# How Is Sound Intensity Affected by Diffraction and Slit Width?

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In summary, diffraction occurs for all types of waves, including sound waves. In this problem, high-frequency sound from a distant source with a wavelength of 9.20 cm passes through a narrow slit 11.5 cm wide. A microphone is placed 35.0 cm directly in front of the center of the slit and is then moved in a direction perpendicular to the line from the center of the slit to point O, the center of the diffraction pattern. The problem asks for the minimal distance from O where the intensity detected by the microphone is zero. To solve this, we use the equations sinθ = (mλ)/a and tanθ = y/L, where a is the width of the slit, L is the
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## Homework Statement

Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 9.20 cm passes through a narrow slit 11.5 cm wide. A microphone is placed 35.0 cm directly in front of the center of the slit. The microphone is then moved in a direction perpendicular to the line from the center of the slit to point O, the center of the diffraction pattern.

At what minimal distance from O will the intensity detected by the microphone be zero?

## Homework Equations

$$sin \theta = \frac{m\lambda}{a}$$

$$tan \theta = \frac{y}{L}$$

$$\lambda = 9.20 cm, a = 11.5 cm, L = 35.0 cm, m = 1$$

## The Attempt at a Solution

$$tan \theta = sin \theta$$

$$\frac{y}{L} = \frac{m\lambda}{a}$$

$$y = \frac{m\lambda L} {a}$$

$$y = \frac{1 \times 9.20 cm \times 35.0 cm}{11.5 cm}$$

y = 28 cm

However, this is not the correct answer. The correct answer is given as y = 46.7 cm. I suspect that my solution didn't work because the angles are not small enough so that the sine and tangent functions can not be set equal to each other. How do you get the correct answer?

Last edited:
I suspect that my solution didn't work because the angles are not small enough so that the sine and tangent functions can not be set equal to each other. How do you get the correct answer?
So try it without this approximation. Based on the geometry of the problem how would you express sinθ? Figure that out and put it in your equation. The math gets a bit messier but it is relatively a straightforward calculation.

## What is diffraction of sound waves?

Diffraction of sound waves is the bending of sound waves as they pass through an opening or around an obstacle. It occurs when sound waves encounter an edge or barrier that is comparable in size to the wavelength of the sound.

## What factors affect the amount of diffraction of sound waves?

The amount of diffraction of sound waves is affected by the wavelength of the sound, the size of the opening or obstacle, and the distance between the sound source and the opening or obstacle. Higher frequencies and smaller openings or obstacles result in less diffraction, while lower frequencies and larger openings or obstacles result in more diffraction.

## How does diffraction of sound waves affect our perception of sound?

Diffraction of sound waves can cause changes in the intensity and direction of sound, which can affect our perception of the sound. It can also cause the sound to reach our ears at slightly different times, resulting in a change in the quality of the sound.

## What are some real-world examples of diffraction of sound waves?

Diffraction of sound waves can be observed in everyday situations, such as when we can hear sound from around a corner or through a partially closed door. It is also commonly seen in concert halls, where the shape and design of the hall can affect the way sound waves diffract and reach the audience.

## How is diffraction of sound waves used in technology and engineering?

Diffraction of sound waves is used in various technologies and engineering applications, such as in the design of loudspeakers and microphones. It is also used in medical imaging techniques, such as ultrasound, where sound waves are diffracted to create images of internal structures in the body.

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