Syllabus Notes Class Schedule Exams & Tests Phys 214, Autumn 2004 Light and Color Single Slit Diffraction Diffraction serves to bend waves whether they are light or sound waves.  Diffraction occurs when a plane wave passes by a sharp edge or barrier such as this illustration of a plane water wave passing through this opening with two sharp barriers (Fig 1.) Figure 1 illustrates a plane wave in a water tank passing through a barrier opening, whose width is comparable to the wavelength of the waves passing through it.  The wavelength is the crest to crest distance seen here as the bright lines.  If the width of the barrier is wider than the wavelength, much of the wave passing through is going to be left unchanged and there will be some rounded portions and some straight portions of the wave. The reason will be obvious shortly.   Sound waves and light waves behave much the same way as the water waves in Fig. 1.  If sound waves did not, we would not be able to hear sounds through a doorway unless we were standing directly in front of the door.  Diffraction effects also play a role in how a speaker must be designed in order to cover a large range of frequencies. It is not just the reflections from walls that allow us to hear around corners but also the diffraction of sound. We will see that the extent of diffraction through a doorway can be large.  Figure 1 does not tell the whole story however.  The wave is bending around both barriers in such a way that each side produces a rounded wave and what you see is the superposition of these two waves.  Figure 2 illustrates sound passing through a doorway.  The figure is looking down from the ceiling. Each side of the doorway produces its own circular wave.  The resultant will then have varying amplitudes across the doorway and beyond. The largest amplitude will be found where the the path lengths are the same thus allowing the maxima and minima in the waves to overlap exactly.  This will always occur on the line that bisects the doorway.  The loudness relationship can be determined by doing a geometrical construction of what is occurring at the doorway.  This is illustrated in Figure 3 below.  If we choose a particular direction off the center line that bisects the door way (here the horizontal dark line in Fig. 2), we can identify an angle θ shown in Figure 2 as the dashed line in order to figure out where the minimum sound intensity location will be. In Figure 3a, we see the door again from above but now there are several points across the doorway that appear to emit sound.  The model that allows us to do this is the Huygens model of a wave,that is, each point can be said to act as a point source of sound.  In Figure 3b, one can see the wave constructions that will occur in the direction θ.  We choose an angle θ, such that the distance marked by the bracket at the bottom door edge, point 5,  is exactly 1 wavelength from the top edge, point 1 in the figure. When we look at the wave that is starting at point 3, the midpoint of the door opening, it is exactly ½ λ off from that wave at the bottom of the doorway, point 5. The crests match the troughs and they will cancel each other out. Likewise, the waves that are leaving from point 4 above will be exactly cancelled by those leaving point 2.  For each point between 5 and 3, there will be a corresponding point between points 3 and 1 that will exactly cancel the sound intensity.  Only the wave from point 1 will not be cancelled.  As the angle is increased further, we will end up with waves that are not going to be cancelled completely again until the bracketed area is equal to 2λ.  The relationship for finding the first minimum in sound intensity for a slit will then be sin θ = (λ/W), where W is the width of the opening.  This is useful for slits with uniform width and what it tells us is the width of the area that will have sound. Beyond that the sound decreases substantially in amplitude. Subsequent minima occur where sinθ = n (λ/W). The intensity distribution can be seen in Fig. 24-10 in Giancoli, on page 734. The equation for determining the minimum for a circular opening is slightly different because the width of the opening varies.  Without derivation, this equation is sin θ = 1.22 (λ/D), Where D is the diameter of the hole.  The extent of the diffraction depends on the ratio of the wavelength to the opening of the hole.  From the above equations we can see that θ will be small when λ/D is small.  This means little diffraction will occur and the sound will be going straight out the opening.  However, if the opening is small relative to the wavelength, the diffraction will increase because λ/D will increase.  In a stereo loudspeaker, a wide dispersion is desirable.    The reason speakers have two or three different openings for use for the low, mid and high frequencies is to eliminate the effect we see on Figure 4 at the left.  The good stereospeakers are broken down into a woofer for the low frequency sounds, a larger speaker, than for the tweeter for the high frequency sounds.  Some of the speakers can be quite small and use sound reflection off walls to help disperse the sounds rather than dispersion.     Example: A 1500-Hz sound and a 8500-Hz sound each emerge from a loudspeaker through a circular opening whose diameter is 0.30 m.  Assuming that the speed of sound in air is 343 m/s, find the diffraction angle θ for each sound. λ1500 = 0.23 m, and λ8500 = 0.04 m found from v = λ f. Using the diffraction equation above for a circular opening, sin θ = 1.22 (0.23m/0.30m)  for 1500-Hz sound yields a θ of 70 deg, And sin θ = 1.22 (0.04m/0.30m) for 8800-Hz sound yields a θ of 9.2 deg. This is substantially different and the reason for having woofers and tweeters. The difference in these two sounds is illustrated in Figure 4. BACK TO TOP  Last Updated: 12/01/04 Contact the instructor at: chopelas@u.washington.edu