
• Rewrite the previous equation:
odd number
2L 
 n 2
2
(for in-phase waves)
• If the waves are to be exactly out of phase so that
there is fully destructive interference, the path
length 2·L must cause either no
additional phase difference or a phase
difference of 1, 2, 3, …
wavelengths. Only then will
the net phase difference be an
odd number of half-wavelengths.
• For a dark film, we must have:
2  L  integer  
(for out-of-phase waves)
where the wavelength is the wavelength λn2 in the
medium containing 2·L.
2  L  integer  n 2
• Remembering that the greater the index of
refraction of a medium, the smaller the wavelength
of light in that medium to rewrite the wavelength of

ray r2 inside the film:
n 2 
n2
where λ is the wavelength of
the incident light in a vacuum (air).
•
• Substituting into
odd number
2L 
 n 2
2
and replacing “odd number/2”: 2  L   m  1   



2  n2
for m = 0, 1, 2, . . . (maxima for bright film in air)
• Similarly, with m replacing “integer”:
2L  m 

n2
for m = 0, 1, 2, . . .
(minima for dark film in air)
• For a given film thickness L, these equations tell us
the wavelengths of light for which the film appears
bright and dark, respectively, one wavelength for
each value of m.
• Intermediate thicknesses give intermediate
brightnesses.
• These equations also tell us the thicknesses of the
films that appear bright and dark in that light,
respectively, one thickness for each value of m.
• Intermediate thicknesses give intermediate
brightnesses.
• Summary: if the film has thickness L, the light is at
normal incidence and has wavelength λ in the film:
▫ If neither or both of the reflected waves from the two
surfaces have a half-cycle reflection phase shift, the
conditions for constructive and destructive
interference are:
▫ Constructive (no relative phase shift):
2·L = m·λ where m = 0, 1, 2, …
▫ Destructive (no relative phase shift):
2·L = (m + ½)·λ where m = 0, 1, 2, …
▫ If one of the two waves has a half-cycle reflection
phase shift, the conditions for constructive and
destructive interference are reversed:
▫ Constructive (half-cycle relative phase shift):
2·L = (m + ½)·λ where m = 0, 1, 2, …
▫ Destructive (half-cycle relative phase shift):
2·L = m·λ where m = 0, 1, 2, …
Summary: Interference in Thin Films
Normal incidence:
• Constructive reflection, no phase shift
2·t = m·λ, m = 0, 1, 2, 3, ...
• Destructive reflection
2·t = (m+½) ·λ, m = 0, 1, 2, 3, ...
• λ: Light wavelength in the film
• λο: Light wavelength in air
λ = λο/n
Phase Shift at Interface
• When na<nb, a phase shift of π, or ½·λ, occurs.
• Constructive reflection: 2·L = (m+½)·λ, m=0, 1, 2, 3…
• Destructive reflection: 2·L = m·λ, m=0, 1, 2, 3...
• If the film has thickness L, the light is at normal
incidence and has wavelength λ in the film; if
neither or both of the reflected waves from the two
surfaces have a half-cycle reflection phase shift:
1. Constructive interference (no relative phase shift):
2·L = m·λ
2. Destructive interference (no relative phase shift)
(no relative phase shift):
2·L = (m + ½)·λ
• If the film has thickness L, the light is at normal
incidence and has wavelength λ in the film; if one of
the two waves has a half-cycle reflection phase
shift:
1. Constructive interference (half-cycle phase shift):
2·L = (m + ½)·λ
2. Destructive interference (half-cycle phase shift)
(no relative phase shift):
2·L = m·λ
Interference in Thin Films
Equation
1 phase
reversal
0 or 2 phase
reversals
2·L = (m +
½)·
constructive
destructive
destructive
constructive
2·L = m·
```