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Hp 15 win

1. 1. How to Use This Presentation <ul><li>To View the presentation as a slideshow with effects </li></ul><ul><li>select “View” on the menu bar and click on “Slide Show.” </li></ul><ul><li>To advance through the presentation, click the right-arrow key or the space bar. </li></ul><ul><li>From the resources slide, click on any resource to see a presentation for that resource. </li></ul><ul><li>From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. </li></ul><ul><li>You may exit the slide show at any time by pressing the Esc key. </li></ul>
2. 2. Chapter Presentation Transparencies Sample Problems Visual Concepts Standardized Test Prep Resources Resources
3. 3. Table of Contents <ul><li>Section 1 Interference </li></ul><ul><li>Section 2 Diffraction </li></ul><ul><li>Section 3 Lasers </li></ul>Interference and Diffraction Chapter 15
4. 4. Objectives <ul><li>Describe how light waves interfere with each other to produce bright and dark fringes. </li></ul><ul><li>Identify the conditions required for interference to occur. </li></ul><ul><li>Predict the location of interference fringes using the equation for double-slit interference. </li></ul>Section 1 Interference Chapter 15
5. 5. Combining Light Waves <ul><li>Interference takes place only between waves with the same wavelength. A light source that has a single wavelength is called monochromatic . </li></ul><ul><li>In constructive interference, component waves combine to form a resultant wave with the same wavelength but with an amplitude that is greater than the either of the individual component waves. </li></ul><ul><li>In the case of destructive interference, the resultant amplitude is less than the amplitude of the larger component wave. </li></ul>Section 1 Interference Chapter 15
6. 6. Interference Between Transverse Waves Chapter 15 Section 1 Interference
7. 7. Combining Light Waves, continued <ul><li>Waves must have a constant phase difference for interference to be observed. </li></ul><ul><li>Coherence is the correlation between the phases of two or more waves. </li></ul><ul><ul><li>Sources of light for which the phase difference is constant are said to be coherent. </li></ul></ul><ul><ul><li>Sources of light for which the phase difference is not constant are said to be incoherent. </li></ul></ul>Section 1 Interference Chapter 15
8. 8. Incoherent and Coherent Light Chapter 15 Section 3 Lasers
9. 9. Combining Light Waves Chapter 15 Section 1 Interference
10. 10. Demonstrating Interference <ul><li>Interference can be demonstrated by passing light through two narrow parallel slits. </li></ul><ul><li>If monochromatic light is used, the light from the two slits produces a series of bright and dark parallel bands, or fringes, on a viewing screen. </li></ul>Section 1 Interference Chapter 15
11. 11. Conditions for Interference of Light Waves Chapter 15 Section 1 Interference
12. 12. Demonstrating Interference, continued <ul><li>The location of interference fringes can be predicted. </li></ul><ul><li>The path difference is the difference in the distance traveled by two beams when they are scattered in the same direction from different points. </li></ul><ul><li>The path difference equals d sin  . </li></ul>Section 1 Interference Chapter 15
13. 13. Interference Arising from Two Slits Chapter 15 Section 1 Interference
14. 14. Demonstrating Interference, continued <ul><li>The number assigned to interference fringes with respect to the central bright fringe is called the order number. The order number is represented by the symbol m . </li></ul><ul><li>The central bright fringe at q = 0 ( m = 0) is called the zeroth-order maximum, or the central maximum . </li></ul><ul><li>The first maximum on either side of the central maximum ( m = 1) is called the first-order maximum. </li></ul>Section 1 Interference Chapter 15
15. 15. Demonstrating Interference, continued <ul><li>Equation for constructive interference </li></ul><ul><ul><li>d sin  = ± m  m = 0, 1, 2, 3, … </li></ul></ul><ul><ul><li>The path difference between two waves = </li></ul></ul><ul><ul><li>an integer multiple of the wavelength </li></ul></ul><ul><li>Equation for destructive interference </li></ul><ul><ul><li>d sin  = ± (m + 1/2 )  m = 0, 1, 2, 3, … </li></ul></ul><ul><ul><li>The path difference between two waves = </li></ul></ul><ul><ul><li>an odd number of half wavelength </li></ul></ul>Section 1 Interference Chapter 15
16. 16. Sample Problem <ul><li>Interference </li></ul><ul><li>The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is measured on a viewing screen at an angle of 2.15º from the central maximum. Determine the wavelength of the light. </li></ul>Section 1 Interference Chapter 15
17. 17. Sample Problem, continued <ul><li>Interference </li></ul><ul><li>1. Define </li></ul><ul><li>Given: d = 3.0  10 –5 m </li></ul><ul><ul><li>m = 2 </li></ul></ul><ul><ul><li>  = 2.15º </li></ul></ul><ul><li>Unknown:   = ? </li></ul><ul><li>Diagram: </li></ul>Section 1 Interference Chapter 15
18. 18. Sample Problem, continued <ul><li>Interference </li></ul><ul><li>2. Plan </li></ul><ul><li>Choose an equation or situation: Use the equation for constructive interference. </li></ul><ul><ul><li>d sin  = m  </li></ul></ul><ul><li>Rearrange the equation to isolate the unknown: </li></ul>Section 1 Interference Chapter 15
19. 19. Sample Problem, continued <ul><li>Interference </li></ul><ul><li>3. Calculate </li></ul><ul><li>Substitute the values into the equation and solve: </li></ul>Section 1 Interference Chapter 15
20. 20. Sample Problem, continued <ul><li>Interference </li></ul><ul><li>4. Evaluate </li></ul><ul><li>This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellow-green color. </li></ul>Section 1 Interference Chapter 15
21. 21. Objectives <ul><li>Describe how light waves bend around obstacles and produce bright and dark fringes. </li></ul><ul><li>Calculate the positions of fringes for a diffraction grating. </li></ul><ul><li>Describe how diffraction determines an optical instrument’s ability to resolve images. </li></ul>Section 2 Diffraction Chapter 15
22. 22. The Bending of Light Waves <ul><li>Diffraction is a change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge. </li></ul><ul><li>Light waves form a diffraction pattern by passing around an obstacle or bending through a slit and interfering with each other. </li></ul><ul><li>Wavelets (as in Huygens’ principle) in a wave front interfere with each other. </li></ul>Section 2 Diffraction Chapter 15
23. 23. Destructive Interference in Single-Slit Diffraction Chapter 15 Section 2 Diffraction
24. 24. The Bending of Light Waves, continued <ul><li>In a diffraction pattern, the central maximum is twice as wide as the secondary maxima. </li></ul><ul><li>Light diffracted by an obstacle also produces a pattern. </li></ul>Section 2 Diffraction Chapter 15
25. 25. Diffraction Gratings <ul><li>A diffraction grating uses diffraction and interference to disperse light into its component colors. </li></ul><ul><li>The position of a maximum depends on the separation of the slits in the grating, d, the order of the maximum m ,, and the wavelength of the light,  . </li></ul><ul><ul><li>d sin  = ± m  m = 0, 1, 2, 3, … </li></ul></ul>Section 2 Diffraction Chapter 15
26. 26. Constructive Interference by a Diffraction Grating Chapter 15 Section 2 Diffraction
27. 27. Sample Problem <ul><li>Diffraction Gratings </li></ul><ul><li>Monochromatic light from a helium-neon laser (  = 632.8 nm) shines at a right angle to the surface of a diffraction grating that contains 150 500 lines/m. Find the angles at which one would observe the first-order and second-order maxima. </li></ul>Section 2 Diffraction Chapter 15
28. 28. Sample Problem, continued <ul><li>Diffraction Gratings </li></ul><ul><li>Define </li></ul><ul><li>Given:   = 632.8 nm = 6.328  10 –7 m </li></ul><ul><ul><li>m = 1 and 2 </li></ul></ul>Section 2 Diffraction Chapter 15 Unknown:   = ?  2 = ?
29. 29. Sample Problem, continued <ul><li>Diffraction Gratings </li></ul><ul><li>Define, continued </li></ul><ul><li>Diagram: </li></ul>Section 2 Diffraction Chapter 15
30. 30. Sample Problem, continued <ul><li>Diffraction Gratings </li></ul><ul><li>2. Plan </li></ul><ul><li>Choose an equation or situation: Use the equation for a diffraction grating. </li></ul><ul><ul><li>d sin  = ± m  </li></ul></ul><ul><li>Rearrange the equation to isolate the unknown: </li></ul>Section 2 Diffraction Chapter 15
31. 31. Sample Problem, continued <ul><li>Diffraction Gratings </li></ul><ul><li>3. Calculate </li></ul><ul><li>Substitute the values into the equation and solve: </li></ul><ul><ul><li>For the first-order maximum, m = 1: </li></ul></ul>Section 2 Diffraction Chapter 15
32. 32. Sample Problem, continued <ul><li>Diffraction Gratings </li></ul><ul><li>3. Calculate, continued </li></ul><ul><li>For m = 2: </li></ul>Section 2 Diffraction Chapter 15
33. 33. Sample Problem, continued <ul><li>Diffraction Gratings </li></ul><ul><li>4. Evaluate </li></ul><ul><li>The second-order maximum is spread slightly more than twice as far 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  = 0.9524). </li></ul>Section 2 Diffraction Chapter 15
34. 34. Function of a Spectrometer Chapter 15 Section 2 Diffraction
35. 35. Diffraction and Instrument Resolution <ul><li>The ability of an optical system to distinguish between closely spaced objects is limited by the wave nature of light. </li></ul><ul><li>Resolving power is the ability of an optical instrument to form separate images of two objects that are close together. </li></ul><ul><li>Resolution depends on wavelength and aperture width. For a circular aperture of diameter D : </li></ul>Section 2 Diffraction Chapter 15
36. 36. Resolution of Two Light Sources Chapter 15 Section 2 Diffraction
37. 37. Objectives <ul><li>Describe the properties of laser light. </li></ul><ul><li>Explain how laser light has particular advantages in certain applications. </li></ul>Section 3 Lasers Chapter 15
38. 38. Lasers and Coherence <ul><li>A laser is a device that produces coherent light at a single wavelength. </li></ul><ul><li>The word laser is an acronym of “ l ight a mplification by s timulated e mission of r adiation.” </li></ul><ul><li>Lasers transform other forms of energy into coherent light. </li></ul>Section 3 Lasers Chapter 15
39. 39. Comparing Incoherent and Coherent Light Chapter 15 Section 3 Lasers
40. 40. Laser Chapter 15 Section 3 Lasers
41. 41. Applications of Lasers <ul><li>Lasers are used to measure distances with great precision. </li></ul><ul><li>Compact disc and DVD players use lasers to read digital data on these discs. </li></ul><ul><li>Lasers have many applications in medicine. </li></ul><ul><ul><li>Eye surgery </li></ul></ul><ul><ul><li>Tumor removal </li></ul></ul><ul><ul><li>Scar removal </li></ul></ul>Section 3 Lasers Chapter 15
42. 42. Components of a Compact Disc Player Chapter 15 Section 3 Lasers
43. 43. Multiple Choice <ul><li>1. In the equations for interference, what does the term d represent? </li></ul><ul><li>A. the distance from the midpoint between the two slits to the viewing screen </li></ul><ul><li>B. the distance between the two slits through which a light wave passes </li></ul><ul><li>C. the distance between two bright interference fringes </li></ul><ul><li>D. the distance between two dark interference fringes </li></ul>Standardized Test Prep Chapter 15
44. 44. Multiple Choice, continued <ul><li>1. In the equations for interference, what does the term d represent? </li></ul><ul><li>A. the distance from the midpoint between the two slits to the viewing screen </li></ul><ul><li>B. the distance between the two slits through which a light wave passes </li></ul><ul><li>C. the distance between two bright interference fringes </li></ul><ul><li>D. the distance between two dark interference fringes </li></ul>Standardized Test Prep Chapter 15
45. 45. Multiple Choice, continued <ul><li>2. Which of the following must be true for two waves with identical amplitudes and wavelengths to undergo complete destructive interference? </li></ul><ul><li>F. The waves must be in phase at all times. </li></ul><ul><li>G. The waves must be 90º out of phase at all times. </li></ul><ul><li>H. The waves must be 180º out of phase at all times. </li></ul><ul><li>J. The waves must be 270º out of phase at all times. </li></ul>Standardized Test Prep Chapter 15
46. 46. Multiple Choice, continued <ul><li>2. Which of the following must be true for two waves with identical amplitudes and wavelengths to undergo complete destructive interference? </li></ul><ul><li>F. The waves must be in phase at all times. </li></ul><ul><li>G. The waves must be 90º out of phase at all times. </li></ul><ul><li>H. The waves must be 180º out of phase at all times. </li></ul><ul><li>J. The waves must be 270º out of phase at all times. </li></ul>Standardized Test Prep Chapter 15
47. 47. Multiple Choice, continued <ul><li>3. Which equation correctly describes the condition for observing the third dark fringe in an interference pattern? </li></ul><ul><li>A. d sin  =  /2 </li></ul><ul><li>B. d sin  = 3  /2 </li></ul><ul><li>C. d sin  = 5  /2 </li></ul><ul><li>D. d sin  = 3  </li></ul>Standardized Test Prep Chapter 15
48. 48. Multiple Choice, continued <ul><li>3. Which equation correctly describes the condition for observing the third dark fringe in an interference pattern? </li></ul><ul><li>A. d sin  =  /2 </li></ul><ul><li>B. d sin  = 3  /2 </li></ul><ul><li>C. d sin  = 5  /2 </li></ul><ul><li>D. d sin  = 3  </li></ul>Standardized Test Prep Chapter 15
49. 49. Multiple Choice, continued <ul><li>4. Why is the diffraction of sound easier to observe than the diffraction of visible light? </li></ul><ul><li>F. Sound waves are easier to detect than visible light waves. </li></ul><ul><li>G. Sound waves have longer wavelengths than visible light waves and so bend more around barriers. </li></ul><ul><li>H. Sound waves are longitudinal waves, which diffract more than transverse waves. </li></ul><ul><li>J. Sound waves have greater amplitude than visible light waves. </li></ul>Standardized Test Prep Chapter 15
50. 50. Multiple Choice, continued <ul><li>4. Why is the diffraction of sound easier to observe than the diffraction of visible light? </li></ul><ul><li>F. Sound waves are easier to detect than visible light waves. </li></ul><ul><li>G. Sound waves have longer wavelengths than visible light waves and so bend more around barriers. </li></ul><ul><li>H. Sound waves are longitudinal waves, which diffract more than transverse waves. </li></ul><ul><li>J. Sound waves have greater amplitude than visible light waves. </li></ul>Standardized Test Prep Chapter 15
51. 51. Multiple Choice, continued <ul><li>5. Monochromatic infrared waves with a wavelength of 750 nm pass through two narrow slits. If the slits are 25 µm apart, at what angle will the fourth order bright fringe appear on a viewing screen? </li></ul><ul><li>A. 4.3º </li></ul><ul><li>B. 6.0º </li></ul><ul><li>C. 6.9º </li></ul><ul><li>D. 7.8º </li></ul>Standardized Test Prep Chapter 15
52. 52. Multiple Choice, continued <ul><li>5. Monochromatic infrared waves with a wavelength of 750 nm pass through two narrow slits. If the slits are 25 µm apart, at what angle will the fourth order bright fringe appear on a viewing screen? </li></ul><ul><li>A. 4.3º </li></ul><ul><li>B. 6.0º </li></ul><ul><li>C. 6.9º </li></ul><ul><li>D. 7.8º </li></ul>Standardized Test Prep Chapter 15
53. 53. Multiple Choice, continued <ul><li>6. Monochromatic light with a wavelength of 640 nm passes through a diffraction grating that has 5.0  10 4 lines/m. A bright line on a screen appears at an angle of 11.1º from the central bright fringe.What is the order of this bright line? </li></ul><ul><li>F. m = 2 </li></ul><ul><li>G. m = 4 </li></ul><ul><li>H. m = 6 </li></ul><ul><li>J. m = 8 </li></ul>Standardized Test Prep Chapter 15
54. 54. Multiple Choice, continued <ul><li>6. Monochromatic light with a wavelength of 640 nm passes through a diffraction grating that has 5.0  10 4 lines/m. A bright line on a screen appears at an angle of 11.1º from the central bright fringe.What is the order of this bright line? </li></ul><ul><li>F. m = 2 </li></ul><ul><li>G. m = 4 </li></ul><ul><li>H. m = 6 </li></ul><ul><li>J. m = 8 </li></ul>Standardized Test Prep Chapter 15
55. 55. Multiple Choice, continued <ul><li>7. For observing the same object, how many times better is the resolution of the telescope shown on the left in the figure below than that of the telescope shown on the right? </li></ul><ul><li>A. 4 </li></ul><ul><li>B. 2 </li></ul><ul><li>C. 1/2 </li></ul><ul><li>D. 1/4 </li></ul>Standardized Test Prep Chapter 15
56. 56. Multiple Choice, continued <ul><li>7. For observing the same object, how many times better is the resolution of the telescope shown on the left in the figure below than that of the telescope shown on the right? </li></ul><ul><li>A. 4 </li></ul><ul><li>B. 2 </li></ul><ul><li>C. 1/2 </li></ul><ul><li>D. 1/4 </li></ul>Standardized Test Prep Chapter 15
57. 57. Multiple Choice, continued <ul><li>8. What steps should you employ to design a telescope with a high degree of resolution? </li></ul><ul><li>F. Widen the aperture, or design the telescope to detect light of short wavelength. </li></ul><ul><li>G. Narrow the aperture, or design the telescope to detect light of short wavelength. </li></ul><ul><li>H. Widen the aperture, or design the telescope to detect light of long wavelength. </li></ul><ul><li>J. Narrow the aperture, or design the telescope to detect light of long wavelength. </li></ul>Standardized Test Prep Chapter 15
58. 58. Multiple Choice, continued <ul><li>8. What steps should you employ to design a telescope with a high degree of resolution? </li></ul><ul><li>F. Widen the aperture, or design the telescope to detect light of short wavelength. </li></ul><ul><li>G. Narrow the aperture, or design the telescope to detect light of short wavelength. </li></ul><ul><li>H. Widen the aperture, or design the telescope to detect light of long wavelength. </li></ul><ul><li>J. Narrow the aperture, or design the telescope to detect light of long wavelength. </li></ul>Standardized Test Prep Chapter 15
59. 59. Multiple Choice, continued <ul><li>9. What is the property of a laser called that causes coherent light to be emitted? </li></ul><ul><li>A. population inversion </li></ul><ul><li>B. light amplification </li></ul><ul><li>C. monochromaticity </li></ul><ul><li>D. stimulated emission </li></ul>Standardized Test Prep Chapter 15
60. 60. Multiple Choice, continued <ul><li>9. What is the property of a laser called that causes coherent light to be emitted? </li></ul><ul><li>A. population inversion </li></ul><ul><li>B. light amplification </li></ul><ul><li>C. monochromaticity </li></ul><ul><li>D. stimulated emission </li></ul>Standardized Test Prep Chapter 15
61. 61. Multiple Choice, continued <ul><li>10. Which of the following is not an essential component of a laser? </li></ul><ul><li>F. a partially transparent mirror </li></ul><ul><li>G. a fully reflecting mirror </li></ul><ul><li>H. a converging lens </li></ul><ul><li>J. an active medium </li></ul>Standardized Test Prep Chapter 15
62. 62. Multiple Choice, continued <ul><li>10. Which of the following is not an essential component of a laser? </li></ul><ul><li>F. a partially transparent mirror </li></ul><ul><li>G. a fully reflecting mirror </li></ul><ul><li>H. a converging lens </li></ul><ul><li>J. an active medium </li></ul>Standardized Test Prep Chapter 15
63. 63. Short Response <ul><li>11. Why is laser light useful for the purposes of making astronomical measurements and surveying? </li></ul>Standardized Test Prep Chapter 15
64. 64. Short Response, continued <ul><li>11. Why is laser light useful for the purposes of making astronomical measurements and surveying? </li></ul><ul><li>Answer: The beam does not spread out much or lose intensity over long distances. </li></ul>Standardized Test Prep Chapter 15
65. 65. Short Response, continued <ul><li>12. A diffraction grating used in a spectrometer causes the third-order maximum of blue light with a wavelength of 490 nm to form at an angle of 6.33º from the central maximum ( m = 0). What is the separation between the lines of the grating? </li></ul>Standardized Test Prep Chapter 15
66. 66. Short Response, continued <ul><li>12. A diffraction grating used in a spectrometer causes the third-order maximum of blue light with a wavelength of 490 nm to form at an angle of 6.33º from the central maximum ( m = 0). What is the separation between the lines of the grating? </li></ul><ul><li>Answer: 7.5  10 4 lines/m = 750 lines/cm </li></ul>Standardized Test Prep Chapter 15
67. 67. Short Response, continued <ul><li>13. Telescopes that orbit Earth provide better images of distant objects because orbiting telescopes are more able to operate near their theoretical resolution than telescopes on Earth. The orbiting telescopes needed to provide high resolution in the visible part of the spectrum are much larger than the orbiting telescopes that provide similar images in the ultraviolet and X-ray portion of the spectrum. Explain why the sizes must vary. </li></ul>Standardized Test Prep Chapter 15
68. 68. Short Response, continued <ul><li>13. (See previous slide for question.) </li></ul><ul><li>Answer: The resolving power of a telescope depends on the ratio of the wavelength to the diameter of the aperture. Telescopes using longer wavelength radiation (visible light) must be larger than those using shorter wavelengths (ultraviolet, X ray) to achieve the same resolving power. </li></ul>Standardized Test Prep Chapter 15
69. 69. Extended Response <ul><li>14. Radio signals often reflect from objects and recombine at a distance. Suppose you are moving in a direction perpendicular to a radio signal source and its reflected signal. How would interference between these two signals sound on a radio receiver? </li></ul>Standardized Test Prep Chapter 15
70. 70. Extended Response, continued <ul><li>14. (See previous slide for question.) </li></ul><ul><li>Answer: The interference pattern for radio signals would “appear” on a radio receiver as an alternating increase in signal intensity followed by a loss of intensity (heard as static or “white noise”). </li></ul>Standardized Test Prep Chapter 15
71. 71. Extended Response, continued <ul><li>Base your answers to questions 15–17 on the information below. In each problem, show all of your work. </li></ul><ul><li>A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe. </li></ul><ul><li>15. What is the wave-length of the light? </li></ul>Standardized Test Prep Chapter 15
72. 72. Extended Response, continued <ul><li>Base your answers to questions 15–17 on the information below. In each problem, show all of your work. </li></ul><ul><li>A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe. </li></ul><ul><li>15. What is the wave-length of the light? </li></ul><ul><li>Answer: 589 nm </li></ul>Standardized Test Prep Chapter 15
73. 73. Extended Response, continued <ul><li>Base your answers to questions 15–17 on the information below. In each problem, show all of your work. </li></ul><ul><li>A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe. </li></ul><ul><li>16. At what angle would the third-order ( m = 3) bright fringe appear? </li></ul>Standardized Test Prep Chapter 15
74. 74. Extended Response, continued <ul><li>Base your answers to questions 15–17 on the information below. In each problem, show all of your work. </li></ul><ul><li>A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe. </li></ul><ul><li>16. At what angle would the third-order ( m = 3) bright fringe appear? </li></ul><ul><li>Answer: 6.77º </li></ul>Standardized Test Prep Chapter 15
75. 75. Extended Response, continued <ul><li>Base your answers to questions 15–17 on the information below. In each problem, show all of your work. </li></ul><ul><li>A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe. </li></ul><ul><li>17. At what angle would the third-order ( m = 3) dark fringe appear? </li></ul>Standardized Test Prep Chapter 15
76. 76. Extended Response, continued <ul><li>Base your answers to questions 15–17 on the information below. In each problem, show all of your work. </li></ul><ul><li>A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe. </li></ul><ul><li>17. At what angle would the third-order ( m = 3) dark fringe appear? </li></ul><ul><li>Answer: 7.90º </li></ul>Standardized Test Prep Chapter 15