SPH3UW/SPH4UI
Notes
Unit 10.4 Interference in Thin Films
Page 1 of 10
Physics Tool box

For reflected light in thin films, destructive interference occurs when the film
thickness has a thickness of 0,

2
, , ,...
4 4 4
For transmitted light in thin films, destructive interference occurs when the
 3 5
, , ,...
4 4 4
For transmitted light in thin films, constructive interference occurs when the
film has a thickness of 0,

3
,...
2
 3 5
film thickness has a thickness of

, ,
For reflected light in thin films, constructive interference occurs when the
film has a thickness of



2
, ,
3
,...
2
Air wedges can be used to determine the thickness of very small objects

.
 2t 
through the relationship x  L 
You have most likely noticed the swirling colours in a soap bubble, or on spilled gasoline on
the road. These effects are produced through optical interference when light is reflected or
transmitted through a thin film.
Consider a thin film like a soap bubble. When light waves strike the outer surface of the
bubble, some of the light is reflected, and some is refracted into the film. A similar process
will occur when the light approaches the inside edge of the bubble, some is reflected and
some is refracted.
As a result, two rays are reflected to the eye of the observer: one (ray 1) from the top
surface and another (ray 2) from the bottom.
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Unit 10.4 Interference in Thin Films
Page 2 of 10
Recall that when waves pass into a slower medium (bubble), the partially reflected waves are
inverted. Also when the transition is from a slower medium to a fast medium, they are not
reflected. Thus transmitted waves are never inverted.
Since both waves originate from the same source, they a initially in phase. Now ray 1 will be
inverted when it is reflected, whereas ray 2 will not. Because the film has a very thin
thickness ( t   ), the extra distance travelled by ray 2 is negligible, and thus the two rays
being 180 out of phase, interfere destructively.
For this reason, the top of a vertical soap film (very thin) appears black.
Now what happens if the soap film is a little thicker. In this case, ray 2 will have an
appreciable path difference in comparison with ray 1. if the thickness t of the film is
path difference is 2 

4
or

4
, the

, thus yielding a 180 phase delay. Thus bringing the two rays
2
back in phase again (constructive interference occurs, and a bright area is observed)
SPH3UW/SPH4UI
Unit 10.4 Interference in Thin Films
When the thickness of the film is
Page 3 of 10

, thus the path difference is  , and the two reflected
2
waves are out of phase and there is destructive interference.
As the thickness increases, the constructive, destructive interference will occurs.

For reflected light in thin films, destructive interference occurs when the film thickness
has a thickness of 0,


2
, ,
3
,... (thus dark areas)
2
For reflected light in thin films, constructive interference occurs when the film has a
thickness of
 3 5
, , ,... (thus bright areas).
4 4 4
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Unit 10.4 Interference in Thin Films
Page 4 of 10
A quantitative approach
Consider a film of uniform thickness t and index of refraction n, as shown in Figure above.
Now assume that the light rays traveling in air are nearly normal (perpendicular) to the two
surfaces of the film. To determine whether the reflected rays interfere constructively or
destructively, we have to consider the following facts:
A wave traveling from a medium of index of refraction n1 toward a medium of index of
refraction n2 undergoes a 180° phase change (becomes inverted) upon reflection
whenever n2>n1 and undergoes no phase change at all if n2<n1.
 The wavelength of light changes when inside the film and this light must travel a
distance of 2t before it again exits the film. We must consider the number of
wavelengths of the light  film that fit into this travelled distance (2t).

 The wavelength of light air is changed when it enters a medium whose index of
refraction is n. The new wavelength of light in that medium n,
n 
n
is determined by:
air
n
Technique: Determine phase change for ray 1 and ray 2 and calculateif their difference is a
multiple of a wavelength (constructive interference), or a multiple of ½
wavelength (destructive interference).
   2  1

2t 
  reflection phase shift of ray 2+    reflection phase shift of ray 1
2 

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Unit 10.4 Interference in Thin Films
Page 5 of 10
Example
Blue light ( = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167 nm) floating on
top of water (nwater = 1.3). Will Ray 1 and Ray 2 produce constructive or destructive
interference?
Solution
Ray 1
First we will determine the phase change of ray 1,
1
Since the glass has a higher index of refraction than air, we have ray 1 being inverted.
1
2
1  
Ray 2
Now the phase change for ray 2. Which has two components to consider, reflection, and
distance travelled.
Since the index of refraction for water is less than the index of refraction for water, ray 2 will
not have ant phase change.
The wavelength of our blue light (500nm)changes when it enters the glass
glass 
air
nglass

500nm
 333.3nm
1.5
The number of wavelengths this is equivalent to is:
2t
glass

2 167nm 
333.3nm
1
Therefore ray 2 will have a phase change of reflection phase change + distance phase
change
 2  reflection 
 0  1
2t
glass
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Unit 10.4 Interference in Thin Films
Page 6 of 10
We need now only consider the difference in the phase changes between ray 1 and ray 2
   2  1
1
 1  
2
1
 
2
Since the final phase change between ray 1 and ray 2 is
1
 , we will have destructive
2
interference.
Example
Light travels from air (n=1/0) to a film (n=1.40) to water (n=1.33). If the film is 1020 nm
thick and the wavelength of light is 476 nm in air, will constructive or destructive interference
occur?
Solution:
Ray 1
1
2
1  
Ray 2
 film 
air
n film
 2  0 
 0 

476nm
 340nm
1.4
2t
 film

2 1020nm 

340nm
 6
Therefore
  2  1
1
 6  
2
1
5 
2
This gives us destructive interference.
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Unit 10.4 Interference in Thin Films
Page 7 of 10
Example
A camera lens (n=1.50) is coated with a film of magnesium fluoride (n=1.25). What is the
minimum thickness of the film required to minimize reflected light of wavelength 550 nm?
Solution
We need the combined Ray 1 and Ray 2 to have destructive interference. Since we want the
minimum thickness, we need

1
.
2
Ray 1
1
2
1  
Ray 2
magnesium 
1
2
2   
air
nmagnesium
2t
magnesium

550nm
 440nm
1.25

Therefore:
   2  1
1

2
1
 1 
2t
 
  
magnesium   2 
2
2t
1

2 magnesium
t
magnesium
4
440nm

4
 110nm
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Unit 10.4 Interference in Thin Films
Page 8 of 10
Air Wedge
Air wedges can be used to determine the thickness of very small objects through the

.
 2t 
relationship x  L 
Example
a) An air wedge between two microscope slides, 10.0 cm long and separated at one end by a
paper of thickness of 0.082 mm, is illuminated with blue light of wavelength 475 nm. What
is the spacing of the dark fringes in the interference patter reflected from the air wedge?
b) How would the spacing change if the wedge was filled with water (n=1.33)?
Solution:
a)

x  L  
 2t 
 4.75 105 cm 

 10.0cm 
 2 8.2 103 cm 


2
 2.90 10 cm


2
The spacing between the dark fringes in air is 2.90 10 cm
b) We first need to find the wavelength of the light in water
 nair
 nwater

 air

4.75 105 cm

1.33
 3.57 105 cm
water  
Now for the spacing distances
SPH3UW/SPH4UI
Unit 10.4 Interference in Thin Films

x  L  
 2t 
 3.57 105 cm 

 10.0cm 
 2 8.2 103 cm 


2
 2.18 10 cm


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SPH3UW/SPH4UI
Unit 10.4 Interference in Thin Films
Extra Notes and Comments
Page 10 of
10