# Thin-Film Optical Coatings

1. Apr 29, 2009

### pphy427

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 30, 2009

### DD31

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

Constructive Thickness = $$\lambda$$m / 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.

3. May 1, 2009

### pphy427

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.

4. May 1, 2009

### DD31

I see what you mean. Got it to work; I used m=2. Just takes a little guess-and-check.

Thanks

5. May 1, 2009

### cashmoney805

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