
#1
Apr1310, 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. 



#2
Apr1310, 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:
r_{1} 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). r_{2} 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 r_{1}. 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 r_{2} reflects off of the other edge. In that case, I would want the phase of r_{2} 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. 



#3
Apr1310, 04:06 PM

P: 25

Okay. Did a bit more thinking on this:
r_{1} 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 r_{2} 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 r_{2} untouched...but in this case they have the same shift, so if we have a thickness that leaves the phase of r_{2} 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 r_{2}. 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) = 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. 



#4
Apr1310, 04:28 PM

P: 129

Anti reflective coating, finding the 'correct thickness'To remember this: Integer number of wavelengths > Doesn't change the initial phase difference Integer number of halfwavelengths > Switches the initial phase difference (in phase to out of phase, vice versa) 


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