W9_parallel_resonance.eps

Week 13: Interference and Diffraction 455

Permutation δ 12 δ 23 |∆δ|

n 1 < n 2 < n 3 π π 0

n 1 > n 2 > n 3 0 0 0

n 1 < n 2 > n 3 π 0 π

n 1 > n 2 < n 3 0 π π

Table 8: Relative phase shift introduced between the wave reflected off of then 1 →n 2 interface
and the transmitted wave reflected off of then 2 →n 3 interface. Note that in the first two cases
(smoothly increasing or decreasingn) there is no net phase shift withn 2 “in the middle”. In the
second two cases, the index of refraction of the thin film medium is either higher than that of its
neighbers or lower, but not in the middle.

13.11.3: No Relative Phase Shift from Surface Reflections

Consider the case whereδ 12 =δ 23 = 0 orπ. In both of these cases there is norelative

phase shift due to the reflections. Either both waves flip (and hence accumulate phase

difference only due to the path difference) or neither wave flips (ditto). Either way, the

totalrelative phase shiftδis just due to the path difference:

δ=k 2 (2d) =

2 πn 2

λ

(2d) =

4 πn 2 d

λ

(1093)

We can now use our simple heuristic rules for max’s and min’s: If the path difference

is an integer number of wavelengthsλ 2 in the thin film, then we expect the two waves

to recombine in phase and while the resultant amplitude may not betwiceeither of the

two waves, it will certainly be larger than either one alone. Similarly, if itis an odd-half

integer number of wavelengths in the film, we expect the waves to beexactly out of phase

and to maximally cancel. We’ll summarize this as:

2 d = mλ 2 =m

λ

n 2

m= 0, 1 , 2 ... maxima (1094)

2 d =

2 m+ 1

2 λ^2 =

(2m+ 1)

2

λ

n 2 m= 0,^1 ,^2 ... minima (1095)

Of course, this is only heuristic. The “correct” way to arrive at thesame place is to setδ

to 0, 2 π, 4 π...for constructive interference and toπ, 3 π, 5 π...for destructive interference.

It is left as a fairly simple (and hopefully by now, familiar) exercise for the student to

show that if you do this, you arrive precisely at our heuristic rules.

13.11.4: A Relative Phase Shift ofπfrom Surface Reflections

Consider the cases whereeitherδ 12 or δ 23 isπand the other is 0. In both of these

cases thereisa relative phase shift due to the reflections. One of the two waves flips

(and hence “suddenly” accumulate an additional phase ofπand the other does not. No

matter which wave flips thetotalrelative phase shiftδmust add or subtract this relative

phase to the one from the path difference:

δ=k 2 (2d) =

2 πn 2

λ

(2d)±π=

4 πn 2 d

λ

±π (1096)

Note that the sign we get differ depending on which one flipped. However, we don’t

really care which sign we get. This is because sin(θ+π) = sin(θ−π) =−sin(theta), so

we can simply move aπwith either sign to whatever side of the equals sign that seems

convenient to us. In order to get the best correspondance with our heuristic rules, we