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2. Resources Resources Chapter Presentation Visual Concepts Sample Problems Transparencies Standardized Test Prep

3. Chapter 15 Interference and Diffraction Table of Contents Section 1 Interference Section 2 Diffraction Section 3 Lasers

4. Section 1 Interference Chapter 15 Objectives • Describehow light waves interfere with each other to produce bright and dark fringes. • Identifythe conditions required for interference to occur. • Predictthe location of interference fringes using the equation for double-slit interference.

5. Section 1 Interference Chapter 16 Combining Light Waves • Interference takes place only between waves with the same wavelength. A light source that has a single wavelength is called monochromatic. • Inconstructive 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. • In the case ofdestructive interference,the resultant amplitude is less than the amplitude of the larger component wave.

6. Section 1 Interference Chapter 16 Interference Between Transverse Waves

7. Section 1 Interference Chapter 15 Combining Light Waves, continued • Waves must have a constant phase difference for interference to be observed. • Coherenceis the correlation between the phases of two or more waves. • Sources of light for which the phase difference is constant are said to be coherent. • Sources of light for which the phase difference is not constant are said to be incoherent.

8. Section 3 Lasers Chapter 15 Incoherent and Coherent Light

9. Section 1 Interference Chapter 15 Combining Light Waves

10. Section 1 Interference Chapter 15 Demonstrating Interference • Interference can be demonstrated by passing light through two narrow parallel slits. • If monochromatic light is used, the light from the two slits produces a series of bright and dark parallel bands, orfringes,on a viewing screen.

11. Section 1 Interference Chapter 15 Conditions for Interference of Light Waves

12. Section 1 Interference Chapter 15 Demonstrating Interference, continued • The location of interference fringes can be predicted. • Thepath differenceis the difference in the distance traveled by two beams when they are scattered in the same direction from different points. • The path difference equals dsinq.

13. Section 1 Interference Chapter 15 Interference Arising from Two Slits

14. Section 1 Interference Chapter 15 Demonstrating Interference, continued • 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. • The central bright fringe at q = 0 (m = 0) is called thezeroth-order maximum,or thecentral maximum. • The first maximum on either side of the central maximum (m = 1) is called thefirst-order maximum.

15. Section 1 Interference Chapter 15 Demonstrating Interference, continued • Equation for constructive interference d sin q = ±ml m = 0, 1, 2, 3, … The path difference between two waves = an integer multiple of the wavelength • Equation for destructive interference d sin q = ±(m + 1/2)l m = 0, 1, 2, 3, … The path difference between two waves = an odd number of half wavelength

16. Section 1 Interference Chapter 15 Sample Problem Interference 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.

17. Section 1 Interference Chapter 15 Sample Problem, continued Interference 1. Define Given:d = 3.0  10–5 m m = 2 q= 2.15º Unknown:l= ? Diagram:

18. Section 1 Interference Chapter 15 Sample Problem, continued Interference 2. Plan Choose an equation or situation: Use the equation for constructive interference. d sin q = ml Rearrange the equation to isolate the unknown:

19. Section 1 Interference Chapter 15 Sample Problem, continued Interference 3. Calculate Substitute the values into the equation and solve:

20. Section 1 Interference Chapter 15 Sample Problem, continued Interference 4. Evaluate This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellow-green color.

21. Section 2 Diffraction Chapter 15 Objectives • Describehow light waves bend around obstacles and produce bright and dark fringes. • Calculatethe positions of fringes for a diffraction grating. • Describehow diffraction determines an optical instrument’s ability to resolve images.

22. Section 2 Diffraction Chapter 15 The Bending of Light Waves • Diffraction isa change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge. • Light waves form adiffraction pattern(are formed) by passing around an obstacle or bending through a slit and interfering with each other. • Wavelets (as in Huygens’ principle) in a wave front interfere with each other.

23. Section 2 Diffraction Chapter 15 Destructive Interference in Single-Slit Diffraction

24. Section 2 Diffraction Chapter 15 The Bending of Light Waves, continued • In a diffraction pattern, the central maximum is twice as wide as the secondary maxima. • Light diffracted by an obstacle also produces a pattern.

25. Section 2 Diffraction Chapter 15 Diffraction Gratings • Adiffraction grating uses diffraction and interference to disperse light into its component colors. • 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, l. d sin q = ±ml m = 0, 1, 2, 3, …

26. Section 2 Diffraction Chapter 15 Constructive Interference by a Diffraction Grating

27. Section 2 Diffraction Chapter 15 Sample Problem Diffraction Gratings Monochromatic light from a helium-neon laser (l = 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.

28. Section 2 Diffraction Chapter 15 Sample Problem, continued Diffraction Gratings • Define Given:l= 632.8 nm = 6.328  10–7 m m = 1 and 2 Unknown:q1 = ? q2 = ?

29. Section 2 Diffraction Chapter 15 Sample Problem, continued Diffraction Gratings • Define, continued Diagram:

30. Section 2 Diffraction Chapter 15 Sample Problem, continued Diffraction Gratings 2. Plan Choose an equation or situation: Use the equation for a diffraction grating. d sin q = ±ml Rearrange the equation to isolate the unknown:

31. Section 2 Diffraction Chapter 15 Sample Problem, continued Diffraction Gratings 3. Calculate Substitute the values into the equation and solve: For the first-order maximum, m = 1:

32. Section 2 Diffraction Chapter 15 Sample Problem, continued Diffraction Gratings 3. Calculate, continued For m = 2:

33. Section 2 Diffraction Chapter 15 Sample Problem, continued Diffraction Gratings 4. Evaluate 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 q = 0.9524).

34. Section 2 Diffraction Chapter 15 Function of a Spectrometer

35. Section 2 Diffraction Chapter 15 Diffraction and Instrument Resolution • The ability of an optical system to distinguish between closely spaced objects is limited by the wave nature of light. • Resolving power is the ability of an optical instrument to form separate images of two objects that are close together. • Resolution depends on wavelength and aperture width. For a circular aperture of diameter D:

36. Section 2 Diffraction Chapter 15 Resolution of Two Light Sources

37. Section 3 Lasers Chapter 15 Objectives • Describethe properties of laser light. • Explainhow laser light has particular advantages in certain applications.

38. Section 3 Lasers Chapter 15 Lasers and Coherence • Alaseris a device that produces coherent light at a single wavelength. • The word laser is an acronym of “light amplification by stimulated emission of radiation.” • Lasers transform other forms of energy into coherent light.

39. Section 3 Lasers Chapter 15 Comparing Incoherent and Coherent Light

40. Section 3 Lasers Chapter 15 Laser

41. Section 3 Lasers Chapter 15 Applications of Lasers • Lasers are used tomeasure distanceswith great precision. • Compact disc and DVD players use lasers to read digital data on these discs. • Lasers have many applications in medicine. • Eye surgery • Tumor removal • Scar removal

42. Section 3 Lasers Chapter 15 Components of a Compact Disc Player

43. Chapter 15 Standardized Test Prep Multiple Choice 1. In the equations for interference, what does the term d represent? A. the distance from the midpoint between the two slits to the viewing screen B. the distance between the two slits through which a light wave passes C. the distance between two bright interference fringes D. the distance between two dark interference fringes

44. Chapter 15 Standardized Test Prep Multiple Choice, continued 1. In the equations for interference, what does the term d represent? A. the distance from the midpoint between the two slits to the viewing screen B. the distance between the two slits through which a light wave passes C. the distance between two bright interference fringes D. the distance between two dark interference fringes

45. Chapter 15 Standardized Test Prep Multiple Choice, continued 2. Which of the following must be true for two waves with identical amplitudes and wavelengths to undergo complete destructive interference? F. The waves must be in phase at all times. G. The waves must be 90º out of phase at all times. H. The waves must be 180º out of phase at all times. J. The waves must be 270º out of phase at all times.

46. Chapter 15 Standardized Test Prep Multiple Choice, continued 2. Which of the following must be true for two waves with identical amplitudes and wavelengths to undergo complete destructive interference? F. The waves must be in phase at all times. G. The waves must be 90º out of phase at all times. H. The waves must be 180º out of phase at all times. J. The waves must be 270º out of phase at all times.

47. Chapter 15 Standardized Test Prep Multiple Choice, continued 3. Which equation correctly describes the condition for observing the third dark fringe in an interference pattern? A.dsin q = l/2 B.dsin q = 3l/2 C.dsin q = 5l/2 D.dsin q = 3l

48. Chapter 15 Standardized Test Prep Multiple Choice, continued 3. Which equation correctly describes the condition for observing the third dark fringe in an interference pattern? A.dsin q = l/2 B. dsin q = 3l/2 C.dsin q = 5l/2 D.dsin q = 3l

49. Chapter 15 Standardized Test Prep Multiple Choice, continued 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.

50. Chapter 15 Standardized Test Prep Multiple Choice, continued 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.