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Thin film interference problem

  • Thread starter rose427
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Homework Statement


You are working with the mineral fluorite (CaF2, n=1.43) and have a sample that is coated with a layer of liquid 158 nm thick. For various wavelengths of visible light incident normal to the surface of the liquid, you observe very strong reflection for green light (λ = 510 nm), essentially zero reflection for red light (λ = 750 nm), and intermediate levels of reflection for all wavelengths between the red and green extremes. What is the index of refraction of the liquid?

Homework Equations


2nt = mλ
2nt = (m+1/2)λ

The Attempt at a Solution


The index of refraction of the liquid can either be greater or smaller than that of mineral fluorite.
First I assumed the first case was true. Because only one phase shift occurred, the constructive condition for green light and destructive condition for red light respectively are:
2nt = (m+1/2)λgreen
2nt = mλred
I derived both equations to solve for m and got
2nt/λgreen - 1/2 = 2nt/λred
And solving this gave me n = 2.52
I did the same thing with second case which gave me n = -2.52.
However my answer is incorrect. Can someone please point out what I did wrong?
Thanks a lot.
 

Answers and Replies

  • #2
Charles Link
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For the second case, you should have ## 2nt=m \lambda_{green} ## and ## 2nt=(m-\frac{1}{2}) \lambda_{red} ##. ## \\ ## (Edit: I'm not completely satisfied with this=let me study it further...)## \\ ## Additional Edit: (Upon further study) ## \\ ## Your method of solution generates a slightly incorrect answer. By setting the ## m's ## equal in the solution without actually putting in an integer number for ## m ##, the result is a slight inaccuracy. ## \\ ## By inspection of the first two equations that you have, you could conclude that ## m=1 ##. (Trying ## m=0 ## doesn't work, and ## m=2 ## starts to generate some values for ## n ## that are too high). It is important to use the integer value of ## m ##, rather than a non-integer value that results from your calculations. (Edit: I subsequently removed a couple of calculations that I did, that you need to do to solve for ## n ##, using ## m=1 ## with either or both of your first two equations. The numbers are slightly different than ## n=2.52 ##...The homework helpers are not allowed to give you the answer.) If you used the ## n=2.52 ##, the ## m ## you compute would be ## m= 1.06 ## instead of ## m=1 ##, if my arithmetic is correct. ## \\ ## Suggestion: Use your method to estimate ## n ##, but subsequently compute ## m ##, and then select the closest integer to it. Then proceed to calculate ## n ## using the integer ## m ##. If you don't get something close to an integer for ## m ##, then you know there is an inconsistency. ## \\ ## And for the second case, (under the assumption that ## n<1.43 ##), I think you also get ## n=2.52 ## using your method. In that case though, solving for ## m ## gives ## m=1.56 ## which is totally off. ## \\ ## Additional comment: This type of observation using two wavelengths for the approximate reflection maximum and minimum is not a highly precise method of computing the index of refraction, but I have to believe that using the ## n ## computed from the ## m=1 ## value is somewhat more legitimate than the value generated for ## n ## that is based on the ## m's ## being equal, but computes to ## m=1.06 ##.
 
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  • #3
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For the second case, you should have 2nt=mλgreen2nt=mλgreen 2nt=m \lambda_{green} and 2nt=(m−12)λred2nt=(m−12)λred 2nt=(m-\frac{1}{2}) \lambda_{red} . \\
Can you please explain why the destructive condition for red light is 2nt = (m-1/2)λ and not 2nt = (m+1/2)λ?
 
  • #4
Charles Link
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The equation for red light destructive interference for the second case is ## 2nt =(l +\frac{1}{2}) \lambda_{red} ## where ## l ## is an integer, not necessarily equal to ## m ##. In this case, since ## \lambda_{red}> \lambda_{green} ##, the likely value for ## l ## is ## l= m-1 ##, although this case was found not to work regardless, since it gives a non-integer value for ## m ##.
 
  • #5
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That makes lots of sense now. Thanks a lot!
 

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