Register to reply

Anti reflective coating, finding the 'correct thickness'

by Glorzifen
Tags: refraction
Share this thread:
Glorzifen
#1
Apr13-10, 01:10 PM
P: 25
1. The problem statement, all variables and given/known data
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)?

2. Relevant equations
[tex]\lambda[/tex] = 2nd/(m+1/2)

3. 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.
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Glorzifen
#2
Apr13-10, 03:21 PM
P: 25
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[tex]\lambda[/tex]/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
#3
Apr13-10, 04:06 PM
P: 25
Okay. Did a bit more thinking on this:

r1 is shifted by [tex]\lambda[/tex]/2 (or pi, still not entirely sure how those two equate. The only formula I have concerning them is k=2pi/[tex]\lambda[/tex]...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 [tex]\lambda[/tex]/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 = [tex]\lambda[/tex](1/2) <-- the [tex]\lambda[/tex] we want is the wavelength of the light once its going through the coating though, so:
2L = [tex]\lambda[/tex]0/n (1/2)
L = t = [tex]\lambda[/tex]0/2n (1/2)
t = [tex]\lambda[/tex]0/4n
= 501nm/4(1.39)
= 92Ám (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
#4
Apr13-10, 04:28 PM
P: 129
Anti reflective coating, finding the 'correct thickness'

Quote Quote by Glorzifen View Post
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.

Quote Quote by Glorzifen View Post
k[tex]\lambda[/tex]/2n. Right?
If k is 2pi/lambda, then that's not the correct equation. If k is some integer then that's correct.

Quote Quote by Glorzifen View Post
The only formula I have concerning them is k=2pi/[tex]\lambda[/tex]...where does the k go?)
Again you shouldn't be using k like this, probably some notation confusion going on here.

Quote Quote by Glorzifen View Post
Anyway, since there is this issue going to the coating --> glass (low to high again) there is ANOTHER phase shift in r2 by [tex]\lambda[/tex]/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)


Register to reply

Related Discussions
Anti-Reflective coatings General Physics 17
Stealth aircraft coating thickness Introductory Physics Homework 1
Reflective Coating for Golf Balls Computing & Technology 12
Waves: Superposition - Thickness of a Reflective coating Introductory Physics Homework 3
Anti-reflective coating optimum index of refraction Introductory Physics Homework 2