91Ó°ÊÓ

(a) As Section \(17.3\) discusses, high-frequency sound waves exhibit less diffraction than low-frequency sound waves do. However, even high-frequency 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 (wave length \(=580 \mathrm{~nm}\), 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

The speed of sound is given as \(343 \text{ m/s}\) and the highest frequent sound is \(2.0 \times 10^4 \text{ Hz}\). Use the equation \(\lambda = \frac{v}{f}\) to find the wavelength: \[ \lambda = \frac{343 \text{ m/s}}{2.0 \times 10^4 \text{ Hz}} = 0.01715 \text{ m} \]

02

Identify the Angle for the First Minimum in Sound Diffraction

For the first minimum in a single-slit diffraction pattern, use \(a \sin \theta = m \lambda\), where \(a\) is the width of the slit (0.91 m), \(m=1\) for the first minimum, and \(\lambda\) is the wavelength of sound calculated in the previous step. Solve for \(\theta\): \[0.91 \sin \theta = 1 \times 0.01715\]\[\sin \theta = \frac{0.01715}{0.91} = 0.0188\]\[\theta = \arcsin(0.0188) = 1.08^{\circ} \]

03

Calculate the Required Width for Light Doorway

Using the same angle \(\theta = 1.08^{\circ}\) for yellow light with wavelength \(580 \text{ nm} = 580 \times 10^{-9} \text{ m}\), apply the same formula for the first dark fringe: \(a \sin \theta = m \lambda\) (\(m=1\)). Solve for \(a\):\[a \sin(1.08^{\circ}) = 580 \times 10^{-9}\]\[a = \frac{580 \times 10^{-9} \text{ m}}{\sin(1.08^{\circ})} = 3.08 \text{ cm}\]

04

Convert Doorway Width to Meters

Since we calculated the width of the hypothetical opening for light in centimeters, convert it to meters for consistency. \[3.08 \text{ cm} = 0.0308 \text{ m}\]

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Key Concepts

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

Sound Waves

Sound waves are an intriguing form of energy. They travel through air and other media as vibrations, and their nature allows them to be heard by our ears.
The frequency of a sound wave is measured in Hertz (Hz) and determines the pitch. High-frequency sound waves have high pitches, whereas low-frequency waves produce deeper sounds.

• Sound travels at different speeds depending on the medium. In air, it travels at approximately 343 meters per second (m/s).
• The wavelength of sound relates inversely to its frequency through the formula: \( \lambda = \frac{v}{f} \). Here, \( \lambda \) stands for wavelength, \( v \) for speed, and \( f \) for frequency.
• waves

Sound diffraction occurs when waves encounter obstacles and bend around them. At low frequencies, it is more pronounced, making it easier to hear sounds around corners.
When sound waves travel through openings, like doorways, the size and frequency of the waves determine the diffraction pattern. This pattern includes central maxima (where sound is strongest) and minima (where sound is weakest), causing variations in sound intensity.

Light Waves

Light waves differ from sound waves in several ways. They are an electromagnetic phenomenon, allowing them to travel through a vacuum without needing a medium. Light waves are incredibly fast, traveling at a speed of about \( 3 \times 10^8 \) meters per second in a vacuum.

• Light waves behave like both waves and particles, a concept known as wave-particle duality.
• The visible spectrum ranges from violet (with short wavelengths) to red (with long wavelengths). Yellow light, for example, has a wavelength of about 580 nanometers (nm).

When light passes through narrow openings, such as a doorway, it forms a diffraction pattern similar to sound. However, due to the much smaller wavelengths compared to sound waves, the diffraction effects are less noticeable.
The angle where the first minimum (dark fringe) occurs is determined using the same principles of diffraction that apply to sound, but because of the smaller wavelength, the angle changes accordingly.

Single-Slit Diffraction Pattern

The single-slit diffraction pattern is a fundamental concept in physics that describes how waves interact with a narrow opening.
It applies to both sound and light waves and helps us understand how waves can bend around obstacles or spread out after passing through slits.

• The diffraction pattern consists of a central maximum where the wave intensity is the strongest, surrounded by alternating minima and maxima (darker and lighter fringes).
• At minima, the waves interfere destructively, canceling each other out, producing a dark fringe.

To find the 'first minimum' directly to the side of the central maximum, we use the equation: - \(a \sin \theta = m \lambda\), - where \(a\) is the slit width, \(\theta\) is the diffraction angle, \(m\) is an integer representing the fringe order, and \(\lambda\) is the wavelength of the wave. For sound waves with low frequencies, the slit can be larger, such as a typical doorway, and still show noticeable diffraction. For light, due to its shorter wavelength, the opening needs to be proportionally smaller to create visible diffraction patterns. This explains why light needs a much narrower slit—calculated to be about 3.08 cm for the given yellow light—to produce similar diffraction as the sound waves in our example.