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Antireflection Coating Problem

by superspartan9
Tags: antireflection, coating
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superspartan9
#1
Oct9-11, 08:08 PM
P: 23
1. The problem statement, all variables and given/known data
"A glass lens (nglass = 1.52) has an antireflection coating of MgF2 (n = 1.38). (a) For 517-nm light, what minimum thickness of the coating will cause reflection rays R2 and R4 to interfere destructively assuming normal incidence? (b) Interference will also occur between forward moving rays R1 and R3. What minimum thickness will cause these rays to interfere constructively?

We are given a figure (attached as temp.jpg).


2. Relevant equations

Δphase = π*(2k + 1) for destructive, Δphase = 2kπ for constructive, Δphase = (2*thickness*n / λ) * 2π + π for part (a) because the rays go from the air to the back wall of the coating an reflect back, part (b) has got me stumped

3. The attempt at a solution

Ok, so attempting to solve part (a), I've solved the equations for Δphase (setting them equal to each other), I got 2k = ((4 * 1.38) / (517 e -9)) * thickness. Now I don't know where to go from here. As far as part (b), I don't understand what the rays are actually doing according to the figure.
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ehild
#2
Oct10-11, 12:38 AM
HW Helper
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P: 10,534
The figure is not shown.


ehild
superspartan9
#3
Oct10-11, 08:20 AM
P: 23
Weird, I thought I attached it to the original post. Regardless here it is.
Attached Thumbnails
temp.jpg  

superspartan9
#4
Oct10-11, 11:25 AM
P: 23
Antireflection Coating Problem

Ok, I solved for part (a), and it makes sense. Now going to part (b), I face a problem concerning the thickness of the glass and how it would play into the phase change.

So far, I've concluded that, using constructive interference equation, Δphase = 2kπ = [(thickness * n) / λ] * 2π + 2π + [(thickness or distance traveled in glass * nglass) / λ] * 2π

I get this equation through analyzing the three phase changing elements in the problem. Firstly, the phase change through the film, which only goes through once hence it's just 1*thickness of the film. Secondly, since the light goes from n = 1 to n = 1.38 to n = 1.52, there are going to be two π shifts as the light goes from low n to high n. Lastly and the strange part, the phase change in the glass. I originally thought there would be no change in phase here, but that yielded the same equation as part (a), so unless the point of this part is to show that the same thickness of film will reflect destructive waves from the glass and let constructive waves through, I don't know where to go.
ehild
#5
Oct10-11, 11:42 AM
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P: 10,534
No need to calculate with the phase change in the glass. In the reflected light, the directly reflected ray interferes with the one reflected from the back boundary of the film and traversing the film twice. Both reflected rays change phase upon reflection as they are reflected from a higher refractive index material. The phase changes (both pi) cancel, and the phase difference is simply (4pi/lambda)*n* (thickness).

In the transmitted light, one ray travels through directly, the other reflects back from the back interface of the MgF2 layer and changes phase by pi; then goes back to the air+layer interface, reflects back (with no phase change this time) and goes through the layer again. So you have a ray which went through the layer once without reflection and the other which went through the layer three times and reflected twice, once from a higher index material, once with a lower index one. What is the phase difference between these two rays then?

ehild
superspartan9
#6
Oct10-11, 12:07 PM
P: 23
[(thickness * n * 2∏) / λ] - ([(3 * thickness * n * 2∏) / λ] + ∏) = 2k∏ ?
superspartan9
#7
Oct10-11, 12:09 PM
P: 23
Actually, I'd probably want to switch the two around to get a positive phase change, right?
ehild
#8
Oct10-11, 03:51 PM
HW Helper
Thanks
P: 10,534
Correct, and make it positive

ehild


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