Oil & Water Film: Min Thickness of 313nm

AI Thread Summary
The discussion focuses on calculating the minimum thickness of a thin oil film over water, where the film appears red at a wavelength of 626 nm. The correct formula used is t = (m + 0.5)λ / 2n, leading to a minimum thickness of 313 nm when m is set to zero. Additionally, a separate problem regarding a magnesium fluoride film on a camera lens reveals that the calculated wavelength of 297.54 nm is not in the visible spectrum, indicating a potential error in the problem's parameters. The refractive indices of magnesium fluoride and glass were discussed, suggesting that the original problem statement may have inaccuracies. The conversation highlights the importance of correctly identifying phase changes and ensuring calculations yield results within the visible range.
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Homework Statement


A thin film of oil (n = 1.50) is spread over a puddle of water (n = 1.33). In a region where the film looks red from directly above (λ = 626 nm), what is the minimum possible thickness of the film?

Homework Equations


2nt = (m+0.5)λ
t = (m+0.5)λ/2n

The Attempt at a Solution


t = (1+0.5)(626E-9m)/2(1.50)
t = 313 nm

Anyone please correct the solution.
 
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You are looking for the minimum thickness. Did you choose the correct value of the integer m to get the minimum thickness?
 
I think it should be zero
 
Right.
 
thanks
 

Homework Statement


A camera lens (n = 1.29) is coated with a thin film of magnesium fluoride (n = 1.71) of thickness 87.0 nm. What wavelength in the visible spectrum is most strongly transmitted through the film?

Homework Equations


2nt = mλ
λ = 2nt/m

The Attempt at a Solution


λ = 2(1.71)(87E-9m)/(1)
= 297.54 nm

Solution is not correct.
 
You failed to consider possible phase changes upon reflection.
 
I think you have the phase change due to reflection accounted for. (You want maximum transmission and hence minimum reflection.) But is your answer in the visible part of the spectrum?
 
No its not
 
  • #10
I might be overlooking something, but it doesn't appear to me that you can get an answer in the visible spectrum.
 
  • #11
TSny said:
I think you have the phase change due to reflection accounted for.
D'oh! You're right. :sorry:
 
  • #12
Oops, I merged the two threads after receiving a report that they were duplicate threads. I see now that they have slight differences in the problem statements. Sorry about that! :smile:
 
  • #13
A web search indicates that magnesium fluoride has an index of refraction of about 1.4. Glass (camera lens) is usually about 1.5. For these values you can get an answer in the visible.
 
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  • #14
Looks like the problem statement (for the visible light question) was mistaken. Where is the problem from?
 
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