Chapter 28 - Chapter 28 5. Picture the Problem: The figure...

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Chapter 28 5. Picture the Problem : The figure shows two students listening to two coherent sound sources. Both students hear constructive interference. Strategy: Since student A is equidistant from both speakers the difference in path lengths from each speaker is zero. For student B to also hear constructive interference (and for the wavelength to be a maximum) the path length difference between him and the two speakers must be equal to one wavelength. Calculate the distance of student B from each speaker. Set the wavelength equal to the difference in distances. Divide the speed of sound by the wavelength to calculate the frequency. Solution: 1. Calculate the distance to the far speaker: 2. Subtract the distances: 3. Divide the speed of sound by the wavelength: Insight: The next higher frequency at which constructive interference occurs is This frequency corresponds to a path length difference of 2 λ for the two waves. 14. Picture the Problem : The first dark fringe in Young’s two-slit experiment occurs at θ = 0.29 ° as shown in the figure. Strategy: Solve equation 28-2 for the wavelength with m = 1 for the ratio of the slit separation ( d ) to the wavelength ( λ ). Solution: Calculate the ratio of slit spacing to wavelength: Insight: If this experiment had been done with red light ( λ = 656 nm), the slits would be 65 μ m apart. 17. Picture the Problem : The figure shows interference pattern for light with wavelength 632.8 nm on a screen that is 1.40 m from the double slits. The separation distance between the minima m and ( m + 4) is 23.0 mm. Strategy: Since 23.0 mm is much smaller than the 1.40 m distance to the screen, use the small angle approximations . Use equation 28-2 to calculate the angles for the minima of order m and ( m + 4). Use equation 28-3 to calculate the location of the minima. Set the difference equal to the 23.0
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Chapter 28 - Chapter 28 5. Picture the Problem: The figure...

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