Wize High School Grade 12 Physics Textbook > Wave Nature of Light

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\degree90° 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_2n1​>n2​ and n2<n3n_2<n_3n2​<n3​.

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The path difference comes from the different lengths traveled by the two rays. The thickness ddd of the thin film is traveled by the second ray twice. Therefore we have Δx=2d\Delta x=2dΔx=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}} \ } Δϕ=kfilm​Δx=kfilm​⋅2d=λfilm​4πd​ ​

Δx\Delta xΔx is the path difference between the two interfering rays
kfilmk_{film}kfilm​ is the wave number of light in the thin film
λfilm\lambda_{film}λfilm​ is the wavelength of light in the thin film
ddd  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:
When ray #1 reflects off the n2n_2n2​ material into the n1n_1n1​material, there is no phase shift since we have a soft boundary condition.
When ray #2 reflects off then3n_3n3​ material into the n2n_2n2​ material, there is a π\bcfi{\pi}π phase shift since we have a hard boundary condition.
Therefore we have a net phase shift of π\bct{\pi}π 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)2π(m+21​) for m=0,±1,±2, ...m=0,\pm1, \pm2, \ ...m=0,±1,±2, ... . This can also be written as πm\pi mπm for m=±1,±3, ...m=\pm1, \pm3, \ ...m=±1,±3, ... (odd only).

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

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Destructive interference:

The phase shift is 2πm2\pi m2πm for m=0,±1,±2, ...m=0,\pm1, \pm2, \ ...m=0,±1,±2, ....

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

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Example: Oil Film on Water

Visible light is incident on a beaker of water (n=1.33n=1.33n=1.33) with a thin film of oil on top (n=1.5n=1.5n=1.5, d=650d = 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\piπ 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\piπ 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}4d=m⋅nλ​   →   λ=m4dn​      with    m=1,3,5, ...m=1, 3, 5 , \ ...m=1,3,5, ...

The possible wavelengths are given by:

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

Part b)

Destructive interference:

2d=m⋅λn   →   λ=2dnm2d=m\cdot\dfrac{\lambda}{n} \ \ \ \to \ \ \ \lambda=\dfrac{2dn}{m}2d=m⋅nλ​   →   λ=m2dn​   with  m=1,2,3 ...m=1,2, 3\ ...m=1,2,3 ...

The possible wavelengths are given by:

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

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Practice: Hanging Film

A thin film of thickness ddd and refractive index n=1.43n=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=300λ1​=300nm and λ1=500\lambda_1=500λ1​=500nm both reflect with maximum intensity?

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