There are many panels on both the walls and the ceilings of a concert hall that attempt to cancel the reflection of sound waves to prevent the "echo" from reaching the audience.  We will discuss the design of a simple sound cancelling panel or tile later in this chapter. To do a good job of cancelling all of the sound of an orchestra at all positions within the hall is impossible. [br][br]To do an excellent job while making it look artistic and allowing comfortable seating and adequate capacity for spectators is both a scientific and an artistic challenge. The one shown above is a spectacular example.

The morpho butterfly is a beautiful blue color, and yet has no blue pigment in its wings.  The blue color is a product of interference that we will learn about in this chapter.  The same is true of the deep-green-colored head of a male Mallard duck, the iridescence of beetle shells, and many other beautiful creatures in nature.  While oil slicks are not a nice occurrence, the color of them as well as the color of soap bubbles, is also due to interference.[br][br]In the realm of technology, interference allows us to encode movies on Blu-Ray discs, allows us to measure distances to tiny fractions of nanometer and to determine crystal structures.  After reading this chapter, expect to understand such concepts.[br]            [br][u]Wave Speed Induced Phase Difference        [br][/u][list][/list]As a first case of wave interference, consider a laser beam that's both monochromatic and coherent (both defined last chapter) that is incident on the top edge of a rectangular-prism-shaped crystal such that the beam is split 50/50.[br][br]While this is not the usual way of beam splitting and recombining used in optics labs, it is instructive nonetheless for learning concepts.  Assume half of the beam passes through the material of the crystal while the other half passes parallel, but just outside the crystal.  Afterwards, when the two beams recombine into a single one, what will happen to the intensity?  The black beam below is the one that splits into two equal amplitude parts as shown below.[br][br]The beam portion traveling inside the crystal will slow down by a factor of the refractive index of the material, or [i]v=c/n.[/i]  This means the wave inside will physically fall behind the wave that is traveling faster outside the crystal.  But another equivalent way to view the same problem is to imagine the waves frozen in time as illustrated below, and to count the phase that fits in each wave - the one inside and the one outside the crystal.  The difference will allow us to find the resulting wave.  To work out the details of this problem,[br] let me point out that there are two equivalent ways to view this problem.  1. We can solve for the temporal (time) delay, or the extra duration of time that the wave inside the material needs to cover the distance corresponding to the length of the crystal as compared withthe wave traveling in air, and then use that to find relative phase.  This would be considered solving in the temporal domain.[br]2. We can solve for the amount of phase that fits into the space corresponding to the length of the crystal and compare that to the amount of phase fitting in the same length in air, and then find the relative phase.  This is a solution in the spatial domain.[br][br]Either way of solving this problem should lead to the same result if we don't make any mistakes, so let's do it both ways.[br]

[u]Method 1 - Temporal domain:[/u] The wave traveling outside the crystal will travel at v=c if we neglect[br] the slight difference between the refractive index of air versus vacuum (which is ok to do here).  It will cover the length of the crystal L in a time [math]t=L/c[/math].  The wave inside the crystal is slowed by a factor of [math]n_{crystal}[/math].  Its travel time will be [math]t=nL/c[/math].  The difference in these two travel times will give rise to a relative phase shift between the two waves of [math]\delta = \omega \Delta t[/math].  Plugging in [math]\Delta t =(n-1)L/c[/math], we find [math]\delta = \omega L(n-1)/c[/math].[br][br][u]Method 2 - Spatial domain:[/u] Think of the waves as long and continuous.  In this spatial view of the problem imagine them frozen in time.  Our job is to count how much phase fits in the wave outside the crystal versus inside the crystal and then get the difference.  Outside the crystal, we find [math]\phi = k_0L[/math], and inside [math]k=nk_0[/math] (more phase fits where the wavelength is shorter).  So inside [math]\phi = nk_0L[/math].  The difference is [math]\delta = (n-1)k_0L[/math].  [br][br]We can see that the two results agree if [math]\omega/c = k_0[/math] for a light wave, and we know it does from our earlier traveling wave discussions.[br][br]If you use numbers in a problem like this you will notice that the phase difference between the waves is very large in radians.  In order for our predictions to be true, we'd have to be sure that our light source has a long enough coherence length, which as you might recall from last chapter can be related to phase by [math]\delta=kL_{coherence}[/math].   

It is very common to find a wave - either in nature or laboratory - that splits while the parts travel along different paths, and then recombine.  A simple situation where this arises is when sound in concert halls needs to be cancelled.  Why cancel the sound in a concert hall?  The quality of the acoustics in an interior space is generally very bad if the walls and ceiling produce echos, or reflect sound.  In places like classrooms we aren't as discerning as in concert halls, nor are the typical frequency ranges as broad or the sound intensities as high.  So in a classroom it generally suffices to use porous ceiling tiles and perhaps to carpet the floors or even the walls to absorb reflected sound, or echos.  The absorption is based on the ability of the material to vibrate at frequencies that match the sound they intend to absorb.  While this is slightly simplified, it's conceptually true.  If you are interested in learning more about sound isolation and control, there are lots of concepts in addition to these that you can read about online.[br][br]One of the techniques used in concert halls is phase cancelling tiles and surfaces.  The idea is that if we cause the sound to travel along different paths and then recombine, if we do it carefully, the recombined waves can cancel by having a phase difference of [math]\delta = \pi[/math].[br]       [u]     [/u][br][list][/list][u]Sound Cancelling Tiles[/u][br]Let's look at the details of a tiled ceiling that's designed for sound cancellation.  Assume there is a front surface and a recessed back surface.  For now we will assume that they are equal areas.  So 1/2 of the sound wave will bounce off the front and 1/2 off the back surface of recess that has a depth [i]d[/i].  Since the wave going to the rear surface must travel both in a distance [i]d[/i] and back out by a distance [i]d[/i], the total path length difference of the waves is [i]2d[/i].  So [math]\Delta L = 2d[/math].  We can always find a phase difference that's induced from a path length difference by using [math]\delta = k\Delta L[/math].  In some cases (not here) we just have to be careful to use the correct wave number [math]k[/math], since it can change in different media as in the case with light where the wavelength decreases by a factor of the refractive index [i]n[/i] inside a material.  Naturally[i] k[/i] would increase by the same factor since [math]k=\frac{2\pi}{\lambda}[/math].  In the present case we need not worry about that, and get [math]\delta = k(2d)[/math].  We want the sound waves to cancel, so obviously we need [math]\delta = \pi[/math] so that the phasors are in direct opposition to one another.  Setting the phase to [/math]\pi[/math] gives us [math]2kd=\pi[/math].  So the depth of tile recess depends on the desired wave number we wish to cancel.  By substituting for either wavelength or frequency, we can write this in terms of either one.  We will need to know separately the speed of sound in air.  [br][br]The speed of sound in air depends on temperature.  A linear approximation of that dependence that works rather well if we don't expect accuracy too far away from ambient temperatures is[br][br][center][math]v=(331+0.6T) m/s,[/math][/center] [br]where T is measured in Celcius.  This gives [i]343m/s[/i] at [i]20 C[/i] which is maybe a bit cool for a concert hall, but what I'll use here nonetheless.[br][list][/list][br][i][b][u]EXAMPLE:[/u][br][br]Suppose the goal is to cancel sound at a frequency of 500Hz.  We know that [math]f=v/\lambda[/math], and that [math]k=\frac{2\pi}{\lambda}[/math].  Combining gives us [math]k=\frac{2\pi f}{v}[/math].  This can be plugged in directly to our phase relation above which was [math]2kd=\pi[/math].  Solving for [math]d[/math] and plugging in gives us [math]d=\frac{\pi v}{2(2\pi f)} = \frac{v}{4f}[/math].  Plugging in numbers indicates that the tile recess depth should be [math]d=0.17m = 17cm.[/math][/b][/i][br]            [br][br]          [br][br]