Physics Solution Manual for 1100 and 2101

This half wavelength must be combined with the extra

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Unformatted text preview: 7.7, sin θ = mλ / d , the angle for the m = 1 bright fringe of red light is greater than that for yellow light, since red has a longer wavelength. 19. (b) As the number of lines per centimeter increases, the separation d between adjacent slits becomes smaller. The angle θ that specifies a principal maxima of a diffraction grating is given by sin θ = mλ / d , where λ is the wavelength and m = 0, 1, 2, 3, … (Equation 27.7). For fixed values of m and the wavelength, this relation shows that as d decreases, θ increases. For a fixed screen location, this means that the distance between two adjacent principal maxima also increases. 20. θ = 9.4º Chapter 27 Problems 1431 CHAPTER 27 INTERFERENCE AND THE WAVE NATURE OF LIGHT PROBLEMS 1. SSM REASONING AND SOLUTION To decide whether constructive or destructive interference occurs, we need to determine the wavelength of the wave. For electromagnetic waves, Equation 16.1 can be written f λ = c , so that c 3.00 × 10 8 m λ= = = 5.60 × 10 2 m f 536 × 10 3 Hz Since the two wave sources are in phase, constructive interference occurs when the path difference is an integer number of wavelengths, and destructive interference occurs when the path difference is an odd number of half wavelengths. We find that the path difference is 8.12 km – 7.00 km = 1.12 km. The number of wavelengths in this path difference is (1.12 × 103 m ) / (5.60 × 102 m ) = 2.00 . Therefore, 2. constructive interference occurs . REASONING In order for the two rays to interfere constructively and thereby form a bright interference fringe, the difference ∆l between their path lengths must be an integral multiple m of the wavelength λ of the light: ∆l = mλ (1) In Equation (1), m can take on any integral value (m = 0, 1, 2, 3, …). In this case, the rays meet at the eight-order bright fringe, so we have that m = 8. SOLUTION Solving Equation (1) for λ, and substituting m = 8, we obtain λ= ∆l 4.57 × 10−6 m = = 5.71× 10−7 m m 8 Using the equivalence 1 nm = 10−9 m, we convert this result to nanometers: ( ) 101−nmm = 571 nm 9 λ = 5.71× 10−7 m 1432 INTERFERENCE AND THE WAVE NATURE OF LIGHT 3. REASONING Let l1 and l2 be the distances from source 1 and source 2, respectively. For constructive interference the condition is l2 – l1 = mλ, where λ is the wavelength and m = 0, 1 1, 2, 3, …. For destructive interference the condition is l2 – l1 = (m + 2 )λ, where m = 0, 1, 2, 3, …. The fact that the wavelength is greater than the separation between the sources will allow us to decide which type of interference we have and to locate the two places in question. SOLUTION a. Since the places we seek lie on the line between the two sources and since the separation between the sources is 4.00 m, we know that l2 – l1 cannot have a value greater than 4.00 m. Therefore, with a wavelength of 5.00 m, the only way constructive interference can occur is at a place where m = 0. But this would mean that l2 = l1, and there is only one place where this is true, namely, at the midpoint between the sources. But we know that there are two places. As a result, we conclude that destructive interference is occurring . b. For destructive interference, the possible values for the integer m are m = 0, 1, 2, 3, …, and we must decide which to use. The condition for destructive interference is 1 l2 – l1 = (m + 2 )λ. Since λ = 5.00 m, all values of m greater than or equal to one result in l2 – l1 being greater than 4.00 m, which we already know is not possible. Therefore, we conclude that m = 0, and the condition for destructive interference becomes l2 – l1 = 1 2 λ. We also know that l2 + l1 = 4.00 m, since the separation between the sources is 4.00 m. In other words, we have 1 l 2 − l1 = 2 λ = 1 2 5 = b.00 m g 2.50 m and l 2 + l 1 = 4 .00 m Adding these two equations gives 2l2 = 6.50 m or l2 = 3.25 m Since l2 + l1 = 4.00m, it follows that l1 = 4.00 m – l2 = 4.00 m – 3.25 m = 0.75 m These results indicate that the two places are 3.25 m and 0.75 m from one of the sources. 4. REASONING The loudspeakers are in-phase sources of identical sound waves. The waves from one speaker travel a distance l1 in reaching point A and the waves from the second speaker travel a distance l2. The condition that leads to constructive interference is l2 – l1 = mλ, where λ is the wavelength of the waves and m = 0, 1, 2, 3 …. In other words, Chapter 27 Problems 1433 the two distances are the same or differ by an integer number of wavelengths. Point A is the midpoint of a side of the square, so that the distances l1 and l2 are the same, and constructive interference occurs. As you walk toward the corner, the waves from one of the sources travels a greater distance in reaching you than does the other wave. This difference in the distances increases until it reaches one half of a wavelength, at which spot destructive interference occurs and you hear no sound. As you walk on, the difference in distances continues to increase, and you gradually hear a louder and louder sound. Ultimately, at the corner, the difference in distances becomes one wavelength, constructive interference occurs, and you hear a maximally loud sound. The general condition that leads to constructive interference is l 2 − l 1 = m λ , where m = 0, 1, 2, 3 … There are many...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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