CHAPTER 15 Instruction and Intervention Support

Interference and Diffraction

1 Core Instruction

Chapter Resources

■■ The Teacher’s Edition wrap has extensive teaching support for every lesson, including Misconception Alerts, Teach from Visuals, Demonstrations, Teaching Tips, Differentiated Instruction, and Problem Solving. ■■ Several additional presentation tools are available online for every chapter, including Animated Physics features, Interactive Whiteboard activities, and Visual Concepts.

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■■ Textbook: Interference ■■ Demonstrations: Interference in Sound Waves • Interference 15.1 Animated Physics: Diffraction in a Ripple Tank • How Distance Traveled Affects Interference • Visual Concepts: Combining Light Waves • Thin-Film Interference Interference Arising from Two Slits ■■ Lab: Double-Slit Interference (Open Inquiry) Teaching Visuals: Conditions for Interference of Light Waves • Comparison of Waves in Phase and 180˚ out of Phase • Path Difference for Light Waves from Two Slits • Position of Higher-Order Interference Fringes PowerPresentations

■■ Textbook: Diffraction ■■ Demonstrations: Waves Bending Around Corners • Diffraction 15.2 Visual Concepts: Diffraction • Maxima and Minima in and Interference by a Single Slit • Light Diffraction by an Diffraction Patterns • Function of a Spectrometer Obstacle: Poisson Spot • Effect of Slit Size on Diffraction Teaching Visuals: Diffraction of Light with Decreasing Patterns • Multiple-Slit Diffraction Slit Width • Constructive Interference by a ■■ Lab: Diffraction (Core Skill) Diffraction Grating • Function and Use of a Diffraction Grating in a Spectrometer • and more PowerPresentations

■■ Textbook: Lasers ■■ Demonstrations: Dancing Light • Interference in Laser Light 15.3 Visual Concepts: Comparing Noncoherent and Coherent Light • Laser • Function of a Compact Disc • Comparing Digital and Analog Signals • and more Teaching Visuals: Operation of a Laser • Components of a Compact Disc Player • Wave Fronts from Noncoherent and Coherent Light Sources PowerPresentations

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2 Support and Intervention 3 Specialized Support

■■ Study Guide ■■ Chapter Summary Audio Files Concept Maps ■■ Differentiated Instruction: Inclusion, Below Level, and ■■ Scientific Reasoning Skill Builder English Learners (TE wrap) Interactive Demos Sample Problem Set I Sample Problem Set II

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Interference and Diffraction 516B CHAPTER 15 The streaks of colored light you see coming from a compact disc resemble the colors that appear when white light passes through a prism. However, the compact disc Chapter Overview does not separate light by means of refraction. Instead, the light Section 1 identifies the conditions waves undergo interference. required for interference to occur and shows how to calculate the location of bright and dark fringes in double- slit interference. Section 2 describes how diffracted light waves interfere, shows how to calculate the position of fringes produced by a diffraction grating, and discusses the resolving power of optical instruments. Section 3 describes how a laser produces coherent light and explores applications of lasers. C15CHO002a

About the Image The colors visible on the discs’ surfaces are characteristic of the light used in the photograph. Have students verify the ability of a compact disc to separate light into its particular spectrum by reflecting light from different sources off a compact disc’s surface. For example, sunlight is separated into all of the visible colors, while most fluorescent lighting produces only a few colors. ©Andrew Douglas/Masterfile (br) ©Don Farrall/Photodisc/Getty Images Lab 516Preview

The following investigations support the Thin-Film Interference

Untitled-295concepts 516 presented in this chapter. Waves Bending Around Corners 5/20/2011 7:54:11 AM Lab Diffraction and Interference by a Single Slit Diffraction (Core Skill) Light Diffraction by an Obstacle: Poisson Spot Double-Slit Interference (Open Inquiry) Effect of Slit Size on Diffraction Patterns Multiple-Slit Diffraction DemonstrationS Dancing Light Interference in Sound Waves Interference in Laser Light Interference in a Ripple Tank How Distance Traveled Affects Interference

516 Chapter 15 CHAPTER 15

Interference Focus and Motivate Activate Prior Knowledge and Diffraction Knowledge to Review • The superposition principle: When two SECTION 1 mechanical waves pass through the Interference same space at the same time, their SECTION 2 displacements at each point add. Diffraction • When two waves with the same SECTION 3 frequency and amplitude overlap, the Lasers resulting wave has the same frequency as the individual waves. If the waves Why It Matters are in phase, the resultant wave has Devices called diffraction twice their amplitude. If they are 180° gratings use the principle of interference to separate out of phase, the amplitudes cancel. light into its component wavelengths. Diffraction Items to Probe gratings are used in • Preconceptions about waves: Ask instruments called spectrometers, which are students what happens when two used to study the waves (A and B) travel toward each chemical composition and temperature of stars. other on a string and meet at a point where A’s displacement is 4 cm up and B’s displacement is 3 cm up. The displacement at that point is 7 cm up. What if A is 4 cm up and B is 3 cm ONLINE Physics down? The displacement is 1 cm up. HMDScience.com

ONLINE LAB Diffraction PREMIUM Double-Slit Interference CONTENT Physics HMDScience.com

Diffraction ©Andrew Douglas/Masterfile (br) ©Don Farrall/Photodisc/Getty Images Why It Matters 517 Connecting to History satisfactory. By the early 1800s, the wave model began to replace the particle model because it Untitled-295 517 Isaac Newton is famous for his accomplishments 5/20/2011 7:54:17 AM in mechanics, especially the laws of gravitation could explain more phenomena about light. The and of motion. During his lifetime, however, he wave model explains not only reflection and was most famous for his work on light. Newton refraction but also diffraction and interference. thought of light as a beam of tiny particles. With After briefly explaining this history to this model, reflection was explained as particles students, ask them if they can think of other bouncing off a surface; refraction was explained scientific models that have changed over time. as caused by forces from the surface acting on Why do scientific models change? What makes light particles. As more became known about one model better than another? light, Newton’s particle model became less

Interference and Diffraction 517 SECTION 1 SECTION 1

Objectives Plan and Prepare Describe how light waves Interference interfere with each other to produce bright and dark fringes. Preview Vocabulary Key Terms Identify the conditions required coherence path difference order number Visual Vocabulary Ask students to for interference to occur. visualize the iridescence of materials such Predict the location of Combining Light Waves as mother-of-pearl or the inside of an interference fringes using the You have probably noticed the bands of color that form on the surface of abalone seashell (show pictures or equation for double-slit a soap bubble, as shown in Figure 1.1. Unlike the colors that appear when examples if possible). Bands of color are interference. light passes through a refracting substance, these colors are the result of formed because each wavelength is light waves combining with each other. reinforced at a different position. This is known as interference. Explain that Interference takes place only between waves with the same wavelength. iridescence is caused by interference, and FIGURE 1.1 To understand how light waves combine with each other, let us review relate it to the other vocabulary from the Interference on a Soap how other kinds of waves combine. If two waves with identical wave- chapter. Bubble Light waves interfere lengths interact, they combine to form a resultant wave. This resultant to form bands of color on a soap wave has the same wavelength as the component waves, but according to bubble’s surface. the superposition principle, its displacement at any instant equals the sum of the displacements of the component waves. The resultant wave is Teach the consequence of the interference between the two waves. Figure 1.2 can be used to describe pairs of mechanical waves or electromagnetic waves with the same wavelength. A light source that has a single wavelength is called monochromatic, which means single Demonstration colored. In the case of constructive interference, the component waves combine to form a resultant wave with the same wavelength but with an Interference in Sound Waves amplitude that is greater than the amplitude of either of the individual component waves. For light, the result of constructive interference is light Purpose Introduce students to interfer- that is brighter than the light from the contributing waves. In the case of ence patterns using sound waves from destructive interference, the resultant amplitude is less than the amplitude two coherent sources. of the larger component wave. For light, the result of destructive interfer- ence is dimmer light or dark spots. Materials sine-wave generator, amplifier, two speakers, tape measures, overhead

projector with grid transparency FIGURE 1.2

Procedure Connect the generator and Wave Interference Two waves can interfere (a) constructively or (b) the speakers to the amplifier. Place the First wave speakers about 3 m apart so that both destructively. In interference, energy is not Second wave lost but is instead redistributed. Resultant wave face the class. Have students stand in (a)

rows perpendicular to a line joining First wave the two speakers. To reduce the Resultant wave Second wave effect of echoes from the walls, have (b)

students cover the ear that is opposite (cl) ©Ian Mckinnell/Getty Images (br) ©GIPhotoStock/Photo Inc. Researchers, the speakers. Differentiated518 Chapter 15 Instruction Set the generator to a fre­quency of HRW • Holt Physics PH99PE-C16-001-001-A about 440 Hz. Turn on the generator and Below Level amplifier, and adjust the two speakers to Untitled-296Students 518 may have trouble remembering the 5/20/2011 7:55:08 AM equal intensity. Have students slowly walk result produced by the two different types of forward in their rows and listen to the interference. Students can remember that intensity of the sound. Tell them to stand destructive interference produces a dimmer still when they find a location where the light or dark spots because destructive, dim, and sound is at a minimum. Have students dark all begin with "d." If they can then recall measure the distance between each point that constructive and destructive are opposites, where sound intensity is at a minimum. they will be able to remember that constructive Using these data, have students draw the interference produces brighter light. destructive interference fringes on the

518 Chapter 15 FIGURE 1.3 Comparison of Waves In Phase transparency. When all of the fringe and 180° Out of Phase (a) (b) positions are recorded, turn the over- (a) The features of two waves in phase completely match, whereas (b) they are head projector on and have students opposite each other in waves that are note that they were standing in places 180° out of phase. where the superposition of waves resulted in destructive interference.

HRW • Holt Physics PH99PE-C16-001-002-A HRW • Holt Physics Teaching Tip Waves must have a constant phase difference for interference PH99PE-C16-001-003-A to be observed. Point out to students that waves can For two waves to produce a stable interference pattern, the phases of the also interfere in an intermediate way in individual waves must remain unchanged relative to one another. If the which the relative phase between the crest of one wave overlaps the crest of another wave, as in Figure 1.3(a), the two have a phase difference of 0° and are said to be in phase. If the crest of waves is not exactly 0° or 180° as it is in one wave overlaps the trough of the other wave, as in Figure 1.3(b), the two Figure 1.2 and Figure 1.3. Ask students to waves have a phase difference of 180° and are said to be out of phase. draw waves of the same wavelength that Waves are said to have coherence when the phase difference between coherence the correlation between are 45°, 90°, and 135° out of phase. What two waves is constant and the waves do not shift relative to each other as the phases of two or more waves time passes. Sources of such waves are said to be coherent. do the resultant waves look like for each When two light bulbs are placed side by side, no interference is of these three cases? observed. The reason is that the light waves from one bulb are emitted independently of the waves from the other bulb. Random changes occurring in the light from one bulb do not necessarily occur in the light from the other bulb. Thus, the phase difference between the light waves Demonstration from the two bulbs is not constant. The light waves still interfere, but the conditions for the interference change with each phase change, and Interference in a Ripple Tank therefore, no single interference pattern is observed. Light sources of this type are said to be incoherent. FIGURE 1.4 Purpose Demonstrate interference Monochromatic Light patterns using two types of point Interference An interference sources. Demonstrating Interference pattern consists of alternating light Materials ripple tank, ripple generator, Interference in light waves from two sources can be demonstrated in the and dark fringes. following way. Light from a single source is passed through a narrow slit two pencils, overhead projector or light and then through two narrow parallel slits. The slits serve as a pair of source, screen coherent light sources because the waves emerging from them come from the same source. Any random change in the light emitted by the source Procedure Touch the water in the tank will occur in the two separate beams at the same time. with a pencil point several times to If monochromatic light is used, the light from the two slits produces a generate a wave. Let students observe series of bright and dark parallel bands, or fringes, on a distant viewing the wave pattern. Repeat the procedure screen, as shown in Figure 1.4. When the light from the two slits arrives at a point on the viewing screen where constructive interference occurs, a with two pencils 20 cm apart so that the bright fringe appears at that location. When the light from the two slits two waves vary in frequency and phase. combines destructively at a point on the viewing screen, a dark fringe Explain that the pencils act as point appears at that location. (cl) (cl) ©Ian Mckinnell/Getty Images (br) ©GIPhotoStock/Photo Inc. Researchers, sources but that they generate waves of Interference and Diffraction 519 different phase or wavelength, or both. Use the ripple generator to create two Below Level coherent sources. Point out that the speakers in the previous Demonstration Untitled-296 519 Below-level students may believe that coherent 5/20/2011 7:55:10 AM were also coherent sources. Explain that waves are merely waves traveling at the same when the sources are coherent, they frequency. Waves with a constant phase differ- produce a stable interference pattern. ence are also coherent. So, while this means that the waves have the same frequency, there is more than just frequency to coherence: phase differ- ence must also be considered. Phase difference is the difference, expressed in t (time) between two waves having the same frequency and referenced to the same point in time.

Interference and Diffraction 519 FIGURE 1.5 When a white-light source is used to observe interference, the situa- tion becomes more complicated. The reason is that white light includes White Light Interference waves of many wavelengths. An example of a white-light interference When waves of white light from Teach continued pattern is shown in Figure 1.5. The interference pattern is stable or well two coherent sources interfere, the defined at positions where there is constructive interference between pattern is indistinct because different light waves of the same wavelength. This explains the color bands on colors interfere constructively and either side of the center band of white light. destructively at different positions. Demonstration Figure 1.6 shows some of the ways that two coherent waves leaving the slits can combine at the viewing screen. When the waves arrive at the HOw Distance Traveled Affects central point of the screen, as in Figure 1.6(a), they have traveled equal distances. Thus, they arrive in phase at the center of the screen, constructive Interference interference occurs, and a bright fringe forms at that location. Purpose Demonstrate that for con- When the two light waves combine at a specific point off the center of structive interference the difference the screen, as in Figure 1.6(b), the wave from the more distant slit must between the distances traveled by two travel one wavelength farther than the wave from the nearer slit. Because coherent wave fronts is equal to a whole the second wave has traveled exactly one wavelength farther than the first wave, the two waves are in phase when they combine at the screen. number times the wavelength. Constructive interference therefore occurs, and a second bright fringe Materials sine-wave generator, appears on the screen. amplifier, two speakers, tape measures If the waves meet midway between the locations of the two bright fringes, as in Figure 1.6(c), the first wave travels half a wavelength farther Procedure Set up the equipment as in than the second wave. In this case, the trough of the first wave overlaps the first Demonstration, and ask the crest of the second wave, giving rise to destructive interference. Consequently, a dark fringe appears on the viewing screen between the students to return to their positions in bright fringes. line. Set the generator to a frequency of about 440 Hz, and ask students to step FIGURE 1.6 Conditions for Interference of Light Waves Bright area at forward and back until they find a place center of screen (a) When both waves of light travel the same distance l where the sound is at a maximum. Have ( l ), they arrive at the screen in phase and interfere 1 1 l2 + l3 + 1 Center Center 2 constructively. (b) If the difference between the distances of screen them measure the distance from their Slits Slits of screen Slits traveled by the light from each source equals a whole location to each of the speakers and Bright Dark wavelength (λ), the waves still interfere constructively. l1 l3 l area area calculate the difference in distance (path (c) If the distances traveled by the light differ by a half 2 difference). Given the sound’s frequency wavelength, the waves interfere destructively. and speed (330 m/s), ask students to calculate its wavelength. 0.75 m Ask (a) Bright area atBright area at students to compare the wavelength to center of screencenter of screen

the path difference, then record the l1 l1 l2 + l2 + l3 + 1 l3 +Center1 Center Center Center 2 2 results on the chalkboard. of screen of screen Slits Slits Slits Slits of screen Slitsof screen Slits HRW • Holt Physics HRW • Holt Physics HRW • Holt Physics PH99PE-C16-001-008-A PH99PE-C16-001-009-A PH99PE-C16-001-010-A Bright Bright Dark Dark Increase the frequency to 660 Hz l1 l1 area area l3 l3 area area (λ = 0.50 m), and repeat the demonstra- l2 l2 tion. Have students notice that they are now standing closer to each other. Have students examine all the results (b) (c)

and note that although each of them has Inc. Researchers, (tl) Taylor/Photo ©David detected a different wave crest, the path Differentiated520 Chapter 15 Instruction difference between the speakers and the HRW • Holt PhysicsHRW • Holt Physics HRW • Holt PhysicsHRW • Holt Physics HRW • Holt PhysicsHRW • Holt Physics PH99PE-C16-001-008-APH99PE-C16-001-008-A PH99PE-C16-001-009-APH99PE-C16-001-009-A PH99PE-C16-001-010-APH99PE-C16-001-010-A wave crest equals a whole number Inclusion multiplied by the wavelength in all cases. Untitled-296Students 520 with color blindness may have 5/20/2011 7:55:12 AM difficulty seeing the color differences in Figure 1.5. Modify this figure by labeling each strip of color with its name. The objective here is not necessarily to determine the different colors represented. Rather, it is to express that there is a range of colors.

520 Chapter 15 Predicting the location of interference fringes. FIGURE 1.7 Consider two narrow slits separated by a distance d, as shown in Path Difference for Light TEACH FROM VISUALS Figure 1.7, and through which two coherent, monochromatic light waves, Waves from Two Slits l and l , pass and are projected onto a screen. If the distance from the 1 2 The path difference for two light FIGURE 1.7 Point out that lines l and l slits to the screen is very large compared with the distance between the waves equals d sin θ. In order to 1 2 slits, then l1 and l2 are nearly parallel. As a result of this approximation, l1 emphasize the path difference, the should intersect for interference θ and l2 make the same angle, , with the horizontal dotted lines that are figure is not drawn to scale. to occur. They are represented as perpendicular to the slits. Angle θ also indicates the position where waves combine with respect to the central point of the screen. parallel only as an approximation so The difference in the distance traveled by the two waves is called their that the path difference can be l1 path difference. Study the right triangle shown in Figure 1.7, and note that found mathematically. the path difference between the two waves is equal to d sin θ. Note carefully that the value for the path difference varies with angle θ and that Ask If d = 0.50 mm and θ = 0.30°, how each value of θ defines a specific position on the screen. large is the path difference? How many d l2 The value of the path difference determines whether the two waves wavelengths does this equal for green are in or out of phase when they arrive at the viewing screen. If the path light with a wavelength of 520 nm? difference is either zero or some whole-number multiple of the wave- dsin length, the two waves are in phase, and constructive interference results. Answer: d sin θ = 2600 nm, or The condition for bright fringes (constructive interference) is given by: 5 wavelengths of the light

Equation for Constructive Interference HRW • Holt Physics d sin θ = ±mλ m = 0, 1, 2, 3, . . . PH99PE-C16-001-012A path difference the difference in the Demonstration the path difference between two waves = distance traveled by two beams when an integer multiple of the wavelength they are scattered in the same direction from different points Thin-Film Interference Purpose Demonstrate that wavelength In this equation, m is the order number of the fringe. The central bright order number the number assigned to interference fringes with respect to affects the position of interference fringe at θ = 0 (m = 0) is called the zeroth-order maximum, or the central the central bright fringe maximum; the first maximum on either side of the central maximum, fringes. which occurs when m = 1, is called the first-order maximum, and so forth. Materials bottles of soap solution; 1 Similarly, when the path difference is an odd multiple of __ λ, the two 2 overhead or slide projector; red, green, waves arriving at the screen are 180° out of phase, giving rise to destruc- tive interference. The condition for dark fringes is given by the following and blue cellophane equation: Procedure Use the soap solution to blow bubbles for the students to Equation for Destructive Interference observe. Explain that the colors seen on d sin θ = ± m + _ 1 λ m = 0, 1, 2, 3, . . . the soap bubble result from interference ( 2 ) between two reflected waves. One is the path difference between two waves = an odd number of half wavelengths reflected from the bubble’s outer surface and the other from its inner

1 __ surface. Because the soap bubble is very If m = 0 in this equation, the path difference is ± 2 λ, which is the condition required for the first dark fringe on either side of the bright thin, the path difference for these waves 3 central maximum. Likewise, if m = 1, the path difference is ± __ λ, which 2 is so small that interference can be is the condition for the second dark fringe on each side of the central maximum, and so forth. observed. Have students note how the (tl) ©David Taylor/Photo Researchers, Inc. Researchers, (tl) Taylor/Photo ©David bands of color swirl over the bubbles’ Problem Solving Interference and Diffraction 521 surfaces. Cover the light source with one of the pieces of cellophane, and have Deconstructing Problems students repeat the demonstration. This time, have students note the widths of Untitled-296 521 When working with problems that require 5/20/2011 7:55:13 AM predicting location of interference fringes, the alternating monochromatic and students should first look for the order black bands. After using all three number (m). When m is zero or a whole number cellophane covers, point out that the multiple of the wavelength, they should use the fringe width in each pattern varies with constructive interference angle. If m is an odd color (wavelength). number multiple of half the wavelength, the students know they are working with destruc- tive interference. Once students determine wavelength, they can then determine color.

Interference and Diffraction 521 FIGURE 1.8 A representation of the interference pattern formed by Position of Higher-Order Interference Fringes double-slit interference is shown in Figure 1.8. The Teach continued The higher-order (m = 1, 2) maxima appear on either side of numbers indicate the two maxima (the plural of maxi- the central maximum (m = 0). m mum) that form on either side of the central (zeroth- 2 order) maximum. The darkest areas indicate the posi- TEACH FROM VISUALS tions of the dark fringes, or minima (the plural of 1 minimum), that also appear in the pattern. FIGURE 1.8 Have students notice that 0 Because the separation between interference fringes the dark and bright fringes form a –1 varies for light of different wavelengths, double-slit symmetrical pattern. –2 interference provides a method of measuring the wavelength of light. In fact, this technique was used to Viewing screen Ask How many zeroth-order bright make the first meas urement of the wavelength of light. lines are there? How many dark ones? PREMIUM CONTENT Answer: one; two HRW • Holt Physics PH99PE-C16-001-013A Interference Interactive Demo HMDScience.com Sample Problem A The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is measured on Classroom Practice a viewing screen at an angle of 2.15° from the central maximum. Interference Determine the wavelength of the light. The distance between two slits is ANALYZE Given: d = 3.0 × 10-5 m m = 2 θ = 2.15° 0.0050 mm. Find the angles of the Unknown: λ = ? zeroth-, first-, second-, third-, and Diagram: Second-order fourth-order bright fringes of bright fringe interference produced with light (m = 2) d = 0.030 mm Zeroth-order with a wavelength of 550 nm. = 2.15˚ bright fringe (m = 0)

Answer: 0°, 6.3°, 13°, 19°, 26° Diagram not to scale When monochromatic light falls on two PLAN Choose an equation or situation: slits with a separation of 0.010 mm, the Use the equation for constructiveHRW • Holt Physics interference. PH99PE-C16-001-014A zeroth-order dark fringes are observed d sin θ = mλ at a 2.0° angle. Find the wavelength. Rearrange the equation to isolate the unknown: Answer: 7.0 ×​ 10 ​−7​ m = 7.0 × 10 2 nm _d sin θ λ = m

SOLVE Substitute the values into the equation and solve: (3.0 × 10 -5 m)(sin 2.15°) λ = ___ Calculator Solution 2 Because the minimum number of -7 2 significant figures for the data is two, the λ = 5.6 × 10 m = 5.6 × 10 nm calculator answer 5.627366 × 10-7 should be rounded to two significant figures. λ = 5.6 × 10 2 nm

CHECK YOUR This wavelength of light is in the visible spectrum. The wavelength WORK corresponds to light of a yellow-green color.

Continued Problem522 Chapter Solving 15 Deconstructing Problems Untitled-296Point 522 out to students that when they 5/20/2011 7:55:14 AM see the keyword central maximum, this is a clue to plug their givens into the constructive interference equation to solve for their unknowns.

522 Chapter 15 Interference (continued) PROBLEM guide A 1. Lasers are devices that can emit light at a specific wavelength. A double-slit Use this guide to assign problems. interference experiment is performed with blue-green light from an argon-gas laser. The separation between the slits is 0.50 mm, and the first-order maximum of SE = Student Edition Textbook the interference pattern is at an angle of 0.059° from the center of the pattern. PW = Sample Problem Set I (online) What is the wavelength of argon laser light? PB = Sample Problem Set II (online) 2. Light falls on a double slit with slit separation of 2.02 × 10 -6 m, and the first bright fringe is seen at an angle of 16.5° relative to the central maximum. Find the Solving for: wavelength of the light. λ SE Sample, 1–2; 3. A pair of narrow parallel slits separated by a distance of 0.250 mm is illuminated Ch. Rvw. 9–11, 29 by the green component from a mercury vapor lamp (λ = 546.1 nm). Calculate the PW 3–4 angle from the central maximum to the first bright fringe on either side of the central maximum. PB 4–6

4. Using the data from item 2, determine the angle between the central maximum θ SE 3–4 and the second dark fringe in the interference pattern. PW Sample, 1–2 PB 7–10 d SE Ch. Rvw. 28 PW 5–7 PB Sample, 1–3 SECTION 1 FORMATIVE ASSESSMENT *Challenging Problem Reviewing Main Ideas 1. What is the necessary condition for a path length difference between two waves that interfere constructively? Destructively? Answers 2. If white light is used instead of monochromatic light to demonstrate interference, how does the interference pattern change? Practice A −7 2 3. If the distance between two slits is 0.0550 mm, find the angle between the 1. 5.1 × ​10 ​ ​ m = 5.1 × 10 nm first-order and second-order bright fringes for yellow light with a wave- 2. 574 nm length of 605 nm. 3. 0.125° Interpreting Graphics 4. 25.2° 4. Two radio antennas simultaneously FIGURE 1.9 transmit identical signals with a wave- Two Radio Antennas length of 3.35 m, as shown in Figure 1.9. A radio sev eral miles away in a car Assess and Reteach traveling parallel to the straight line Antenna = 3.35 m Car between the antennas receives the signals. If the second maximum is d Assess Use the Formative Assessment = 1.28 located at an angle of 1.28° north of the on this page to evaluate student central maximum for the interfering signals, what is the distance, d, between mastery of the section. the two antennas? Reteach For students who need HRW • Holt Physics PH99PE-C16-001-015A additional instruction, download the Answers to Section Assessment Interference and Diffraction 523 Section Study Guide. Response to Intervention To reassess 1. a difference of an integral number of students’ mastery, use the Section Quiz,

Untitled-296 523 wavelengths; a difference of an odd 5/20/2011 7:55:15 AM available to print or to take directly integral number of half wavelengths online at HMDScience.com. 2. It becomes blurred, and the bright fringes are made up of narrow, colored bands. 3. 0.63° 4. 3.00 × 10 2 m

Interference and Diffraction 523

Plan andPrepare Demonstration 524 path by anobstacle iscalleddiffraction. the divergence ofawave from itsinitial if thiswere alightwave. Point outthat light andshadow that would beformed they observe andto describe areas of students to sketch thepatterns that illustrate Huygens’ principle.Ask waves that appear to atstart the corners beyond thebarrier. Explainthat the “shadow” ofquietwater extending Let studentsexamine theedges ofthe and turnonthestraight-wave generator. Procedure light source, screen generator, barrier, overhead projector or Materials in arippletank. Purpose Wa concepts discussed inthechapter. definition ofdiffraction andtheother directions. Relate thisroot to the refers to lightbreaking upinto different diffringere, for “break into pieces,” which the term diffraction from theLatin verb ized by Francesco Grimaldi.Hecoined diffraction oflightwere firstcharacter- Latin Word Origins Preview Vocabulary SECTION Teach ves BendingAr Chapter 15 Demonstrate wave diffraction ripple tank, straight-waveripple tank, Place thebarrierintank, 2 The effects of ound C orners Differentiated Instruction Untitled-297 524 occurs when waves are deflectedasthey go wave travels parallel to Refraction anobject. shadow oftheslit.Diffraction means that a a smallholeandhasflared-out geometric Diffraction occurs whenawave goes through terms denote awave changing itsangle. the difference between theconcepts. Both students may have difficultyunderstanding diffraction refers to lightbending,below-level Since for theinstruction bothrefraction and Below Le vel diffract, around objects. A property of all waves is that they bend, or Water Waves and Diffraction diffraction obstacle, anopening,oredge a wavewhentheencountersan 524 FIGURE 2.1 images. instrument’s ability to resolve an determines optical Describe how diffraction fringes for a diffraction grating. Calculate the positions of bright and dark fringes. around obstacles and produce Describe how light waves bend SECTION 2 Chapter 15 achangeinthedirectionof Objectives The Bending of Light Waves diffraction Key Terms Diffraction bright rectangle clearly marked with onthescreen. edges But iftheslit is placed between a distant light source and a screen, the light produces a is more) or mm (1 slit wide a When edges. bysharp or obstacles, around initial direction oftravel iscalled shadowed. be This divergence oflightthat from wouldotherwise its result isthat light deviates from astraight-line path andenters theregion asapoint source oflight,serves thewaves spread out from theslits. The wave front can treated be asapoint source ofwaves. Because each slit understood using Huygens’s principle, which states that any point ona ofthelightSome to theright bends through andleftasitpasses each slit. on thescreen assharply shadows. defined But nothappen. thisdoes The rest ofthescreen dark. wouldbe The oftheslits edges wouldappear oflight thinstrips two wheresee each slit andthesource were up. lined interference pattern inthedouble-slit demonstration. Instead, you would observed. easily be to small too is bend they amount wavelengths, the short Figure 2.1. Figure in shown barriers the as such obstacles, around waves water bend that sound waves are able to bend around the corner. In a similar fashion, is reason The corner,person. the aroundthe talking see cannot you but is who someone hear can you building, a of corner the near stand you If In general, diffraction occurs when waves pass through small openings, The bending oflight through asitpasses slits can each ofthetwo be If light traveled instraight lines, an you able wouldnotbe to observe Light waves can also bend around obstacles, but because of their their of because but obstacles, around bend wavesalso can Light same directionsame itwas before. wave changes, anditnolonger travels inthe through asubstance. Thedirection ofthe resolving power diffraction. 5/20/2011 7:56:20AM

©Fundamental Photographs, New York

(tr) ©Houghton Mifflin Harcourt gradually narrowed, the light eventually begins to spread out and pro- FIGURE 2.2 duce a diffraction pattern, such as that shown in Figure 2.2. Like the interference fringes in the double-slit demonstration, this pattern of light Diffraction of Light with and dark bands arises from the combination of light waves. Decreasing Slit Width Demonstration Diffraction becomes more evident as the width of the slit is narrowed. Diffraction and Interference Wavelets in a wave front interfere with each other. Slit width by a Single Slit Diffraction patterns resemble interference patterns because they also Wide Narrow result from constructive and destructive interference. In the case of Purpose Demonstrate diffraction and interference, it is assumed that the slits behave as point sources of light. single-slit interference in a ripple tank. For diffraction, the actual width of a single slit is considered. Materials ripple tank, straight-wave According to Huygens’s principle, each portion of a slit acts as a source of waves. Thus, light from one portion of the slit can interfere with generator, two straight barriers, over- light from another portion of the slit. The resultant intensity of the head projector or light source, screen diffracted light on the screen depends on the angle, θ, through which the light is diffracted. Procedure Place two barriers about To understand the single-slit diffraction pattern, consider Figure 2.3(a), 10 cm from the straight- wave generator, which shows an incoming plane wave passing through a slit of width a. and leave a wide opening between Each point (or, more accurately, each infinitely thin slit) within the wide slit them. Ask students to predict how these is a source of Huygens wavelets. The figure is simplified by showing only five among this infinite number of sources. As with double-slit interfer- obstacles will affect the wave; then, ence, the viewing screen is assumed to be so far from the slit that the rays generate the wave to confirm their emerging from the slit are nearly parallel. At the viewing screen’s midpoint, prediction. Ask what will change when all rays from the slit travel the same distance, so a bright fringe appears. the width of the opening decreases. The wavelets from the five sources can also interfere destructively Have students note that diffraction when they arrive at the screen, as shown in Figure 2.3(b). When the extra distance traveled by the wave originating at point 3 is half a wavelength becomes more evident as the width of longer than the wave from point 1, these two waves interfere destructively the slit narrows. at the screen. At the same time, the wave from point 5 travels half a wavelength farther than the wave from point 3, so these waves also Tell students that you will increase interfere destructively. With all pairs of points interfering destructively, the wavelength by decreasing the wave this point on the screen is dark. generator’s frequency until interference For angles other than those at which destructive interference completely patterns appear. (Increase the slit’s width, occurs, some of the light waves remain un canceled. At these angles light appears on the screen as part of a bright band. The brightest band appears if necessary.) Explain that each portion in the pattern’s center, while the bands to either side are much dimmer. of the slit acts as a point source. Ask students under what conditions they FIGURE 2.3 might expect light to produce similar Destructive Interference in (a) (b) Single-Slit Diffraction diffraction and interference patterns. (a) By treating the light coming through when the slit is wider—but not vastly the slit as a line of infinitely thin sources 1 1 wider—than the wavelength along the slit’s width, one can determine 2 2 (b) the conditions at which destructive a 3 a 3 interference occurs between the waves 4 1 from the upper half of the slit and the 5 Central 4 2 bright fringe waves from the lower half. Incident 5 wave

Viewing screen ©Fundamental Photographs, ©Fundamental York New Photographs, (tr) ©Houghton Mifflin Harcourt

HRW • Holt Physics Interference and Diffraction 525 Problem Solving PH99PE-C16-002-005A HRW • Holt Physics Alternative ApproachesPH99PE-C16-002-004A3. Draw the tangent of AB (and label it A1B1). Draw all circles in a forward direction. Untitled-297 525 Here is a step-by-step method for proving the 5/20/2011 7:56:21 AM Huygens’ Principle: 1. Take few points having values 1, 2, 3 ... on the wavefront AB. These points are considered the source of secondary wavelets. 2. Use the secondary wavelets to make a radius vt at time (t). Now every point should be taken in the center and circles of the radii vt should be drawn.

Interference and Diffraction 525

Demonstration 526 producing a brightspot. interference will occur at this point of light, however, predicts constructive shadow would be dark. The wave theory stream thecenter ofparticles, ofthe would if light were composed of a with no bending around obstacles, as it light. Iflighttravels instraight-line paths crucial in confirming the wave theory of shadow. Explain that this experiment was the bright spot at the center of the edges ofthehead. Have studentsnote shadow dueto lightbendingaround the expect an interference pattern in the the pinhead; others may correctly students may expect a dark shadow of they to expect see onthescreen. Some Ask students to describe the pattern beam is larger a little than the pinhead). laser beam (far enough away that the and place the pinhead in the path of the Procedure from thestudents. CAUTION clay, screen Materials obstacle. diffracted lightaround theedge ofan of lightproduced by interference of Purpose Obs Light Dffra Teach continued t a Chapter 15 cle: Poiss Demonstrate thebrightspot laser, pinwitharound head, Direct the laserbeamaway Place the pin tip in the clay, ction b on Spot y an Problem Solving Untitled-297 526 Huygens’s wavelet (at each pointinspace). totaling phases andamplitudesofeach form ofadiffraction pattern isfigured by have a series of minimums and maximums. The amplitudes. Thisiswhydiffraction patterns zero andthecombined total oftheindividual (a.k.a. interference). Thistotal canbebetween and amplitudesoftheindividualwaves comes from addingtheirrelative phases When waves are combined, theirsum Take ItFur similar to that of a diffraction grating. into its component colors in a manner as wide maxima. as the secondary In a diffraction pattern, the central maximum is twice Diffraction Pattern from a Single Slit on a CD Constructive Interference 526 FIGURE 2.5 FIGURE 2.4 FIGURE 2.6 Shadow of a Washer and inside the washer. edge. Note the dark and light stripes both around the washer the washer’s shadow when light is diffracted at the washer’s Chapter 15 Compact discs disperse light ther A diffraction pattern forms in Diffraction Gratings parallel slits. Gratings are made by spacedlinesonapiece equally ruling glass prism. Atransmission grating consists ofmany spaced equally lightdisperse into itscomponentsimilar to aneffect that colors with ofa to eithertransmit orreflect light, diffraction uses andinterference to a diffraction grating. Adiffraction grating, which can constructed be Figure 2.6. disc’s surface, causing you a“rainbow” to see ofcolor, asshownin wavelength of light atrespect angle can seen aparticular with be to the ing light, theorientation ofthedisc,andlight’s wavelength. Each interference constructive goes directions. incertain consist entirely ofreflecting material, light so reflected from themunder- much that light ofthedisc asthethinportions separate them. Theseareas spacedrows. closely forms Theserows ofdata donotreflect nearly as information (alternating reflecting pitsandsmooth onthedisc surfaces) disc, streaks ofcolorare visible. Thesestreaks appear because thedigital You have probably that noticed ifwhite light isincident onacompact This phenomenonhas put to been practical called ina device use This interference constructive onthedirection depends oftheincom- glass isrequired thepattern. to observe pared thewavelength with ofthelight, andamagnifying around oftheshadow. theedge The washer islarge com- ofbright anddarka series bands oflight that continue shadow ofawasher. The pattern consists oftheshadow and Figure 2.5 minima. (called by a series of narrower, less intense secondary bands intense central band—the broad, a of consists slit single a through passing light The diffraction pattern that results from monochromatic Light diffracted by an obstacle also produces a pattern. about twice as wide asthenext brightest aswide about maximum. twice that thecentral isquite brightabitbrighter fringe and lie approximately halfway thedark between fringes. Note points at which maximum interference constructive occurs An example ofsuch apattern isshownin Diffraction occurs around ofallobjects. theedges secondary maximasecondary shows thediffraction pattern that appears inthe ), and a series of dark bands, or central maximum Figure 2.4. —flanked —flanked The 5/20/2011 7:56:23AM

(bl) ©Stockbyte/Getty Images; (t) ©Houghton Mifflin Harcourt; (cl) © Omikron/Photo Researchers, Inc.

(tr) ©amana images inc./Alamy FIGURE 2.7 Constructive Interference by a Diffraction Grating Conceptual Challenge Light of a single wavelength passes through each of the slits of a Demonstration diffraction grating to constructively interfere at a particular angle θ. Spiked Stars Photographs of stars always show spikes Effect of Slit Size on extending from the stars. Given Diffraction grating Lens P that the aperture of Diffraction Patterns a camera’s rect- Purpose Demonstrate that light angular shutter diffraction is more evident in narrow slits. has straight d edges, explain Materials two razor blades, clear glass Screen how diffraction or plastic plate, unfrosted colored light accounts for the spikes. bulb, adhesive tape Radio Diffraction Procedure Tape the two razor blades Visible light waves are not observed diffracting around flat on the plate with one blade’s edge d buildings or other obstacles. nearly touching the edge of the other However, radio waves can dsin blade. In a dark room, have students be detected around buildings or mountains, even when stand 3 to 6 m away from the light and the transmitter is not visible. look through the slit to observe the Ex plain why diffraction is more diffraction pattern. Ask how the of glass using a diamond cutting point driven by an elaborate machinePHYSICS evident for radio waves than for called a ruling engine. Replicas are then made by pouring liquid Spec.plastic Number on PH 99visible PE C16-002-009-Alight. distance from the first dark line to the Boston Graphics, Inc. the grating and then peeling it off once it has set. This plastic grating617.523.1333 is bright central fringe would vary as the then fastened to a flat piece of glass or plastic for support. slit’s width decreases. The distance Figure 2.7 shows a schematic diagram of a section of a diffraction increases. Point out that the central grating. A monochromatic plane wave is incoming from the left, normal to fringe is wider and much brighter than the plane of the grating. The waves that emerge nearly parallel from the grating are brought together at a point P on the screen by the lens. The the others. Explain that diffraction intensity of the pattern on the screen is the result of the combined effects gratings have many very narrow slits, of interference and diffraction. Each slit produces diffraction, and the allowing the observer to see the first diffracted beams in turn interfere with one another to produce the pattern. bright fringe from each slit. (The other For some arbitrary angle, θ, measured from the original direction of travel of the wave, the waves must travel different path lengths before fringes are less evident.) reaching point P on the screen. Note that the path difference between waves from any two adjacent slits is d sin θ. If this path difference equals one wavelength or some integral multiple of a wavelength, waves from all slits will be in phase at P, and a bright line will be observed. The condi- tion for bright line formation at angle θ is therefore given by the equation Answers for constructive interference: Conceptual Challenge

d sin θ = ±mλ m = 0, 1, 2, 3, . . . 1. The edges of the camera shutter diffract light, causing secondary This equation can be used to calculate the wavelength of light if you maxima to appear on the sides of know the grating spacing and the angle of deviation. The integer m is the order number for the bright lines of a given wavelength. If the incident the star’s image. These maxima give radiation contains several wavelengths, each wavelength deviates by a the appearance of light spikes specific angle, which can be determined from the equation. (bl) ©Stockbyte/Getty Images; (t) ©Houghton Mifflin Harcourt; (cl) © Omikron/Photo Inc. Researchers, (tr) ©amana images inc./Alamy around the star.

Interference and Diffraction 527 2. The wavelength of a radio wave is much longer than that of visible light. Deconstructing Problems Untitled-297 527 In order to determine the specific angle at 5/20/2011 7:56:25 AM which each wavelength deviates, rearrange the equation to solve for the unknown. d sin θ = ±mλ

±mλ sin θ = _​ ​ d

θ = sin-1 _​ ±m ​λ d

Interference and Diffraction 527 FIGURE 2.8 Maxima from a Diffraction Grating Teach continued Light is dispersed by a diffraction grating. The angle of deviation for the first-order maximum is smaller for blue light than for yellow light.

The Language of Diffraction grating Physics As in double-slit interference, d represents the distance between two adjacent slits. Bright lines occur for special values of θ that exist when the path difference from adjacent slits is a whole number of wavelengths Second order First order Zeroth order First order Second order (m = –2) (m = –1) (m = 0) (m = 1) (m = 2) (m = 0, 1, 2, 3, and so forth).

FIGURE 2.9 Note in Figure 2.8 that all wavelengths combine at θ = 0, which corre- sponds to m = 0. This is called the zeroth-order maximum. ThePHYSICS first-order Spectrometer The spectrometer maximum, corresponding to m = 1, is observed at an angle thatSpec. satisfies Number PH 99 PE C16-002-010-A Demonstration uses a grating to disperse the light Boston Graphics, Inc. the relationship sin θ = λ/d. The second-order maximum, corresponding from a source. 617.523.1333 to m = 2, is observed at an angle where sin θ = 2λ/d. Multiple-Slit Diffraction Light rays Telescope The sharpness of the principal maxima and the broad range of the dark Purpose Demonstrate patterns formed Slit areas depend on the number of lines in a grating. The number of lines per by a diffraction grating and the effect of unit length in a grating is the inverse of the line separation d. For example, different grating line separations. a grating ruled with 5000 lines/cm has a slit spacing, d, equal to the inverse of this number; hence, d = (1/5000) cm = 2 × 10-4 cm. The greater the Materials laser, two optical gratings number of lines per unit length in a grating, the less separation between Source with different grating constants, screen the slits and the farther spread apart the individual wavelengths of light. Lens Grating Diffraction gratings are frequently used in devices called spectrometers, CAUTION Avoid directing the laser which separate the light from a source into its monochromatic compo- beam toward the students. nents. A diagram of the basic components of a spectrometer is shown in Procedure Shine the laser beam onto Figure 2.9. The light to be analyzed passes through a slit and is formed into a parallel beam by a lens. The light then passes through the grating. The the screen, and place the optical grating diffracted light leaves the grating at angles that satisfy the diffraction Spec. Number PH 99 PE C16-002-011-A directly in front of the beam. Have 617.523.1333grating equation. A telescope with a calibrated scale is used to observe the students note the bright central fringe first-order maxima and to measure the angles at which they appear. From these measurements, the wavelengths of the light can be determined and (the zeroth-order maximum) and the the chemical composition of the light source can be identified. An exam- first-order maxima, one on each side of ple of a spectrum produced by a spectrometer is shown in Figure 2.10. the zeroth-order maximum. Spectrometers are used in astronomy to study the chemical compositions and temperatures of stars, interstellar gas clouds, and galaxies. Replace the first optical grating with FIGURE 2.10 the second. Ask students why there are differences in the separation between Spectrum of Mercury Vapor The light from mercury vapor is passed the zeroth-order and first-order maxima through a diffraction grating, producing produced by each grating. Smaller line the spectrum shown. spacing produces greater separation 528 Chapter 15 between the zeroth-order and first- Differentiated Instruction order maxima. Inclusion mount it at a 60° angle to the bottom of the Remind students that diffraction is Untitled-297 528 box. At the top, cut a hole for the CD to be 5/20/2011 7:56:26 AM greatest when the size of the openings Provide an opportunity for tactile learners to experience how a spectroscope works by viewed. Cover any holes in the box or around and the wavelength of waves are of the the CD with tape. Point the slit toward light same order of magnitude. building one in class from common household equipment: a CD, a cereal box, duct tape, thick sources and observe. paper, and something with which to cut the box. The cereal box will feature a narrow slit that produces a beam of light, and the CD will separate light into colors. Construct a slit (.2 to .3 mm) on one side of the box using duct tape and some thick paper. Cut the CD in half and

528 Chapter 15 PREMIUM CONTENT

Diffraction Gratings Interactive Demo HMDScience.com Sample Problem B Monochromatic light from a helium-neon Classroom Practice laser (λ = 632.8 nm) shines at a right angle to the surface of a Diffraction Gratings diffraction grating that contains 150 500 lines/m. Find the angles at which one would observe the first-order and second-order maxima. Monochromatic light shines at the surface of a diffraction grating with -7 ANALYZE Given: λ = 632.8 nm = 6.328 × 10 m m = 1 and 2 5.0 × 103 lines/cm. The first-order 1 1 d = __ = _ m maximum is observed at a 15° angle. _lines 150 500 150 500 m Find the wavelength. Unknown: θ = ? θ = ? 1 2 Answer: 5.2 × 102 nm Diagram: Second-order maximum (m = 2) Find the first-order and the second- order angles of diffraction observed through a 1.00 × 104 lines/cm diffraction Grating First-order θ maximum grating with light of wavelengths 400.0 2 (m = 1) λ = 632.8 nm θ1 nm, 500.0 nm, and 600.0 nm.

1 Screen Answers: 400.0 nm: θ1 = 23.6°, θ2 = 53.1° d = 150 500 m 500.0 nm: θ1 = 30.0°,

θ2 = 90.0° (does not occur) PLAN Choose an equation or situation: Use the equation for a diffraction grating. 600.0 nm: θ1 = 36.9°, d sin θ = ±mλ sinθ2 = 1.2 (does not occur) Rearrange the equation to isolate the unknown:

θ = sin-1 _mλ Light of 400.0 nm wavelength is shined ( d ) on a 5.0 × 103 lines/cm grating. How many diffraction lines can be observed? SOLVE Substitute the values into the equation and solve: For the first-order maximum, m = 1: Answer: five (m = 0, 1, 2, 3, 4; The m = 5 –7 fringe cannot be seen because it is at an θ = sin-1 _λ = sin-1 __6.328 × 10 m 1 ( ) d _1 m angle of 90°.) ( 150 500 )

θ1 = 5.465° Calculator Solution For m = 2: Because the minimum number of significant figures for the data is four, the calculator θ = sin-1 _2λ answers 5.464926226 and 10.98037754 2 ( d ) should be rounded to four significant figures. -7 -1 2(6.328__ × 10 m) θ2 = sin _1 m ( 150 500 )

θ2 = 10.98°

Continued Problem Solving Interference and Diffraction 529 Deconstructing Problems Untitled-297 529 Students should scan diffraction grating 5/20/2011 7:56:27 AM problems for keywords to determine the problem’s unknowns and givens. The phrase “calculate the angular separation” indicates that students will be given two wavelengths or two orders of maxima. They should plug in the wavelength and other data to obtain the angle (θ) for each case, and then find the difference between the angles.

θ2-θ1 = separation

Interference and Diffraction 529 Diffraction Gratings (continued)

CHECK YOUR The second-order maximum is spread slightly more than twice as far Teach continued WORK from the center as the first-order maximum. This diffraction grating does not have high dispersion, and it can produce spectral lines up to the tenth-order maxima (where sin θ = 0.9524). PROBLEM guide B Use this guide to assign problems. 1. A diffraction grating with 5.000 × 103 lines/cm is used to examine the sodium SE = Student Edition Textbook spectrum. Calculate the angular separation of the two closely spaced yellow lines PW = Sample Problem Set I (online) of sodium (588.995 nm and 589.592 nm) in each of the first three orders.

PB = Sample Problem Set II (online) 2. A diffraction grating with 4525 lines/cm is illuminated by direct sunlight. The Solving for: first-order solar spectrum is spread out on a white screen hanging on a wall opposite the grating. θ SE Sample, 1–2; a. At what angle does the first-order maximum for blue light with a wavelength of Ch. Rvw. 9–21 422 nm appear? PW 4–5 b. At what angle does the first-order maximum for red light with a wavelength of 655 nm appear? PB 4–6 λ PW Sample, 1–2 3. A grating with 1555 lines/cm is illuminated with light of wavelength 565 nm. What is the highest-order number that can be observed with this grating? (Hint: PB 7–10 Remember that sin θ can never be greater than 1 for a diffraction grating.)

d SE 5*; Ch. Rvw. 28, 30 4. Repeat item 3 for a diffraction grating with 15 550 lines/cm that is illuminated with PW 3 light of wavelength 565 nm.

PB 3–4 5. A diffraction grating is calibrated by using the 546.1 nm line of mercury vapor. The m SE 3–4 first-order maximum is found at an angle of 21.2°. Calculate the number of lines per centimeter on this grating. PW 6–7 PB Sample, 1–2 *Challenging Problem Diffraction and Instrument Resolution FIGURE 2.11 The ability of an optical system, such as a microscope or a Limits of an Optical System Each of two distant telescope, to distinguish between closely spaced objects is Answers point sources produces a diffraction pattern. limited by the wave nature of light. To understand this limitation, consider Figure 2.11, which shows two light Practice B sources far from a narrow slit. The sources can be taken as 1. 0.02°, 0.04°, 0.11° two point sources that are not coherent. For example, they could be two distant stars that appear close to each other in 2. a. 1.0° Source 1 the night sky. b. 17.2° If no diffraction occurred, you would observe two distinct Source 2 bright spots (or images) on the screen at the far right. 3. 11 However, because of diffraction, each source is shown to 4. 1 have a bright central region flanked by weaker 3 bright and dark rings. What is observed on the screen is the 5. 6.62 × 10 lines/cm resultant from the superposition of two diffraction patterns, Slit Screen one from each source. (br) ©Courtesy University School of of Birmingham Physics Astronomy, &

530 Chapter 15HRW • Holt Physics Problem SolvingPH99PE-C16-002-014-A Deconstructing Problems When students they are asked to calculate order number, they should solve for m. Untitled-297When 530 students are asked, “At what angle does 5/20/2011 7:56:28 AM the first-order maximum for blue light with a d sin θ = ±mλ wavelength of ____ nm appear?” they should ±dsinθ use the provided wavelength and other givens m = ​ _ ​ λ in the problem and rearrange the problem to solve for the angle (θ). d sin θ = ±mλ

±mλ sin θ = _​ ​ d

530 Chapter 15 Resolution depends on wavelength and aperture width. If the two sources are separated so that their central maxima do not overlap, as in Figure 2.12, their images can just be distinguished and are The Language of said to be barely resolved. To achieve high resolution or resolving power, resolving power the ability of an the angle between the resolved objects, θ, should be as small as possible optical instrument to form separate Physics images of two objects that are close Figure 2.11. as shown in The shorter the wavelength of the incoming light together Point out that the equation for resolving or the wider the opening, or aperture, through which the light passes, the power is known as the Rayleigh criterion. smaller the angle of resolution, θ, will be and the greater the resolving power will be. For visible-light telescopes, the aperture width, D, is This equation gives the angle θ in approximately equal to the diameter of the mirror or lens. The equation radians. One rad equals (180/π)°. For to determine the limiting angle of resolution in radians for an optical example, for light with a wavelength of instrument with a circular aperture is as follows: 500.0 nm, a lens that is 0.10 m in diam- -6 θ = 1.22 _λ eter has a limiting angle of 6.1 × 10 rad, D or (3.5 × 10-4)°. This may seem like a very The constant 1.22 comes from the derivation of the equation for small angle, but for viewing the surface circular apertures and is absent for long slits. Note that one radian equals 6 (180/π)°. The equation indicates that for light with a short wavelength, of Mars (nearly 75 × 10 km away), it such as an X ray, a small aperture is sufficient for high resolution. On the corresponds to a distance of about other hand, if the wavelength of the light is long, as in the case of a radio 460 km. This means a telescope with wave, the aperture must be large in order to resolve distant objects. This is one reason why radio telescopes have large dishlike antennas. such a lens would show two mountains Yet, even with their large sizes, radio telescopes cannot resolve sources separated by a distance of less than as easily as visible-light telescopes resolve visible-light sources. At the 460 km as one blurred mountain. shortest radio wavelength (1 mm), the largest single antenna for a radio telescope—the 305 m dish at Arecibo, Puerto Rico—has a resolution angle of 4 × 10-6 rad. The same resolution angle can be obtained for the longest visible light waves (700 nm) by an optical telescope with a 21 cm mirror.

FIGURE 2.12

Resolution Two point sources are barely resolved if the central maxima of their diffraction patterns do not overlap.

HRW • Holt Physics PH99PE-C16-002-015bA (br) ©Courtesy University School of of Birmingham Physics Astronomy, & Differentiated Instruction Interference and Diffraction 531 English Learners Untitled-297 531 English Learners may be familiar with the term 5/20/2011 7:56:29 AM resolution, meaning “a decision” or “a formal expression.” The text implies a scientific definition for resolution (“the ability of an optical system, such as a microscope or telescope to distinguish between closely spaced objects”) but does not explicitly label it. To avoid textual confusion, point out the scientific definition for the term before moving on to discuss resolving power.

Interference and Diffraction 531

Assess andReteach 532 online at HMD available to printorto take directly students’ mastery, use Quiz, theSection Response to Intervention StudyGuide. Section downloadadditional instruction, the Reteach mastery ofthesection. on thispage to evaluate student Assess wavelength ofred light. thanthe about 10000timesshorter hundredths ofananometer. Thisis microscopes typicallyismeasured in wavelength of electrons used in electron with visiblelightbecause theeffective smaller detailsincells thanispossible microscopes allow usto distinguish wavelength Point outthat electron blue, because ithasashorter resolutionbetter withamicroscope. Ask whichcolor, blueorred, gives a Teaching Tip Teach continued

Chapter 15 Use theFormative Assessment For studentswhoneed Science.com. To reassess Answers to Assessment Section Untitled-297 532

4. 2. 5. 3. 1. 532 a. than thelight’s wavelength. Asthehole sharp shadow when the hole is much wider The image ofthepointsource castsa diffracted more. orange light;Longer wavelengths are of thewavelength ofvisiblelight. a humanhair;Itssizeisclosest to thesize increases as the width of the slit decreases. The widthofthecentral maximum b. FIGURE 2.13 arranged to have the resolving power of a 36 km wide radio telescope. distant resolution radio for sources.observing The antennas can be Large Array in New Mex ico are used together to provide improved Combining Many Combining TelescopesMany 5.89 ×​ 24.7° 6. 5. 4. 3. 2. 1. SECTION 2 Chapter 15 ReviewingMainIdeas ultraviolet radiation rather than visible light? Explain. Would easier itbe to resolve nearby ifyou objects themusing detected projected onto ascreen asthehole’s isreduced. width single hole. what Discuss happens to theimage of thepoint source A point source oflight isinside acontainer that isopaque except fora why. Explain Would orange light orbluelight produce diffraction awider pattern? apple, lead, orahuman apencil your hair? answer. Explain wouldproduceWhich object themost distinct diffraction pattern: an diffraction pattern oftheslit ismade asthewidth smaller. thechange ofthecentral inwidth Describe maximum ofthesingle-slit b. a. first-order maximum at anangle of12.07°. Light through passes adiffraction grating a 3550lines/cm andforms with

Critical Thinking At what maximum thesecond appear? angle will What isthewavelength ofthelight? 10 ​ −7 ​ m(589 nm) FORMATIVE ASSESSMENT The 27 antennas at the Very

6. power isgreater for wavelengths. short length thanvisiblelight,andresolving yes; Ultraviolet wave lighthasashorter - central fringe. of darkfringes oneithersideofthe bright tively interfere, producing a regular pattern widen. Lightwaves from theholedestruc- the hole’s edges, causingtheimage to narrows, the light waves bend more around telescope. Anexample ofthisisshownin that functionlikeso amuch will they larger waves, onecan combine several radio telescopes resolving power. to thetelescope’s isclose telescope theoretical conditions, theactual resolving power ofthe operates inthevacuum ofspace. Under these largely quality becausesuperior thetelescope images from theHubble Space Telescope are of which blurthelight from inspace. objects The constantly moving layers ofairintheatmosphere, by islimited optical onEarth the telescopes kilometers. radio had telescope adiameter ofseveral radio “images” isthesameif asitwouldbe that each antenna receives, theresolution ofthe line and computers are to process used the signals Figure 2.13. To compensate resolution forthepoor ofradio It shouldthat noted be theresolving power for Iftheradio antennas are arranged ina 5/20/2011 7:56:30AM

©Phototake/Getty Images SECTION 3 SECTION 3

Objectives Lasers Describe the properties of laser Plan and Prepare light. Key Term Explain how laser light has Preview Vocabulary laser particular advantages in certain applications. Scientific Meanings Explain that Lasers and Coherence scientific words sometimes originate from acronyms. Laser is one such word. At this point, you are familiar with electromagnetic radiation that is produced by glowing, or incandescent, light sources. This includes light It comes from the acronym LASER from light bulbs, candle flames, or the sun. You may have seen another (Light Amplification by Simulated form of light that is very different from the light produced by incandes- Emission of Radiation). The verb lase, cent sources. The light produced by a laser has unique properties that laser a device that produces coherent make it very useful for many applications. light at a single wavelength which means “to produce coherent To understand how laser light is different from conventional light, light,” also comes from this acronym. consider the light produced by an incandescent light bulb, as shown in Relate the acronym to the definition Figure 3.1. When electric charges move through the filament, electromag- of laser. netic waves are emitted in the form of visible light. In a typical light bulb, there are variations in the structure of the filament and in the way charges move through it. As a result, electromagnetic waves are emitted at different times from different parts of the filament. These waves have different intensities and move in different directions. The light also covers Teach a wide range of the electromagnetic spectrum because it includes light of different wavelengths. Because so many different wavelengths exist, and because the light is changing almost constantly, the light produced is Demonstration incoherent. That is, the component waves do not maintain a constant phase difference at all times. The wave fronts of incoherent light are like Did YOU Know? the wave fronts that result when rain falls on the surface of a pond. No The light from an ordinary electric lamp Dancing Light two wave fronts are caused by the same event, and they therefore do not undergoes about 100 million (108) Purpose Demonstrate characteristics of random changes every second. produce a stable interference pattern. a laser beam.

FIGURE 3.1 Materials cassette-tape or CD player with an open speaker, clear plastic wrap, Wave Fronts from Incoherent and Coherent Light Sources Waves from an reflective polyester film, laser, light pen incoherent light source (right) have changing phase CAUTION Avoid directing the laser relationships, while waves from a coherent light source (left) have constant phase relationships. beam toward students. Procedure Before students arrive, pull the clear wrap tightly around the speaker, and secure it with tape. Tape Coherent light Coherent light the reflective film to the clear wrap in front of the speaker. In class, direct the laser beam so that it is reflected off the

Incoherent light Incoherent light reflective film attached to the speaker and onto the ceiling. Play the music, Interference and Diffraction 533 and darken the room. Replace the laser C16-003-001-ADifferentiated Instruction C16-003-001-A with a light pen, and have students Inclusion compare the two beams. Have them note that laser light is emitted in one Untitled-298 533 Visually-impaired students may require a 5/20/2011 7:57:48 AM narrow beam in one direction and is modification of Figure 3.1. To demonstrate the almost monochromatic. concepts illustrated by Figure 3.1, use a tactile substitution that students can feel, such as textured strings arranged in incoherent and coherent patterns.

Interference and Diffraction 533 Lasers, on the other hand, typically produce a narrow beam of coher- ent light. The waves emitted by a laser are in phase, and they do not shift relative to each other as time progresses. Because all the waves are in Teach continued phase, they interfere constructively at all points. The individual waves effectively behave like a single wave with a very large amplitude. In addition, the light produced by a laser is monochromatic, so all the waves have exactly the same wavelength. As a result of these properties, the Demonstration intensity, or brightness, of laser light can be made much greater than that of incoherent light. For light, intensity is a meas ure of the energy trans- Interference in Laser Light ferred per unit time over a given area. Purpose Demonstrate interference of Lasers transform energy into coherent light. light waves with the same wavelength. Did YOU Know? A laser is a device that converts light, electrical energy, or chemical Materials laser, light pen, projection The word laser is an acronym energy into coherent light. There are a variety of different types of lasers, screen, opaque slide with two slits (a word made from the first letters of but they all have some common features. They all use a substance called several words) that stands for “light the active medium to which energy is added to produce coherent light. (to make one, spray a layer of paint amplification by stimulated emission The active medium can be a solid, liquid, or gas. The composition of the on a glass plate and use a pin to etch of radiation.” active medium determines the wavelength of the light produced by the two parallel slits about 1 mm apart laser. through the paint) The basic operation of a laser is shown in Figure 3.2. When high-energy light or electrical or chemical energy is added to the active medium, as in CAUTION Avoid directing the light Figure 3.2(a), the atoms in the active medium absorb some of the energy. from the laser at the students. Procedure Shine the laser through two FIGURE 3.2 slits and project the beam toward the Operation of a Laser Atoms or molecules with added energy screen. Point out that each slit acts as a (a) Atoms or molecules in the active medium of a laser absorb point source of light. Have students energy from an external source. examine the bands of brightness and darkness. Explain that the areas of darkness correspond to the nodal lines Active Mirror Mirror Energy input observed in the experiments with sound medium (partially transparent) and water waves in the first two (b) When a spontaneously emitted Demonstrations of this chapter. Stimulated emission Spontaneously emitted light light wave interacts with an atom, Repeat the experiment with a light it may cause the atom to emit an Before stimulated pen. Explain that although the light from identical light wave. emission the two slits is nearly coherent, white After stimulated light contains many wavelengths. emission Although these interfere with each other, the pattern is blurred. (c) Stimulated emission increases Coherent light output the amount of coherent light in the of laser active medium, and the coherent waves behave as a single wave. ©Philippe Plailly/Eurelios/Photo Inc. Researchers, Differentiated534 Chapter 15 Instruction Pre-AP Ask students to determine the wavelengths for the following colors and order them from Untitled-298Lasers 534 are concentrated light sources, which 5/20/2011 7:57:48 AM means they can damage the human eye. The not harmful to the retina to most harmful: coherence of laser light combined with the ultraviolet, near ultraviolet, red, green, blue, focusing mechanism of the eye causes laser indigo. red, green, blue, indigo, near ultraviolet, light to be concentrated into an an extremely ultraviolet small spot on the retina. Wavelengths in the 400 to 1400 nm range can do permanent damage to the retina. Waves with shorter wavelengths cause more damage than waves with longer wavelengths.

534 Chapter 15 Untitled-298 535

©Philippe Plailly/Eurelios/Photo Researchers, Inc. Applications of Lasers Earth-to-moon distance about to be Earth-to-moon 3.84×10 takes to travel andback, to themoon scientists have measured the missions. By oflight knowing thespeed andmeasuring thetimelight were onthemoon’s placed by surface astronauts theApollo during by thelaser. slightly transparent, which allows theintense coherent light emitted to be active medium, more itbecomes andmore intense. ofthemirrors One is even whentheheight changes by only afewcentimeters. from processes. geological measurements forthese can Lasers used be repeated measurements to record changes intheheight ofEarth’s crust as shownin medium, where stimulatethey the emission of more coherent light waves, the endsofmaterial return coherent these light waves into theactive of thetube, producing more stimulated emission asitgoes. Mirrors on sides oftheglass tube. However, ofthelight some moves along thelength pointed at distant reflectors andthereflected light can detected. be result, to meas can alaser used be ure large distances, because itcan be that allthelight waves emitted by have thelaser thesame direction. Asa little spreadingfrom undergoes very distance. alaser with reason One is bulb oreven thelight that by isfocused aparabolic reflector, thelight emerges from asanarrow thelaser beam. Unlike thelight from alight Of theproperties light, oflaser theonethat ismost evident is that it Lasers are used to measure distances with great precision. medical applications. specialized andvery uses ofindustrial variety to awide household uses region are ofthespectrum. Lasers inmany used ways, from common created similar, devices but to masers lasers operate inthemicrowave far infrared to theX-ray region ofthespectrum. Scientists have also There are types, oflaser variety wavelengths awide with ranging from the wave, in shown as initial the wavelength, as same direction the phase, waves light and as with energy excess their release to atoms energized wave other cause initial can wave,light a of form this the in energy its releases spontaneously atom one wavelength. wavessame light release the After to of induced be can atoms netic radiation return whenthey to theiroriginal, lower energy states. atomsexcited thenrelease theirexcessofelectromag- energy intheform energy state, theatom can excited be to ahigher energy state. These “Atomic Physics.” to anatom energy When isadded that isat alower You learn that will atoms at exist different states inthechapter energy A pulseoflight isdirected toward oneofseveral 0.25m points onthemoon’s distance. theEarth-to-moon to determine surface strides. Group Brepresents laser light.Students about, walking inalldirections with different Students onthissideoftheroom should mill incoherent by lightemitted alightbulb. the room. Group Arepresents waves of into two groups, withonegroup ineach halfof Divide theroom inhalfandseparate students outincoherentact lightandcoherent light. To appeal to kinestheticlearners, have students Inclusion Most ofthelight produced by stimulated emission escapes out the As shownin atoms, the excited to wavelength applied is certain a of light When Figure 3.2(c). Figure 3.3, Figure3.2(b). Asthelight through back passes andforth the astronomers direct light laser at particular This process is called called is process This 5 km. Geologists use use km. Geologists stimulated emission stimulated 2 reflectors that in sync witheach other. should lineupandmarch direction, inthesame . Interference andDiffraction FIGURE 3.3 380 000 km. 3 centimeters out of a total of trip yields a distance accurate to lunar missions. Timing of the laser’s Moon left behind by US and Russian d’Azur is aimed at mirrors on the laser de at la the Côte l’Observatoire Distance to the Moon The 5/20/2011 7:57:49AM 535 emit light. something triggers theexcited atoms to atoms remain inanexcited state until spontaneously. Inlasers, most ofthe period oftime,butthey emitlight higher level ofenergy for alonger excited atoms are ableto remain at a light. Inphosphorescent paint,the immediately lose theenergy by emitting become excited (by absorbing energy) that in ­ fluo way lasers work. Make sure they realize and phosphorescence are related to the Students may thinkthat fluorescence Misconception Alert! rescent lamps, any atoms that Interference andDiffraction 535 WHY IT MATTERS Teach continued Digital Video Why It Matters Digital video Players Players CDs, DVDs, and high definition video

discs, such as Blu-ray discs, all fall into a n interesting application of the laser is the digital class of digital media called optical video disc (DVD) player. In a DVD player, light storage devices. A from a laser is directed through a series of optics toward a video disc on which the music or data have The laser typically used for a CD been digitally recorded. The DVD player “reads” the data These binary, digital data in a DVD are stored as a series player has a wavelength of 780 nm, in the way the laser light is reflected from the disc. of pits and smooth areas (called lands) on the surface of the disc. The series of pits and lands is recorded starting at which is light in the near-infrared region. In digital recording, a video signal is sampled at regular the center of the disc and spiraling outward along in A DVD player laser has a wavelength of intervals of time. Each sampling is converted to an tracks the DVD. These tracks are just 320 nm wide and spaced electrical signal, which in turn is converted into a series of 650 nm, which is visible as red light. A 740 nm apart. If you could stretch the data track of a DVD binary numbers. Binary numbers consist only of zeros and Blu-ray player laser has a wavelength of out, it would be 12 km long! ones. The binary numbers are coded to contain information 405 nm, which in actuality is more violet about the signal, including the sound and image, as well When a DVD is played, the laser light is reflected off than blue. All of these lasers are diode as the speed of the motor that rotates the disc. This this series of pits and lands into a detector. In fact, the lasers. They are semiconductor devices process is called analog-to-digital (a-d) conversion. depth of the pit is chosen so that destructive interference occurs when the laser transitions from a pit to a land or that produce low-intensity, coherent from a land to a pit. The detector records the changes in light when they carry current. light reflection between the pits and lands as ones and smooth areas as zeros—binary data that are then converted back to the analog signal you see as video or

Disc hear as music. This step is called digital-to-analog (d-a) Pit Land conversion, and the analog signal can then be amplified Lens to the television set and speaker system. A DVD drive on your computer works in much the same way. Data from a computer are already in a digital Lens format, so no a-d or d-a conversion is needed. You may wonder how a DVD-recordable (DVD-R) disc is different. These discs don’t have any pits and lands at all. Instead, they have a layer of light-sensitive dye sandwiched between a smooth reflective metal, usually Prism Photoelectrical cell aluminum, and clear plastic. A DVD-R drive has an additional laser, about 10 times more powerful than a DVD reading laser, that writes the digital data along the Laser tracks of the DVD-R disc. When the writing laser shines on the light-sensitive dye, the dye turns dark and Light from a laser is directed toward the creates nonreflecting areas along the track. This process surface of the disc. Smooth parts of the disc creates the digital pattern that behaves like the pits and reflect the light back to the photoelectrical cell. lands, which a standard DVD player can read. (tr) © Bailey-Cooper Photography/Alamy Images

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536 Chapter 15 (tr) © Bailey-Cooper Photography/Alamy Images Untitled-298 537 Answers to Assessment Section

central ofthecornea to cause flatter. portion itto become to correct nearsightedness used can be also by focusing onthe thebeam tiny burned holetointhetissue, be which relieves thepressure. Lasers eventual blindness. Focusing at drainage alaser theclogged allows a port great. untreated, Left glaucoma and can lead to damage oftheopticnerve glaucoma, aconditioninwhich thefluidpressure theeyeistoo within effectiveatbe treating lesions oftheretina, inside theeye. eye—the cornea andlens—without damaging them. Therefore, can lasers lengths can oflasers pass through structures certain at thefront ofthe conventional surgical methods. For wave- specific example, very some massive performing surgery. without As aresult, surgeons can stop internal orremove bleeding tumors through anddirectedinserted anorifice to internal structures. body trapped canbeam be also inanoptical endoscope, fiber which can be thereby reducing loss anddecreasing blood Alaser ofinfection. therisk from coagulates also vessels, thelaser blood inthenewly opened blood water contained inthecells. advantage One isthat ofalaser theenergy canlasers cut through muscle tissue by heating andevaporating the carbon having dioxide lasers awavelength of10µm. Carbon dioxide water can vaporizedby be high-intensity infrared light produced by tissues are protected. responds to thewavelength oflight inthelaser, used but otherbody ofbirthmarkstypes affecting surrounding without tissues. The scar tissue For example, to lighten can lasers used be orremove scars and certain fact that tissues body absorb specific different wavelengths light. oflaser areLasers formany used also medical procedures by making ofthe use Lasers have applications many in medicine. 2. 3. 1. 3. 2. 1. SECTION 3 Lasers areLasers forothereyesurgeries, to used correct including surgery to treat used can be Lasers also tissues that cannot reached be by Many medical applications take oflasers advantage ofthefactthat different components withdifferent does notspread outvery muchinto Laser lightisnearly monochromatic, so it energy source. in theactive mediumby anexternal from theenergy that isaddedto anatom no; Thesecond lightwave isobtained in phase). (they are coherent andcontinuously relative to each other astimeprogresses Waves by emitted alaser donotshift ReviewingMainIdeas knowledge ofrefraction to explain why. light instead ofwhite light isused to transmit thesignal. Apply your thin strands ofextremely pure glass. In fiber-optic these systems, laser Fiber-optic systems transmit light by means ofinternal reflection within your answer. Explain ofenergy conservation? principle that isidentical to thefirst. thisgaining wave Does ofasecond violate the The process ofstimulated emission involves producing wave asecond same wavelength but are notcoherent? How light does from differfrom alaser light waves whose allhave the

Critical Thinking FORMATIVE ASSESSMENT absent over distances). short equipment (that is,dispersionisnearly and thetransmission and receiving wavelengths asitpasses between thefiber Interference andDiffraction name and price to the cash register. system, which returns the product’s transmitted to the store’s computer number.inventory This information is codes that represent the product’s the bar code reproduces the binary reflected from the bars and spaces of these products are scanned, laser light codes found on many products. When is also the basis for the reading of bar information stored on a compact disc The principle behind reading the Did Know? YOUKnow? 5/20/2011 7:57:53AM 537 Assess andReteach online at HMDScience.com. available to printorto take directly students’ mastery, use Quiz, theSection Response to Intervention StudyGuide. Section downloadadditional instruction, the Reteach mastery ofthesection. on thispage to evaluate student Assess them report ononeofthese topics. tion abouthow lasers are used andhave star. Have studentsfindmore informa- Telescope discovered anultraviolet laser instars.In1996,theHubbleSpaceaction Astronomers have even found laser saws, andmoonquake detectors. They are used asbloodless scalpels, industrial, andscientific applications. Lasers now have medical,military, lasers, X-ray lasers, andtunablelasers. lasers, CO lasers, including semiconductor (diode) Point outthat there are manytypesof T eaching Tip Use theFormative Assessment For studentswhoneed 2 lasers, ruby lasers, microwave Interference andDiffraction To reassess 537

Careers in Physics CAREERS IN PHYSICS Laser Surgeon Dr. Shawn Wong practices ophthalmology Laser Surgeon in Austin, Texas, and specializes in LASIK (laser-assisted in situ keratomileusis) surgery to correct vision problems. In his aser surgery combines two fields—eye care and spare time, Dr. Wong enjoys bicycle racing engineering—to give perfect vision to people who and flying model airplanes. L otherwise need glasses or contacts. To learn more about this career, read the interview with ophthalmologist Dr. Wong particularly likes the Dr. L. Shawn Wong, who runs a laser center in Austin, Texas. problem-solving aspects of his work. He Dr. Wong makes measurements of the also enjoys “not only making a living, What sort of education helped you become eye in preparation for laser surgery. a laser surgeon? but helping to solve people’s problems Besides using my medical school training, I use a lot of along the way. Laser eye surgery is a very engineering in my work; physics and math courses are very How does your work relate to the physics of interference and diffraction? dynamic field—what we can’t do now is helpful. In high school, even in junior high, having a love of math and science is extremely helpful. Measuring diffraction and interference is part of every aspect just around the corner.” of what we do. The approach is based on doing many small Who helped you find your career path? things correctly. Applying small physics principles in the right Of all my teachers, my junior high earth science teacher order can solve very big problems. made the biggest impression on me. What I learned in those What advice would you give to somebody classes I actually still use today: problem solving. who is considering a career in laser Interestingly, I work in the town where I grew up; a lot of my surgery? former teachers are my patients today. My education didn’t start in medical school; it started by asking questions as a kid. You need a genuine love of taking What makes laser surgery interesting on complex problems. A background in physics and math is to you? extremely helpful. Technol ogy in It’s nice to be able to help people. Unlike glasses and medicine is based on contacts, laser surgery is not a correction; it’s a cure. When engineering. you are improving people’s vision, everybody in the room gets to see the results. I don’t need to tell patients they’re Being well rounded will help doing well—they can tell. you get into medical school —and get out, too. You have to What is the nature of your work? be comfortable doing the A typical patient is somebody born with poor vision. We make science, but you also these patients undergo a lot of formal diagnostic testing and have to be comfortable informal screening to be sure they are good candidates. dealing with people. Lasers are used for diagnosing as well as treating. Laser tolerances are extremely small—we’re talking in terms of submicrons, the individual cells of the eye. What is the favorite thing about your job? What part would you most like to change? Shawn Wong My favorite thing is making people visually free. I would like to be able to solve an even wider range of problems. We can’t solve everything.

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538 Chapter 15 Chapter summary CHAPTER 15 Summary Teaching Tip SECTION 1 Interference KEY TERMS Because double-slit interference can • Light waves with the same wavelength and constant phase differences coherence be a difficult subject to understand, interfere with each other to produce light and dark interference patterns. path difference students may find it helpful to write • In double-slit interference, the position of a bright fringe requires that the order number path difference between two interfering point sources be equal to a whole an expository essay with diagrams, number of wavelengths. explaining how an interference pattern • In double-slit interference, the position of a dark fringe requires that the path difference between two interfering point sources be equal to an odd is created. Essays should include a number of half wavelengths. discussion of diffraction as it relates to double-slit interference.

SECTION 2 Diffraction KEY TERMS

• Light waves form a diffraction pattern by passing around an obstacle or diffraction bending through a slit and interfering with each other. resolving power • The position of a maximum in a pattern created by a diffraction grating depends on the separation of the slits in the grating, the order of the maximum, and the wavelength of the light.

SECTION 3 Lasers KEY TERM

• A laser is a device that transforms energy into a beam of coherent laser monochromatic light.

VARIABLE SYMBOLS

Quantities Units

λ wavelength m meters

θ angle from the center of an ° degrees interference pattern Problem Solving See Appendix D: Equations for a summary d slit separation m meters of the equations introduced in this chapter. If you need more problem-solving practice, m order number (unitless) see Appendix I: Additional Problems.

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Interference and Diffraction 539 chapter review CHAPTER 15 Review Answers 1. The amplitude of the resultant wave Interference PRACTICE PROBLEMS is twice the amplitude of either REVIEWING MAIN IDEAS For problems 9–11, see Sample Problem A.

interfering wave, so a bright 1. What happens if two light waves with the same 9. Light falls on two slits spaced 0.33 mm apart. If fringe forms; The amplitude of the amplitude interfere constructively? What happens if the angle between the first dark fringe and the resultant wave is zero, so a dark they interfere destructively? central maximum is 0.055°, what is the wavelength of the light? fringe forms. 2. Interference in sound is recognized by differences in volume; how is interference in light recognized? 10. A sodium-vapor street lamp produces light that is 2. by differences in brightness nearly monochromatic. If the light shines on a 3. A double-slit interference experiment is performed 3. λ < λ , so interference wooden door in which there are two straight, parallel blue light red light with red light and then again with blue light. In what cracks, an interference pattern will form on a distant fringes form at smaller angles for ways do the two interference patterns differ? (Hint: wall behind the door. The slits have a separation of blue light. Consider the difference in wavelength for the two 0.3096 mm, and the second-order maximum occurs colors of light.) 4. d , m, θ at an angle of 0.218° from the central maximum. 4. What data would you need to collect to correctly Determine the following quantities: 5. θ would decrease because λ is calculate the wavelength of light in a double-slit a. the wavelength of the light shorter in water. interference experiment? b. the angle of the third-order maximum c. the angle of the fourth-order maximum 6. no; because the light from the stars CONCEPTUAL QUESTIONS 11. All but two gaps within a set of venetian blinds have is not coherent been blocked off to create a double-slit system. These 5. If a double-slit experiment were performed underwa- gaps are separated by a distance of 3.2 cm. Infrared 7. No interference is observed ter, how would the observed interference pattern be radiation is then passed through the two gaps in the affected? (Hint: Consider how light changes in a because the two waves have blinds. If the angle between the central and the medium with a higher index of refraction.) different wavelengths. second-order maxima in the interference pattern is 0.56°, what is the wavelength of the radiation? 8. The separation of the fringes 6. Because of their great distance from us, stars are essentially point sources of light. If two stars were decreases because d and sin θ are near each other in the sky, would the light from inversely proportional. them produce an interference pattern? Explain Diffraction your answer. 9. 630 nm REVIEWING MAIN IDEAS 7. Assume that white light is provided by a single source 12. Why does light produce a pattern similar to an 10. a. 589 nm in a double-slit experiment. Describe the interference interference pattern when it passes through a single pattern if one slit is covered with a red filter and the b. 0.327° slit? other slit is covered with a blue filter. c. 0.436° 13. How does the width of the central region of a 8. An interference pattern is formed by using green light single-slit diffraction pattern change as the 11. 160 µm and an apparatus in which the two slits can move. If wavelength of the light increases? 12. At certain angles the diffracted the slits are moved farther apart, will the separation of the bright fringes in the pattern decrease, increase, 14. Why is white light separated into a spectrum of colors waves from different parts of the slit or remain unchanged? Why? when it is passed through a diffraction grating? destructively interfere with each 15. Why might orbiting telescopes be problematic for the other, forming dark fringes like those radio portion of the electromagnetic spectrum? observed in an interference pattern. 13. It increases as the light with longer wavelength is diffracted more.

14. Light of a particular wavelength ©Photo Inc. Researchers, interferes constructively at a 540 Chapter 15 particular angle. 15. The larger an orbiting telescope is, the more expensive it is to place in Untitled-293 540 5/20/2011 7:53:17 AM orbit. In order to have adequate resolution, an orbiting radio tele- scope would have to be very large.

540 Chapter 15 CHAPTER REVIEW

CONCEPTUAL QUESTIONS 20. If light with a wavelength of 353 nm is passed through 16. The sin θ is inversely proportional the diffraction grating with 795 slits/cm, find the 16. Monochromatic light shines through two different angle at which one would observe the second-order to d, which is the reciprocal of the diffraction gratings. The second grating produces a maximum. pattern in which the first-order and second-order number of lines per centimeter. maxima are more widely spread apart. Use this 21. By attaching a diffraction-grating spectroscope to an The grating that spreads the information to tell if there are more or fewer lines per astronomical telescope, one can measure the spectral pattern the most has the most lines centimeter in the second grating than in the first. lines from a star and determine the star’s chemical composition. Assume the grating has 3661 lines/cm. per centimeter. 17. Why is the resolving power of your eye better at night a. If the wavelengths of the star’s light are 478.5 nm, 17. The pupil is larger at night, so the than during the day? 647.4 nm, and 696.4 nm, what are the angles at angle of resolution is smaller and 18. Globular clusters, such as the one shown below, are which the first-order spectral lines occur? spherical groupings of stars that form a ring around b. At what angles are these lines found in the resolving power is greater. second-order spectrum? the Milky Way galaxy. Because there can be millions 18. easier: a, c, d of stars in a single cluster and because they are distant, resolving individual stars within the cluster is harder: b a challenge. Of the following conditions, which would Lasers 19. 3.22° make it easier to resolve the component stars? Which REVIEWING MAIN IDEAS would make it more difficult? 20. 3.22° 22. What properties does laser light have that are not 21. a. 10.09°, 13.71°, 14.77° found in the light used to light your home? b. 20.51°, 28.30°, 30.66° 23. Laser light is commonly used to demonstrate double-slit interference. Explain why laser light is 22. Laser light is coherent and mono- preferable to light from other sources for observing chromatic. interference. 23. Light must be coherent for an 24. Give two examples in which the uniform direction of laser light is advantageous. Give two examples interference pattern to form. An in which the high intensity of laser light is interference pattern is well defined advantageous. with monochromatic light. 25. Laser light is often linearly polarized. How would you 24. Answers may include distance show that this statement is true? a. The number of stars per unit volume is half as measurements, compact disc players, great. and fiber-optic communications; b. The cluster is twice as far away. Mixed Review Answers may include laser surgery c. The cluster is observed in the ultraviolet REVIEWING MAIN IDEAS portion instead of in the visible region of the and fiber-optic communications. electromagnetic spectrum. 26. The 546.1 nm line in mercury is measured at an angle 25. by rotating a polarizing sheet in front d. The telescope’s mirror or lens is twice as wide. of 81.0° in the third-order spectrum of a diffraction grating. Calculate the number of lines per centimeter of the laser light—if the intensity of PRACTICE PROBLEMS for the grating. the transmitted light varies, the light

For problems 19–21, see Sample Problem B. 27. Recall from your study of heat and entropy that the is polarized entropy of a system is a measure of that system’s 26. 6030 lines/cm 19. Light with a wavelength of 707 nm is passed through disorder. Why is it appropriate to describe a laser as a diffraction grating with 795 slits/cm. Find the angle an entropy-reducing device? 27. Energy (light, for example) added to at which one would observe the first-order maximum. the active medium’s atoms is not highly ordered. This light travels in all directions, is incoherent, and is not

©Photo Researchers, ©Photo Inc. Researchers, monochromatic. The laser converts a

Chapter Review 541 portion of this energy into a beam of more coherent, monochromatic light.

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Interference and Diffraction 541 chapter review CHAPTER REVIEW

28. A double-slit interference experiment is performed 30. Visible light from an incandescent light bulb ranges 28. 2.41 × 10–4 m using blue light from a hydrogen discharge tube from 400.0 nm to 700.0 nm. When this light is focused 29. 432.0 nm (λ = 486 nm). The fifth-order bright fringe in the on a diffraction grating, the entire first-order spec- interference pattern is 0.578° from the central trum is seen, but none of the second-order spectrum 30. 8.000 × 10–7 m maxi mum. How far apart are the two slits separated? is seen. What is the maximum spacing between lines –3 on this grating? 31. 1.93 × 10 mm = 3 λ; 29. A beam containing light of wavelengths λ1 and λ2 a maximum passes through a set of parallel slits. In the interfer- 31. In an arrangement to demonstrate double-slit ence pattern, the fourth bright line of the λ1 light interference, λ = 643 nm, θ = 0.737°, and occurs at the same position as the fifth bright line of d = 0.150 mm. For light from the two slits interfering

the λ2 light. If λ1 is known to be 540.0 nm, what is the at this angle, what is the path difference both in value of λ2? millimeters and in terms of the number of wave- Alternative lengths? Will the interference correspond to a Assessment Answers maximum, a minimum, or an intermediate condition? 1. To represent constant wavelength, the distances between lines should not exceed the lines’ thickness. ALTERNATIVE ASSESSMENT 4. Thomas Young’s 1803 experiment provided crucial Students can create the same effects evidence for the wave nature of light, but it was met with strong opposition in England until Augustin by overlapping drawings on com- 1. Simulate interference patterns. Use a computer to draw concentric circles at regular distances to represent Fresnel presented his wave theory of light to the puter screens. waves traveling from a point source. Photocopy the French Academy of Sciences in 1819. Research the lives and careers of these two scientists. Create a 2. One hole should produce concentric page onto two transparencies, and lay them on an overhead projector. Vary the distances between “source presentation about one of them. The presentation can diffraction rings; two holes should points,” and observe how these variations affect be in the form of a report, poster, short video, or produce a striped pattern. Suggested interference patterns. Design transparencies with computer presentation. improvements will vary. thicker lines with larger separations to explore the effect 5. Research waves that surround you, including those of wavelength on interference. used in commercial, medicinal, and industrial applica- Students should use path difference 3. 2. Investigate the effect of slit separation on interference tions. Interpret how the waves’ characteristics and and refraction to predict wavelengths. patterns. Wrap a flashlight or a pen light tightly with behaviors make them useful. For example, investigate what kinds of waves are used in medical procedures 4. Presentations will vary but should tinfoil and make pinholes in the foil. First, record the pattern you see on a screen a few inches away with one such as MRI and ultrasound. What are their wave- indicate the source material used. hole; then, do the same with two holes. How does the lengths? Research how lasers are used in medicine. Young (1773–1829) was a child distance between the holes affect the distance between How are they used in industry? Prepare a poster or the bright parts of the pattern? Draw schematic chart describing your findings, and present it to prodigy. Fresnel (1788–1827) was eight diagrams of your observations, and compare them with the class. years old before he could read. the results of double-slit interference. How would you 5. Student presentations will vary. improve your equipment? Findings should be explained clearly. 3. Soap bubbles exhibit different colors because light that is reflected from the outer layer of the soap film interferes with light that is refracted and then re flected from the inner layer of the soap film. Given a refractive index of n = 1.35 and thicknesses ranging from 600 nm to 1000 nm for a soap film, can you predict the colors of a bubble? Test your answer by making soap bubbles and observing the order in which the colors appear. Can you tell the thickness of a soap bubble from its colors? Organize your findings into a chart, or create a computer program to predict the thicknesses of a bubble based on the wavelengths of light it reflects.

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542 Chapter 15 CHAPTER REVIEW CHAPTER REVIEW

Double-Slit Experiment

One of the classic experiments that demonstrate the wave In this equation, d is the slit separation, θ is the fringe nature of light is the double-slit experiment. In this experiment, angle, m is the order number, and λ is the wavelength light from a single source is passed through a narrow slit and of the incident wave. Typically, only the first few fringes then through two narrow parallel slits. When the light appears (m = 0, 1, 2, 3) are bright enough to see. on a viewing screen behind the slits, you see a pattern of In this graphing calculator activity, you will calculate a table of alternating bright and dark fringes corresponding to fringe angles. By analyzing this table, you will gain a better constructive and destructive interference of the light. understanding of the relationship between fringe angles, As you studied earlier in the chapter, the bright fringes are wavelength, and slit separation. described by the following equation. Go online to HMDScience.com to find this graphing calculator activity. d sin θ = ± mλ

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Interference and Diffraction 543 STANDARDS-BASED ASSESSMENT Standards-Based Assessment Answers 1. B MULTIPLE CHOICE 5. Monochromatic infrared waves with a wavelength of 750 nm pass through two narrow slits. If the slits 2. H 1. In the equations for interference, what does the are 25 µm apart, at what angle will the fourth-order 3. C term d represent? bright fringe appear on a viewing screen? A. the distance from the midpoint between the two A. 4.3° 4. G slits to the viewing screen B. 6.0° 5. C B. the distance between the two slits through which C. 6.9° 6. H a light wave passes D. 7.8° C. the distance between two bright interference 7. B fringes 6. Monochromatic light with a wavelength of 640 nm D. the distance between two dark interference passes through a diffraction grating that has 4 fringes 5.0 × 10 lines/m. A bright line on a screen appears at an angle of 11.1° from the central bright fringe. 2. Which of the following must be true for two waves What is the order of this bright line? with identical amplitudes and wavelengths to F. m = 2 undergo complete destructive interference? G. m = 4 F. The waves must be in phase at all times. H. m = 6 G. The waves must be 90° out of phase at all times. J. m = 8 H. The waves must be 180° out of phase at all times. J. The waves must be 270° out of phase at all times. 7. For observing the same object, how many times better is the resolution of the telescope shown on 3. Which equation correctly describes the condition the left in the figure below than that of the telescope for observing the third dark fringe in an interference shown on the right? pattern? A. 4 A. d sin θ = λ/2 B. 2 B. d sin θ = 3λ/2 C. _1 C. d sin θ = 5λ/2 2 D. d sin θ = 3λ D. _1 4 4. Why is the diffraction of sound easier to observe than the diffraction of visible light? F. Sound waves are easier to detect than visible light waves. G. Sound waves have longer wavelengths than visible light waves and so bend more around barriers. H. Sound waves are longitudinal waves, which diffract more than transverse waves. J. Sound waves have greater amplitude than visible light waves.

Area of mirror = 80 m2 Area of mirror = 20 m2

C15TEP001A

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544 Chapter 15 TEST PREP

8. F 8. What steps should you employ to design a telescope EXTENDED RESPONSE 9. D with a high degree of resolution? 10. H 14. Radio signals often reflect from objects and F. Widen the aperture, or design the telescope to 11. The beam does not spread out much detect light of short wavelength. recombine at a distance. Suppose you are moving G. Narrow the aperture, or design the telescope to in a direction perpendicular to a radio signal or lose intensity over long distances. detect light of short wavelength. source and its reflected signal. How would 12. 7.5 × 104 lines/m = 750 lines/cm H. Widen the aperture, or design the telescope to interference between these two signals sound on a radio receiver? 13. The resolving power of a telescope detect light of long wavelength. depends on the ratio of the J. Narrow the aperture, or design the telescope to detect light of long wavelength. wavelength to the diameter of the aperture. Telescopes using 9. What is the property of a laser called that causes longer wavelength radiation (visible coherent light to be emitted? Original radio signal A. different intensities light) must be larger than those B. light amplification using shorter wavelengths C. monochromaticity (ultraviolet, X ray) to achieve the D. stimulated emission same resolving power. 10. Which of the following is not an essential compo- Observer 14. The interference pattern for radio nent of a laser? Reflected radio signal signals would “appear” on a radio F. a partially transparent mirror receiver as an alternating increase G. a fully reflecting mirror in signal intensity followed by a H. a converging lens loss of intensity (heard as static or J. an active medium “white noise”). SHORT RESPONSE 15. 589 nm Base your answers to questions 15–17 on the information below. 16. 6.77° 11. Why is laser light useful for the purposes of making In each problem, show all of your work. astronomical measurements and surveying? 17. 7.90° C15TEP002A A double-slit apparatus for demonstrating interference 12. A diffraction grating used in a spectrometer causes is constructed so that the slits are separated by 15.0 µm. the third-order maximum of blue light with a A first-order fringe for constructive interference appears wavelength of 490 nm to form at an angle of 6.33° at an angle of 2.25° from the zeroth-order (central) from the central maximum (m = 0). What is the fringe. ruling of the grating in lines/cm? 15. What is the wavelength of the light? 13. Telescopes that orbit Earth provide better images of distant objects because orbiting telescopes are more 16. At what angle would the third-order (m = 3) bright able to operate near their theoretical resolution than fringe appear? telescopes on Earth. The orbiting telescopes needed to provide high resolution in the visible part of the 17. At what angle would the third-order (m = 3) dark spectrum are much larger than the orbiting tele- fringe appear? scopes that provide similar images in the ultraviolet and X-ray portion of the spectrum. Explain why the sizes must vary.

11 12 1 Test Tip 10 2 9 3 Be sure that angles in all calculations 8 4 involving trigonometric functions are 7 6 5 computed in the proper units (degrees or radians).

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Interference and Diffraction 545