How to Calculate Optimal Thickness for Thin-Film Optical Coatings

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Homework Help Overview

The problem involves determining the optimal thickness of a MgF2 coating for dichroic glass to achieve specific colors for transmitted and reflected light. The relevant wavelengths are 692 nm for red and 519 nm for blue, with the refractive index of the coating given as n=1.39.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the need for both constructive and destructive interference to achieve the desired optical effects. There is exploration of the formulas for calculating thickness based on interference conditions, with attempts to relate the wavelengths and the refractive index.

Discussion Status

Some participants have shared their attempts to solve the problem, including rephrasing the relevant formulas and exploring different values for m. There is a recognition of the complexity involved in balancing the conditions for both colors, and some participants express uncertainty about their approaches.

Contextual Notes

Participants note the challenge of finding integer ratios for m values that satisfy both conditions, and there is mention of the phase change due to the refractive index, which adds complexity to the calculations.

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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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