W9_parallel_resonance.eps

456 Week 13: Interference and Diffraction

should probably use the minus sign no matter which one flipped (which Ijust proved

that we can do):

δ=k 2 (2d) =^2 πn^2

λ

(2d)−π=^4 πn^2 d

λ

−π (1097)

That will let us move it over onto the same side as the otherπ’s with a plus sign later.

The heuristic rules for max’s and min’s, are nowexactly the oppositeof the ones above:

2 d =^2 m+ 1

2

λ 2 =(2m+ 1)

2

λ

n 2

m= 0, 1 , 2 ... maxima (1098)

2 d = mλ 2 =m

λ

n 2 m= 0,^1 ,^2 ... minima (1099)

This is because the extra phase shift ofπor minus sign in the wave corresponds to

exactlyhalf of a wavelength path difference in the medium, just enough to make the

two rules swap places. In words, if the path difference contains an odd-half integer

number of wavelengths in the medium, the phase shift ofπat the surface contributes

the equivalent of another half wavelength and the waves will recombine constructively

inphase. Similarly, if the path difference in the medium contains an integer number of

wavelengths, the extra phase shift puts them back exactly out ofphase for (maximally)

destructive interference and a minimum.

Again, the “correct” way to arrive at this heuristic is to setδto 0, 2 π, 4 π...for constructive

interference and toπ, 3 π, 5 π...for destructive interference. The extra factor ofπis there,

ready to be moved to the other side with whatever sign that pleasesyou. Again, a

diligent student should verify that this leads straight to the heuristic rules.

13.11.5: The Limits ofVeryThin Films

The occurrence of discrete phase shifts ofπupon reflection from none, one, or both

surfaces has one easily observable consequence. Averythin film, one that is much

thinner than a wavelength (d≪λ) will havenophase shift from path difference, as the

film isn’t thick enough. The only shifts that matter, then, are thosethat arise from the

inversions reflecting off of a higher-ninterface. There are as before only two combinations

that matter – norelativereflection shift or a relative reflection shift of±π.

In the former case (two shifts or no shift’s, norelativeshift), light reflected from the

upper and lower surface emerge in phasefor all wavelengths! The surface becomes shiny

white, even mirror-like.

In the latter case (one shift in either order), light comes off of the surfaces almost exactly

out of phase for all wavelengths, and destructive interference results. Light is not reflected

from the surface; it becomes extremelytransparent.

Whether or not you know it, you have probably observed concreteexamples of both of

these limits. For example, a drop of oil or gasoline that falls onto a rainpuddle over

black pavement instantly spreads out and forms a thin film. We have all seen the initial

rainbow swirl of strange “metallic” colors, followed by the surface becoming shiny and

grey. What one is seeing is the oil forming a layer on top of water with the order of

indices of refractionnair< noil< nwater.

A second “experiment” – one that is greatly enjoyed by physics students the world over,

including very young ones – is to blow soap bubbles^137. All of us are familiar with the

swirl of colors seen in the reflections from these spherical balls of thin soap film, and at

this point you should understand that colors are the results of theenhancement of some

(^137) That’s right, this is anassignment! Go down to the store and get a bottle of bubble soap in any size that suits
you. Blow bubbles, the bigger the better, ideally on a still,quiet, warm day where you get good ‘hang time’...