《波的干扰》课件.ppt

1、 Lecture 2:InterferenceS3S2PIncident wave(wavelength l)yLddS12dOverview:lInterference of Sound waveslTwo-Slit Interference of LightlPhasorslMultiple-Slit InterferencelWhen two waves are present at the same point in space and time they lead to interference.For the single w w caselAdd amplitudes(e.g.,

2、pressures or electric fields).lWhat we observe however is Intensity(absorbed power).I=A2Stereo speakers:Listener:y2=A1cos(kx-w wt+)y1=A1cos(kx-w wt)121y+y=2A cos(/2)cos(kxt/2)w A =0:waves add“in phase”(“constructive”)I=|2 A1|2=4|A1|2=4I1 =p=p:waves add“out of phase”(“destructive”)I=|2 A1*0|2=0In thi

3、s lecture we confine ourselves to waves with the same wavelengths.The two waves at this point are“out of phase”.Their phase difference depends on the path difference d d r2-r1 relative to wavelength l l.r2r1The relative phase of(two or more)waves also depends on the relative distances to the sources

4、:=2p(d/l)=2p(d/l)=360 =360(d/l)(d/l)ld=p2=“#wavelengths out of phase”(for in-phase sources)d d I I0 0l/4l/4l/2l/2l l Path difference Phase differenceA=2A1cos(/2)d d I I0 0 4I 4I1 1l/4l/4 2I 2I1 1l/2l/2 0 0l l 4I 4I1 1d d 0 00 02A2A1 1l/4l/4p/2 p/2 2A2A1 1l/2l/2p p 0 0l l2p2p 2A 2A1 1 Path difference

5、 Phase differenceThe two waves at this point are“out of phase”.Their phase difference depends on the path difference d d r2-r1 relative to wavelength l l.r2r1A=2A1cos(/2)The relative phase of(two or more)waves also depends on the relative distances to the sources:=2p(d/l)=2p(d/l)=360 =360(d/l)(d/l)l

6、d=p2=“#wavelengths out of phase”(for in-phase sources)Path difference Phase differenceA=2A1cos(/2)Here we use equal intensities.What is the spatial average intensity?Plot here as a function of.0 l 2l0 l 2l3l 4l 5l3l 4l 5l A=2A1cos(/2)I=4A12cos2(/2)0 2p0 2p 4p 6p 8p 10p 4p 6p 8p 10pd dConstructive In

7、terferenceDestructive Interference2A14A12Iave=4I1*0.5=2I1For equal intensities.Procedure:1)Compute path-length difference:d d=r2-r1=2)Compute wavelength:l l=3)Compute phase difference(in degrees):=4)Write a formula for the resultant amplitude:5)Compute the resultant intensity,I=Sound velocity:v=330

8、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 the intensity I at the listener:=2p(d/l)=2p(d/l)with d=d=r2 r1ld= path-length difference:d d=r2-r1=1 m2)Compute wavelength:l l=v/f=(330 m/s)/(900 Hz)=0.367 m3)Compute phase d

9、ifference(in degrees):=360 (d(d/l)=l)=360(1/0.367)=981 4)Write a formula for the resultant amplitude:A=2A1cos(/2),A1=I15)Compute the resultant intensity,I=4 I1cos2(/2)=4(1 W/m2)(0.655)2 =1.72W/m2Sound velocity:v=330 m/sr2r13 m4 m=2p(d/l)=2p(d/l)with d=d=r2 r1ld= speaker alone produces intensity I1=1

10、 W/m2 at the listener,and f=900 Hz.Drive the speakers in phase.Compute the intensity I at the listener: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.increaser2r1What happens to the intensity

11、 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.increaser2r1f decreases l l increases d/l d/l decreases decreases I decreases(see figure)I 0 360 720 900 =981 lThe resultant intensity of two equal-intensity waves of the sa

12、me wavelength at the same point in space is:I=4 I1cos2(/2)lFor nonequal intensities,the maximum and minimum intensities areImax=|A1+A2|2 Imin=|A1-A2|2lThe 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 b

13、oth.The difference due to path lengths iswith d=r2 r1=2p(d/l)=2p(d/l)What happens when a plane wave meets a small aperture?Answer:The result depends on the ratio of the wavelength l l to the size of the aperture a:Huygens principle(1678)All points on wavefront are point sources for spherical seconda

14、ry wavelets with speed,frequency equal to initial wave.Wavefront at t=0 Wavefront at time tl l aSimilar to a wave from a point source.“Diffraction”:Interference of waves from objects or aperturesl l d.S1S2ObserverLightdl lr2r1ObserverSoundl ldlAssume 2 sources radiating in phase:d=d=dsinq qWhen obse

15、rver distance slit spacing (r d):=2p(d/l)=2p(=2p(d/l)=2p(d sinq/l)/l)Normal to dq qdd dObserverrq qdd dm=0,1,2,.d d=dsinq q=ml lConstructiveInterferenced d=dsinq q=(m+1/2)l lDestructiveInterference Basic result:m=0m=1m=2m=-1m=-2q qq q=sin-1(ml/l/d)“lines”of constructive interference:dqqI2l2l/dl l/d0

16、-l 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 tanq qy m(l/l/d)Ly (m+1/2)(l/l/d)LdqY Lq qI2l2lL/dl lL/d0-l lL/dLThe slit-spacing d is often large compared to l l,so that q q is small.Then we can use the small an

17、gle approximations to simplify our results:y=L tan q q L q q(in radians)m=0,1,2,.q q m(l/l/d)ConstructiveInterference:q q (m+1/2)(l/l/d)DestructiveInterference:For small angles:(q q l l q q is smalld sinq qi=mi l l d q qi q qi mi(l/l/d)D Dy L(q2 q1)L(2 1)(l/d)=Ll/d=(2 m)(0.663 mm)/125 mm=0.01 mS1S2D

18、 Dy1.What is the spacing D Dy between fringe maxima on a screen 2m away?a.1 m mmb.1 mm c.1 cm 2.If we increase the spacing between the slits,what will happen to D Dy?a.decrease b.stay the samec.increase3.If we instead use a green laser(smaller l l),D Dy will?a.decrease b.stay the samec.increaseA las

19、er of wavelength 633 nm is incident on two slits separated by 0.125 mm.S1S2D DySince D Dy 1/d,the spacing decreases.Note:This is a general phenomenon the“far-field”interference pattern varies inversely with slit dimensions.Since D Dy l l,the spacing decreases.lWe now want to introduce a new way of s

20、olving interference problems,using phasors to represent the interfering amplitudes(this will make it easier to solve other problems later on).A1 A A1coscos11AAA=cos21A=Now =/2=/2 =2cos21AA To get the intensity,we simply square this amplitude:=2cos4221AI=2cos421II where I1=A12 is the intensity when o

21、nly one slit is open This is identical to our previous result!Represent a wave by a vector with magnitude(A1)and direction().One wave has =0.More generally,if the phasors have different amplitudes A and B,C2=A2+B2+2AB cos Here is the external angle.ACBlPlot the phasor diagram for different :A1A1A=45

22、8/ld=A1A1A=904/ld=A1A1A=1358/3ld=A1A1=1802/ld=A1A1A=360 ld=A1A1A=00=dA1A1A=2258/5ld=A1A1A=2704/3ld=A1A1A=3158/7ld=I04I102p p-2p pq*l l/d-l l/dy(l(l/d)L-(l l/d)LSlits Demo(Small-angle approx.assumed here)What changes if we increase the number of slits(e.g.,N=3,4,1000?)lFirst look at the“principle max

23、ima”:lIf slit 1 and 2 are in phase with each other,than slit 3 will also be in phase.Conclusion:Position of“principle interference maxima”are the same!(i.e.,d sinq q=m l l)S3S2PIncident wave(wavelength l l)yLdS1d2dA1A1A1A What about amplitude of principle maxima?Draw phasor diagram:A1A1A1Atot=3 A1 I

24、tot=9 I1 For N slitsItot=N2 I1 lFor other directions,simple geometry can tell us the resultant amplitude A in terms of A1 and .lUse this to find interference minimaNote:phasor angle is with respect to adjacent slit!Used in HWs etc1.In 2-slit interference,the first minimum corresponds to =p=p.For 3-s

25、lits,we have a secondary maximum at =p =p(see diagram).What is the intensity of this secondary maximum?(Hint:Use phasors.)(a)=p p/2(b)=2p p/3(c)=3p p/4 2.What value of corresponds to the first zero of the 3-slit interference pattern?(a)I1(b)1.5 I1(c)3 I1100100590g()x1010 x 0 02p2p2p2pI0 09I1 =?I I=?

26、(a)=p p/2(b)=2p p/3(c)=3p p/4 3.What value of corresponds to the first zero of the 4-slit interference pattern?1.In 2-slit interference,the first minimum corresponds to =p=p.For 3-slits,we have a secondary maximum at =p =p(see diagram).What is the intensity of this secondary maximum?(Hint:Use phasor

27、s.)(a)I1(b)1.5 I1(c)3 I1100100590g()x1010 x 0 02p2p2p2pI0 09I1 =?I I=?What does the phasor diagram look like?Two of the three phasors cancel,leaving only one I1A1=(a)=p p/2(b)=2p p/3(c)=3p p/4 2.What value of corresponds to the first zero of the 3-slit interference pattern?1.In 2-slit interference,t

28、he first minimum corresponds to =p=p.For 3-slits,we have a secondary maximum at =p =p(see diagram).What is the intensity of this secondary maximum?(Hint:Use phasors.)(a)=p p/2(b)=2p p/3(c)=3p p/4 2.What value of corresponds to the first zero of the 3-slit interference pattern?(a)I1(b)1.5 I1(c)3 I110

29、0100590g()x1010 x 0 02p2p2p2pI0 09I1 =?I I=?A=p/2=p/2 No.A is not zero.=2p/3p/3 Yes!Equilateral triangle gives A=0.=3p/4p/4 No,triangle does not close.A(a)=p p/2(b)=2p p/3(c)=3p p/4 2.What value of corresponds to the first zero of the 3-slit interference pattern?100100590g()x1010 x 0 02p2p2p2pI0 09I

30、1 =?I I=?(a)=p p/2(b)=2p p/3(c)=3p p/4 3.What value of corresponds to the first zero of the 4-slit interference pattern?=p/2=p/2 Yes.Square gives A=0.=2p/3p/3 No.With 4 slits we start to“wrap around”again(A=A1)=3p/4p/4 No.A 0 A For N slits the first zero is at 2p p/N.The positions of the principal m

31、axima of the intensity patterns always occur at =0,2p p,4p,.p,.is the phase between adjacent slits (i.e.,dsinq q=ml l,m=0,1,2,).The principal maxima become taller and narrower as N increases.The intensity of a principal maximum is equal to N2 times the maximum intensity from one slit.The width of a

32、principal maximum goes as 1/N.The#of zeroes between adjacent principal maxima is equal to N-1.The#of secondary maxima between adjacent principal maxima is N-2.02p2pI016I1N=402p2pI025I1N=502p2pI09I1N=31001001020250h5()x1010 xl/d 0 l/dq100100590g()x1010 xql/d 0 l/d10010010160h()x1010 xl/d 0 l/dqSlits Demo Laser Demo