Thin film interference occurs when light reflects off of and transmits through thin films. Constructive and destructive interference results from the phase differences between light waves reflected at different film interfaces. Constructive interference produces brighter, more intense colors when reflected light waves are in phase. Destructive interference produces less intense or no color when reflected light waves are out of phase. The conditions for constructive and destructive interference, including film thickness and wavelength, can be determined using thin film interference equations.

1. Learning Goal: Determine conditions for constructive and destructive
interference of light reflected from/ transmitted through thin films
!
By Jennifer Whetter

2. THIN FILM INTERFERENCE
Thin film: A film with a thickness of the same magnitude
as the λ of light involved (in nm)
How it works
White light is incident on a surface
A portion of the light is reflected, and the rest is
transmitted through the medium
At each successive interface (where two different
mediums meet), a portion of light is reflected, and the
rest is transmitted
The light reflected from the interfaces will interfere
The pattern of light that results from this interference
appears as either light and dark bands or as colorful
bands
Interface 1
Interface 2
White Light

3. CONSTRUCTIVE
DESTRUCTIVE
• When the reflected light waves A and
B are in phase, the crests and troughs
reinforce each other
• Resultant wave, C, has a greater
amplitude than individual waves,
resulting in more intense color
• Mathematically, two reflected waves
must be shifted by an integer number
of wavelengths
• 2dcosθ=mλ/n where m=1,2,3,..
• When the reflected light waves A and B
are out of phase, the resulting wave is
attenuated
• The decrease in amplitude results in a
less intense resultant color, or the lack of
color
• Mathematically, the path difference must
be an odd number of half λ
• 2dcosθ=(m+1/2)λ/n where
m=0,1,2,3…
} dsinθ }dsinθ
d: thickness; θ: angle with respect to the normal; λ: wavelength in air/ vacuum; n: refractive index of the film

4. HOWTO APPLYTHE CONCEPTS
Both aforementioned equations assume both rays undergo a soft or hard reflection
If one undergoes a soft reflection, and the other a soft reflection, there is an additional reflection of π
radians and the equations are reversed
The type of interference that occurs depends on:
The path length difference
Whether the reflected waves are out of phase depends on the extra distance (through the film and
back) that the second ray must travel before rejoining the first ray
A path length difference equal to an odd number of half wavelengths results in a phase shift of π radians
Reflection
A wave is reflected π radians out of phase when it tries to enter a medium with a lower speed of light
(the refractive index of current medium is less than the refractive index of medium it is trying to enter)

5. TIPS
You have to be very careful to account for whether a phase
shift occurs at an interface where reflection is taking place
Wavelengths (and hence interference patterns) often
depend on whether or not a phase shift occurs at both, one,
or none of the interfaces

6. PRACTICE PROBLEM
A butterfly’s wings are made up of
two thin layers of a transparent
substance with a refraction index
of 1.56 called keratin.When
viewed in daylight, a portion of the
wings reflects blue light of
wavelength 475nm. Estimate the
minimum thickness of this section
of the film assuming m=1.
Morpho menelaus

7. SOLUTION
Understanding the problem
As the refractive index of keratin is greater than the
refractive index (n) of air, ray 2 will be reflected out of
phase with ray 1 (with a phase inversion of 1/2 a λ)
At the second interface, the ray is traveling from keratin to
air (higher to lower refractive index), and ray 5 will be in
phase with ray 1
Since one reflection is 1/2 λ and the other is 0 λ, the net
phase inversion, Φ is 1/2 λ
Since bright blue is observed, you are dealing with
constructive interference therefore you use the formula:
path difference=mλ where m=1,2,3

8. Solving the problem
Find the wavelength of blue light in keratin
The λ of blue light is 475nm in air
Therefore, λkeratin=λair/nkeratin=475/1.56=304nm
Equation for thin film interference
path difference=2t+Φ where m=1,2,3,…
mλ =2t+1/2λ where t is thickness
2t=mλkeratin-1/2λkeratin
t=1/2(m-1/2)(304)
t=1/2(1-1/2)(304)
t=1/4(304)
t=76nm
SOLUTION CONTINUED
minimum non-zero
thickness of keratin film

9. REFERENCES
http://www.webexhibits.org/causesofcolor/15E.html
http://physics.bu.edu/py106/notes/Thinfilm.html
http://www.tufts.edu/as/tampl/projects/micro_rs/theory.html
http://dev.physicslab.org/Document.aspx?
doctype=3&filename=PhysicalOptics_ThinFilmInterference.xml