What is the Index of Refraction of the Liquid in Thin Film Interference?

[rose427]

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)λ_green
2nt = mλ_red
I derived both equations to solve for m and got
2nt/λ_green - 1/2 = 2nt/λ_red
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.

 

[Charles Link]

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 $m=1.06$.

 

[rose427]

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?

 

[Charles Link]

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 $m$.

 

[rose427]

That makes lots of sense now. Thanks a lot!