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Chapter 9 Interference and di®raction

This ¯le contains the \Interference and di®raction" chapter of a potential book on Waves, designed

for college sophomores. In this chapter we'll study what happens when waves from two or more sources exist at a given point in space. In the case of two waves, the total wave at the given point is the sum of the two waves. The waves can add constructively if they are in phase, or destructively if they are out of phase, or something inbetween for other phases. In the general case of many waves, we need to add them all up, which involves keeping track of what all the phases are. The results in this chapter basically boil down to (as we'll see) getting a handle on the phases and adding them up properly. We won't need to worry about various other wave topics, such as dispersion, polarization, and so on; it pretty much all comes down to phases. The results in this chapter apply to any kind of wave, but for convenience we'll generally work in terms of electromagnetic waves. The outline of this chapter is as follows. In Section 9.1 we do the warm-up case of two waves interfering. The setup consists of a plane wave passing through two very narrow (much narrower than the wavelength of the wave) slits in a wall, and these two slits may be considered to be the two sources. We will calculate the interference pattern on a screen that is located far away. We'll be concerned with this \far-¯eld" limit for most of this chapter, with the exception of Section 9.5. In Section 9.2 we solve the general case of interference fromNnarrow slits. In addition to showing how the phases can be added algebraically, we show how they can be added in an extremely informative geometric manner. In Section 9.3 we switch gears from the case of many narrow slits to the case of one wide slit. The word \di®raction" is used to describe the interference pattern that results from a slit with non- negligible width. We will see, however, that this still technically falls into the category ofN narrow slits, because one wide slit can be considered to be a collection of a large (in¯nite) number of narrow slits. In section 9.4 we combine the results of the two previous sections and calculate the interference pattern fromNwide slits. Finally, in Section 9.5 we drop the assumption that the screen is far away from the slit(s) and discuss \near-¯eld" interference and di®raction. This case is a bit more complicated, but fortunately there is still a nice geometric way of seeing how things behave. This involves a very interesting mathematical curve known as theCornu spriral. 1

2CHAPTER 9. INTERFERENCE AND DIFFRACTION

9.1 Two-slit interference

Consider a plane wave moving toward a wall, and assume that the wavefronts are parallel to

the wall, as shown in Fig.1. If you want, you can consider this plane wave to be generated plane wave

wall

Figure 1

by a point source that is located a very large distance to the left of the wall. Let there be two very small slits in the wall that let the wave through. (We'll see in Section 9.3 that by \very small," we mean that the height is much smaller than the wavelength.) We're assuming that the slits are essentially in¯nite in length in the direction perpendicular to the page. So they are very wide but very squat rectangles. Fig. 2 shows a head-on view from wall wallslits (view from distant source) wall extends infinitely in both directions wall wall extends infinitely in both directions

Figure 2

the far-away point source. By Huygens' principle we can consider each slit to be the source of acylindricallypropa- gating wave. It is a cylindrical (and not spherical) wave because the wave has no dependence in the direction perpendicular to the page, due to the fact that it is generated by a line source (the slit). If we had a point source instead of a line source, then we would end up with a standard spherically propagating wave. The reason why we're using a line source is so that we can ignore the coordinate perpendicular to the page. However, having said this, the fact that we have a cylindrical wave instead of a spherical wave will be largely irrelevant in this chapter. The main di®erence is that the amplitude of a cylindrical wave falls o® like 1=p r(see Section [to be added] in Chapter 7) instead of the usual 1=rfor a spherical wave. But for reasons that we will see, we can usually ignore this dependence. In the end, since we're ignoring the coordinate perpendicular to the page, we can consider the setup to be a planer one (in the plane of the page) and e®ectively think of the line source as a point source (namely, the point on the line that lies in the page) that happens to produce a wave whose amplitude falls o® like 1=p r(although this fact won't be very important). The important thing to note about our setup is that the two sources arein phasedue to the assumption that the wavefronts are parallel to the wall.

1Note that instead of this

setup with the incoming plane wave and the slits in a wall, we could of course simply have two actual sources that are in phase. But it is sometimes di±cult to generate two waves that are exactly in phase. Our setup with the slits makes this automatically be the case. As the two waves propagate outward from the slits, they will interfere. There will be constructive interference at places where the two waves are in phase (where the pathlengths from the two slits di®er by an integral multiple of the wavelength). And there will be destructive interference at places where the two waves are 180

±out of phase (where the

pathlengths from the two slits di®er by an odd multiple of half of the wavelength). For example, there is constructive interference at pointAin Fig.3 and destructive interference A B

Figure 3

at pointB. What is the interference pattern on a screen that is located very far to the right of the wall? Assume that the screen is parallel to the wall. The setup is shown in Fig. 4. The distance between the slits isd, the distance to the screen isD, the lengths of the two paths to a given pointParer1andr2, andµis the angle that the line toPmakes with the normal to the wall and screen. The distancexfromPto the midpoint of the screen is then x=Dtanµ. 1

Problem 9.1 shows how things are modi¯ed if the wavefronts aren't parallel to the wall. This is done in

the context of theN-slit setup in Section 9.2. The modi¯cation turns out to be a trivial one.

9.1. TWO-SLIT INTERFERENCE3wallscreen

P d xS1r 1 r 2 S 2 D qessentially parallel

Figure 4

In ¯nding the interference pattern on the screen, we will work in the so-calledfar-¯eld limit where the screen is very far away. (We'll discuss thenear-¯eldcase in Section 9.5.)2 The quantitative de¯nition of the far-¯eld limit isDÀd. This assumption thatDis much larger thandleads to two important facts. ²IfDÀd, then we can say that the two pathlengthsr1andr2in Fig. 4 are essentially equal in amultiplicativesense. That is, the ratior1=r2is essentially equal to 1. This follows from the fact that the additive di®erencejr1¡r2jis negligible compared with r

1andr2(becausejr1¡r2jcan't be any larger thand, which we are assuming is

negligible compared withD, which itself is less thanr1andr2). Thisr1=r2¼1 fact then tells us that the amplitudes of the two waves at pointPfrom the two slits are essentially equal (because the amplitudes are proportional to 1=p r, although the exact power dependence here isn't important). ²IfDÀd, then we can say that ther1andr2paths in Fig. 4 are essentially parallel, and so they make essentially the same angle (namelyµ) with the normal. The parallel nature of the paths then allows us to easily calculate theadditivedi®erence between the pathlengths. A closeup of Fig. 4 near the slits is shown in Fig. 5. The di®erence to screen far away dS 1 S 2 qq d sinq

Figure 5

in the pathlengths is obtained by dropping the perpendicular line as shown, so we see that the di®erencer2¡r1equalsdsinµ. The phase di®erence between the two waves is then k(r2¡r1) =kdsinµ=2¼ dsinµ= 2¼¢dsinµ :(1) In short,dsinµ=¸is the fraction of a cycle that the longer path is ahead of the shorter path. Remark:We found above thatr1is essentially equal tor2in amultiplicativesense, but not in

anadditivesense. Let's be a little more explicit about this. Let²be de¯ned as the di®erence,

²´r2¡r1. Thenr2=r1+², and sor2=r1= 1 +²=r1. Sincer1> D, the second term here is less than²=D. As we mentioned above, this quantity is negligible because²can't be larger thand, and because we're assumingDÀd. We therefore conclude thatr2=r1¼1. In other words,r1¼r2in a multiplicative sense. This then implies that the amplitudes of the two waves are essentially equal.

However, the phase di®erence equalsk(r2¡r1) = 2¼(r2¡r1)=¸= 2¼²=¸. So if²is of the

same order as the wavelength, then the phase di®erence isn't negligible. Sor2isnotequal tor1 2 The fancier terminology for these two cases comes from the people who did pioneering work in them:

theFraunhoferlimit for far-¯eld, and theFresnellimit for near-¯eld. The correct pronunciation of \Fresnel"

appears to be fray-NELL, although many people say freh-NELL.

4CHAPTER 9. INTERFERENCE AND DIFFRACTION

in anadditivesense. To sum up, the multiplicative comparison ofr2andr1(which is relevant for the amplitudes) involves the comparison of²andD, and we know that²=Dis negligible in the far-¯eld limit. But the additive comparison ofr2andr1(which is relevant for the phases) involves the comparison of²and¸, and²may very well be of the same order as¸.| Having found the phase di®erence in Eq. (1), we can now ¯nd the total value of the wave at pointP. LetAPbe the common amplitude of each of the two waves atP. Then up to an overall phase that depends on when we pick thet= 0 time, the total (complex) wave at

Pequals

E tot(P) =APei(kr1¡!t)+APei(kr2¡!t) =AP(eikr1+eikr2)e¡i!t:(2) Our goal is to ¯nd the amplitude of the total wave, because that (or rather the square of it) yields the intensity of the total wave at pointP. We can ¯nd the amplitude by factoring out the average of the two phases in the wave, as follows. E tot(P) =AP³ eik(r1¡r2)=2+e¡ik(r1¡r2)=2´ eik(r1+r2)=2e¡i!t = 2APcosµk(r1¡r2) 2 e i¡ k(r1+r2)=2¡!t¢ = 2APcosµkdsinµ 2 e i¡ k(r1+r2)=2¡!t¢ ;(3) where we have usedk(r2¡r1) =kdsinµfrom Eq. (1). The amplitude is the coe±cient of the exponential term, so we see that the total amplitude atPis A tot(P) = 2APcosµkdsinµ 2

¡!Atot(µ) = 2A(µ)cosµkdsinµ

2 ;(4) where we have rewrittenAPasA(µ), andAtot(P) asAtot(µ), to emphasize the dependence onµ. Note that the amplitude atµ= 0 is 2A(0)cos(0) = 2A(0). Therefore, A tot(µ) A tot(0)=A(µ)

A(0)cosµkdsinµ

2 (5) The intensity is proportional to the square of the amplitude, which gives I tot(µ) I tot(0)=A(µ)2

A(0)2cos2µkdsinµ

2 (6) Since the amplitude of a cylindrically propagating wave is proportional to 1=p r, we have

A(µ)

A(0)=1=p

r(µ) 1=p r(0)=s r(0) r(µ)=s D

D=cosµ=p

cosµ:(7)

Therefore,

I tot(µ) I tot(0)= cosµcos2µkdsinµ 2 = cosµcos2µ¼dsinµ :(8)

9.1. TWO-SLIT INTERFERENCE5

This result holds for all values ofµ, even ones that approach 90±. The only approximation we've made so far is the far-¯eld one, which allows us to say that (1) the amplitudes of the waves from the two slits are essentially equal, and (2) the two paths are essentially parallel. The far-¯eld approximation has nothing to do with the angleµ. If we want to writeItotin terms of the distancexfrom the midpoint of the screen, instead ofµ, then we can use cosµ=D=p x

2+D2and sinµ=x=p

2+D2. This gives

I tot(x) I tot(0)=D p x

2+D2cos2µxkd

2 p x D p x

2+D2cos2µx¼d

p x :(9)

Plots ofItot(x)=Itot(0) are shown in Fig. 6, fordvalues of (:01)¸, (0:5)¸, 5¸, and 50¸.I(x)/I(0)I(x)/I(0)

I(x)/I(0)I(x)/I(0)

(d = 50λ)(d = 5λ)(d = 0.5 λ)(d = .01 λ) x/D x/D x/Dx/D- - 42024
0.2 0.4 0.6 0.8 1.0 42024
0.2 0.4 0.6 0.8 1.0 42024
0.2 0.4 0.6 0.8 1.0 42024
0.2 0.4 0.6 0.8 1.0

Figure 6

As you can see from the ¯rst plot, ifdis much smaller than¸, the interference pattern isn't too exciting, because the two paths are essentially in phase with each other. The most they can be out of phase is whenµ!90±(equivalently,x! 1), in which case the pathlength di®erence is simplyb= (:01)¸, which is only 1% of a full phase. Since we have d¿¸in the ¯rst plot, the cosine-squared term in Eq. (9) is essentially equal to 1, so the curve reduces to a plot of the functionD=p x

2+D2. It decays to zero for the simple

intuitive reason that the farther we get away from the slit, the smaller the amplitude isquotesdbs_dbs1.pdfusesText_1

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Chapter 9

Interference and di®raction