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Chapter 15 Section 1 Interference Preview • Objectives • Combining Light Waves • Demonstrating Interference • Sample Problem © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Objectives • Describe how light waves interfere with each other to produce bright and dark fringes. • Identify the conditions required for interference to occur. • Predict the location of interference fringes using the equation for double-slit interference. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Combining Light Waves • Interference takes place only between waves with the same wavelength. A light source that has a single wavelength is called monochromatic. • 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. • In the case of destructive interference, the resultant amplitude is less than the amplitude of the larger component wave. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Interference Between Transverse Waves © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Combining Light Waves, continued • Waves must have a constant phase difference for interference to be observed. • Coherence is 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. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Combining Light Waves Click below to watch the Visual Concept. Visual Concept © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference 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, or fringes, on a viewing screen. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Conditions for Interference of Light Waves © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Demonstrating Interference, continued • The location of interference fringes can be predicted. • The path difference is the difference in the distance traveled by two beams when they are scattered in the same direction from different points. • The path difference equals dsin. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Interference Arising from Two Slits Click below to watch the Visual Concept. Visual Concept © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference 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 the zeroth-order maximum, or the central maximum. • The first maximum on either side of the central maximum (m = 1) is called the first-order maximum. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Demonstrating Interference, continued • Equation for constructive interference d sin = ±m m = 0, 1, 2, 3, … The path difference between two waves = an integer multiple of the wavelength • Equation for destructive interference d sin = ±(m + 1/2) m = 0, 1, 2, 3, … The path difference between two waves = an odd number of half wavelength © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference 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. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Sample Problem, continued Interference 1. Define Given: d = 3.0 10–5 m m=2 = 2.15º Unknown: =? Diagram: © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Sample Problem, continued Interference 2. Plan Choose an equation or situation: Use the equation for constructive interference. d sin = m Rearrange the equation to isolate the unknown: d sin m © Houghton Mifflin Harcourt Publishing Company Section 1 Interference Chapter 15 Sample Problem, continued Interference 3. Calculate Substitute the values into the equation and solve: 3.0 10 –5 m sin2.15º 2 5.6 10 –7 m 5.6 102 nm 5.6 102 nm © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 1 Interference Sample Problem, continued Interference 4. Evaluate This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellowgreen color. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Preview • Objectives • The Bending of Light Waves • Diffraction Gratings • Sample Problem • Diffraction and Instrument Resolution © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Objectives • Describe how light waves bend around obstacles and produce bright and dark fringes. • Calculate the positions of fringes for a diffraction grating. • Describe how diffraction determines an optical instrument’s ability to resolve images. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction The Bending of Light Waves • Diffraction is a change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge. • Light waves form a diffraction pattern 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. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Destructive Interference in Single-Slit Diffraction © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction 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. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Diffraction Gratings • A diffraction 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, . d sin = ±m m = 0, 1, 2, 3, … © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Constructive Interference by a Diffraction Grating © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Sample Problem Diffraction Gratings 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. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings 1. Define Given: = 632.8 nm = 6.328 10–7 m m = 1 and 2 1 1 d m lines 150 500 150 500 m Unknown: 1 = ? © Houghton Mifflin Harcourt Publishing Company 2 = ? Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings 1. Define, continued Diagram: © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings 2. Plan Choose an equation or situation: Use the equation for a diffraction grating. d sin = ±m Rearrange the equation to isolate the unknown: m sin d –1 © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings 3. Calculate Substitute the values into the equation and solve: For the first-order maximum, m = 1: –7 6.328 10 m –1 –1 1 sin sin 1 d m 150 500 1 5.465º © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Sample Problem, continued Diffraction Gratings 3. Calculate, continued For m = 2: –1 2 2 sin d –7 2 6.328 10 m –1 2 sin 1 m 150 500 2 10.98º © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction 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 = 0.9524). © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction Function of a Spectrometer Click below to watch the Visual Concept. Visual Concept © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 2 Diffraction 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: 1.22 © Houghton Mifflin Harcourt Publishing Company D Chapter 15 Section 2 Diffraction Resolution of Two Light Sources © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Preview • Objectives • Lasers and Coherence • Applications of Lasers © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Objectives • Describe the properties of laser light. • Explain how laser light has particular advantages in certain applications. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Lasers and Coherence • A laser is 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. © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Comparing Incoherent and Coherent Light Click below to watch the Visual Concept. Visual Concept © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Laser Click below to watch the Visual Concept. Visual Concept © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Applications of Lasers • Lasers are used to measure distances with 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 © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Components of a Compact Disc Player © Houghton Mifflin Harcourt Publishing Company Chapter 15 Section 3 Lasers Incoherent and Coherent Light © Houghton Mifflin Harcourt Publishing Company