How to Calculate Optimal Thickness for Thin-Film Optical Coatings

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SUMMARY

The optimal thickness for a MgF2 coating (n=1.39) to achieve red transmission at 692 nm and blue reflection at 519 nm can be calculated using the formulas for constructive and destructive interference. The constructive thickness is given by the equation d = λm / (2n) and the destructive thickness by d = λ(m - 0.5) / (2n). By setting m=2 for blue reflection, the calculations yield the necessary thickness to achieve the desired optical effects. The phase change due to the refractive index must also be considered, particularly for the blue maximum, which modifies the equation to d = λ(m + 0.5) / (2n).

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Homework Statement


A jewelry maker has asked your glass studio to produce a sheet of dichroic glass that will appear red (wavelength=692 nm) for transmitted light and blue (wavelength=519 nm) for reflected light. If you use a MgF2 coating (n=1.39), how thick should the coating be.


Homework Equations



for constructive interference: lambda = (2nd)/m
for destructive interference: lambda = (2nd)/(m-.5)


The Attempt at a Solution


Both constructive and destructive interference of the reflected waves are required here, at different wavelengths. I think we need to minimize the red for reflection and maximize the blue. I really don't know how to make sense of this problem. I'm sorry I don't have a better attempt. I will really appreciate any help you can offer. Thanks so much!
 
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I'm having the same issue with this same problem. Just to rephrase the formulas,

Constructive Thickness = \lambdam / 2*n
Destructive Thickness = \lambda(m-.5) / 2*n

I figured the same; that we wanted to maximize blue reflection and minimize red. So far, I've tried solving for the value that makes the red destructive, and tried a few multiples of it (in other words, a few m values) in comparison to the blue construcive in hope I'd find an integer match (ie a 5/4 or something ratio so I can figure the m's), but no luck on that. I've only got one crack at it left, so I want to make sure I get it.
 
ok, I've been working at this one and i think it's way easier than i thought it was.

we know that the film is the same in both conditions so if we solve both the constructive and destructive formulas for "d" we can find a corresponding "m" value. whether you use the situation where blue is constructive and red is destructive or vice versa, you'll eventually get the same answer.
 
I see what you mean. Got it to work; I used m=2. Just takes a little guess-and-check.

Thanks
 
DD31 said:
I'm having the same issue with this same problem. Just to rephrase the formulas,

Constructive Thickness = \lambdam / 2*n
Destructive Thickness = \lambda(m-.5) / 2*n

I figured the same; that we wanted to maximize blue reflection and minimize red. So far, I've tried solving for the value that makes the red destructive, and tried a few multiples of it (in other words, a few m values) in comparison to the blue construcive in hope I'd find an integer match (ie a 5/4 or something ratio so I can figure the m's), but no luck on that. I've only got one crack at it left, so I want to make sure I get it.
However, you have to take into account the phase change due to n=1.54. So in this case, the equation for blue max is t= \lambda(m+.5) / 2*n
 

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