《波的干扰》课件

1、 Lecture 2: InterferenceS3S2PIncident wave(wavelength l)yLddS12dOverview:Interference of Sound wavesTwo-Slit Interference of LightPhasorsMultiple-Slit Interference Lecture 2: InterfereInterference of Waves (from last lecture):When two waves are present at the same point in space and time they lead t

2、o interference. For the single w caseAdd amplitudes (e.g., pressures or electric fields).What we observe however is Intensity (absorbed power). I = A2Stereo speakers: Listener:y2 = A1cos(kx - t + )y1 = A1cos(kx - t) = 0:waves add “in phase” (“constructive”) I = |2 A1|2 = 4|A1|2 = 4I1 = p:waves add “

3、out of phase” (“destructive”) I = |2 A1*0|2 = 0In this lecture we confine ourselves to waves with the same wavelengths.Interference of Waves (from laSpatial InterferenceThe two waves at this point are “out of phase”. Their phase difference f depends on the path difference d r2 - r1 relative to wavel

4、ength l.r2r1The relative phase of (two or more) waves also depends on the relative distances to the sources:f = 2p(d/l) = 360(d/l)= “ # wavelengths out of phase” (for in-phase sources)df I0l/4l/2l Path difference Phase differenceA = 2A1cos(f/2)Spatial InterferenceThe two wadf I0 4I1l/4 2I1l/2 0l 4I1

5、df002A1l/4p/2 2A1l/2p 0l2p 2A1 Path difference Phase differenceSpatial InterferenceThe two waves at this point are “out of phase”. Their phase difference f depends on the path difference d r2 - r1 relative to wavelength l.r2r1A = 2A1cos(f/2)The relative phase of (two or more) waves also depends on t

6、he relative distances to the sources:f = 2p(d/l) = 360(d/l)= “ # wavelengths out of phase” (for in-phase sources) Path difference Phase differenceA = 2A1cos(f/2)Here we use equal intensities.df Idf Amplitude vs. Intensity (for 2 interfering waves) cos(f/2) cos2(f/2)What is the spatial average intens

7、ity? Plot here as a function of f.0 l 2l3l 4l 5lfA = 2A1cos(f/2)I = 4A12cos2(f/2)f0 2p 4p 6p 8p 10pdConstructive InterferenceDestructive Interference2A14A12Iave = 4I1*0.5 = 2I1For equal intensities.Amplitude vs. Intensity (for 2Procedure:1) Compute path-length difference: d = r2 - r1 =2) Compute wav

8、elength: l =3) Compute phase difference (in degrees): f =4) Write a formula for the resultant amplitude:5) Compute the resultant intensity, I =Sound velocity: v = 330 m/sr2r13 m4 mEach speaker alone produces intensity I1 = 1 W/m2 at the listener, and f = 900 Hz. Drive the speakers in phase. Compute

9、the intensity I at the listener:f = 2p(d/l)with d = r2 r1/interferenceSound wave example:Procedure:Sound velocity: Procedure:1) Compute path-length difference: d = r2 - r1 = 1 m2) Compute wavelength: l = v/f = (330 m/s)/(900 Hz) = 0.367 m3) Compute phase difference (in degrees): f = 360 (d/l) = 360(

10、1/0.367) = 9814) Write a formula for the resultant amplitude: A = 2A1cos(f/2), A1 = I15) Compute the resultant intensity, I = 4 I1cos2(f/2) = 4 (1 W/m2 ) (0.655)2 =1.72W/m2Sound velocity: v = 330 m/sr2r13 m4 mf = 2p(d/l)with d = r2 r1/interferenceSound wave example:Each speaker alone produces intens

11、ity I1 = 1 W/m2 at the listener, and f = 900 Hz. Drive the speakers in phase. Compute the intensity I at the listener:Procedure:Sound velocity: What happens to the intensity at the listener if we decrease the frequency f by a small amount? (Recall the phase shift was 981.)a. decrease b. stay the sam

12、ec. increaseACT 2: Speaker interferencer2r1What happens to the intensity What happens to the intensity at the listener if we decrease the frequency f by a small amount? (Recall the phase shift was 981.)a. decrease b. stay the samec. increaseACT 2: Speaker interferencer2r1f decreases l increases d/l

13、decreases f decreases I decreases (see figure)If0 360 720 900f = 981What happens to the intensity SummaryThe resultant intensity of two equal-intensity waves of the same wavelength at the same point in space is: I = 4 I1cos2(f/2)For nonequal intensities, the maximum and minimum intensities areImax =

14、 |A1 + A2|2 Imin = |A1 - A2|2The phase difference between the two waves may be due to a difference in their source phases or a difference in the path lengths to the observer, or both. The difference due to path lengths iswith d = r2 r1f = 2p(d/l)Summarywith d = r2 r1f = 2p(What happens when a plane

15、wave meets a small aperture?Answer: The result depends on the ratio of the wavelength l to the size of the aperture a :Huygens principle (1678) All points on wavefront are point sources for spherical secondary wavelets with speed, frequency equal to initial wave.Wavefront at t=0 Wavefront at time tl

16、 aSimilar to a wave from a point source.“Diffraction”: Interference of waves from objects or aperturesl d.S1S2ObserverLightdlr2r1ObserverSoundldIts just like sound waves!In Simple formula for the path difference, d , when the observer is far from sources.Assume 2 sources radiating in phase:d = dsinq

17、When observer distance slit spacing (r d) :f = 2p(d/l) = 2p(d sinq/l)Normal to dqddObserverrqddSimple formula for the path diTwo-Slit Interferencem = 0, 1, 2,.d = dsinq = mlConstructiveInterferenced = dsinq = (m + 1/2)lDestructiveInterference Basic result:m=0m=1m=2m=-1m=-2qq = sin-1(ml/d) “lines” of

18、 constructive interference:dqqI2l/dl/d0-l/dr Usually we care about the linear (as opposed to angular) displacement y of the pattern(because our screens are often flat):Lyy = L tanqTwo-Slit Interferencem = 0, 1Two-Slit Interference, small angles:y m(l/d)Ly (m + 1/2)(l/d)LdqY LqI2lL/dlL/d0-lL/dLThe sl

19、it-spacing d is often large compared to l, so that q is small.Then we can use the small angle approximations to simplify our results:y = L tan q L q (in radians)m = 0, 1, 2,.q m(l/d)ConstructiveInterference:q (m + 1/2)(l/d)DestructiveInterference:For small angles: (q l q is smalld sinqi = mi l d qi

20、qi mi (l/d) Dy L(q2 q1) L(2 1) (l/d) = Ll/d = (2 m)(0.663 mm)/125 mm = 0.01 mS1S2Dy1. What is the spacing Dy betw1. What is the spacing Dy between fringe maxima on a screen 2m away? a.1 mmb. 1 mm c. 1 cm 2. If we increase the spacing between the slits, what will happen to Dy?a.decrease b. stay the s

21、amec. increase3. If we instead use a green laser (smaller l), Dy will?a.decrease b. stay the samec. increaseACT 3: 2-slit interferenceA laser of wavelength 633 nm is incident on two slits separated by 0.125 mm. S1S2DySince Dy 1/d, the spacing decreases. Note: This is a general phenomenon the “far-fi

22、eld” interference pattern varies inversely with slit dimensions.Since Dy l, the spacing decreases. 1. What is the spacing Dy betwPhasorsWe now want to introduce a new way of solving interference problems, using phasors to represent the interfering amplitudes (this will make it easier to solve other

23、problems later on).A1AA1Now To get the intensity, we simply square this amplitude: where I1= A12 is the intensity when only one slit is open This is identical to our previous result !Represent a wave by a vector with magnitude (A1) and direction (f). One wave has f = 0.More generally, if the phasors

24、 have different amplitudes A and B,C2 = A2 + B2 + 2AB cos fHere f is the external angle.ACBPhasorsWe now want to introducPhasors for 2-SlitsPlot the phasor diagram for different f:A1A1fAf=45A1A1fAf=90A1A1fAf=135A1A1f=180fA1A1Af=360 A1A1Af=0A1A1fAf=225A1A1fAf=270A1A1fAf=315I04I1f02p-2pq*l/d-l/dy(l/d)

25、L-(l/d)LSlits Demo(Small-angle approx. assumed here)Phasors for 2-SlitsPlot the phMulti-Slit InterferenceWhat changes if we increase the number of slits (e.g., N = 3, 4, 1000?)First look at the “principle maxima”:If slit 1 and 2 are in phase with each other, than slit 3 will also be in phase.Conclus

26、ion: Position of “principle interference maxima” are the same!(i.e., d sinq = m l)S3S2PIncident wave(wavelength l)yLdS1d2dA1fA1fA1A What about amplitude of principle maxima? Draw phasor diagram:A1A1A1Atot = 3 A1 Itot = 9 I1 For N slitsItot = N2 I1 For other directions, simple geometry can tell us th

27、e resultant amplitude A in terms of A1 and f.Use this to find interference minimaNote: phasor angle is with respect to adjacent slit!Used in HWs etcMulti-Slit InterferenceWhat chAct 41. In 2-slit interference, the first minimum corresponds to f = p. For 3-slits, we have a secondary maximum at f = p

28、(see diagram). What is the intensity of this secondary maximum? (Hint: Use phasors.)(a) f=p/2 (b) f=2p/3 (c) f=3p/4 2. What value of f corresponds to the first zero of the 3-slit interference pattern?(a) I1 (b) 1.5 I1 (c) 3 I1f02p-2pI09I1f =? I =? (a) f=p/2 (b) f=2p/3 (c) f=3p/4 3. What value of f c

29、orresponds to the first zero of the 4-slit interference pattern?Act 41. In 2-slit interferencAct 41. In 2-slit interference, the first minimum corresponds to f = p. For 3-slits, we have a secondary maximum at f = p (see diagram). What is the intensity of this secondary maximum? (Hint: Use phasors.)(

30、a) I1 (b) 1.5 I1 (c) 3 I1f02p-2pI09I1f =? I =? What does the phasor diagram look like?Two of the three phasors cancel, leaving only one I1A1 =(a) f=p/2 (b) f=2p/3 (c) f=3p/4 2. What value of f corresponds to the first zero of the 3-slit interference pattern?Act 41. In 2-slit interferencAct 41. In 2-

31、slit interference, the first minimum corresponds to f = p. For 3-slits, we have a secondary maximum at f = p (see diagram). What is the intensity of this secondary maximum? (Hint: Use phasors.)(a) f=p/2 (b) f=2p/3 (c) f=3p/4 2. What value of f corresponds to the first zero of the 3-slit interference pattern?(a) I1 (b) 1.5 I1 (c) 3 I1f02p-2pI09I1f =? I =? A f=p/2fNo. A is not zero. f=2p/3 f Yes! Equilateral triangle gives A = 0. f=3p/4 fNo, triangle does not close. A Act 41. In 2-slit interferencAct 4(a) f=p/2 (b) f=2p/3 (c) f=3p/4 2. What value of f corresp