91Ó°ÊÓ

(a) As Section 17.3 discusses, high-frequency sound waves exhibit less diffraction than low-frequency sound waves do. However, even highfrequency sound waves exhibit much more diffraction under normal circumstances than do light waves that pass through the same opening. The highest frequency that a healthy ear can typically hear is \(2.0 \times 10^{4} \mathrm{Hz}\) Assume that a sound wave with this frequency travels at \(343 \mathrm{m} / \mathrm{s}\) and passes through a doorway that has a width of \(0.91 \mathrm{m}\). Determine the angle that locates the first minimum to either side of the central maximum in the diffraction pattern for the sound. This minimum is equivalent to the first dark fringe in a single-slit diffraction pattern for light. (b) Suppose that yellow light (wavelength \(=580 \mathrm{nm}\) in vacuum) passes through a doorway and that the first dark fringe in its diffraction pattern is located at the angle determined in part (a). How wide would this hypothetical doorway have to be?

Step by step solution

01

Calculate the Wavelength of Sound

To find the angle of the first diffraction minimum, we start by calculating the wavelength of the sound using the formula:\[ \lambda = \frac{v}{f} \]where \( v = 343 \, \mathrm{m/s} \) is the speed of sound, and \( f = 2.0 \times 10^4 \, \mathrm{Hz} \) is the frequency of the sound. Substituting the values:\[ \lambda = \frac{343}{2.0 \times 10^4} = 0.01715 \, \mathrm{m} \]

02

Apply the Single-Slit Diffraction Formula for Sound

The angle \( \theta \) of the first diffraction minimum can be found using the condition for single-slit diffraction:\[ a \sin(\theta) = m \lambda \]where \( a = 0.91 \, \mathrm{m} \) is the doorway width, \( m = 1 \) (for the first minimum), and \( \lambda = 0.01715 \, \mathrm{m} \). Solving for \( \theta \):\[ 0.91 \sin(\theta) = 0.01715 \]\[ \sin(\theta) = \frac{0.01715}{0.91} \approx 0.01885 \]\[ \theta = \arcsin(0.01885) \approx 1.08^\circ \]

03

Determine the Width for Yellow Light

Given the angle \( \theta = 1.08^\circ \) and the wavelength for yellow light \( \lambda = 580 \, \mathrm{nm} = 580 \times 10^{-9} \, \mathrm{m} \), we use the same diffraction condition:\[ a \sin(\theta) = m \lambda \]Substitute \( \theta \) and \( \lambda \) into the equation and solve for \( a \): \[ a \sin(1.08^\circ) = 580 \times 10^{-9} \]\[ a \times 0.01885 = 580 \times 10^{-9} \]\[ a = \frac{580 \times 10^{-9}}{0.01885} \approx 3.08 \times 10^{-5} \, \mathrm{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

single-slit diffraction

Single-slit diffraction describes how waves spread out after passing through a narrow opening. Imagine water waves passing through a small gap; they spread out on the other side, forming a pattern of alternating light and dark bands. Sound waves do the same when they encounter obstacles or apertures that are comparable in size to their wavelength. In this scenario, we are considering sound waves passing through a doorway.

The concept relies on wave interference, where waves either combine to amplify or cancel each other. This creates a diffraction pattern with maxima and minima (bright and dark fringes). For the first minimum or dark fringe, the angle can be calculated. The formula to use is:

• \[ a \sin(\theta) = m \lambda \] where \( a \) is the slit width (or doorway width), \( \theta \) is the diffraction angle, \( m \) is the order of the minimum (1 for the first minimum), and \( \lambda \) is the wavelength of the wave.

For sound, determining the angle of the first minimum helps us understand how sound bends around corners into shadow regions.

frequency of sound

The frequency of a sound wave is the number of vibrations or cycles per second and is measured in Hertz (Hz). It determines the pitch of the sound; higher frequencies correspond to higher pitches. Humans generally hear sounds in the range of 20 Hz to 20,000 Hz.

In this exercise, the frequency of interest is \( 2 \times 10^4 \, \mathrm{Hz} \), which is at the upper limit of human hearing. This is considered a high frequency sound. High-frequency sounds tend to diffract less than low-frequency sounds because their wavelengths are shorter. However, when they pass through a narrow slit or small opening, they can still show diffraction effects.

Understanding frequency is crucial because it connects to how sound behaves when interacting with objects, affecting phenomena like diffraction and interference patterns.

wavelength calculation

The wavelength of a sound wave is the distance between successive crests or troughs of the wave. It determines how "spread out" the sound is and is inversely related to the frequency.

To calculate the wavelength \( \lambda \) of a sound wave, use the formula:

• \[ \lambda = \frac{v}{f} \] where \( v \) is the speed of sound (provided as 343 m/s here) and \( f \) is the frequency (\( 2 \times 10^4 \, \mathrm{Hz} \)).

For our given values, the calculation provides:

• \[ \lambda = \frac{343}{2 \times 10^4} = 0.01715 \, \mathrm{m} \]

This result gives the sound wave's wavelength, essential for predicting how it will diffract. Shorter wavelengths, as seen with higher frequencies, tend to bend less as they pass through openings.

diffraction pattern

A diffraction pattern emerges when waves encounter an obstacle or slit. It is a series of light and dark bands that illustrate the wave interference effect. In a single-slit set-up, the diffraction pattern results from the interaction of waves spreading out from various points along the slit.

For sound waves, especially high-frequency ones like 20,000 Hz, the diffraction pattern reveals how sound moves around obstacles. The first dark fringe or minimum in a diffraction pattern is a key detail—it is where sound waves destructively interfere, cancelling each other's effect. In this exercise, identifying the angle corresponding to the first minimum helps visualize how sound propagates through the doorway.

Visualizing this pattern can help in understanding phenomena such as "hearing around corners," where even if you can't see a source, you can still hear it. This is due to the way sound waves bend, forming a diffraction pattern that extends into regions otherwise blocked by obstacles.