![]() ![]() It is a phenomenon that occurs when sound waves encounter an obstacle or an opening that is comparable in size to the wavelength of the sound wave. This process causes the sound waves to change direction and spread out, resulting in the sound being heard in areas that would otherwise be in the shadow of the obstacle. Sound diffraction is a fundamental concept in the field of physical acoustics, which studies the properties and behavior of sound waves. It is governed by the principles of wave propagation and interference. One of the key principles that explains sound diffraction is the Huygens-Fresnel principle, which states that every point on a wavefront can be considered as a source of secondary wavelets that spread out in all directions. Diffraction of Sound Waves Definitionĭiffraction of sound waves can be understood by considering the interaction between the wavefront and the obstacle or opening. When a sound wave encounters an obstacle, such as a wall or a solid object, it causes the wavefront to bend around the obstacle. This bending of the wavefront is due to the wavefront interacting with the edges of the obstacle, causing changes in the phase of the wave. The diffraction of sound waves is influenced by several factors, including the frequency and wavelength of the sound wave, the size and shape of the obstacle or opening, and the distance between the sound source and the obstacle. These factors determine the extent of diffraction and the resulting diffraction patterns. In the case of a narrow opening, such as a doorway or a window, the sound waves diffract and spread out in a fan-like pattern. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License. Use the information below to generate a citation. Then you must include on every digital page view the following attribution: If you are redistributing all or part of this book in a digital format, ![]() Then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a print format, Want to cite, share, or modify this book? This book uses the This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Since the arc subtends an angle ϕ ϕ at the center of the circle, To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of Figure 4.7. In solving that problem, you will find that they are less than, but very close to, ϕ = 3 π, 5 π, 7 π, … rad. The exact values of ϕ ϕ for the maxima are investigated in Exercise 4.120. As a result, E 1 E 1 and E 2 E 2 turn out to be slightly larger for arcs that have not quite curled through 3 π 3 π rad and 5 π 5 π rad, respectively. Since the total length of the arc of the phasor diagram is always N Δ E 0, N Δ E 0, the radius of the arc decreases as ϕ ϕ increases. These two maxima actually correspond to values of ϕ ϕ slightly less than 3 π 3 π rad and 5 π 5 π rad. The proof is left as an exercise for the student ( Exercise 4.119). In part (e), the phasors have rotated through ϕ = 5 π ϕ = 5 π rad, corresponding to 2.5 rotations around a circle of diameter E 2 E 2 and arc length N Δ E 0. The amplitude of the phasor for each Huygens wavelet is Δ E 0, Δ E 0, the amplitude of the resultant phasor is E, and the phase difference between the wavelets from the first and the last sources is The phasor diagram for the waves arriving at the point whose angular position is θ θ is shown in Figure 4.7. This distance is equivalent to a phase difference of ( 2 π a / λ N ) sin θ. If we consider that there are N Huygens sources across the slit shown in Figure 4.4, with each source separated by a distance a/N from its adjacent neighbors, the path difference between waves from adjacent sources reaching the arbitrary point P on the screen is ( a / N ) sin θ. To calculate the intensity of the diffraction pattern, we follow the phasor method used for calculations with ac circuits in Alternating-Current Circuits. Calculate the intensity relative to the central maximum of an arbitrary point on the screen.Calculate the intensity relative to the central maximum of the single-slit diffraction peaks.By the end of this section, you will be able to: ![]()
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