Anti reflective coating, finding the 'correct thickness'

[Glorzifen]

Homework Statement

To reduce reflections from glass lenses (n ≈ 1.5), the glass surfaces
are coated with a thin layer of magnesium fluoride (n ≈ 1.39).
What is the correct thickness of the coating for green light (510 nm
vacuum wavelength)?

Homework Equations

\lambda = 2nd/(m+1/2)

The Attempt at a Solution

I don't really know where to start. It says "correct thickness"...what defines the correct thickness? Is it when a certain value is 0? Or when a certain amount of light gets through? I'm confused as to where to start because I'm not sure what this means...if anyone could give me a hint, that would be awesome. Thanks.

 

[Glorzifen]

I've been studying this a bit - is the correct thickness they're looking for one that gives off destructive interference? Normally, to get destructive interference...I would do this:

r₁ has a phase shift of pi when its reflected off of the coating (in my notes that is also written as lambda/2...not sure why that is). r₂ continues on through the coating and reflects off the other edge, travels back through the coating and out into the air where it interferes with r₁.

In a previous example I did, there was air on the other side of the thin film and so there was no phase shift when r₂ reflects off of the other edge. In that case, I would want the phase of r₂ to remain unchanged or changed only by k\lambda/2n. Right?

But in this case, there is an additional phase shift off the outer edge because the light is going from the coating to the glass (n=1.39 to n=1.5). I'm confused as how to fit that into my solution.

Any help would be appreciated.

 

[Glorzifen]

Okay. Did a bit more thinking on this:

r₁ is shifted by \lambda/2 (or pi, still not entirely sure how those two equate. The only formula I have concerning them is k=2pi/\lambda...where does the k go?)

Anyway, since there is this issue going to the coating --> glass (low to high again) there is ANOTHER phase shift in r₂ by \lambda/2. Normally there wouldn't be the same shift so we could just have a thickness that leaves the phase of r₂ untouched...but in this case they have the same shift, so if we have a thickness that leaves the phase of r₂ untouched...they will be the same and therefore constructive interference will occur. Right?

So then we want to have a thickness that causes a slight shift in the phase of r₂. The smallest shift we can do is by 1/2 a wavelength...so:

2L = \lambda(1/2) <-- the \lambda we want is the wavelength of the light once its going through the coating though, so:
2L = \lambda₀/n (1/2)
L = t = \lambda₀/2n (1/2)
t = \lambda₀/4n
= 501nm/4(1.39)
= 92µm (the correct answer)

That is the right answer, but there are a few questions I put in there along the way. I want to make sure I'm understanding this pretty comprehensively so if there is anything anyone wants to clarify for me I would be grateful.

 

[AtticusFinch]

r₁ has a phase shift of pi when its reflected off of the coating (in my notes that is also written as lambda/2...not sure why that is).

It's because if you phase shift any sine or cosine function by pi radians you move some integer multiple of lambda/2.

k\lambda/2n. Right?

If k is 2pi/lambda, then that's not the correct equation. If k is some integer then that's correct.

The only formula I have concerning them is k=2pi/\lambda...where does the k go?)

Again you shouldn't be using k like this, probably some notation confusion going on here.

Anyway, since there is this issue going to the coating --> glass (low to high again) there is ANOTHER phase shift in r₂ by \lambda/2. Normally there wouldn't be the same shift so we could just have a thickness that leaves the phase of r₂ untouched...but in this case they have the same shift, so if we have a thickness that leaves the phase of r₂ untouched...they will be the same and therefore constructive interference will occur. Right?

Yes. If you shift anything by a integer number of wavelengths the waves will still line up like they were the same wave (different from shifting by integer number of half wavelengths).

To remember this:

Integer number of wavelengths --> Doesn't change the initial phase difference
Integer number of half-wavelengths --> Switches the initial phase difference (in phase to out of phase, vice versa)