Thickness of AntiReflection Coating for destructive interference

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Homework Help Overview

The discussion revolves around the thickness of an anti-reflection coating made of magnesium fluoride (MgF2) applied to glass, focusing on achieving destructive interference to minimize light reflection. The problem involves understanding the conditions for destructive interference and calculating the necessary thickness of the coating for light of a specific wavelength.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between path length differences and phase differences in the context of destructive interference. There is discussion about the correct application of equations related to phase difference and the significance of the variable m in determining the thickness of the coating.

Discussion Status

Participants are actively questioning the assumptions regarding the value of m and its implications for the thickness of the coating. There is a recognition of the need for clarity on the phase difference required for destructive interference, with some guidance provided on the correct interpretation of the equations involved.

Contextual Notes

Participants are navigating the complexities of phase differences in interference phenomena, particularly focusing on the conditions for the first instance of destructive interference and the implications of using different values for m.

smithisize
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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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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.
 
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?
 
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.
 
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?
 
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.
 
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
 

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