光学光学光学光学 (5).pdf

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1、-153-Wave Optics Interference，Interferometer，Coherence Diffraction-Fourier Optics Fundamental principle:Superposition of waves Central problem:(the kernel question)Phase distribution over a wave front.We shall first look at interference(Hecht chap7.1-7.2,Zhaos Chap2 section 1-4),focusing on the prin

2、ciple of superposition illustrating the importance of phase relation in interference,and the evaluation of phase distribution over a wavefront,the using Youngs Experiment as example.Later we shall look into some interferometer devices(Zhaos,chap3;Hecht 9.3,9.4,9.6)and the important concept of cohere

3、nce.Chapter 5 Interference,Interferometer and Coherence Lets start this important chapter with some examples to illustrate the phenomena we call interference,which is the hallmark behavior of wave.-154-In both cases,the total intensity(energy)from contribution of two paths is not simple summation of

4、 the energy of each path:12III+,such phenomenon is called interference,and we shall see it arises from superposition of waves(not the superposition of irradiance or intensity).Whether the superposition of waves displays interference or not depends on the coherence among component waves.5-1 Superposi

5、tion of Waves This is a fundamental postulate like that superposition of forces and the fields(here the wave is just E-M field anyway).The math view of the -155-principle of superposition is originated from the wave equation discussed earlier.22221uUVt=If the v is independent of U then:iiiUaU=is the

6、 solution of the wave equation provided iU is the solution,ia is a coefficient.The field(or waves)shall be added to get the total field(wave),and the system response to the field will be treated linear as long as the field is not too strong,this is the domain of linear optics.5-1-1 Superposition pri

7、nciple in linear optics A general light wave can be written as:()0ipi tEE ee=where 0E the amplitude;()p the spatial phase(including the initial phase);t the oscillation with time or temporal phase.For plane wave:0()pk r=+?spherical wave:0()pkr=+However,common detectors only detect the irradiance(or

8、intensity)of the light,which we know:220TIEE=Since we are interested in the relative irradiance in most cases,and we shall let:(as if chosen proper unit)220IEE=(5-1)-156-Our common sense sometimes misleadingly makes us to perceive that in the cases of multiple light sources,the resultant irradiance

9、is iiII=This is the case for illumination by incoherent sources such as light bulbs,candles,etc.However,iiII=is generally wrong,as demonstrated by the two examples at the beginning of the chapter.It is the summation of the field E,not I that is physically correct,as stated in the principle of superp

10、osition:Suppose we have many different field sources contributing to a point p,then from superposition principle,the total field at p is:()0()jjipitpjjjjEEpE ee=?(5-2)We shall first study simpler case where different components have same frequency and in such casej=,the intensity can be easily compu

11、ted with(5-1):22000|()()2cos()()jkjjkjkjkjjkjIEEEEEEpp=+(5-4)sintan()cosojjjojjjEpE=(5-5)(5-4)is just(5-3),a result of superposition;the(5-5)can be relatively easily derived by phasor method.-158-The phasor method in superposition of waves usually offers an intuitive and speedy solution of waves wit

12、h same frequency and parallel E0.5-1-3 Standing Waves (Hecht,7.1.4)An interesting puzzle:recalling the reflection by a surface at normal incident angle in discussion of Fresnel equations:In the example above when tinn(external reflection),the reflected lights pE and B?field are shown in the figure.F

13、rom the Fresnel equation we know ipE and rpE are reversed in direction(so,irB B?are in same direction,since irk andk?reversed,andEBk for traveling -159-E-M waves).So it seems if we superpose the incident and reflecting waves,irEEE=+?diminishes,while irBBB=+?increases.Recall that for traveling waves,

14、EcB=there seems existing an apparent paradox.The reason is that the superposition of two traveling waves travelling in opposite directions will result in a standing wave,not a traveling wave.The general solution to the wave equation takes form of 12(,)()()x tC f xvtC g xvt=+So when two waves travel

15、in opposite direction,the summation also satisfies the wave equation.For this case,the resultant wave will be a standing wave.(I follow the Hechts derivation and use sines for wave;if you use cosines or Euler formula,same results)For the wave propagates along-x:0sin()IIIEEkxt=+The reflected wave pro

16、pagates along+x:0sin()RRREEkxt=+The Reflecting Mirror at x=0,the boundary condition at x=0,requires 0IREEE=+=Lets consider the simple case where r=1 or00IREE=,we can set 0I=(which is arbitrary anyway,we always have the freedom to set the initial phase of ONE wave as zero),and the(0)00IRRxEE=+=and th

17、e superposition of wave is:00sin()sin()2sincosEEkxtkxtEkxt=+=(5-6)-160-(5-6)is the expression for a standing wave.It is standing in a sense that the spatial distribution sinkx,does not move with time(the nodes and the antinodes are fixed in space,in contrast to the traveling wave);the amplitude chan

18、ges with time as coswt(see figure 7.10,7.11 in Hechts book and Fig 7.14 shows a setup to verify the existence of the standing wave).For an open end standing wave(i.e.only has boundary condition at one end,such as waves reflected by one mirror as shown):,the resulting standing wavesincoskxt,there is

19、no restriction on the value of k,it can take any values.Another important case is the light field in a cavity:such as the two mirror cavity separated by L.The boundary condition would be(0)0E xxL=,the resultant stable wave is still in the form ofsincoskxt,but with restriction on k.where 22,0,1,2(57)

20、mLmmk=This is directly from the boundary condition E(x=L)=0klm=.The reason that I wrote it in form of(5-7)is that it represents a clear physical -161-picture,as I shall demonstrate below.From the superposition of wave,the field inside the cavity bounded by M1,M2 is the superposition of the incident

21、wave and the reflected ones(multiple times)by M1and M2.Another equivalent way to look at this 2-Mirror,L length cavity is taking it as circle with 2L periphery,the waves travels along in circle.The only stable ones are the waves that after one round passage overlap exactly with the original ones i.e

22、.02,0,1,2_2LmmornLm=in terms of Optical Path length.These waves are standing waves;for those do not satisfy(5-7)the superposition will result a null field(E=0).For length=L cavity,the standing waves of first 3 lowest m(called modes)are:The standing waves have form of(5-6),and then the expression for

23、 the B -162-field can be derived from Maxwell equation:cossinBEBkxtt Thus what are nodes for the E field(sinkx=0),are antinodes for B(coskx=1);what are antinodes for E are nodes for B.This is the relation for,E B?of standing waves,and this explains the puzzle stated at the beginning of this section.

24、Not only in the light(such as light field in a cavity,which is fundamental in laser),standing wave also plays crucial roles in other waves,such as sound waves.Most music instruments are based on it.Also if you ever tried to entertain your friends by inhaling helium(He)gas,your sound would be much hi

25、gher pitched(high frequency).Can you explain this using what you have learned?5-1-4 The addition of waves of different frequency (Hecht,7.2)(a)Beats Simple case:two waves with different frequency,travel in same direction,parallel E field,same 0E?10=+20=10kkk=+20kkk=(true for+2min121 212122()OOIIII I

26、EEII=+=p1E1232E3E-171-The interference is essentially a redistribution of energy,creating peaks and valleys(maximum and minimum)in intensity over the space,but the average irradiance o I over the space is I1+I2,since space=0.Interference does not violate the conservation of energy.(b)Contrast(visibi

27、lity)maxminmaxminIIII+For the two sources case,substitute Imax Imin 1 2122 I III=+(5-13)012III+0(1cos)II=+(5-14)is a measure the contrast between maximum and minimum in the intensity distribution in interference.Larger means larger contrast,easier to observe interference.1 if Imin 0 0 if Imin Imax F

28、or the two coherent sources,1(or Imin 0)if the amplitude of the two sources are equal:E01=E02(I1=I2).-172-5-2-2.Conditions for Coherence The sources are considered coherent if:(1)Phase term 1cos()cos()0t TTtpp dtT+=i.e.have stationary phase difference()()()jiPPP=(2)Same frequency(3)Have parallel E?c

29、omponents.The above conditions are needed for nonzero cross terms in intensity formula.The sources satisfying the above conditions are called Coherent(means well-correlated)For condition(1),(P)consists two parts:()()()jijiPPP=+()()jiPP is the phase difference due to space distribution(such as 12()kr

30、r?for plane wave,or 12()k rrfor spherical wave)and ji is the phase difference due to initial phase.There should also be time dependent partt,but under same frequency condition,that is cancelled.The spatial part,for fixed arrangement of sources and observer,00()()jiPP usually has fixed relations;ji=w

31、ill however depend on the nature of the light source,it could be random,and the stationary(P)requires that has to be stable too(stable means do not -173-change with time,at least within the time interval of the measurement).For condition(2),if we have different frequencies in hand,it effectively int

32、roduces a time varying phase factor into(P),()t),and time average may become 0 For condition(3),to guarantee that0ijEE?,it is obvious necessary to have parallel component.In the superposition of two perpendicular vectors,E1,E2,E=E1+E2 ,clearly 22212EEE=+no cross term,no interference.The above are id

33、eal coherent conditions which may not be satisfied rigorously in practice,since the light sources are only quasi-monochromatic,the phases cannot be stationary infinitely.The modified Coherent Condition:(1)nearly same frequency(2)()iiP+is stationary over the detection period(3)E?has parallel componen

34、t Now the phase difference is:()pt=+.The conditions(1),(2)that make nonzero cross term will depend on the detection period,(the time that intensity is averaged).During the time period of average,if 2 or 2,the time varying of frequency difference is -174-small and the average will not be zero,that is

35、 the reason for small frequency difference as require by(1);()p the phase difference due to spatial difference usually can be maintained at fixed value by fix the OPL the light travels so that will not be a problem here;depends on the nature of the light source,it has to be stationary within detecti

36、on period.The detection period has to be longer than the time resolution(response time)of the detectors rest(it is the smallest time interval allowed by the detector to make meaningful measurement),i.e.rest(rest10-1s for eye,10-9s for PMT the photo multiplier tube)In the visible light region,the tim

37、e period of optical oscillation is on the order of141521010Ts=,andT.Thus the average I=,the cost term is averaged out in the process,that is the reason we have20IE For condition(1)with different frequency components add up,we have seen that will result in modulation of the envelope of amplitude.(wav

38、e packet or beat),the amplitude envelope varies approximately with cos()t,for small ,2,then within the detection time cos()t varies little and will not average to zero.For condition(2),the stability of the phase difference,the i term,is -175-depending on the Coherent time 0 of the source.For convent

39、ional thermal sources,12801010 s;for lasers,6501010 s.For 0 cos()0P=,no interference,the condition 2 fails.For 0 cos()0P ,i.e.within the detection process,the sources are somewhat coherent.5-3 Wavefront Splitting Interference 5-3-1 Interference of Two Point Sources Q1.Q2 is two point sources,radiate

40、 spherical waves,with same amplitude and equal initial phase,same frequency.Then the field from such point sources is in spherical wave form:iikri tioiAEeer+=121 2()2cosI PIII I=+This is just(5-12)given before.2210111()AI PAr=,2220222()AIPAr=For 12,r rd,d is separation between Q1,Q2.-176-12II or 12A

41、AA=Then 222()2(1cos)4cos2I PAA=+=（515）To evaluate:12122()()()PPrr=（516）So for every points in the field,by knowing the phase difference from(5-16),we know the intensity by(5-15)or(5-12).We shall concentrate on the maxima and minima of the interference pattern Then it is easy to see,，for 12120,1,2,()

42、max1()0,1,2,()min2rrmmI P isimumrrmmI P isimum=+=The contour of points Ps satisfying the above relations are hyperbola surfaces,with Q1,Q2 at their foci(S1,S2 in the Fig 9.3 in Hechts):The interference pattern on an observing screen would generally be hyperbolic curves,i.e.-177-Under paraxial condit

43、ion,i.e.222,xyD,then the curves are approximately close to straight lines.5-3-2 Youngs Experiment Paraxial condition:22222,dDXYD Lets first treat the ideal case,where S,S1,S2 are treated as point sources(As we shall see later,that the sizes of S consisting of incoherent sources will give vise to spa

44、tial Coherence;the sizes of S1,S2 will lead to the problem of diffraction).The cleverness of Youngs setup is to eliminate the effect of random phase fluctuation in conventional incoherent light sources.For two incoherent light sources,the(P),phase difference at P,consists of -178-()P+,where ,the ini

45、tial phase difference randomly change,and cos0T=,no interference.In the Youngs Experiment,such random initial phase change is cancelled out.i.e.:1()1111()i kssi tsAE SAeeASS+=2()2222()i kssi tsAE SA eeASS+=11()111()i k SSri tSAEPeer+=22()222()i k SSri tSAEPeer+=Though is a random number,(P)is not de

46、pending on it,i.e.S1,S2 are coherent.Using the figure above,we can prove 22(,)4cos()2kdI x yAxD=A is the amplitude at point P by S1 or S2(they are equal under paraxial condition).12()()Pk rrk L=(Here,we take12SSSS=)12sinmmLrrddforxD=mdxLxDD=-179-()(5 17)kdPxD=From(5-12):121 2()2cos()kdI PIII IxD=+(5

47、-18).under paraxial condition,Then ,r0 is the center of to P.(5-19)The interference pattern by double slits is something like:The spacing between the adjacent maxima(or minima)is given by:,and this will give:12II22dD21120()AIIr12S S21()4 cos()2kdI PIxD,0,1,max1();0 1min2ththdLxmmmimaDmmmnima=+=2kdxD

48、=Dxd=-180-is the angle spanned by the center of receiving screen with S1,S2.5-3-3 A more Strict and Systematic Treatment(1)Wavefront distribution(Zhaos P148-158)As stated earlier,the key question in interference is to evaluate then we have to know,the general expression for phase distribution of a p

49、oint source over space.Previously,we state that wavefront is a constant phase surface,here we generalize it referring to any surface in the field,and we are interested in the field distribution E(P)across this surface.Any light sources can be visualize as consisting of many point sources,so the basi

50、c question would be,what is E(P)on a wavefront in a field created by point sources.O,Q,are point sources on(x,y)plane,what are the field distribution dD=()()()jiPPP=()iP-181-on(x,y)plane?Two planes are separated by distance Z.For multiple light sources,we are going to divide them into axial points a