Chapter 7

Experiment 5:Diffraction of Light Waves

WARNINGThis experiment will employ Class III(a) lasers as a coherent, monochromatic lightsource. The student must read and understand the laser safety instructions onpage 90 before attending this week’s laboratory.

7.1 Introduction

In this lab the phenomenon of diffraction will be explored. Diffraction is interference of awave with itself. According to Huygen’s Principle waves propagate such that each pointreached by a wavefront acts as a new wave source. The sum of the secondary waves emittedfrom all points on the wavefront propagate the wave forward. Interference between secondarywaves emitted from different parts of the wave front can cause waves to bend around cornersand cause intensity fluctuations much like interference patterns from separate sources. Someof these effects were touched in the previous lab on interference.

General InformationWe will observe the diffraction of light waves; however, diffraction occurs when anywave propagates around obstacles. Diffraction and the wave nature of particles leadsto Heisenberg’s Uncertainty Principle and to the very reason why atoms are the sizesthat they are.

In this lab the intensity patterns generated by monochromatic (laser) light passingthrough a single thin slit, a circular aperture, and around an opaque circle will be predictedand experimentally verified. The intensity distributions of monochromatic light diffracted

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CHAPTER 7: EXPERIMENT 5

a

Plane Wave

x

y

m = 0

m = 1

m = 2

m = 3

m = 4

m = -1

m = -2

m = -3

m = -4

Diffraction

Pattern

Figure 7.1: A sketch of a plane wave incident upon a single slit being diffracted and theresulting intensity distribution. The intensity is on a logarithmic scale; since our eyes havelogarithmic response, this much more closely matches the apparent brightness of the respectvedots. Keep in mind that this week we are concerned with the dark intensity MINIMA andthat m = 0 is NOT a minimum.

from the described objects are based on:

-7 -5 -3 -1 1 3 5 70

0.2

0.4

0.6

0.8

1

Intensity vs. Position

y (cm)

Figure 7.2: A graphical plot of an exam-ple diffraction intensity distribution for asingle slit.

1. the Superposition Principle,

2. the wave nature of light Disturbance:

(a) Amplitude: A = A0 sin(ωt+ ϕ)(b) Intensity: I ∝ (∑A)2,

3. Huygen’s Principle - Light propagates insuch a way that each point reached bythe wave acts as a point source of a newlight wave. The superposition of all thesewaves represents the propagation of thelight wave.

All calculations are based on the assumptionthat the distance, x, between the slit and theviewing screen is much larger than the slit widtha:, i.e. x >> a. These results also apply toplane waves incident on the obstruction. This

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particular case is called Fraunhöfer scattering. Plane waves result from a laser in thiscase, but a point source very far away or at the focus of a lens also give plane waves. Thecalculations of this type of scattering are much simpler than the Fresnel scattering where thedistant point source constraint is removed. Fresnel scattering assumes spherical wave-fronts.

CheckpointWhat is the difference between Fraunhofer and Fresnel scattering?

CheckpointAre the eyes sensitive to the amplitude, to the phase, or to the intensity of light? Arethe eyes’ response linear, logarithmic, or something else?

7.2 The Experiment

Figure 7.3: A photograph of Pasco’sOS-8523 optics single slit set. We areinterested in the single slits at the topright and the circular apertures at thetop left.

We will observe the diffraction patterns from slitshaving several widths and apertures having differentdiameters. Each will be illuminated with a laser’scoherent light and the resulting pattern will be char-acterized.

7.2.1 Diffraction From a Single Slit

A narrow slit of infinite length and width a isilluminated by a plane wave (laser beam) as shownin Figure 7.1. The intensity distribution observed (ona screen) at an angle θ with respect to the incidentdirection is given by Equation (7.1). This relation isderived in detail in the appendix and every studentmust make an effort to go through its derivation. Themathematics used to calculate this relation are verysimple. The contributions from the field at each smallarea of the slit to the field at a point on the screen areadded together by integration. Squaring this resultand disregarding sinusoidal fluctuations in time givesthe intensity. The main difficulty in the calculation

is determining the relative phase of each small contribution. Figure 7.2 shows the expected

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shape of this distribution; mathematics can describe this distribution,

I(θ)I(0) =

(sinαα

)2with α = πa

λsin(θ), (7.1)

where a is the width of the slit, λ is the wavelength of the light, θ is the angle between theoptical axis and the propagation direction of the scattered light, I(θ) is the intensity of lightscattered to angle θ, and I(0) is the transmitted light intensity on the optical axis. Whenθ = 0, α = 0, and the relative intensity is undefined (0/0); however, when θ is very small andyet not zero, the relative intensity is very close to 1. In fact, as θ gets closer to 0, this ratioof intensities gets closer to 1. Using calculus we describe these situations using the limit asα approaches 0,

limα→0

sinαα

= 1.

Although 0/0 might be anything, in this particular case we might think that the ratio iseffectively 1; this is the basis for the approximation sin θ ≈ θ when θ << 1 radian. Certainly,the intensity of the light on the axis is defined, is measurable, and is quite close to theintensity near the optical axis. The numerator of this ratio can also be zero when thedenominator is nonzero. In these cases the intensity is predicted and observed to vanish.These intensity minima occur when 0 = sinα for each time

πm = α and mλ = a sin θ for m = ±1,±2, . . . (7.2)

Figure 7.4: A sketch of our apparatus showing the view screen, the sample slide, the laser,and the laser adjustments. Our beam needs to be horizontal and at the level of the disk’scenter.

We might note the similarity between this relation and the formula for interferencemaxima; we must avoid confusing the points that for diffraction these directions are intensityminima and for diffraction the optical axis (m = 0) is an exceptional maximum instead ofan expected minimum.

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This relation is satisfied for integer values of m. Increasing values of m give minima atcorrespondingly larger angles. The first minimum will be found for m = 1, the second form = 2 and so forth. If is less than one for all values of θ, there will be no minima, i.e. whenthe size of the aperture is smaller than a wavelength (a < λ). This indicates that diffractionis most strongly caused by protuberances with sizes that are about the same dimension as awavelength.

Four single slits (along with some double slits) are on a disk shown in Figure 7.3 photo-graph of Pasco’s OS-8523 optics single slit set. We are interested in the single slits at thetop right and the circular apertures at the top left.. This situation is similar to the onediagrammed in Figure 7.1. To observe diffraction from a single slit, align the laser beamparallel to the table, at the height of the center of the disk, as shown in Figure 7.4. When theslit is perpendicular to the beam, the reflected light will re-enter the laser. The diffractionpattern you are expected to observe is shown in Figure 7.5.

Figure 7.5: Observed Diffraction Pattern. The pattern observed by one’s eyes doesnot die off as quickly in intensity as one expects when comparing the observed pattern withthe calculated intensity profile given by Equation (7.1) and shown in Figure 7.2. This isbecause the bright laser light saturates the eye. Thus the center and nearby fringes seem tovary slightly in size but all appear to be the same brightness.

Observe on the screen the different patterns generated by all of the single slits on thismask. Note the characteristics of the pattern and the slit width for each in your Data. Is itpossible from our cursory observations that mλ = a sin θ? If a quick and cheap observationcan contradict this equation, then we need not spend more money and time.

Calculate the width of one of the four single slits. This quantity can be calculated fromEquation (7.2) using measurements of the locations of the intensity minima. The wavelengthof the HeNe laser is 6328Å, (1Å= 10−10 m). The quantity to be determined experimentallyis sin θ. This can be done using trigonometry as shown in Figure 7.6.

Measure the slit width using several intensity minima of the diffraction pattern. Place along strip of fresh tape on the screen and carefully mark all of the dark spots in the fringes.Be sure to record the manufacturer’s specified slit width, a. Avoid disturbing the laser, theslit, and the screen until all of the minima are marked and the distance, x, between theslit and the screen is measured. Is your screen perpendicular to your optical axis? Removethe slit and circle the undeflected laser spot. Ideally, this spot is exactly the center of thediffraction pattern. Make a table in your Data and record the order numbers, m, and thedistances, ym, between the optical axis and the dark spots. It is more accurate to measure

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the distance 2 ym between the two mth minima on opposite sides of the axis; you might wishto avoid this additional complication. Note the units in the column headings and the errorsin the table body.

Enter your data table into Vernier Software’s Ga3 graphical analysis program. A suitablesetup file for Ga3 can be downloaded from the lab’s website at

http://groups.physics.northwestern.edu/lab/diffraction.html

Otherwise, change the column names and units to represent your data. Add a calculatedcolumn to hold your measured slit widths. Enter the column name and appropriate unitsand type

“m” * 632.8 * x / (c * “ym”)

into the Equation edit control. Use the “Column” button to obtain the quoted values; chooseyour column names instead of mine. Instead of “x” type the number of cm, m, etc. betweenyour screen and slit; use the same units for x as for ym so that they cancel in the above ratio.What units remain to be the units of a? If you like, you can use the value “c” to convertthese units to nm (c=1), µm (c=1000), mm (c=1000000), or m (c = 109). Be sure to usethe correct units in the column heading to represent your numbers. Plot your slit widthson the y-axis and the order number on the x-axis. Click the smallest number on the y-axisand change the range to include 0. Does it look like all of your data are the same withinyour error? If so, draw a box around your data points and Analyze/Statistics. Move theparameters box off of your data points.

Review Section 2.6 and specify your best estimate of slit width as a = (a± sa)U.

x

Ry

Figure 7.6: A schematic diagram of our experiment showing how we can determine thescattering angle, θ, by measuring the distance from the optical axis, y, and the distancebetween the slits and the screen, x.

Checkpoint

The relation of sin2 α/α2 for α = 0 is an undefined expression of 0/0. What is thelimit of this relation for the limit α→ 0?

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CheckpointWhich relation gives the position of the diffraction minima for a slit of width a,illuminated with plane wave light of wavelength λ?

CheckpointIf you want to sharpen up a beam spot by inserting a narrow vertical slit into thebeam, will the beam spot get more and more narrow as you close the slit? Explain.

7.2.2 Diffraction by a Circular Aperture

For a circular hole of diameter, a, the diffraction pattern of light with wavelength λ consistsof concentric rings, which are analogous to the bands which we obtained for the single slits.The pattern for this intensity distribution can be calculated in the same way as for the singleslit (see Appendix), but because the aperture here is circular, it is more convenient to usecylindrical coordinates (z, ρ, φ). The superposition principle requires us to integrate over adisk, and the result is a Bessel function. The condition for observing a minimum of intensityis found from the zeroes of the Bessel function, 0 = J1(πkm), and the minima are at anglesθ such that

sin θ = kmλ

a. (7.3)

Table 7.1m km

1 1.2202 2.2333 3.2384 4.2415 5.2436 6.2447 7.2458 8.2459 9.246

The first several proportionality constants, km, are displayed inTable 7.1. These are slightly larger than the corresponding integers.

The disk provided with the apparatus has circular openings of twodifferent diameters. Put the apertures in the laser beam one at atime and observe how the diameter of the dark rings depends on theaperture diameter. Record your observations in your Data and choose theconfiguration which gives the best diffraction pattern for detailed study.Having the aperture perpendicular to the beam and in the center of thebeam gives the best results. Having the screen perpendicular to the beamwill avoid stretching the circles in the projection into ellipses.

Calculate the diameter of the aperture. To make this measurementaccurately, make the distance between the object and the screen large; you may choose touse the wall as your screen. First, measure the distance x between the view screen and thecircular aperture. Without disturbing this distance, carefully mark the diameters of at least5-6 dark rings in the pattern. A circular ring scale is provided for those who wish to use aphone camera; align the bright central spot on the scale’s black center and take a photograph.The ring diameters can then be read from the photograph; do not use a ruler because thephotograph changes the scale. Alternately, white paper can be marked on opposite sides of

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the dark rings and the ruler can be used to find the diameters between the dots. Please donot mark on the scales. The angle in Equation (7.3) is between the optical axis (center ofthe central spot) and the dark ring. A little geometry and trigonometry should reveal that

sin θm ≈ tan θ = Rm

x= (2Rm)

2x (7.4)

as shown in Figure 7.6. If the dark circle is irregular, use a typical value instead of thelargest or smallest radius and consider its shape when estimating δRm. If you cannot get agood pattern, mark the slit set’s position with one of the elevator blocks and ask your TAfor assistance; your aperture might need cleaned. Don’t forget to include your units anduncertainties.

Option 1: Statistics of Aperture Diameter

Measure the diameters of at least 5 dark rings and use each with the respective km fromTable 7.1 to compute a measurement for the aperture diameter. Plot the aperture diameterson the y-axis and decide whether all predictions are equal or whether there is a dependenceon m. Compute the statistics and report the diameter, a = a± sa.

Option 2: Measure the km

The roundoff error in k1 is limited to ±0.0005 and you have estimated uncertainties for xand R1. The uncertainty in our wavelength is much smaller than these so estimate the errorin aperture diameter using

a ≈ 1.22λxR1

and δa = a

√√√√(δR1

R1

)2

+(0.0005

1.220

)2. (7.5)

Measure the value of the constant km in front of the ratio λa

in Equation (7.3) forthe second order minimum, m = 2, and the third order minimum, m = 3. Do this byusing as aperture diameter a the value previously obtained above and by measuring sin θcorresponding to the second minimum and then the third minimum,

km ≈aRm

λxand δkm = km

√√√√(δaa

)2

+(δRm

Rm

)2

.

In measuring the constant km you are determining the zeroes of “the first Bessel functionof the first kind.” Bessel functions come in two kinds, Jm and Km, and they solve mathproblems with cylindrical symmetry.

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CheckpointBeams of particles act like waves with very short wavelengths when scattered byprotuberances such as other particles. If a target of Pb and one of C are bombardedwith a beam of protons, which target will show the sharpest diffraction pattern, thelarge size Pb target or the small C one? (Hint: The pattern due to an aperture isidentical to the pattern caused by its photographic negative: a disk in empty space.)

CheckpointWhat size object will generate an observable diffraction pattern if placed in the pathof light with wavelength λ?

Checkpoint

What is responsible for the factor 1.22 in Equation (7.3)?

7.2.3 The Poisson Spot

Figure 7.7: A sketch illustrating the optical beam line used to observe the Poisson spot inthe center of a sphere’s shadow.

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Historical AsideIn 1818 Augustin-Jean Fresnel entered a competition sponsored by the French Academyto try to clear up some outstanding questions about light. His paper was on a wavetheory of light in which he showed that the refraction and reflection of light andall of their characteristics could be explained if we only allowed light to be treatedas a wave. His theory even explained the results of Young’s double-slit experiment.This was about 30 years after Sir Isaac Newton presented a corpuscle theory of lightin which light particles moved from a source to a destination but could be bent bymaterial surfaces.

Everyone knows to hide behind a rock or a tree if someone is throwing stones or shootingbullets at him since the projectiles will either be stopped by the obstacle or pass harmlesslyby. One is not so safe from waves in a swimming pool, however, because waves will bendaround obstacles having the size of their wavelengths or smaller.

Historical AsideMany scientists at the meeting were having problems with Fresnel’s paper becauseeveryone can see that objects cast shadows. Simeon Denis Poisson, one of the judges,was particularly disturbed by Fresnel’s paper and he used wave theory to show that iflight is to be described as a wave phenomenon, then a bright spot would be visible atthe center of the shadow of a circular opaque obstacle, a result which he felt provedthe absurdity of the wave theory of light.

This surprising prediction, fashioned by Poisson as the death blow to wave theory, wasalmost immediately verified experimentally by Dominique Arago, another member of thejudges committee. The spot actually exists! The spot is still called the Poisson spot (morerecently “Poisson-Arago spot” has come into fashion) despite Argo’s having discovered it. Iguess this just goes to show that some of our mistakes will never be lived down. . .

Historical AsideFresnel won the competition, but it would require James Clerk Maxwell (sixty yearslater) to locate the medium in which the waves of light propagate.

In less than 60 seconds you can now settle a controversy that has preoccupied the mindsof the brightest philosophers and scientists for centuries. Is light described as a stream ofparticles or as a wave phenomenon? In other words, does the Poisson spot exist?

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Observe the Poisson Spot

Use the concave (diverging) lens included with the apparatus to expand the beam slightly.Place the circular obstacle in the beam, (see Figure 7.7), and observe on the screen, at thecenter of the shadow generated by this obstacle, a bright spot – the Poisson spot! By varyingthe obstacle’s position between the lens and screen, you can optimize the intensity of thePoisson spot sort of like focusing the Fresnel zone plate.

Record your observations in your Data. Can you explain the Poisson spot as constructiveinterference from the thin ring of light wave-fronts that just barely miss the sphere? Whatmust be true for all of this light to interfere constructively at the location of the spot?

CheckpointWhat is the Poisson spot? What does the existence of the Poisson spot demonstrate?

7.2.4 A 21st Century View of Light

The distinction between wave and particle relies on two types of experiments. Observationof interference phenomena demonstrates the presence of waves. Experiments which showdiscrete as opposed to continuous changes such as the photoelectric effect demonstrateparticle phenomena (each photon of light knocks one electron off an atom, molecular bond,or metal surface). Light is a wave (we observe interference) and light is also a particlecalled the photon (we observe emission, scattering, and absorption of single photons). Thesetwo descriptions of light are not mutually exclusive. It seems that light propagates andoccupies space like a wave but interacts with electric charge like a particle. The locationwhere the photon interacts with electric charge cannot be known but its probable locationis proportional to the wave’s intensity. The electromagnetic field propagates as a wave andits square is the intensity. This is called the “wave - particle duality.” The strangest part,perhaps, is that atomic-scale particles also exhibit the wave - particle duality.

CheckpointWhat is our present view of light? Is light a wave in some medium or a stream ofparticles?

7.3 Analysis

We have measured a slit width and the slit manufacturer measured the width of the same slit.The same is true for one aperture diameter. Use the strategy in Section 2.9.1 to determine

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whether our measurements and Pasco’s measurements agree. Keep Pasco’s 1% tolerancespecifications in mind.

What properties of our apparatus and its environment (NOT already included in σ) mightaffect the data that we have gathered? For each, estimate how much it might have influencedour results. (Communicate with complete sentences.) Consider all of your observations (notjust the numerical, statistical analysis above) and proffer a brief assessment about whetherlight is best represented as a wave or a stream of particles and whether we can rely uponour equations to make further predictions.

7.4 Conclusions

We are now ready to state simply and briefly what our data says about light. We have alreadydetailed the reasons why and should not repeat those reasons here; Conclusions are merelya series of results that our data supports. Always communicate with complete sentences tointroduce valid equations (equations are equations NOT sentences despite the fact that wetend to read them as though they were sentences) and define the symbols (a, λ, θ, . . .) in theequations. We need not argue why we conclude as we do at this point; however, we muststate our conclusions clearly so that our reader can understand them completely withoutreading anything else in our report. Rather than re-write equations in the Conclusions, it ispreferred simply to name them (e.g. “Snell’s law” or “Equation (3)”).

Finally, report any measured quantities that we or other experimenters might find usefulin future work. If we expect that we might re-use our slit or aperture, we might reporttheir measured dimensions. The Bessel functions help us describe many phenomena havingcylindrical symmetry, so the locations of their zeroes is useful information. In each casecommunicate with complete sentences and include the units and errors for the measurements.

What improvements might you suggest? Can you think of any applications for what wehave observed?

7.4.1 References

Sections of this write up were taken from:

• Bruno Gobi, Physics Laboratory: Third Quarter, Northwestern University

• D. Halliday & R. Resnick, Physics Part 2, John Wiley & Sons.

• R. Blum & D. E. Roller, Physics Volume 2, Electricity, Magnetism, and Light, Holden-Day.

• E. Hecht/A. Zajac, Optics, Addison-Wesley Publishing.

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7.5 APPENDIX:Intensity Distribution from a Single Slit

The calculation of the intensity distribution of diffraction phenomena is based upon the su-perposition principle and Huygen’s principle. Light is a wave and if we choose to characterizeit by the magnitude of its electric field vector, E, then it can be represented as,

E = E0 sin(ωt− kx). (7.6)

a

y

P

Figure 7.8: A schematic illustration of aplane wave incident upon a single slit andthe angular coordinate used to locate points ofobservation.

If two waves with responses E1 and E2 reachthe same point P at the same time, thenthe superposition principle gives the totaldisplacement of E as

E0 = E1 + E2. (7.7)

This total displacement depends upon therelative phase ϕ between the two waves.Looking at Figure 7.8, try to imagine whathappens when E2 shifts with respect to E1(this shift is given by the phase ϕ = −kx).By lining up peaks with peaks and valleyswith valleys, one gets a maximum displace-ment of the wave medium, E0 in this case. Inthe opposite case, when one lines up peakswith valleys, the response, E0, becomes zero.The light intensity, I, is related to theamplitude, E0, by the important relation

I = ε0c

2 |E0|2 . (7.8)

Note that amplitudes must be summed first before the square is taken for the intensity,

|E0|2 = |E1 + E2|2 .

Figure 7.8 shows the quantities used to calculate the diffraction from a single slit of width a.Each point in the slit along the y-direction will generate its own wavelet (Huygen’s principle)and each of these wavelets will have the same wavelength, λ, as the incident plane wave. Awavelet generated at point y and reaching point P will disturb the electric field at P by theamount

Ey(t) = E0

asin(ωt+ ϕ(y))

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with ω = 2πT

= 2πf . The phase ϕ(y) can be determined from the path difference

ϕ(y) = k∆ = 2πyλ

sin θ (7.9)

so that the electric field response at P due to the wavelet emanating from y in the slit is

Ey(t) = E0

asin

(ωt+ 2πy

λsin θ

). (7.10)

The superposition principle determines the total amplitude at P to be the sum of thecontributions from every point in the slit. This is obtained by integrating,

E(t) =∫ a/2

−a/2Ey(t) dy =

∫ a/2

−a/2

E0

asin

(ωt+ 2πy

λsin θ

)dy = E0

2α

∫ α

−αsin(ωt+ ϕ) dϕ

= −E0

2α (cos(ωt+ α)− cos(ωt− α)) = E0

(sinαα

)sinωt. (7.11)

-10 -8 -6 -4 -2 0 2 4 6 8 10

-0.3

-0.1

0.1

0.3

0.5

0.7

0.9

y (cm)

Figure 7.9: The amplitude of the electricfield’s oscillations at each position on theviewscreen. The negative ‘amplitudes’ reflect ahalf-period phase shift, but the intensity is thesame in this case as for positive amplitudes.

We used Equation (7.9) to change inte-gration variables from y to ϕ and Equa-tion (7.1) to simplify to α. Figure 7.9 showshow this function varies with position ona view screen. We are interested in thelight intensity; note that the amplitude ofEquation (7.11) is EP = E0

(sinαα

), and the

intensity at P is

I(θ)I(0) =

ε0c2 E

2P

ε0c2 E

20

=(sinα

α

)2, (7.12)

where I(0) is the intensity on the opticalaxis, α = πa

λsin θ, a is the slit width, λ is

the wavelength, and θ is the direction to Pfrom the slide. I(0) is always the maximumintensity of the diffraction pattern. Equa-tion (7.12) will have a value of zero eachtime that sin2 α = 0 unless α = 0 also. Thisoccurs when α = ±mπ for integer m 6= 0and yields Equation (7.2) for predicting thedirections to points with minimum light

intensity from a single slit. The shape of the distribution given by Equation (7.12) is shownin Figure 7.2.

We might note as one final characteristic of diffraction that the intensity maxima are notmidway between the intensity zeroes as we might naively expect. The division by α shifts themaxima toward the optical axis. Toward the optical axis corresponds to division by smallera and a larger intensity. Since dividing zero by a still yields zero, the zeroes are not shifted.

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Calculus allows us to determine that the maximum intensities occur when tanαm = αm.

7.5.1 Diffraction from Double-Slit

A very similar procedure will allow us to analyze the diffraction expected from two identicalslits. Figure 7.10 shows an appropriate coordinate system; one might note the similarity toFigure 7.8 that we used to analyze the single slit. In the double-slit case, we must integrateover the bottom slit where −d

2 −a2 < y < −d

2 + a2 and then add this to the integral over the

top slit where +d2 −

a2 < y < +d

2 + a2 . We will make the same change of variable from y to ϕ

and the same simplification from a to α as we used for the single-slit case. Additionally, we

d

y

a

y = 0

P

Figure 7.10: A schematic diagram illustrating a plane wave incident upon two identicalslits and the angular coordinate used to locate points of observation.

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will use β = πdλ

sin θ to simplify from d to β in the same fashion.

E(t) = E0

a

[∫ − d−a2

− d+a2

sin (ωt+ ϕ(y)) dy +∫ d+a

2

d−a2

sin (ωt+ ϕ(y)) dy]

= E0

2α

[∫ −(β−α)

−(β+α)sin (ωt+ ϕ) dϕ+

∫ (β+α)

(β−α)sin (ωt+ ϕ) dϕ

]

= E0

2α

[− cos (ωt+ ϕ)

]−(β−α)

−(β+α)+ E0

2α

[− cos (ωt+ ϕ)

](β+α)

(β−α)(7.13)

= E0

2α

[cos (ωt− (β + α))− cos (ωt− (β − α))

+ cos (ωt+ (β − α))− cos (ωt+ (β + α))]

= E0

2α[

cos (ωt− α− β) + cos (ωt− α + β)

− cos (ωt+ α− β))− cos (ωt+ α + β)]

= E0

2α[2 cos (ωt− α) cos β − 2 cos (ωt+ α) cos β

]= E0

α

[cos (ωt− α)− cos (ωt+ α)

]cos β

= 2E0sinαα

cos β sin(ωt) (7.14)

We have used two trigonometric identities

cos(A−B) + cos(A+B) = 2 cosA cosBcos(A−B)− cos(A+B) = 2 sinA sinB

the first was used once with A = ωt+ α and B = β and once with A = ωt− α and B = β.The second was used in the last step before a re-arrangement of multiplicands. To extendthis development to N slits, we would find it convenient to utilize the complex exponentialrepresentation sin θ = eiθ− e−iθ

2i .An alternate and less rigorous approach is to consider the net amplitude at P after single-

slit diffraction from each slit individually to be the emission from each slit. Each of these twowaves then travels its individual distance to the screen and they interfere at P. When a planewave is incident on two (or more) identical slits, the amplitude of the electric field responseat P for each slit is given by Equation (7.11). Figure 7.10 shows the quantities used tocalculate the double slit diffraction intensity distribution. Since the slit widths a are equal,we expect the intensity of the waves from them arriving at P to have equal amplitudes,

E1 = E2 = E0

(sinαα

). (7.15)

If we measure the distances between the slits and P from the midpoint between the slits, we

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can determine the relative phases for the two waves

ϕ1 = k

(−∆

2

)= −2π

λ· 1

2d sin θ = −πdλ

sin θ. (7.16)

If we letβ = πd

λsin θ, then ϕ1 = −β and similarly ϕ2 = β. (7.17)

The total electric field from the slits, then, is

E(t) = E1(t) + E2(t) = E0

(sinαα

)[sin(ωt− β) + sin(ωt+ β)]

= 2E0

(sinαα

)cos β sin(ωt) (7.18)

where we have used sinA + sinB = 2 cos A−B2 sin A+B

2 . The intensity distribution is foundto be

I(θ)I(0) =

(sinαα

)2cos2 β, (7.19)

where we note that the electric field amplitude for two slits on the optical axis is twice aslarge as for one slit. We might note that the diffraction from two slits is the same as for oneslit, as a comparison of Equation (7.11) with Equation (7.19) easily shows. The interferencebetween the two slits yields fringes described by cos2 β and the diffraction merely modulatesthe fringes’ peak brightness values. Examination of the definitions of β (Equation (7.19))and α (Equation (7.1)) allows us to note that β must change faster with θ than α; after all,the slit widths cannot exceed the separation; a = d means one big slit twice as wide insteadof two discrete slits.

Checkpoint

The distribution of light from two slits is represented by the product(

sinαα

)2(cos β)2.

Which one of these two terms is called the diffraction term and which one is theinterference term? Which term is responsible for the interference fringes?

General InformationThis product of interference between two point sources and diffraction of one extendedsource is one example of a more general strategy. When linear superposition applies,the response of several identical extended sources can be determined by dividing theproblem into two simpler problems: solve one extended source, solve point sourcesat each of their centers, and multiply the two solutions. This process is called theconvolution of the two solutions.

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