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Wize University Physics Textbook (Master) > Wave Optics

Thin Films

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Thin Film Interference



Imagine a lens made of glass is coated with a thin film. If we assume that the incident angle is very close to 90°90\degree we can neglect refraction and look at the interference of two reflected waves as shown in the picture below.

Let's look at what happens when n1>n2n_1>n_2 and n2<n3n_2<n_3.

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The path difference comes from the different lengths traveled by the two rays. The thickness dd of the thin film is traveled by the second ray twice. Therefore we have Δx=2d\Delta x=2d and the phase difference is given by:

 Δϕ=kfilmΔx=kfilm⋅2d=4πdλfilm \boxed { \ \Delta\phi=k_{film}\Delta x =k_{film} \cdot2d=\dfrac{4\pi d}{\lambda_{film}} \ }

  • Δx\Delta x is the path difference between the two interfering rays
  • kfilmk_{film} is the wave number of light in the thin film
  • λfilm\lambda_{film} is the wavelength of light in the thin film
  • dd is the thickness of the thin film




Wize Concept
You have to consider the potential phase shift of the reflected waves.


Let's look at the phase shifts that are happening for the two rays:
  1. When ray #1 reflects off the n2n_2 material into the n1n_1material, there is no phase shift since we have a soft boundary condition.
  2. When ray #2 reflects off then3n_3 material into the n2n_2 material, there is a π\bcfi{\pi} phase shift since we have a hard boundary condition.
Therefore we have a net phase shift of π\bct{\pi} and we need to swap the constructive / destructive equations once.




Constructive interference:

The phase shift is 2π(m+12)2\pi \bigg(m+\dfrac{1}{2}\bigg) for m=0,±1,±2, ...m=0,\pm1, \pm2, \ ... . This can also be written as πm\pi m for m=±1,±3, ...m=\pm1, \pm3, \ ... (odd only).

Combining this with the equation at the top we get:
 d=mλfilm4\boxed{ \ d=\dfrac{m\lambda_{film}}{4}}
m=1,3,5,  ...m=1,3,5, \ \ ... (odd only)


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Destructive interference:

The phase shift is 2πm2\pi m for m=0,±1,±2, ...m=0,\pm1, \pm2, \ ....

Combining this with the equation at the top we get:
 d=mλfilm2\boxed{ \ d=\dfrac{m\lambda_{film}}{2}}
m=1,2,3, ...m=1, 2,3, \ ...






Watch Out!
In all these equations, the wavelength has to be in the thin film, so it will be  λfilm=λnfilm \boxed{ \ \lambda_{film}=\dfrac{\lambda}{n_{film}} \ } .

Example: Oil Film on Water


Visible light is incident on a beaker of water (n=1.33n=1.33) with a thin film of oil on top (n=1.5n=1.5, d=650d = 650 nm).

a) For which wavelengths will the reflection be the brightest?
b) For which wavelengths will there be no reflected light?

The visible wavelengths range from 400 nm (purple) to 700 nm (red).




→\to The ray that reflects off the layer of oil into air has a phase shift of π\pi since oil has a higher index of refraction (hard boundary condition).
→\to The ray that reflects off the water into oil has no phase shift, because water has a smaller index of refraction (soft boundary condition). So the two reflected beams are out of phase by pi.

Therefore we have a π\pi phase shift overall, and have to swap the equations once.

Part a)


Constructive interference:

4d=m⋅λn   →   λ=4dnm4d=m\cdot\dfrac{\lambda}{n} \ \ \ \to \ \ \ \lambda=\dfrac{4dn}{m} with m=1,3,5, ...m=1, 3, 5 , \ ...

The possible wavelengths are given by:

m=5  :m=5 \ \ : λ=780 nm   ✕\lambda=780\ nm\ \ \ ✕
m=7  :m=7 \ \ : λ=557 nm   ✓\lambda=557\ nm\ \ \ ✓
m=9  :m=9 \ \ : λ=433  nm   ✓\lambda=433\ \ nm\ \ \ ✓
m=11  :m=11 \ \ : λ=355  nm   ✕\lambda=355\ \ nm\ \ \ ✕

Part b)


Destructive interference:

2d=m⋅λn   →   λ=2dnm2d=m\cdot\dfrac{\lambda}{n} \ \ \ \to \ \ \ \lambda=\dfrac{2dn}{m} with m=1,2,3 ...m=1,2, 3\ ...

The possible wavelengths are given by:

m=2  :m=2 \ \ : λ=975 nm   ✕\lambda=975\ nm\ \ \ ✕
m=3  :m=3 \ \ : λ=650 nm   ✓\lambda=650\ nm\ \ \ ✓
m=4  :m=4 \ \ : λ=488  nm   ✓\lambda=488\ \ nm\ \ \ ✓
m=5  :m=5 \ \ : λ=390  nm   ✕\lambda=390\ \ nm\ \ \ ✕

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Practice: Hanging Film


A thin film of thickness dd and refractive index n=1.43n=1.43 is hanging from the ceiling.

a) Write the equations for constructive and destructive interference of the two reflected beams.

b) What is the minimum thickness required so that the two wavelengths λ1=300\lambda_1=300nm and λ1=500\lambda_1=500nm both reflect with maximum intensity?