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Learning Goal: Determine conditions for constructive and destructive
interference of light reflected from/ transmitted through thin films	

!
By Jennifer Whetter
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
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	

• dsinθ=mλ 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	

• dsinθ=(m+0.5)λ where m=1,3, 5,
} dsinθ }dsinθ
Note: dsinθ is the difference in path lengths (extra distance beam A must travel through medium)
HOWTO APPLYTHE CONCEPTS
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)	

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
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
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
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
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

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Thin Film Interference Jennifer Whetter

  • 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 • dsinθ=mλ 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 • dsinθ=(m+0.5)λ where m=1,3, 5, } dsinθ }dsinθ Note: dsinθ is the difference in path lengths (extra distance beam A must travel through medium)
  • 4. HOWTO APPLYTHE CONCEPTS 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) 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
  • 5. 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
  • 6. 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
  • 7. 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