UP 2 Ch 35 Diffraction.pdf

7/25/2019 UP 2 Ch 35 Diffraction.pdf

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1 Understanding Physics: Ch. 35 (Diffraction) Draft 1 9/1/2015 1:13 PM

35 Diffraction

This image shows a pattern of light and dark bands produced when monochromatic light passesthe inner and outer edges of a razor blade. The lines of alternating maximum and minimumintensity are the result of interference, but the interfering waves did not come from light passingthrough two or more slits.

Evidence Is In This ChapterWhere in this chapter can you find experimental evidence and logical reasoning that is used todevelop and support the idea that the pattern of light and dark bands visible in this image of arazor blade is the result of interference of light waves?

Figure 35-1 Discussion pages 3-9 Figure 35-4

Figure 35-5 Discussion pages 12-14


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Introduction

In Chapter 34 we defined diffraction as the flaring of a wave as it emerges from a narrow slit. Inthis chapter we will learn that diffracted light can produce an interference pattern, called adiffraction pattern, in some surprising situations. The opening photo of the razor blade is an

example. There is only one slit present yet interference seems to occur. In addition, there alsoseem to be interference effects at the outer edge. We

Learning Outcomes

Qualitatively and quantitatively describe the location of the dark bands in a single slitdiffraction pattern and their relation to physical parameters including slit width andwavelength. (35-1)

Predict how the intensity versus angular position graph produced by a narrow single slitwill change when physical parameters of the system change and calculate the intensity at

particular locations. (35-2) Qualitatively and quantitatively describe double slit interference patterns for wider slitsas a combination of single slit diffraction and double slit interference. (35-3)

Calculate the dispersion, resolvability and location of bright bands produced by adiffraction grating. (35-4)

Apply Braggs law to determine the atomic spacing in a crystal given information aboutthe x-ray diffraction pattern. (35-5)

Qualitatively and quantitatively describe the diffraction pattern produced by a circularopening or lens and identify the conditions under which two point sources can beresolved. (35-6)

Critical Ideas to Remember

Rays, superposition, interference, coherent, Huygen, plane waves, wavelets,

monochromatic

When a wave passes through a slit it will spread out or flare on the outgoing side of the slit.This is called diffraction. The narrower the slit, the more the wave spreads out. That is, thenarrower the slit, the greater the diffraction. For example, see Fig. 35-2 or 34-8.

35-1 Diffraction with a Single SlitWhen li ght passes through a sli t dif fr action of the light waves can produce a pattern thatreminds us of the in ter ference patterns seen in Chapter 34. Surpri singly, this can happen

even when there is only one sl i t.

When monochromatic light from a distant source, or laser light, passes through a single narrowslit a special pattern is sometimes observed on the viewing screen. Examples are shown in theopening photo of the razor blade and Fig. 35-1. We call such a pattern a diffraction pattern. Adiffraction patternconsists of a broad and very bright central maximumand a number of


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narrower and less bright maxima (called secondaryor sidemaxima) to both sides. In betweenthe maxima are dark bands or minima.The alternating light and dark bands in this pattern looklike those we saw in the interference patterns discussed in Chapter 34. However, we have only asingle slit here so it is not obvious how interference could be occurring. Nonetheless, we willapply our interference model, which is based on the concept of superposition of waves, and see if

it can explain how this pattern arises.

Fig. 35-1 A diffraction pattern produced when monochromatic light passes through a single slit.

In order to apply our interference model to analyze a single slit diffraction pattern letsconsider light waves of wavelength that are diffracted by a single slit of width a that is severaltimes larger than the wavelength of the light. Recall that, according to Huygensprinciple, whenan incoming plane wave hits a narrowslit (a slit with width athat is smaller than the

wavelength) a single new wave or waveletis formed as shown in Fig. 35-2a. What happens ifthe slit is wider than the wavelength of the incoming wave? In this case, a collection ofwavelets form at points along the slit as shown in Fig. 35-2b. The wavelets from different pointswithin the slit travel different path lengths to the viewing screen. Consequently, applying ourinterference model from Chapter 34 leads us to predict that the new waves will produce aninterference pattern of bright and dark fringes that is consistent with the diffraction pattern ofFig. 35-1.

When a plane wave passes through a slit one or more new waves, called wavelets, are generated.If the width of the slit is less than the wavelength of the incoming wave then just one wavelet isformed. However, if the width of the slit is greater than the wavelength of the incoming wave

then several wavelets are formed at points along the slit.

HowDo WeKnow?

Newimageneedalong thelines ofthis butnotidentical


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Figure 35-2: A wave passes through a slit. (a) When the slit is narrow, the extent of diffraction is great. That is,

the wave flares (vertically) a great deal. However, there is only a single wavelet produced on the outgoing side of

the slit. (b) When the slit is wider, diffraction is reduced. (The wave does not flare vertically as much.) In addition,

several wavelets are produced on the outgoing side of the slit.

In order to confirm that it is appropriate to apply an interference model to explain diffractionpatterns, we will develop an expression for the location of fringes using this model and thencompare our theoretical predictions to experimental observations. However, because there canbe many wavelets that interfere to produce a diffraction pattern, diffraction is moremathematically challenging than two slit interference. We get around this complication byfocusing only on finding equationsfor the dark fringes. In order to do this, lets consider lightwaves of wavelength that are diffracted by a single slit of width a that is several times largerthan the wavelength of the light. The slit is in an otherwise solid screenB, as shown in crosssection in Fig. 35-3a. Because the slit is wider than the wavelength, a collection of waveletsform at points along the slit as shown in Fig. 35-2b.

2

HowDo WeKnow?

The wavelets produced from a

single slit superimpose and

interfere. Under the right

conditions this can produce an

interference pattern (the

diffraction pattern) even in the

case of a single slit.


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FIGURE35-3 A slit of width awith a length that extends into and out of the page is located in an otherwise opaquescreen B. Incoming wave fronts are parallel to the screen (a) Wavelets from the top points of two zones of widtha/2 undergo totally destructive interference at point P1 on viewing screen C.(b) For distance to the screenD>>a,we can approximate rays r1and r2as being parallel, at angle to the central axis.

As shown in Fig. 35-3, wavelets from different points within the slit travel different pathlengths to the viewing screen. Consequently, applying our interference model from Chapter 34leads us to predict that the new waves may produce an interference pattern of bright and darkfringes on the viewing screen. We can justify the central bright fringe seen in Fig. 35-1 bynoting that the wavelets from all points in the slit travel about the same distance to reach thecenter of the pattern and thus are in phase there. As for the other bright fringes, we can say onlythat they are approximately halfway between adjacent dark fringes.

To find the dark fringes, we shall use a clever and simplifying strategy that involves pairingup all the rays coming through the slit and then finding what conditions cause the wavelets of therays in each pair to cancel each other. Figure 35-3ashows how we apply this strategy to locatethe first dark fringe, at pointP1. First, we mentally divide the slit into two zones of equal widthsa/2. Then we extend toP1a light ray r1from the top point of the top zone and a light ray r2fromthe top point of the bottom zone. A central axisis drawn from the center of the slit to screen C,andP1is located at an angle to that axis.

The wavelets of the pair of rays r1and r2are in phase within the slit because they originatefrom the same wavefront passing through the slit. However, to produce the first dark fringe they

The wavelength is the

distance between wave

fronts.

Notice that the slit

width a is much larger

than the wavelength.


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must be out of phase by /2 when they reachP1. This phase difference is due to their path lengthdifference, with the wavelet of r2traveling a longer path to reachP1than the wavelet of r1. Todisplay this path length difference, we find a point bon ray r2such that the path length from btoP1 matches the path length of ray r1. Then the path length difference between the two rays is thedistance from the center of the slit to b.

When viewing screen Cis near screenB, as in Fig. 35-3a, the diffraction pattern on Cisdifficult to describe mathematically. However, we can simplify the mathematics considerably ifwe arrange for the distance between the slit and screenDto be much larger than the slit width a.Then we can approximate rays r1and r2as being parallel, at angle to the central axis (Fig. 35-3b). We can also approximate the triangle formed by point b, the top point of the slit, and thecenter point of the slit as being a right triangle with one of the angles inside that triangle being .See Fig. 35-3b. The path length difference between rays r1and r2is then equal to (a/2) sin .

Consequently, for destructive interference we must have sin .2 2

a

We can repeat this analysis for any other pair of rays originating at corresponding points inthe two zones (say, at the midpoints of the zones) and extending to pointP1. Each such pair ofrays has the same path length difference (a/2) sin. Setting this common path length differenceequal to /2 (our condition for the first dark fringe), we have

sin = ,2 2

a

which gives us

sin first minimum for . (35-1)a D a

For slit width aand wavelength , sina tells us the angle of the first dark fringe aboveand (by symmetry) below the central axis.

We find the second dark fringes above and below the central axis as we found the first darkfringes, except that we now divide the slit intofourzones of equal widths. Using the samereasoning as above, we find that

sin = ,4 2

a

which gives us

sin =2 second minimum for (35-2)a D a

for the second minimum.


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We could now continue to locate dark fringes in the diffraction pattern by splitting up theslit into more zones of equal width. We would always choose an even number of zones so thatthe zones (and their waves) could be paired. If we did this, we would find that the dark fringesabove and below the central axis can be located with the following general equation:

sin = for 1,2,3,... . (35-3)a m , m= Angul ar Location of Dark F ri nges in Single Sli t Dif fr action Pattern

Comparisons of sin =a m (Eq. 35-3) with experimental observations confirm that the

expression correctly predicts the location of the dark bands in a diffraction pattern. Thisincreases our confidence that diffraction patterns, even those from single slits, can be correctlyunderstood as an interference phenomenon.

Diffraction is a special case of interference. As is the case with other examples of interference,diffraction patterns result when waves with different phases combine at a given location.

Sometimes people wonder when they should use the word diffraction rather than theword interference. There is no firm answer to this question. In this book, if the combiningwaves originate from a small number of coherent sourcesas in a double-slit experiment withthe slit width amuch less than the wavelength (a


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For most slits we encounter in our everyday lives, the slit width a is muchgreater than thewavelengthfor visible light. In this case, sinwillapproach zero for all physically meaningfulvalues of m. This indicates that all the dark bands are compressed into the point where the slit isprojected onto the viewing screen (the central axis). Because there are bright bands half waybetween each dark band, and all the bands are compressed together, the diffraction pattern

becomes invisible in this case. All we see is a bright spot where the light has passed through theslit. This is what our everyday experience tells us to expect for a wide slit.

As shown in Fig. 35-4, if we narrow the slit while holding the wavelength constant, webegin to increase the angle at which the first dark fringes appear. That is, the extent of thediffraction (the extent of the flaring and the width of the pattern) isgreaterfor a narrowerslit.This is also consistent with our earlier discussions of diffraction. (See Figs. 34-8 and 35-2 forexample.)

However, this trend only continues until the slit width is narrowed to equal the wavelength.At this point, (a =), the expression

sin =m

a

gives sin equal to one for mequal to one. Thistells us that the angular location of the first darkfringes is 90. Since the first dark fringes mark the two edges of the central bright fringe, thatbright fringe would cover even an infinitely long viewing screen. That is, there is no pattern oflight and dark bands visible, only a single, spread out bright spot that decreases in intensitymoving outward from the center.

Lastly, the expression sin =m

a

is undefined if the slit width ais less than the

wavelength. The expression is undefined becausesinwould be greater than one in that caseand that, of course, cannot occur. This mathematical result is consistent with the fact we dontsee single slit diffraction effects when we use vanishingly narrow slits as in Chapter 34. Avanishingly narrow slit acts as a point source of a single wavelet and thus no interference ispossible.

The diffraction (or interference) pattern associated with a single slit is only visible whenthe slit is significantly wider than the wavelength of the light,

but not so wide thata

is approximately zero. Within this range of slit widths,

the central bright spot widens as the slit is narrowed and fewer bands are visible.


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Fig. 35-4(a) Red laser light passes through a narrow slit that is wider than the wavelength of the red light. Theactual diffraction pattern and a graph of intensity versus angular position are shown. For a very wide slit the image is

simply the image of the slit. (b) The slit is narrowed but is still wider than the wavelength of red light. Notice thatthe central maximum widens becoming wider than the slit and the number of side maxima decrease. For a verynarrow slit the image becomes a bright spot covering the entire screen with no dark bands visible.http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html (R. Nave)

Diffraction at an Edge

Diffraction of light is not limited to situations involving light passing through a narrowopening such as a slit. It also occurs when light passes an edge, such as the edges of the razorblade whose diffraction pattern is shown in the opening photo. Note the lines of maxima andminima that run approximately parallel to the edges, at both the inside edges of the blade and the

outside edges. This diffraction pattern occurs because when the light passes the vertical edge atthe left of the razor blade, it flares left and right and undergoes interference, producing thepattern along the left edge.

As we have seen in this section, diffraction finds a ready explanation in the wave theory oflight. However, this theory, originally advanced in the late 1600s by Huygens and used 123 yearslater by Young to explain double-slit interference, was very slow in being adopted. This islargely because it ran counter to Newtons theory that light was a stream of particles.

Newtons view was the prevailing view in French scientific circles of the early 19th century,when Augustin Fresnel was a young military engineer. Fresnel, who believed in the wave theory

of light, submitted a paper to the French Academy of Sciences describing his experiments withlight and his wave-theory explanations of them.

In 1819, the Academy, dominated by supporters of Newton and thinking to challenge thewave point of view, organized a prize competition for an essay on the subject of diffraction.Fresnel won. The Newtonians, however, were neither converted nor silenced. One of them, S. D.Poisson, pointed out the strange result that if Fresnels theories were correct, then light waves

should flare into the shadow region of a sphere as they pass the edge of the sphere, producing a

How DoWe

Know?

Slit width100

Narrower slit

but widthstill >

Actual image on viewing screen.

Graph of intensity vs. angular position.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html


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bright spot at the center of the shadow. The prize committee arranged to have Dominique Argotest Poissonsprediction. He discovered (see Fig. 35-5) that the predictedFresnel bright spot,aswe call it today, was indeed there!*Nothing builds confidence in a theory so much as having oneof its unexpected and counterintuitive predictions verified by experiment.

FIGURE35-5A photograph of the diffraction pattern produced by a disk. Note the concentric diffraction rings andthe Fresnel bright spot at the center of the pattern. This experiment is essentially identical to that arranged by thecommittee testing Fresnels theories, because both the sphere they used and the disk used here have a cross section

with a circular edge.

CheckPoint 35-1We produce a diffraction pattern on a viewing screen by means of a long narrow slit illuminatedby blue light. Does the pattern expand away from the bright center (the maxima and minima shiftaway from the center) or contract toward it if we (a) decrease the slit width ? (b) switch toyellow light?

Example 35-1:White L ight, Red Light

A slit of width ais illuminated by white light (which consists of all the wavelengths in thevisible range).

(a)What slit width should you choose if the first minimum for red light of wavelength 650 nmis to appear at = 5?

(b) For the slit width you determine in (a), what is the wavelength of the light whose first sidediffraction maximum is at 5, thus coinciding with the first minimum for the red light?

SOLUTIONCategorization

This problem is about single-slit diffraction. Therefore, we will apply principles of interferencewhich are fundamentally superposition of waves.

* Since Poisson predicted the spot and Argo discovered it, an alternate name is the Poisson-Argo bright spot.

HowDo WeKnow?


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Interpretation

Here is a picture to help us put the problem and its parameters into perspective:

Figure 35-6 A typical diagram used to help one think about a single-slit diffraction problem with relevantinformation for this problem shown.

Computation

The fundamental physical principles for this problem are:

Light exhibits wave-like properties including interference resulting from thesuperposition of waves with differing phases.

The relative phases of interfering waves determine the resulting wave intensity.

Path length difference (and changes in phase that occur at an interface) determine therelative phase of the waves.

(a) AKey Insight for this problem is that diffraction occurs separately for each wavelengthpresent in the light that passes through the slit. The locations of the minima of the diffractionpattern are given by asin = m(Eq. 35-3). When we set m= 1 (for the first minimum),substitute the given values of and(=650 nm for red light) and rearrange we get a slit width aof

1 650nmsin sin 5

7500nm 7.5 m. (Answer)

ma

(b)Our next Key Insight is to assumethat the first side maximum for any wavelength is abouthalfway between the first and second minima for that wavelength. Those first and secondminima can be located with asin = m(Eq. 35-3) by setting m= 1 and m= 2,respectively. Thus, the first side maximum can be located approximatelyby setting m= m= 1.5. Then Eq. 35-3 becomes

sin = .a m


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We could simply solve this for and substitute known values. Instead, lets insertouralgebraic solution for the slit width from above.

sin

ma

so

sin sinsin

1650nm

1.5

433 nm (Answer)

ma

m m

m

m

Looking at Fig. 33-Xwhich shows the visible spectrum, we see that light of this wavelength isviolet.

Expect and Check

Expect:Diffraction of visible light is not something we encounter in our day-to-day

experiences. One reason for this is that in order for the dark bands of an interference pattern tobe visible, the light must pass through a slit that is on the order of its wavelength. For ouranswer, we should expect the slit width to be larger than (but not too much larger than) thewavelength of red light.

Check:Our answer for the slit width, a= 7.5 m, is very narrow. For comparison, note that afine human hair may be about 100m in diameter. While it is narrow, our answer for the slitwidth is about 10 times larger than the wavelength of the red light (650 nm) which is closeenough to produce a visible diffraction pattern.

35-2 Intensity in a Single-Slit Diffraction PatternAn in terf erence model allows us to corr ectly predict the intensity curve for single sli tdiffraction.

In Section 35-1 we learned how to find the positions of the minima and the maxima in a single-slit diffraction pattern. Now we turn to a more general problem: Finding an expression for theintensityIof the pattern as a function of , the angular position of a point on a viewing screen.

To do this, we divide the slit of Fig. 35-2aintoNzones of equal widths xsmall enough thatwe can assume each zone acts as a source of Huygens wavelets. We wish to superimpose thewavelets arriving at an arbitrary pointPon the viewing screen, at angle to the central axis, so

that we can determine the amplitudeEof the magnitude of the electric field of the resultantwave atP. The intensity of the light atPis then proportional to the square of that amplitude.

To findE, we need the phase relationships among the arriving wavelets. The phasedifference between wavelets from adjacent zones is given by

2

(phase difference) path length difference .


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For pointPat angle , the path length difference between wavelets from adjacent zones is xsin, so the phase difference between wavelets from adjacent zones is

2

sin . (35-4)x

We assume that the wavelets arriving atPall have the same amplitude E. To find theamplitudeEof the resultant wave atP, we add the amplitudes Evia phasors. To do this, weconstruct a diagram ofNphasors, one corresponding to the wavelet from each zone in the slit.

For pointP0at = 0 on the central axis of Fig. 35-2a, 2

sin (Eq. 35-4)x

tells

us that the phase difference between the wavelets is zero; that is, the wavelets all arrive inphase. Figure 35-7ais the corresponding phasor diagram; adjacent phasors represent waveletsfrom adjacent zones and are arranged head to tail. Because there is zero phase differencebetween the wavelets, there is zero angle between each pair of adjacent phasors. The amplitudeEof the net wave atPis the vector-like sum of these phasors. This arrangement of the phasorsturns out to be the one that gives the greatest value for the amplitudeE. We call this valueEmax;that is,Emaxis the value ofEfor = 0.

We next consider a pointPthat is at a small angle to the central axis. The

expression 2

sin (Eq. 35-4)x

now tells us that the phase difference between

wavelets from adjacent zones is no longer zero. Figure 35-7bshows the corresponding phasordiagram; as before, the phasors are arranged head to tail, but now there is an angle betweenadjacent phasors. The amplitudeEat this new point is still the vector sum of the phasors, but it

is smaller than the amplitude in Fig. 35-7a, which means that the intensity of the light is less atthis new pointPthan atP

FIGURE35-7Phasor diagrams for N = 18 phasors, corresponding to the division of a single slit into 18 zones.Resultant amplitudesEare shown for (a) the central maximum at = 0, (b) a point on the screen lying at a smallangle to the central axis, (c) the first minimum, and (d) the first side maximum

If we continue to increase , the angle between adjacent phasors increases, and eventuallythe chain of phasors curls completely around so that the head of the last phasor just reaches thetail of the first phasor (Fig. 35-7c). The amplitudeEis now zero, which means that the intensityof the light is also zero. We have reached the first minimum, or dark fringe, in the diffractionpattern. The first and last phasors now have a phase difference of 2rad, which means that the

HowDo We

Know?


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path length difference between the top and bottom rays through the slit equals one wavelength.Recall that this is the condition we determined for the first diffraction minimum.

As we continue to increase , the angle between adjacent phasors continues to increase,the chain of phasors begins to wrap back on itself, and the resulting coil begins to shrink.

AmplitudeEnow increases until it reaches a maximum value in the arrangement shown in Fig.35-7d. This arrangement corresponds to the first side maximum in the diffraction pattern.

If we increase a bit more, the resulting shrinkage of the coil decreasesE,which means thatthe intensity also decreases. When is increased enough, the head of the last phasor again meetsthe tail of the first phasor. We have then reached the second minimum.

We could continue this qualitative method of determining the maxima and minima of thediffraction pattern but, instead, we shall now turn to a quantitative method.

The expression sin =a m tells us how to locate the minima of the single-slit

diffraction pattern. However, as we see in Fig. 35-2athe intensityIof the pattern varies as afunction . We can quantify these intensity as

2

max

sin, (35-5)I I

1

where sin . (35-6)2

a

I ntensity as a Function of Angle for Single-Sli t Di ff raction

The symbol here is just a convenient connection between the angle that locates a point on theviewing screen and the light intensityIat that point.Imaxis the greatest value of the intensityIin the pattern and occurs at the central maximum (where = 0), and is the phase difference(in radians) between the top and bottom rays from the slit width a.

The expression for intensity we found above (Eq. 35-5) is consistent with experimentalobservations of intensity patterns for diffracted light, including those shown in Fig. 35-4.

CheckPoint 35-2Light of wavelength, 650 nm is used in two separate single-slit diffraction experiments. Fig. 35-8 shows the results as graphs of intensityIversus angular position for the two diffractionpatterns. (a) Are the slit widths larger or smaller than 650 nm? (b) Is the slit wider inexperiment A or experiment B?

HowDo WeKnow?


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Fig. 35-8: CheckPoint 35-2

Example 35-2:

Maxima I ntensiti es

What are the intensities of the first three secondary maxima (side maxima) in the single-slitdiffraction pattern, measured relative to the intensity of the central maximum?

SOLUTION

Categorization

Problems involving interference and diffraction of light waves are fundamentally problemsabout the superposition of light waves. We can use the concept of superposition to develop anexpression for the intensity of a single-slit diffraction pattern. We did so in Section 35-2.

Interpretation

In Section 35-2 we discussed the use of superposition principles and the expression2

max

sin,

I

I

(Eq. 35-5). A graph of this function is shown in Fig. 35-9 which helps us

visualize the relative amplitudes of the side maxima.

A

B


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Fig 35-9: Graph of the relative intensities of (one half of) a single-slit diffraction pattern. This graph contains the

three side maxima as well as the central maximum. It was generated by calculating

2

max

sinI

I

(Eq. 35-5)

for a range of values of .

As in Worked Example 35-1, we will assumethat the maxima lie approximately halfwaybetween the minima.

Computation

From2

max

sinI I

(Eq. 35-5) we find the locations of the secondary maxima are given by

1, for 1,2,3,...,

2m m =

with in radian measure.

Thus, we get2

2

max

1sin

sin 2

, for 1, 2, 3,....1

2

mI

mIm

The first of the secondary maxima occurs for m= 1, and its relative intensity is


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2

2

1

max

2

1sin 1

sin1.52

1 1.51

2

4.5 10 4.5%. (Answer)

I

I

For m= 2 and m= 3 we find that

2 3

max max

1.6% and 0.83%. (Answer)I I

I I

Expect and Check

Expect: From the graph in Fig 35-9, we expect the I1/Imax value to be roughly 0.05. (The othertwo side maxima are harder to estimate because they are so low.) Also, we see that the relativeintensities of the successive maxima decrease very rapidly.

Check: Our first answer of 0.045 matches our expectation quite well. Also, our three answers--4.5%, 1.6%, and 0.83%--exhibit a very clear trend of being both significantly weaker than thecentral maxima and becoming progressively weaker as we move outward from the centralmaxima.

35-3 Diffraction with Two SlitsThe dif fr action pattern associated with double sli ts is a double sli t interference pattern

contained withi n the envelope of a single sli t dif fr action pattern.

When we discussed interference in the double-slit experiments of Chapter 34, we implicitlyassumed that the slits were narrow compared to the wavelength of the light illuminating them;

that is,a


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in which sin (35-8)d

and = sin . (35-9)a

Recall that dis the distance between the centers of the slits and ais the slit width.

Note carefully that the right side of 2

2

max

sincos (Eq. 35-7)I I

is the

product ofImaxand two factors. One factor is associated with interference and one factor isassociated with diffraction. (1) The interference factorcos2is due to the interference between

two slits with slit separation d where sin

d

as above. (2) The diffraction factor2

max

sin(Eq. 35-5)I

is due to diffraction by a single slit of width awhere, as before,

1sin (Eq. 35-6)

2

a

. In other words,

A double-slit interference pattern with slit widths on the order of the wavelength is adouble slit interference pattern limited by a single slit diffraction pattern.

As an example, consider the intensity plot of Fig. 35-10a. This plot is like the intensity plotfor double slit interference that was shown in Fig. 34-15. Figure 35-10a suggests the double-slitinterference pattern that would occur if the slits were infinitely narrow (for a


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FIGURE35-10: Intensity versus angular position plots forreddish-orange incident light (= 623 nm) in (a) Adouble-slit interference experiment with vanishingly narrow slits (b) A single-slit diffraction experiment with a slit

of width a=0.031 mm (a 50 ). (c) A double-slit interference pattern for slits with the same separation as in (a)and widths as in (b) (a 50 ). Note that the first minima of the diffraction pattern of (b) eliminate the double-slit

fringes that would occur near 1.2 in (c).

Figure 35-11a shows an actual intensity pattern in which both double-slit interference andsingle-slit diffraction are evident. As one would predict, if one slit is covered the single-slitdiffraction pattern of Fig. 35-11b results. Note the correspondence between the predictedintensity graphs of Figs. 35-10 and the actual experimental results of Fig. 35-11. In comparingthese figures, bear in mind that 35-11 has been deliberately overexposed to bring out the faintsecondary maxima and that two secondary maxima (rather than one) are shown.

Double slit interference pattern

for vanishingly narrow slits.

Single slit diffraction pattern

produced by a slit with a 50

Double slit interference pattern for slits with the same separation as in (a)but widths as in (b) (a 50 ). The pattern is the double-slit interference

pattern limited by single-slit diffraction effects. The intensity curve for

single slit diffraction, shown in (b), acts as an envelope for the double-slit

interference fringes, shown in (a).

How

Do WeKnow?


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FIGURE35-11(a) The diffraction/interference pattern for a double-slit with slit widths on the order of thewavelength. Here, the interference fringes appear within the bright bands of the single-slit diffraction pattern. Thisexperimental result is consistent with what we predicted based on the expression

2

2

max

sincosI I

(Eq. 35-7) as shown in Fig. 35-10c. (b) The diffraction pattern for a single slit

appears if one of the slits is covered.

CheckPoint 35-3

(a)How would Fig. 35-11b change if one used the same wavelength light but the slit width wasincreased slightly? (b) How would Fig. 35-11a change if one used the same wavelengthlight and slit width but the slit separation was increased slightly?

Example 35-3:

Br ight Fri nges

Lets consider a double slit with an unusually small spacing. Suppose the wavelength of thelight source is 405 nm, the slit separation is 19.44m, and the slit width is 4.050m. Considerthe interference of the light from the two slits and also the diffraction of the light through eachslit.

(a)How many bright interference fringes are within the central peak of the diffractionenvelope?

(b) How many bright fringes are within either of the first side peaks of the diffractionenvelope?

SOLUTION

Categorization

This problem is going to involve bringing together the concepts of interference and diffraction.

Interpretation

(a)

(b)

HowDo WeKnow?


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Heres a sketch of the interference pattern that shows the effects of single-slit diffractionpattern on the two-slit interference pattern. It also highlights the bright interference fringesrelated to the questions.

FIGURE35-12One side of the intensity plot for a two-slit interference experiment; the diffraction envelope isindicated by the dotted curve. The smaller inset shows (vertically expanded) the intensity plot within the first andsecond side peaks of the diffraction envelope.

Computation

The basic ideas are, once again:

Light exhibits wave-like properties including interference and diffraction.


Letsbegin by analyze the two basic mechanisms responsible for the optical pattern producedin the experiment:

Single-sli t diff raction:The Key Insighthere is that the limits of the central peak are the firstminima in the diffraction pattern due to either slit, individually. (See Fig. 35-12.) The angularlocations of those minima are given by Eq. 35-3 (asinm). Lets write this equation asasin= m1, with the subscript 1 referring to the single-slit (one-slit) diffraction. For the firstminima in the diffraction pattern, we substitute m1=1, obtaining

sin = . (35-10)a

Double-sli t in terf erence:A Key Insighthere is that the angular locations of the bright fringesof the double-slit interference pattern are given by Eqs. 34-7 and 34-8 (from Chapter 34),which we can combine to write

2sin = , for 1,2,3,.... (35-11)2d m m Here the subscript 2 refers to the double-slit (two-slit) interference.

Another Key Insightis that we can locate the first diffraction minimum within the double-slitfringe pattern by dividing Eq. 35-11 by Eq. 35-10 (above) and solving for m2. By doing so andthen substituting the given data, we obtain

Wiley: In the main figure: Pleaseremove the numbers 6, 7, 8. They

dont add anything.

In the insert figure: Please remove Imax

and accompanying connecting line. Itis not correct. Also, the horizontaldashed line is superfluous, so pleaseremove that as well.


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2

19.44 m4.8.

4.050 m

dm

a

This calculation tells us that the bright interference fringe for m2= 4 fits into the central peak ofthe single-slit diffraction pattern, but the fringe for m2=5 does not. Within the centraldiffraction peak we have the central bright fringe (m2= 0), and four bright fringes (up to m2=

4) on each side of it. Thus, a total of nine bright fringes of the double-slit interference patternare within the central peak of the diffraction envelope. (Answer)

(b)The Key Insighthere is that the outer limits of the first side diffraction peaks are thesecond diffraction minima, each of which is at the angle given by a sin = m1 with m1 :

sin =2 (35-12)a

Dividing 2sin =d m (Eq. 35-11) by Eq. 35-12, we find

2

2 19.44 m29.6.

4.050 m

dm

a

This tells us that the second diffraction minimum occurs just before the bright interference

fringe for m2= 10 in 2sin =d m

(Eq. 35-11). Within either first side diffraction peak we havethe fringes from m2= 5 to m2= 9 for a total of five bright fringes of the double-slit interferencepattern (shown in the inset of Fig. 35-12).

Expect and Check

Expect:Because single-slit diffraction patter and the two-slit interference pattern are fairlyregular near the center, we would expect the number of bright fringes within two adjacentpeaks of the diffraction envelope to be about the same.

Check:For both (a) and (b) the number of peaks we determined to be on one side of the centerwas the same, five.

Note: In practice, actually seeing the peaks labeled 4, 5, and 9 in Fig. 35-12 might be difficultbecause they lie so close minima of the diffraction envelope.

35-4 Diffraction GratingsScienti sts often use a more elaborate form of the double sli t arrangement we have studied, one

with numerous sli ts, to analyze light and objects that emit and absorb l ight.

One of the most useful tools in the study of light and of objects that emit and absorb light is thediffraction grating.A diffraction gratingis a device that uses interferencephenomena toseperate a beam of light by wavelength. A diffraction grating is a more elaborate form of thedouble-slit arrangement of Fig. 34-11. This device has a much greater numberNof slits, oftencalled rulings,perhaps as many as several thousand per millimeter. An idealized gratingconsisting of only five slits is represented in Fig. 35-13. When monochromatic light is sentthrough the slits, it forms narrow interference fringes that can be analyzed to determine the


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wavelength of the light. (Diffraction gratings can also be opaque surfaces with narrow parallelgrooves arranged like the slits in Fig. 35-13. Light then scatters back from the grooves to forminterference fringes rather than being transmitted through open slits.)

FIGURE35-13An idealized diffraction grating, consisting of only five rulings, that produces an interference patternon a distant viewing screen C.

With monochromatic light incident on a diffraction grating, if we gradually increase thenumber of slits from two to a large numberN, the intensity plot changes from the typical double-slit plot of Fig. 35-10cto a much more complicated one and then eventually to a simple graphlike that shown in Fig. 35-14a. The pattern you would see on a viewing screen usingmonochromatic red light from, say, a helium-neon laser, is shown in Fig. 35-14b. The maximaare now very narrow (and so are called lines); they are separated by relatively wide dark regions.

We use a familiar procedure to find the locations of the bright lines on the viewing screen.We first assume that the screen is far enough from the grating so that the rays reaching aparticular pointPon the screen are approximately parallel when they leave the grating (Fig. 35-

15). Then we apply to each pair of adjacent rulings the same reasoning we used for double-slitinterference. The separation dbetween rulings is called thegrating spacing.(IfNrulings occupya total width w, then d=w/N.) The path length difference between adjacent rays is again d sin(Fig. 35-15), where is the angle from the central axis of the grating (and of the diffractionpattern) to pointP. A line will be located atPif the path length difference between adjacent raysis an integer number of wavelengthsthat is, if

sin = for 0,1,2,... (35-13)d m , m=

Angular l ocation of br ight l ines produced by a diff raction grating

(measur ed relati ve to the central axis)

where is the wavelength of the light. Each integer mrepresents a different line; hence theseintegers can be used to label the lines, as in Fig. 35-14. The integers are then called the ordernumbers,and the lines are called the zeroth-order line (the central line, with m= 0), the first-order line, the second-order line, and so on.


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FIGURE35-14A diffraction grating illuminated with a single wavelength of light. (a) The intensity plot produced bya diffraction grating with a great many rulings consists of narrow peaks, here labeled with their order numbers m.(b)The corresponding bright fringes seen on the screen are called lines and are here also labeled with order numbers m.Lines of the zeroth, first, second, and third orders are shown.

If we rewrite sin = (Eq. 35-13)d m as =sin-1(m/d) we see that, for a given diffraction

grating, the angle from the central axis to any line (say, the third-order line) depends on thewavelength of the light being used. Thus, when light of an unknown wavelength is sent through adiffraction grating, measurements of the angles to the higher-order lines can be usedin sin =d m to determine the wavelength. Even light of several unknown wavelengths can be

distinguished and identified in this way. We cannot do that with a double-slit arrangement eventhough the same equationand wavelength dependence apply there. In double-slit interference,the bright fringes due to different wavelengths overlap too much to be distinguished.

FIGURE35-15The rays from the rulings in a diffraction grating to a distant point P are approximately parallel. Thepath length difference between each two adjacent rays is dsin, where is measured as shown. (The rulings extendinto and out of the page.)

Width of the Lines

A gratings ability to resolve (separate) lines of different wavelengths depends on the width of

the lines. We shall here derive an expression for the half-widthof the central line (the line forwhich m= 0) and then state an expression for the half-widths of the higher-order lines. We


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measure the half-width of the central line as the angle hw from the center of the line at = 0outward to where the line effectively ends and darkness effectively begins with the firstminimum (Fig. 35-16). At such a minimum, theNrays from theNslits of the grating cancel oneanother. (The actual width of the central line is, of course 2(hw), but line widths are usuallycompared via half-widths.)

In Section 35-1 we were also concerned with the cancellation of a great many rays, there dueto diffraction through a single slit. We obtained sin =a m (Eq. 35-3), which, because of the

similarity of the two situations, we can use to find the first minimum here. It tells us that the firstminimum occurs where the path length difference between the top and bottom rays equals . Forsingle-slit diffraction, this difference is asin . For a grating ofNrulings, each separated fromthe next by distance d, the distance between the top and bottom rulings isNd(Fig. 35-17), so thepath length difference between the top and bottom rays here isNdsin hw. Thus, the firstminimum occurs where

hwsin .Nd

FIGURE 35-16The half-width hwof the central line is measured from the center of that line to the adjacentminimum on a plot ofIversus like Fig. 35-14a.

Because hwis small, sin hwhw(in radian measure). Substituting this into

hwsinNd gives the half-width of the central line as

hw (half-width of central line).Nd

This expression can be generalized for any other line as

hw half-width of line at . (35-14)cosNd

Note that for light of a given wavelength and a given ruling separation d, the widths of the linesdecrease with an increase in the numberNof rulings. Thus, of two diffraction gratings, thegrating with the larger value ofNis better able to distinguish between wavelengths because itsdiffraction lines are narrower and so produce less overlap. But the line width of a monochromaticlight beam is determined by the number of slits that the beam encounters. In a diffraction gratingspectrometer, a collimating telescope can be used to illuminate allNslits of the grating.


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FIGURE35-17The top and bottom rulings of a diffraction grating of Nrulings are separated by distanceNd. The topand bottom rays passing through these rulings have a path length difference ofNdsin hw, where hwis the angleto the first minimum. (The angle is here greatly exaggerated for clarity.)

The Diffraction Grating Spectrometer

Diffraction gratings are widely used to determine the wavelengths that are emitted by sources oflight ranging from lamps to stars. Figure 35-18 shows a simplegrating spectroscopein which agrating is used for this purpose. Light from source Sis focused by lensL1on a vertical slit S1placed in the focal plane of lensL2. The light emerging from tube C(called a collimator) is aplane wave and is incident perpendicularly on grating G, where it is diffracted into a diffractionpattern, with the m= 0 order diffracted at angle = 0 along the central axis of the grating.

We can view the diffraction pattern that would appear on a viewing screen at any angle simply by orienting telescope Tin Fig. 35-18 to that angle. LensL3of the telescope then focusesthe light diffracted at angle (and at slightly smaller and larger angles) onto a focal planeFFwithin the telescope. When we look through eyepiece E, we see a magnified view of this focusedimage.

By changing the angle of the telescope, we can examine the entire diffraction pattern. Forany order number other than m= 0, the original light is spread out according to wavelength (orcolor) so that we can determine, with sin = (Eq. 35-13)d m , just what wavelengths are being

emitted by the source. If the source emits a number of discrete wavelengths, what we see as werotate the telescope horizontally through the angles corresponding to an order mis a vertical lineof color for each wavelength, with the shorter-wavelength line at a smaller angle m= 0 than thelonger-wavelength line.

For example, the light emitted by a hydrogen lamp, which contains hydrogen gas, has four

discrete wavelengths in the visible range. If our eyes intercept this light directly, it appears to bewhite. If, instead, we view it through a grating spectroscope, we can distinguish, in severalorders, the lines of the four colors corresponding to these visible wavelengths. (Such lines arecalled emission lines.) Four orders are represented in Fig. 35-19. In the central order (m= 0), thelines corresponding to all four wavelengths are superimposed, giving a single white line at = 0.The colors are separated in the higher orders.


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The third order is not shown in Fig. 35-19 for the sake of clarity; it actually overlaps thesecond and fourth orders. The fourth-order red line is missing because it is not formed by thegrating used here. That is, when we attempt to solve sin = (Eq. 35-13)d m forthe angle for

the red wavelength when m= 4, we find that sin is greater than unity, which is not possible.The fourth order is then said to be incompletefor this grating; it might not be incomplete for a

grating with greater spacing d, which will spread the lines less than in Fig. 35-19. Figure 35-20 isa photograph of the visible emission lines produced by cadmium.

FIGURE35-18A simple type of grating spectroscope used to analyze the wavelengths of light emitted by source S.

FIGURE35-19The zeroth, first, second, and fourth orders of the visible emission lines from hydrogen. Note that thelines are farther apart at greater angles. (The lines are also dimmer and wider, although that is not shown here. Also,the third order line is eliminated for clarity.)

HowDo WeKnow?


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FIGURE35-20The visible emission lines of cadmium, as seen through a grating spectroscope.

FIGURE35-21The fine rulings, each 0.5m wide, on a compact disc function as a diffraction grating. When a smallsource of white light illuminates a disc, the diffracted light forms colored lanes that are the composite of the

diffraction patterns from the rulings.

Dispersion

To be useful in distinguishing wavelengths that are close to each other (as in a gratingspectroscope), a grating must spread apart the diffraction lines associated with the variouswavelengths. The spreading of diffraction lines from various wavelength light is calleddispersion and is defined as

. (35-15)D

Dispersion Defined

Here is the angular separation of two lines whose wavelengths differ by . The greaterDis,the greater is the distance between two emission lines whose wavelengths differ by .

How is the dispersion of a grating related to the line spacing and other characteristics ofthe device? The expression sin =d m (Eq. 35-13) is the expression for the locations of the

lines in the diffraction pattern of a grating. If we regard and as variables and takedifferentials of this equation. We find

cos ,d d m d

where the differentials dand dare placed in parentheses to distinguish them from the productof the center to center slit spacing dand the angle or wavelength . For small enough angles,we can write these differentials as small differences, obtaining

cos , (Eq. 35 16)d m


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or .cos

m

d

Recall that the dispersionDis defined in Eq. 35-15 as D

. Consequently, we find

. (35-17)cos

mD

d

Dispersion of a diff raction grating

Thus, to achieve higher dispersion we must use a grating of smaller grating spacing d and workin a higher order m. Note that the dispersion does not depend on the number of rulings. The SIunit forDis the degree per meter or the radian per meter.

Resolving Power

To resolvelines whose wavelengths are close together we need to make the linesdistinguishable, and so the line should be as narrow as possible. Expressed otherwise, the gratingshould have a high resolving power R,defined as

. (35-18)R

Resolving power defined

Here () is the mean wavelength of two emission lines that can barely be recognized as separate,and is the wavelength difference between them. The greater Ris, the closer two emissionlines can be and still be resolved.

Again, we would like to relate this characteristic to design aspects of the grating that we

can control. Our differential expression cos , (Eq. 35 16)d m from above is anexpression for the locations of the lines in the diffraction pattern formed by a grating. Here isthe small wavelength difference between two waves that are diffracted by the grating, and is

the angular separation between them in the diffraction pattern. If is to be the smallest anglethat will permit the two lines to be resolved, it must (by Rayleighs criterion) be equal to thehalf-width of each line, which is given by

hw Eq. 35 14cosNd


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If we substitute hwas given here for in cos , (Eq. 35 16)d m , we find that

,mN

from which it follows that

.R Nm

. (35-19)R Nm Resolving power of a grating

Consequently, we find that to achieve high resolving power, we must spread out the light beamso it is incident on many rulings (largeNin Eq. 35-19).

Comparing Dispersion and Resolving Power

It is easy to confuse resolving power with dispersion so we take a moment clear to compare andcontrast them. Table 35-1 shows the characteristics of three gratings, all illuminated with light ofwavelength = 589 nm, whose diffracted light is viewed in the first order (m= 1 in

sin = (Eq. 35-13)d m . You should verify that the values of dispersionDand resolving power

Ras given in the table can be calculated with (Eq. 35-17)cos

mD

d and

(Eq. 35-19)R Nm , respectively. (In the calculations forD, you will need to convert radians

per meter to degrees per micrometer.)

TABLE35- 1 Three Gratingsa

Specifications Calculated Values

Grating N d(nm) D(/m) R

A 10 000 2540 13.4 23.2 10 000

B 20 000 2540 13..4 23.2 20 000

C 10 000 1370 25.5 46.3 10 000

aData are for = 589 nm and m= 1.

For the conditions noted in Table 35-1, gratingsAandBhave the same dispersion andAandChave the same resolving power.

Figure 35-22 shows the intensity patterns (also called line shapes) that would be produced bythese gratings for two lines of wavelengths 1and 2, in the vicinity of = 589 nm. GratingB,


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with the higher resolving power, produces narrower lines and thus is capable of distinguishinglines that are much closer together in wavelength than those in the figure. Grating C, with thehigher dispersion, produces the greater angular separation between the lines.

FIGURE35-22:The intensity patterns for light of two wavelengths sent through the gratings of Table 35-1. GratingBhas the highest resolving power and grating Cthe highest dispersion.

CheckPoint 35-4The figure shows lines of different orders produced by a diffraction grating in monochromatic red light. If we switchto monochromatic light of a longer wavelength will the resolving power of the device increase or decrease? Will thedispersion increase or decrease? Describe changes you would expect to see in the line pattern produced ascompared to Fig. 35-23.

Fig. 35-23CheckPoint 35-4.

Example 35-4:

Diff raction Grating

A diffraction grating has 1.26104rulings uniformly spaced over 25.4 mm (so there are 496lines/mm). It is illuminated at normal incidence by yellow light from a sodium vapor lamp.This light contains two closely spaced emission lines (known as the sodium doublet) of

wavelengths 589.00 nm and 589.59 nm.(a) Using the dispersion of the grating, calculate the angular separation between the doubletlines in the first order.

(b) What is the least number of rulings a grating can have and still be able to resolve thesodium doublet in the first order?

SOLUTION

Narrower peaks as

compared to A-indicates

higher resolving power

Greater separation of peaks

as compared to A-indicates

higher dispersion.


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Categorization

This problem involves looking at how the interference pattern from a diffraction gratingdepends on the wavelength of the light.

Interpretation

The geometry of this problem is important, so a picture would certainly help:

Figure 35-24:The light from the sodium source shines (at normal incidence) on the diffraction grating. Theinterference maxima of the two different wavelengths of the doublet are resolved by the grating at two slightlydifferent angles.

Knowns: The width of the grating: w= 25.4 mm

The number of lines (or rulings): N = 1.26104rulings

The wavelengths of each of the sodium doublet emission lines: 1= 589.00 nm and 2=589.59 nm

ComputationThe basic principles are:

Light exhibits wave-like properties.


One Key Insightis that the angular separation between the two lines in the first orderdepends on their wavelength difference and the dispersionDof the grating where thedispersion is defined asD (Eq. 35-15). A second Key Insightis that the dispersionDdepends on the angle and can be evaluated using D = m/dcos (Eq. 35-17). So we need tostart by first determining the location of the first maximum of one of the wavelengths of the

doublet.

The maxima produced by the diffraction grating can be determined with dsin = m(Eq. 35-13). The grating spacing dfor this diffraction grating is

3

4

6

25.4 10 m

1.26 10

2.016 10 m 2016nm.

wd

N


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The first-order maximum corresponds to m= 1. Substituting these values for dand minto dsin = m(Eq. 35-13) and rearranging leads to

-1 -1 1 589.00nmsin sin2016nm

16.99

m

d

Another Key Insightis that we can assume that, in the first order, the two sodium lines occurclose enough to each other for us to evaluateDat the angle = 16.99 we found above for theshorter wavelength line. Then Eq. 35-17 gives the dispersion as

4

1

cos 2016nm cos16.99

5.187 10 rad/nm.

mD

d

From the definition of dispersion,D (Eq. 35-15), we then have

44

5.187 10 rad/nm 589.59nm-589.00nm

3.06 10 rad=0.0175 (Answer)

D

You can show that this result depends on the grating spacing dbut not on the number of rulingsof the grating.

(b) One Key Insight here is that the resolving powerRof a grating in any order mis physicallyset by the number of rulingsNin the grating. According to Eq. 35-19 these parameters arerelated byR=Nm. A second Key Insightis that the least wavelength difference resolvabledepends on the average wavelength and the resolving powerRof the grating, according to thedefinition of the resolving power,R/ (Eq. 35-18).

One last Key Insightis that the sodium doublet to be barely resolved, must be thedifference between their wavelengths, 0.59 nm, and must be their average wavelength,589.30 nm. Putting these ideas together, we find that the least number of rulings for a gratingto resolve the sodium doublet is

589.30nm

999rulings. (Answer)1 0.59nm

RN

m m


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Expect and CheckExpect:We determined the angular separation between the two maxima to be about 0.02o. Thatis big enough that we could simply calculate the angular location of the first maxima of the twowavelengths individually and compare them. We already know that the angular location of theshorter wavelength is 16.99o. So we can predict that the calculation for location of the first

maxima of the other wavelength should be about 16.99o

+ 0.02o

= 17.01o

.Check: If we repeat the calculation for the location of the first maximum of the longerwavelength as we did for the shorter wavelength we get

-1 -1 1 589.59nmsin sin2016nm

17.01 ,

m

d

which is exactly where we anticipated it should be based on our answers to the problem.

Expect: Our answer for the number of rulings on the diffraction grating should be realistic.Typically, diffraction gratings are manufactured with hundreds or thousands of lines. If weexpect our diffraction grating to work, it must at least be in that regime.Check: In (b) we found that our diffraction grating needs about 1000 lines, so that is in theballpark of what we would expect.

35-5 X-Ray DiffractionDi ff raction occurs with electromagnetic waves outside the visible spectrum too. X-ray

dif fr action i s often used to analyze the structur e of crystals.

X rays are electromagnetic radiation whose wavelengths are of the order of 1 (= 0.1 nm = 10 -10m). Compare this with a wavelength of 550 nm (= 5.5 10-7m) at the center of the visiblespectrum. Figure 35-25 shows that x rays are produced when electrons escaping from a heatedfilamentFare accelerated by a potential difference Vand strike a metal target T.

A standard optical diffraction grating cannot be used to discriminate between differentwavelengths in the x-ray wavelength range. For = 1 (= 0.1 nm) and d= 3000 nm, forexample, sin = (Eq. 35-13)d m shows that the first-order maximum occurs at

-1 -1 1 0.1mnsin sin 0.0019 .3000nm

m

d


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FIGURE35-25:X rays are generated when electrons leaving heated filamentF are accelerated through a potentialdifference Vand strike a metal target T. The windowWin the evacuated chamber C is transparent to x rays.

This is too close to the central maximum to be practical. A grating with d= is desirable, but,since x-ray wavelengths are about equal to atomic diameters, such gratings cannot be constructedmechanically.

In 1912, it occurred to German physicist Max von Laue that a crystalline solid, which

consists of a regular array of atoms, might form a natural three-dimensional diffraction gratingfor x rays. The idea is that, in a crystal such as sodium chloride (NaCl), a basic unit of atoms(called the unit cell) repeats itself throughout the array. In NaCl four sodium ions and fourchlorine ions are associated with each unit cell. Figure 35-26arepresents a section through acrystal of NaCl and identifies this basic unit. The unit cell is a cube measuring a0on each side.

When an x-ray beam enters a crystal such as NaCl, x rays arescatteredthat is, redirectedin all directions by the crystal structure. In some directions the scattered waves undergodestructive interference, resulting in intensity minima; in other directions the interference isconstructive, resulting in intensity maxima. This process of scattering and interference is a formof diffraction, although it is unlike the diffraction of light traveling through a slit or past an edge

as we discussed earlier.

HowDo WeKnow?


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FIGURE 35-26:(a) The cubic structure of NaCl, showing the sodium and chlorine ions and a unit cell (shaded). ( b)Incident x rays undergo diffraction by the structure of (a). The x rays are diffracted as if they were reflected by afamily of parallel planes, with the angle of reflection equal to the angle of incidence, both angles measured relativeto the planes (not relative to a normal as in optics). (c) The path length difference between waves effectivelyreflected by two adjacent planes is 2dsin .(d) A different orientation of the incident x rays relative to the structure.A different family of parallel planes now effectively reflects the x rays.

Although the process of diffraction of x rays by a crystal is complicated, the maxima turn outto be in directions as if the x rays were reflected by a family of parallel reflecting planes(orcrystal planes) that extend through the atoms within the crystal and that contain regular arrays ofthe atoms. (The x rays are not actually reflected; we use these fictional planes only to simplifythe analysis of the actual diffraction process.)

Figure 35-26bshows three of the family of planes, with interplanar spacing d, from whichthe incident rays shown are said to reflect. Rays 1, 2, and 3 reflect from the first, second, andthird planes, respectively. At each reflection the angle of incidence and the angle of reflection arerepresented with . Contrary to the custom in optics, these angles are defined relative to the

surfaceof the reflecting plane rather than a normal to that surface. For the situation of Fig. 35-26b, the interplanar spacing happens to be equal to the unit cell dimension a0.

Figure 35-26cshows an edge-on view of reflection from an adjacent pair of planes. Thewaves of rays 1 and 2 arrive at the crystal in phase. After they are reflected, they must again bein phase, because the reflections and the reflecting planes have been defined solely to explain theintensity maxima in the diffraction of x rays by a crystal. Unlike light rays, the x rays havenegligible refraction when entering the crystal; moreover, we do not define an index of refractionfor this situation. Thus, the relative phase between the waves of rays 1 and 2 as they leave thecrystal is set solely by their path length difference. For these rays to be in phase, the path lengthdifference must be equal to an integer multiple of the wavelength of the x rays.

By drawing the dashed perpendiculars in Fig. 35-26c, we find that the path length differenceis 2dsin . In fact, this is true for any pair of adjacent planes in the family of planes representedin Fig. 35-26b. Thus, we have, as the criterion for intensity maxima for x-ray diffraction,

2 sin = , for 1,2,3,... , (35-20)d m m=

Braggs Law for X-Ray Di ff raction of Crystals

where mis the order number of an intensity maximum. This expression which relates theinterplanar spacing dto characteristics of the diffraction pattern is called Braggs law afterBritish physicist W. L. Bragg, who first derived it. (He and his father shared the 1915 Nobel

Prize for their use of x rays to study the structures of crystals.) The angle of incidence in Fig. 35-26 is called aBragg angle.

Regardless of the angle at which x rays enter a crystal, there is always a family of planesfrom which they can be said to reflect so that we can apply Braggs law. In Fig. 35-26d, thecrystal structure has the same orientation as it does in Fig. 35-26a, but the angle at which thebeam enters the structure differs from that shown in Fig. 35-26b. This new angle requires a new


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family of reflecting planes, with a different interplanar spacing dand different Bragg angle , inorder to explain the x-ray diffraction via Braggs law.

Figure 35-27 shows how the interplanar spacing dcan be related to the unit cell dimensiona0. For the particular family of planes shown there, the Pythagorean theorem gives

05 5 ,d a

0or = .5

ad

Figure 35-27 suggests how the dimensions of the unit cell can be found once the interplanarspacing has been measured by means of x-ray diffraction.

FIGURE35-27:A family of planes through the structure of Fig. 35-26a, and a way to relate the edge length a0of aunit cell to the interplanar spacing d

X-ray diffraction is a powerful tool for studying both x-ray spectra and the arrangement of

atoms in crystals. To study spectra, a particular set of crystal planes, having a known spacing d,is chosen. These planes effectively reflect different wavelengths at different angles. A detectorthat can discriminate one angle from another can then be used to determine the wavelength ofradiation reaching it. The crystal itself can be studied with a monochromatic x-ray beam, todetermine not only the spacing of various crystal planes but also the structure of the unit cell.

35-6 Diffraction by a Circular ApertureDi ff raction occurs with a circular opening too and limi ts the abili ty of optical instruments to

distinguish between objects that are close together.

In this last section we consider diffraction by a circular aperture. A circular apertureis acircular opening, such as a circular lens, through which light can pass. In the next chapter,Chapter 36, we will focus on image formation including the images formed by lenses. What isimportant to carry forward to that chapter from this is that diffraction can impact the nature ofthe image formed by lenses.


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Figure 35-28 shows the image of a distant point source of light that has passed through acircular aperture. As you can see, the image is not a point as expected based on the point sourceof light. Instead, the image is a circular disk surrounded by several progressively faintersecondary rings. Comparison of Fig. 35-28 with any of the previous images of diffraction leaveslittle doubt that diffraction occurs with circular openings as well as rectangular slits.

Careful analysis of the diffraction images produced by circular apertures shows that the firstminimum for the diffraction pattern is located by

sin =1.22 . (35-21)d

F irst M inimum in the Diff raction Pattern for a Circular Apertur e

where the angle is the angle from the central axis to any point on that (circular) minimum and dis the diameter of the aperture.

FIGURE 35-28The diffraction pattern of a circular aperture. Note the central maximum and the circular secondarymaxima. The figure has been overexposed to bring out these secondary maxima, which are much less intense thanthe central maximum.

Compare this expression with

sin = first minimum single slit , (Eq. 35-1)

a

which locates the first minimum for a long narrow slit of width a. The main difference is thefactor 1.22, which enters because of the circular shape of the aperture.

HowDo WeKnow?


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Resolvability of Circular Apertures

The fact that images formed by circular lenses can involve diffraction patterns is important whenwe wish to resolve, ordistinguish, two distant point objects whose angular separation is small.Figure 35-29 shows three example cases for two distant point objects (perhaps stars) with small

angular separation. In Figure 35-29a, the objects are not resolved because of diffraction. That is,their diffraction patterns (mainly their central maxima) overlap so much that the two objectscannot be distinguished from a single larger point object. In Fig. 35-29cthe are fully resolved. Inthe middle case, shown in Fig. 35-29b, the two point sources are separated by the minimumamount necessary to distinguish that there are two point sources rather than a larger singlesource. This occurs when the central maximum of the diffraction pattern of one source iscentered on the first minimum of the diffraction pattern of the other. The minimum angularseparation of point sources that allows an observer to detect that there is more than one source is

called the Rayleighs criterionfor resolvability. Fromsin =1.22 (Eq. 35-21)

d

, two objectsthat are barely resolvable by this criterion must have an angular separation Rof

-1

R

1.22sin .

d

Because the angles involved are small, we can replace sin Rwith Rexpressed in radians andwrite:

R 1.22 Rayleighs criterion circular aperture . (35-22)d

FIGURE35-29At the top, the images of two point sources (stars), formed by a lens. At the bottom, representationsof the image intensities. In (a) the angular separation of the sources is too small for them to be distinguished; in ( b)they can be marginally distinguished, and in (c) they are clearly distinguished. Rayleighs criterion is just satisfied in(b), with the central maximum of one diffraction pattern coinciding with the first minimum of the other.


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Rayleighs criterion for resolvability is only an approximation, because resolvability dependson many factors, such as the relative brightness of the sources and their surroundings, turbulencein the air between the sources and the observer, and the functioning of the observers visual

system. Experimental results show that the least angular separation that can actually be resolvedby a person is generally somewhat greater than the value given by Rayleighs criterion

R1.22 (Eq. 35-22)

d

. However, for the sake of calculations here, we shall take

R 1.22d

as being a precise criterion:

If the angular separation between the sources is greater than R,we can resolve the sources; if it is less, we cannot.

If we want to resolve objects of very small angular separation, we have to limit diffraction

effects. If we are using an external lens or lens system such as a magnifying glass or telescope

to make the observation we can do this according to R 1.22 (Eq. 35-22)d

, either by

increasing the lens diameter or by using light of a shorter wavelength. For this reason ultravioletlight is often used with microscope. Because of its shorter wavelength, it permits finer detail tobe examined than would be possible for the same microscope operated with visible light. It turnsout that under certain circumstances, a beam of electrons behaves like a wave. In an electronmicroscopesuch beams may have an effective wavelength that is 10-5of the wavelength ofvisible light. They permit the detailed examination of tiny structures, like that in Fig. 35-30, thatwould be blurred by diffraction if viewed with an optical microscope using visible light.

FIGURE35-30A false-color scanning electron micrograph of red blood cells traveling through an arterial branch.

CheckPoint 35-5Suppose you can barely resolve two red dots, due to diffraction by the pupil of your eye. If weincrease the general illumination around you so that the pupil decreases in diameter, does theresolvability of the dots improve or diminish? Consider only diffraction. (You might experimentto check your answer.)


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Example 35-5:

Circular Lens

A circular lens with diameter d= 5 cm forms images of distant point objects. Consideringdiffraction by the lens, what is the minimum separation distance between two point objects sotheir separate images can be resolved?

SOLUTION

Characterization

This problem focuses on the concept that there are physical limits on resolvability associatedwith diffraction effects.

Interpretation

The limits on resolvability are expressed in the Rayleighs criterion.The wavelength of light isgoing to be relevant here, but is not given. Lets assumea wavelength = 550 nm.

Computation

The basic principle is that lenses have limited angular resolving power. For circular lenses thisis given by the Rayleigh criterion:

1.22

R

d

.

Substituting the given data, we obtain

95

2

1.22

550 101.22 1.3 10 rad.5 10

R

d

m

m


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How Do We Know?

Use evidence and reasoning presented in this chapter to explain how you know that thediffraction pattern (light and dark bands) visible in this image of a razor blade is the result ofinterference of diffracted light waves.

Where to find

Evidence-Based Model Development in Chapter 34

Figure 34-2 Figure 34-3 Figure 34-7 Figure 34-10 Discussion pages 19-20 Figure 34-14 Figure 34-18 Figure 34-19 Discussion page 34


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Conceptual Questions

1 You are conducting a single-slit diffraction experiment with light of wavelength. What appears, on a distant viewingscreen, at a point at which the top and bottom rays through the slit have a path length difference equal to (a) 5and (b)

4.5?2 In a single-slit diffraction experiment, the top and bottom rays through the slit arrive at a certain point on the viewing

screen with a path length difference of 4.0 wavelengths. In a phasor representation like those in Fig35-7,how manyoverlapping circles does the chain of phasors make?

3 For three experiments, Fig. 35-31gives the parameterof Eq.35-9versus angle for two-slit interference using lightof wavelength 500 nm. The slit separations in the three experiments differ. Rank the experiments according to (a) theslit separations and (b) the total number of two-slit interference maxima in the pattern, greatest first. Hint for Part b:Look back at Chapter 34

Figure 35-31 Question3.

4 For three experiments, Fig.35-32gives of Eq. 35-6 versus angle in one-slit diffraction using light of wavelength500 nm. Rank the experiments according to (a) the slit widths and (b) the total number of diffraction minima in the

pattern, greatest first.


5 Figure35-33shows four choices for the rectangular opening of a source of either sound waves or light waves. Thesides have lengths of eitherLor 2L, withLbeing 3.0 times the wavelength of the waves. Rank the openings according

to the extent of (a) left-right spreading and (b) up-down spreading of the waves due to diffraction, greatest first.

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6

7

Light of frequencyfilluminating a long narrow slit produces a diffraction pattern. If we switch to light of frequency1.3f, does the pattern expand away from the center or contract toward the center?

Light of frequencyfilluminating a long narrow slit produces a diffraction pattern. Does the pattern expand or contractif, instead, we submerge the equipment in clear corn syrup?

8 At night many people see rings (called entoptic halos) surrounding bright outdoor lamps in otherwise dark

surroundings. The rings are the first of the side maxima in diffraction patterns produced by structures that are thoughtto be within the cornea (or possibly the lens) of the observer's eye. (The central maxima of such patterns overlap thelamp.) (a) Would a particular ring become smaller or larger if the lamp were switched from blue to red light? (b) If alamp emits white light, is blue or red on the outside edge of the ring?

9 (a) For a given diffraction grating, does the smallest differencein two wavelengths that can be resolved increase,decrease, or remain the same as the wavelength increases? (b) For a given wavelength region (say, around 500 nm), isgreater in the first order or in the third order?

10 Figure35-34shows a red line and a green line of the same order in the pattern produced by a diffraction grating. If weincreased the number of rulings in the gratingsay, by removing tape that had covered the outer half of the rulingswould (a) the half-widths of the lines and (b) the separation of the lines increase, decrease, or remain the same? (c)Would the lines shift to the right, shift to the left, or remain in place?

Figure 35-34 Questions10 and 11.

11 For the situation of Question9and Fig.35-34, if instead we increased the grating spacing, would (a) the half-widths ofthe lines and (b) the separation of the lines increase, decrease, or remain the same? (c) Would the lines shift to theright, shift to the left, or remain in place?

12 (a) Figure35-35ashows the lines produced by diffraction gratingsAandBusing light of the same wavelength; thelines are of the same order and appear at the same angles . Which grating has the greater number of rulings? (b)Figure35-35bshows lines of two orders produced by a single diffraction grating using light of two wavelengths, bothin the red region of the spectrum. Which lines, the left pair or right pair, are in the order with greater m? Is the centerof the diffraction pattern located to the left or to the right in (c) Fig.35-35aand (d) Fig.35-35b?


13 Figure35-36shows the bright fringes that lie within the central diffraction envelope in two double-slit diffractionexperiments using the same wavelength of light. Are (a) the slit width a, (b) the slit separation d, and (c) the ratio d/ain experimentBgreater t

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