Thickness of AntiReflection Coating for destructive interference

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  • #1
smithisize
13
0

Homework Statement


Glass, despite being transparent, still reflects a little bit of light. When making lenses and such, you'd rather not waste any light with stray reflections. Fortunately, you can use destructive interference to suppress those reflections. The picture below shows a piece of glass with a coating of magnesium fluoride (MgF2). The arrows represent incoming light. Some of the light will reflect off the air-MgF2 interface. Some of the light will reflect off the MgF2-glass interface. If you can get these two portions of light to interfere destructively, there won't be any visible reflections.

Use what you know about path length differences and interference to figure out how thick the coating of MgF2 needs to be for an anti-reflective coating for light of wavelength 510 nm. Use the thinnest coating possible. MgF2 has an index of refraction of 1.38 and the glass has an index of 1.56.

Note: the wavelength given is the wavelength of the light in air.

http://img811.imageshack.us/img811/5161/arcoating.jpg

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Homework Equations


r2-r1=d*sinθ = (m + 1/2)λ
V(in material) = c/n
δ = 2*Pi(m + 1/2)


The Attempt at a Solution


Eventually I'll need the wavelength in the glass. So: c=λ(in air)*f → f = c/λ(in air)
V(in glass) = c/n(glass) = λ(in glass) * f →
λ(in air)/n(in glass) = λ(glass)

Did some reading, and my physics book says that the phase difference between the beams is only due to the path length difference, which is 2t where t is thickness of film.
So, basically it said this is true:
2*Pi(m + 1/2) = 2t*2*pi/λ(glass)
So, I solved for t, got this: t = λ(glass)/2

That was wrong.

Correct answer: t = λ(glass)/4. Where is this factor of 1/2 coming from?
 
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  • #2
smithisize said:
Edit: Been reading more. So, given the scenario, the difference in phase between the two reflected waves is only due to their path length difference, which should in fact be 2t.
Good.
Since m is 1 and we're looking for destructive interference, I set 2Pi = (2t*2Pi)/λ(glass) →
t = λ(glass)/2
What must the phase difference be for destructive interference.
 
  • #3
Doc Al said:
Good.

What must the phase difference be for destructive interference.

Shouldn't it be δ= 2*Pi(m+1/2) as I did? Or should it be simply 2*Pi and I'm confusing my equations?
 
  • #4
smithisize said:
Shouldn't it be δ= 2*Pi(m+1/2) as I did?
That's the correct equation, but what does m equal?
Or should it be simply 2*Pi and I'm confusing my equations?
That is what you did and it's incorrect.
 
  • #5
Doc Al said:
That's the correct equation, but what does m equal?

That is what you did and it's incorrect.

Well I thought m should equal 1 because we're looking for the first instance of destructive interference to minimize the thickness of the coating.
Why isn't it 1?
 
  • #6
smithisize said:
Well I thought m should equal 1 because we're looking for the first instance of destructive interference to minimize the thickness of the coating.
Why isn't it 1?
The first instance is when m = 0.

The main thing is that for destructive interference the phase difference has to be an odd multiple of pi, not 2pi.
 
  • #7
Doc Al said:
The first instance is when m = 0.

The main thing is that for destructive interference the phase difference has to be an odd multiple of pi, not 2pi.

Oh ok, that is my mistake, though first instance was m = 1.

And the phase difference being a multiple of pi makes sense, because at 2pi basically there is constructive interference, and half way between 0 and two pi there is destructive, hence the multiples of pi.

Cool, thanks a lot, makes a lot more sense now!

Smith
 

Related to Thickness of AntiReflection Coating for destructive interference

1. What is the purpose of an anti-reflection coating?

An anti-reflection coating is a thin layer of material that is applied to the surface of an optical component, such as a lens or mirror, to reduce the amount of light that is reflected off the surface. This allows more light to pass through the component, resulting in improved image quality and clarity.

2. How does destructive interference play a role in the thickness of an anti-reflection coating?

Destructive interference occurs when two waves of equal amplitude and opposite phase cancel each other out. In the case of an anti-reflection coating, the thickness of the coating is designed to create destructive interference between the reflected light waves and the incident light waves, resulting in minimal reflection and maximum transmission.

3. What factors determine the optimal thickness of an anti-reflection coating?

The optimal thickness of an anti-reflection coating depends on several factors, including the refractive index of the coating material, the refractive index of the substrate material, and the wavelength of the incident light. It is also influenced by the desired level of reflection reduction and the angle of incidence of the light.

4. Can the thickness of an anti-reflection coating be calculated mathematically?

Yes, the thickness of an anti-reflection coating can be calculated using the equation: t = (λ / 4) * (n1 + n2) where t is the thickness of the coating, λ is the wavelength of the incident light, and n1 and n2 are the refractive indices of the coating and substrate materials, respectively.

5. Are there any limitations to the thickness of an anti-reflection coating?

There are limitations to the thickness of an anti-reflection coating, as a coating that is too thick can lead to unwanted reflections and reduced transmission. Additionally, the thickness of the coating must be within the range of the available coating materials, which may vary depending on the application and desired wavelength range.

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