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Home , Arago spot, Diffraction, Superposition principle

Chapter 7

Experiment 5: Diffraction of

WARNING This experiment will employ Class III(a) as a coherent, monochromatic light source. The student must read and understand the safety instructions on page 90 before attending this week’s laboratory.

7.1 Introduction

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

General Information We will observe the diffraction of light waves; however, diffraction occurs when any wave propagates around obstacles. Diffraction and the wave nature of particles leads to Heisenberg’s and to the very reason why atoms are the sizes that they are.

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

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m = 4

m = 3

m = 2

= 1 y m

= 0 a x m

m = -1

m = -2 m = -3 Pattern

= -4 m

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

from the described objects are based on:

1. the , Intensity vs. Position 2. the wave nature of light Disturbance: 1

(a) : A = A0 sin(ωt + ϕ) 0.8 (b) Intensity: I ∝ (P A)2, 0.6 3. Huygen’s Principle - Light propagates in such a way that each point reached by 0.4 the wave acts as a point source of a new light wave. The superposition of all these 0.2 waves represents the propagation of the 0 light wave. -7 -5 -3 -1 1 3 5 7 y (cm) All calculations are based on the assumption that the distance, x, between the slit and the Figure 7.2: A graphical plot of an exam- viewing screen is much larger than the slit width ple diffraction intensity distribution for a a:, i.e. x >> a. These results also apply to single slit. plane waves incident on the obstruction. This

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

Checkpoint What is the difference between Fraunhofer and Fresnel scattering?

Checkpoint Are the eyes sensitive to the amplitude, to the , or to the intensity of light? Are the eyes’ response linear, logarithmic, or something else?

7.2 The Experiment

We will observe the diffraction patterns from slits having several widths and having different diameters. Each will be illuminated with a laser’s coherent 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 is illuminated by a plane wave (laser beam) as shown in Figure 7.1. The intensity distribution observed (on a screen) at an angle θ with respect to the incident direction is given by Equation (7.1). This relation is derived in detail in the appendix and every student Figure 7.3: A photograph of Pasco’s must make an effort to go through its derivation. The OS-8523 single slit set. We are mathematics used to calculate this relation are very interested in the single slits at the top simple. The contributions from the field at each small right and the circular apertures at the area of the slit to the field at a point on the screen are top left. added together by integration. Squaring this result and disregarding sinusoidal fluctuations in time gives the 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) sin α2 πa = with α = sin(θ), (7.1) I(0) α λ

where a is the width of the slit, λ is the of the light, θ is the angle between the optical axis and the propagation direction of the scattered light, I(θ) is the intensity of light scattered to angle θ, and I(0) is the transmitted light intensity on the optical axis. When θ = 0, α = 0, and the relative intensity is undefined (0/0); however, when θ is very small and yet not zero, the relative intensity is very close to 1. In fact, as θ gets closer to 0, this ratio of intensities gets closer to 1. Using calculus we describe these situations using the limit as α approaches 0, sin α lim = 1. α→0 α Although 0/0 might be anything, in this particular case we might think that the ratio is effectively 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 the intensity near the optical axis. The numerator of this ratio can also be zero when the denominator is nonzero. In these cases the intensity is predicted and observed to vanish. These intensity minima occur when 0 = sin α 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’s center.

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

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This relation is satisfied for integer values of m. Increasing values of m give minima at correspondingly larger angles. The first minimum will be found for m = 1, the second for m = 2 and so forth. If is less than one for all values of θ, there will be no minima, i.e. when the size of the aperture is smaller than a wavelength (a < λ). This indicates that diffraction is most strongly caused by protuberances with sizes that are about the same dimension as a wavelength. 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 the top right and the circular apertures at the top left.. This situation is similar to the one diagrammed in Figure 7.1. To observe diffraction from a single slit, align the laser beam parallel to the table, at the height of the center of the disk, as shown in Figure 7.4. When the slit is perpendicular to the beam, the reflected light will re-enter the laser. The diffraction pattern you are expected to observe is shown in Figure 7.5.

Figure 7.5: Observed Diffraction Pattern. The pattern observed by one’s eyes does not die off as quickly in intensity as one expects when comparing the observed pattern with the calculated intensity profile given by Equation (7.1) and shown in Figure 7.2. This is because the bright laser light saturates the . Thus the center and nearby fringes seem to vary 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 this mask. Note the characteristics of the pattern and the slit width for each in your Data. Is it possible from our cursory observations that mλ = a sin θ? If a quick and cheap observation can 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 from Equation (7.2) using measurements of the locations of the intensity minima. The wavelength of the HeNe laser is 6328 Å, (1 Å= 10−10 m). The quantity to be determined experimentally is 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 a long 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, the slit, and the screen until all of the minima are marked and the distance, x, between the slit and the screen is measured. Is your screen perpendicular to your optical axis? Remove the slit and circle the undeflected laser spot. Ideally, this spot is exactly the center of the diffraction pattern. Make a table in your Data and record the order numbers, m, and the distances, ym, between the optical axis and the dark spots. It is more accurate to measure

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th the distance 2 ym between the two m minima on opposite sides of the axis; you might wish to avoid this additional complication. Note the units in the column headings and the errors in the table body. Enter your data table into Vernier Software’s Ga3 graphical analysis program. A suitable setup 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 calculated column to hold your measured slit widths. Enter the column name and appropriate units and type

“m” * 632.8 * x / (c * “ym”) into the Equation edit control. Use the “Column” button to obtain the quoted values; choose your column names instead of mine. Instead of “x” type the number of cm, m, etc. between your 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 convert these units to nm (c = 1), µm (c = 1000), mm (c = 1000000), or m (c = 109). Be sure to use the correct units in the column heading to represent your numbers. Plot your slit widths on the y-axis and the order number on the x-axis. Click the smallest number on the y-axis and change the range to include 0. Does it look like all of your data are the same within your error? If so, draw a box around your data points and Analyze/Statistics. Move the parameters box off of your data points.

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

R y

x

Figure 7.6: A schematic diagram of our experiment showing how we can determine the scattering angle, θ, by measuring the distance from the optical axis, y, and the distance between 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 the limit of this relation for the limit α → 0?

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

Checkpoint If you want to sharpen up a beam spot by inserting a narrow vertical slit into the beam, 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 consists of 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 single slit (see Appendix), but because the aperture here is circular, it is more convenient to use cylindrical coordinates (z, ρ, φ). The superposition principle requires us to integrate over a disk, and the result is a Bessel . The condition for observing a minimum of intensity is found from the zeroes of the , 0 = J1(πkm), and the minima are at angles θ such that λ sin θ = k . (7.3) m a

The first several proportionality constants, km, are displayed in Table 7.1 Table 7.1. These are slightly larger than the corresponding integers. m km The disk provided with the apparatus has circular openings of two 1 1.220 different diameters. Put the apertures in the laser beam one at a 2 2.233 time and observe how the diameter of the dark rings depends on the 3 3.238 aperture diameter. Record your observations in your Data and choose the 4 4.241 configuration which gives the best diffraction pattern for detailed study. 5 5.243 Having the aperture perpendicular to the beam and in the center of the 6 6.244 beam gives the best results. Having the screen perpendicular to the beam 7 7.245 will avoid stretching the circles in the projection into ellipses. 8 8.245 Calculate the diameter of the aperture. To make this measurement 9 9.246 accurately, make the distance between the object and the screen large; you may choose to use the wall as your screen. First, measure the distance x between the view screen and the circular aperture. Without disturbing this distance, carefully mark the diameters of at least 5-6 dark rings in the pattern. A circular ring scale is provided for those who wish to use a phone 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 the photograph changes the scale. Alternately, white paper can be marked on opposite sides of

97 CHAPTER 7: EXPERIMENT 5 the dark rings and the ruler can be used to find the diameters between the dots. Please do not mark on the scales. The angle in Equation (7.3) is between the optical axis (center of the central spot) and the dark ring. A little geometry and trigonometry should reveal that

R (2R ) sin θ ≈ tan θ = m = m (7.4) m x 2x as shown in Figure 7.6. If the dark circle is irregular, use a typical value instead of the largest or smallest radius and consider its shape when estimating δRm. If you cannot get a good pattern, mark the slit set’s position with one of the elevator blocks and ask your TA for assistance; your aperture might need cleaned. Don’t forget to include your units and uncertainties.

Option 1: Statistics of Aperture Diameter

Measure the diameters of at least 5 dark rings and use each with the respective km from Table 7.1 to compute a measurement for the aperture diameter. Plot the aperture diameters on the y-axis and decide whether all predictions are equal or whether there is a dependence on 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 x and R1. The uncertainty in our wavelength is much smaller than these so estimate the error in aperture diameter using v u !2 1.22λx u δR 0.00052 a ≈ and δa = at 1 + . (7.5) R1 R1 1.220

λ Measure the value of the constant km in front of the ratio a in Equation (7.3) for the second order minimum, m = 2, and the third order minimum, m = 3. Do this by using as aperture diameter a the value previously obtained above and by measuring sin θ corresponding to the second minimum and then the third minimum, v u !2 !2 aRm u δa δRm km ≈ and δkm = kmt + . λx a Rm

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

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

Checkpoint What size object will generate an diffraction pattern if placed in the path of 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 in the center of a sphere’s shadow.

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

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

Historical Aside Many scientists at the meeting were having problems with Fresnel’s paper because everyone 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 if light is to be described as a wave phenomenon, then a bright spot would be visible at the center of the shadow of a circular opaque obstacle, a result which he felt proved the absurdity of the wave theory of light.

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

Historical Aside Fresnel won the competition, but it would require James Clerk Maxwell (sixty years later) 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 minds of the brightest philosophers and scientists for centuries. Is light described as a stream of particles 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 the center of the shadow generated by this obstacle, a bright spot – the Poisson spot! By varying the obstacle’s position between the lens and screen, you can optimize the intensity of the Poisson spot sort of like focusing the Fresnel . Record your observations in your Data. Can you explain the Poisson spot as constructive interference from the thin ring of light wave-fronts that just barely miss the sphere? What must be true for all of this light to interfere constructively at the location of the spot?

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

7.2.4A 21st Century View of Light

The distinction between wave and particle relies on two types of experiments. Observation of interference phenomena demonstrates the presence of waves. Experiments which show discrete as opposed to continuous changes such as the photoelectric effect demonstrate particle phenomena (each 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 particle called the photon (we observe emission, scattering, and absorption of single ). These two descriptions of light are not mutually exclusive. It seems that light propagates and occupies space like a wave but interacts with like a particle. The location where the photon interacts with electric charge cannot be known but its probable location is proportional to the wave’s intensity. The electromagnetic field propagates as a wave and its 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.

Checkpoint What is our present view of light? Is light a wave in some medium or a stream of particles?

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

101 CHAPTER 7: EXPERIMENT 5 whether our measurements and Pasco’s measurements agree. Keep Pasco’s 1% tolerance specifications in mind. What properties of our apparatus and its environment (NOT already included in σ) might affect the data that we have gathered? For each, estimate how much it might have influenced our results. (Communicate with complete sentences.) Consider all of your observations (not just the numerical, statistical analysis above) and proffer a brief assessment about whether light is best represented as a wave or a stream of particles and whether we can rely upon our 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 already detailed the reasons why and should not repeat those reasons here; Conclusions are merely a series of results that our data supports. Always communicate with complete sentences to introduce valid equations (equations are equations NOT sentences despite the fact that we tend to read them as though they were sentences) and define the symbols (a, λ, θ, . . .) in the equations. We need not argue why we conclude as we do at this point; however, we must state our conclusions clearly so that our reader can understand them completely without reading anything else in our report. Rather than re-write equations in the Conclusions, it is preferred simply to name them (e.g. “Snell’s law” or “Equation (3)”). Finally, report any measured quantities that we or other experimenters might find useful in future work. If we expect that we might re-use our slit or aperture, we might report their measured dimensions. The Bessel functions help us describe many phenomena having cylindrical symmetry, so the locations of their zeroes is useful information. In each case communicate 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 we have observed?

7.4.1 References

Sections of this write up were taken from:

• Bruno Gobi, 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 characterize it by the magnitude of its electric field vector, E, then it can be represented as,

E = E0 sin(ωt − kx). (7.6)

If two waves with responses E1 and E2 reach the same point P at the same time, then the superposition principle gives the total displacement of E as P E0 = E1 + E2. (7.7)

This total displacement depends upon the relative phase ϕ between the two waves. Looking at Figure 7.8, try to imagine what a happens when E2 shifts with respect to E1 y (this shift is given by the phase ϕ = −kx). By lining up peaks with peaks and valleys with valleys, one gets a maximum displace- ment of the wave medium, E0 in this case. In the opposite case, when one lines up peaks Figure 7.8: A schematic illustration of a with valleys, the response, E0, becomes zero. plane wave incident upon a single slit and The light intensity, I, is related to the the angular coordinate used to locate points of amplitude, E0, by the important relation observation. ε c I = 0 |E |2 . (7.8) 2 0 Note that must be summed first before the square is taken for the intensity,

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

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 (Huygen’s principle) and each of these will have the same wavelength, λ, as the incident plane wave. A wavelet generated at point y and reaching point P will disturb the electric field at P by the amount E E (t) = 0 sin(ωt + ϕ(y)) y a

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2π with ω = T = 2πf. The phase ϕ(y) can be determined from the path difference 2πy ϕ(y) = k∆ = sin θ (7.9) λ so that the electric field response at P due to the wavelet emanating from y in the slit is E  2πy  E (t) = 0 sin ωt + sin θ . (7.10) y a λ The superposition principle determines the total amplitude at P to be the sum of the contributions from every point in the slit. This is obtained by integrating,

a/2 a/2 α Z Z E0  2πy  E0 Z E(t) = Ey(t) dy = sin ωt + sin θ dy = sin(ωt + ϕ) dϕ −a/2 −a/2 a λ 2α −α E sin α = − 0 (cos(ωt + α) − cos(ωt − α)) = E sin ωt. (7.11) 2α 0 α

We used Equation (7.9) to change inte- gration variables from y to ϕ and Equa-

0.9 tion (7.1) to simplify to α. Figure 7.9 shows how this function varies with position on 0.7 a view screen. We are interested in the light intensity; note that the amplitude of 0.5  sin α  Equation (7.11) is EP = E0 α , and the 0.3 intensity at P is

I(θ) ε0c E2 sin α2 0.1 = 2 P = , (7.12) I(0) ε0c E2 α -10 -8 -6 -4 -2 0 2 4 6 8 10 2 0 -0.1 where I(0) is the intensity on the optical -0.3 πa axis, α = λ sin θ, a is the slit width, λ is y (cm) the wavelength, and θ is the direction to P from the slide. I(0) is always the maximum Figure 7.9: The amplitude of the electric intensity of the diffraction pattern. Equa- field’s oscillations at each position on the tion (7.12) will have a value of zero each viewscreen. The negative ‘amplitudes’ reflect a time that sin2 α = 0 unless α = 0 also. This half-period phase shift, but the intensity is the occurs when α = ±mπ for integer m 6= 0 same in this case as for positive amplitudes. and yields Equation (7.2) for predicting the directions to points with minimum light intensity from a single slit. The shape of the distribution given by Equation (7.12) is shown in Figure 7.2. We might note as one final characteristic of diffraction that the intensity maxima are not midway between the intensity zeroes as we might naively expect. The division by α shifts the maxima toward the optical axis. Toward the optical axis corresponds to division by smaller a 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 tan αm = αm.

7.5.1 Diffraction from Double-Slit

A very similar procedure will allow us to analyze the diffraction expected from two identical slits. Figure 7.10 shows an appropriate coordinate system; one might note the similarity to Figure 7.8 that we used to analyze the single slit. In the double-slit case, we must integrate d a d a over the bottom slit where − 2 − 2 < y < − 2 + 2 and then add this to the integral over the d a d a top slit where + 2 − 2 < y < + 2 + 2 . 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

P y

d y = 0 a

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

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πd will use β = λ sin θ to simplify from d to β in the same fashion.

" − d−a d+a # E Z 2 Z 2 E(t) = 0 sin (ωt + ϕ(y)) dy + sin (ωt + ϕ(y)) dy d+a d−a a − 2 2 E "Z −(β−α) Z (β+α) # = 0 sin (ωt + ϕ) dϕ + sin (ωt + ϕ) dϕ 2α −(β+α) (β−α) E  −(β−α) E  (β+α) = 0 − cos (ωt + ϕ) + 0 − cos (ωt + ϕ) (7.13) 2α −(β+α) 2α (β−α) E  = 0 cos (ωt − (β + α)) − cos (ωt − (β − α)) 2α  + cos (ωt + (β − α)) − cos (ωt + (β + α))

E h = 0 cos (ωt − α − β) + cos (ωt − α + β) 2α i − cos (ωt + α − β)) − cos (ωt + α + β) E h i = 0 2 cos (ωt − α) cos β − 2 cos (ωt + α) cos β 2α E h i = 0 cos (ωt − α) − cos (ωt + α) cos β α sin α = 2E cos β sin(ωt) (7.14) 0 α We have used two trigonometric identities

cos(A − B) + cos(A + B) = 2 cos A cos B cos(A − B) − cos(A + B) = 2 sin A sin B

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 extend this development to N slits, we would find it convenient to utilize the complex exponential eiθ− e−iθ representation sin θ = 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 two waves then travels its individual distance to the screen and they interfere at P. When a plane wave is incident on two (or more) identical slits, the amplitude of the electric field response at P for each slit is given by Equation (7.11). Figure 7.10 shows the quantities used to calculate 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, sin α E = E = E . (7.15) 1 2 0 α If we measure the distances between the slits and P from the midpoint between the slits, we

106 CHAPTER 7: EXPERIMENT 5 can determine the relative phases for the two waves

∆! 2π 1 πd ϕ = k − = − · d sin θ = − sin θ. (7.16) 1 2 λ 2 λ

If we let πd β = sin θ, then ϕ = −β and similarly ϕ = β. (7.17) λ 1 2 The total electric field from the slits, then, is

sin α E(t) = E (t) + E (t) = E [sin(ωt − β) + sin(ωt + β)] 1 2 0 α sin α = 2E cos β sin(ωt) (7.18) 0 α

A−B A+B where we have used sin A + sin B = 2 cos 2 sin 2 . The intensity distribution is found to be I(θ) sin α2 = cos2 β, (7.19) I(0) α where we note that the electric field amplitude for two slits on the optical axis is twice as large as for one slit. We might note that the diffraction from two slits is the same as for one slit, as a comparison of Equation (7.11) with Equation (7.19) easily shows. The interference between the two slits yields fringes described by cos2 β and the diffraction merely modulates the 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 instead of two discrete slits.

Checkpoint

2  sin α  2 The distribution of light from two slits is represented by the product α (cos β) . Which one of these two terms is called the diffraction term and which one is the interference term? Which term is responsible for the interference fringes?

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

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