# Homework 13 Questions

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1. What are the formulas for constructive vs. destructive interference in thin films? The formulas in the book contradict the formulas given in HW 12 #10 "watch it."

2. For question 9, to find the number of fringes, we said in class to use m=d/a and multiply m by 2 and add 1 to that value. Although, when doing the problem myself, the correct answer was correct when I did not add 1 to 2m.  Can you explain?

From observing the HW problem and looking at the slides, I have found that the formulas of thin films depend on the situation of phase change observed. The book and the slides did seem contradictory at first, but then on further examination, it seemed to be just badly explained.

1. This is with the standard 3 layers having n1, n2, n3 in the order as top,middle, and bottom layer index of refractions:

Case 1: If only one ray reflects 180 degrees of phase:  and n1<n2>n3 or n1>n2<n3.

The formula for constructive interference is 2nt=(m+1/2)*wavelength and destructive interference 2nt=(m)*wavelength.

Case 2: If both rays reflect 180 degrees out of phase: n1<n2<n3 or n1>n2>n3 : Notice: the formulas switch!

The formula for destructive interference is 2nt=(m+1/2)*wavelength and constructive interference 2nt=(m)*wavelength.

I am not 100% certain I wrote that right, but I think it does match up with what I have seen in the slides.

If you look on one of the slides: pg. 8 of the pdf for slides #1203

The first example asks for which equation represents the destructive interference where n1<n2<n3, the answer was number 4: 2t=(1/2)wavelength2 matching up with Case 2 above.

The second example on the same slide asks for the film minimum non-zero thickness to get destructive interference. The wavelength given as 500nm. This time, n1<n2>n3 (1<1.5>1.3) and coincides with Case 1 above. In that case, destructive interference is 2nt=(m)*wavelength and n2 = 1.5. Since t is to be non-zero, the right hand side must be a non-zero number for which the lowest value of m = 1 to be non-zero. The answer then simply becomes t=wavelength/(1.5 * 2) or 500nm/3

Looking at Problem #10 in HW12 - the problem you listed about the oil film. Air has an index n=1.00. Water has an index n=1.33 (the bottom layer). My problem had it with oil having a value of n=1.30 (thin film layer). So once again n1<n2<n3 you apply the Case 2 form. A wavelength that is not reflected is destructive. A wavelength that is most reflected is constructive. Since m is not provided, it needs to be solved for using the two equations of constructive and destructive.

2.

I am pretty sure that if you reread the slide you will find that m=d/a needs to be rounded to the lowest integer before multiplying by 2.

Example: if you get a value such as 4.9 for m, you will need to round down to 4.0 (the maximum integer value) before doubling and adding the single bright fringe in the center at m=0.

I think the same case occurs for part b of this question.

Hope that helps!

answered Dec 11, 2014 by (140 points)
edited Dec 11, 2014 by MC75