(a) Explanation Write the expression for angle of diffraction minima. θ n = sin − 1 ( n λ d ) Here, θ n is the angle of n th minima, λ is the wavelength and d is the slit width. The wavelength is given by, λ = v f Here, v is the speed of sound and f is the frequency. Therefore, the angle of diffraction minima is, θ n = sin − 1 ( n c d f ) Conclusion: Substitute 1 for n , 343 m/s for v , 650 Hz for f and 1.10 m for d to get θ 1 . θ 1 = sin − 1 ( 343 m/s ( 1 (b) To determine The angular directions of diffraction minima.

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

9th Edition

ISBN: 9781305266292

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Chapter 38, Problem 3P

(a)

To determine

The number of minima.

(b)

To determine

The angular directions of diffraction minima.

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Chapter 38, Problem 2P

Chapter 38, Problem 4P

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A steady sound with a frequency of f = 750 Hz is produced by a source located far from an open doorway set in a sound- absorbing wall. The sound waves pass through the w = 1.18 m-wide doorway. (Assume the speed of sound is 343 m/s.) (a) If a person walks parallel to the wall beyond the open doorway, how many diffraction minima will she encounter? (b) What are the angular directions (in degrees) of these diffraction minima? (Enter the magnitudes from smallest to largest starting with the first answer blank. Enter NONE in any remaining answer blanks. Do not enter any duplicate numerical values.) smallest 土土 largest

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Chapter 38 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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Chapter 38 Solutions