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Learning Goal: Determine conditions for constructive and destructive
interference of light reflected from/ transmitted thro...
THIN FILM INTERFERENCE	

Thin film: A film with a thickness of the same magnitude
as the λ of light involved (in nm)	

How i...
CONSTRUCTIVE	

 DESTRUCTIVE
• When the reflected light waves A and
B are in phase, the crests and troughs
reinforce each ot...
HOWTO APPLYTHE CONCEPTS
Both aforementioned equations assume both rays undergo a soft or hard reflection	

If one undergoes...
TIPS
You have to be very careful to account for whether a phase
shift occurs at an interface where reflection is taking pla...
PRACTICE PROBLEM
A butterfly’s wings are made up of
two thin layers of a transparent
substance with a refraction index
of 1...
SOLUTION
Understanding the problem	

As the refractive index of keratin is greater than the
refractive index (n) of air, r...
Solving the problem	

Find the wavelength of blue light in keratin	

The λ of blue light is 475nm in air	

Therefore, λker...
REFERENCES
http://www.webexhibits.org/causesofcolor/15E.html	

http://physics.bu.edu/py106/notes/Thinfilm.html	

http://www...
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Conditions for Constructive and Destructive Thin Film Interference

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Physics Learning Object 2

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Conditions for Constructive and Destructive Thin Film Interference

  1. 1. Learning Goal: Determine conditions for constructive and destructive interference of light reflected from/ transmitted through thin films ! By Jennifer Whetter
  2. 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. 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. 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. 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. 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. 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. 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. 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

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