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THIN FILM INTERFERENCE
LO9
Source: https://figures.boundless.com/16755/full/oil-20slick.jpe
Produces colours of soap bubbles, oil films, peacock
feather and some type of butterflies.
Net reflection — (2) and (6). If they are in PHASE, constructive interference
will make net reflected ray relatively bright....
MINDMAP
Phase changes
𝜆n = 𝜆/n
index of refraction
Wavelength in
a medium
1. 180o (hard reflection): medium with
greater re...
CONCEPT CHECK ☑
• n1= 1.50 , n2= 2.00
• In all cases, determine what happened to reflected light when its
thickness is appr...
SOLUTIONS
• Case A:
( ✔ ) eliminated by destructive
interference
( ) bright by constructive interference
• Case B
( ✔ ) el...
SOLUTIONS
• Case C:
( ) eliminated by destructive interference
( ✔ ) bright by constructive interference
• Case D:
( ) eli...
PROBLEM 1
An anti-reflective coat (n= 1.38) is used on a glass
surface.
a) Find the minimum thickness of the coating that c...
SOLUTION
• Visualize
• Anti-reflective coating = destructive interference
n=1.34
⟷t
SOLUTION
a) 𝜆n = 𝜆/n
𝜆n = 480/1.34 = 358 nm
2t= 𝜆n/2
t = 358/4 = 89.5 nm
b) Next two possible conditions,
2t= 3𝜆n/2 2t= 5𝜆...
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LO9- Thin Film Interference

  1. 1. THIN FILM INTERFERENCE LO9 Source: https://figures.boundless.com/16755/full/oil-20slick.jpe
  2. 2. Produces colours of soap bubbles, oil films, peacock feather and some type of butterflies.
  3. 3. Net reflection — (2) and (6). If they are in PHASE, constructive interference will make net reflected ray relatively bright. Unless, destructive interference with no reflection. Source: Nelson Physics for Scientists and Engineers depend Extra distance travelled by light in coating
  4. 4. MINDMAP Phase changes 𝜆n = 𝜆/n index of refraction Wavelength in a medium 1. 180o (hard reflection): medium with greater refraction index. 2. No change (soft): medium with smaller refraction index. 2t= m𝜆n/2 (constructive interference) —————————— t=(2m+1)𝜆n/4 (destructive interference) m=0,1,2,… thickness
  5. 5. CONCEPT CHECK ☑ • n1= 1.50 , n2= 2.00 • In all cases, determine what happened to reflected light when its thickness is approaching zero. Source: Boston University
  6. 6. SOLUTIONS • Case A: ( ✔ ) eliminated by destructive interference ( ) bright by constructive interference • Case B ( ✔ ) eliminated by destructive interference ( ) bright by constructive interference In both cases, light experienced half wavelength shift due to the inverted wave from the bottom of the surface, therefore create destructive interference.
  7. 7. SOLUTIONS • Case C: ( ) eliminated by destructive interference ( ✔ ) bright by constructive interference • Case D: ( ) eliminated by destructive interference (✔ ) bright by constructive interference In case C, wave is inverted all the way therefore no shift in wavelenght. While in case D, neither wave is inverted.
  8. 8. PROBLEM 1 An anti-reflective coat (n= 1.38) is used on a glass surface. a) Find the minimum thickness of the coating that can be used for blue light (480 nm). b) Find the next two possible thickness.
  9. 9. SOLUTION • Visualize • Anti-reflective coating = destructive interference n=1.34 ⟷t
  10. 10. SOLUTION a) 𝜆n = 𝜆/n 𝜆n = 480/1.34 = 358 nm 2t= 𝜆n/2 t = 358/4 = 89.5 nm b) Next two possible conditions, 2t= 3𝜆n/2 2t= 5𝜆n/2 t = 268.5 nm t= 447.5 nm Note:We need extra phase shift of 180o to create constructive interference therefore we need wall thickness of 𝜆n/4, 3𝜆n/4, 5𝜆n/4, and so on.
  • ShaileshSomani1

    Sep. 7, 2019
  • KomalWaheeed

    Apr. 21, 2018

Getting familiar with Thin Film Interference

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