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Ch 36: Diffraction
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 36: DiffractionProblem 7b
Chapter 36, Problem 7b

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at ±61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. At what other angles do you find no waves hitting the shore?

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Determine the wavelength of the water waves using the relationship between wave speed, frequency, and wavelength. The wave speed \( v \) is given as 15.0 cm/s, and the frequency \( f \) can be calculated from the observation that 75.0 wave crests pass by each minute. Use the formula \( f = \frac{75.0}{60} \) to find the frequency in Hz, and then use \( \lambda = \frac{v}{f} \) to find the wavelength.
Understand the phenomenon of diffraction and interference. The hole in the barrier acts as a single slit, and the pattern observed on the shore is due to the interference of waves. The locations where no waves are observed correspond to destructive interference, which occurs at specific angles \( \theta \) given by the condition \( d \sin \theta = m \lambda \), where \( d \) is the width of the slit (hole), \( \lambda \) is the wavelength, and \( m \) is an integer (\( \pm 1, \pm 2, \pm 3, \dots \)).
From the problem, the first destructive interference occurs at \( \pm 61.3 \) cm from the point directly opposite the hole. Use this information to calculate the angle \( \theta \) for the first minimum. The distance from the hole to the shore is 3.20 m, so \( \tan \theta = \frac{61.3}{320} \). Solve for \( \theta \) and use \( \sin \theta \approx \tan \theta \) for small angles.
Once \( \sin \theta \) for the first minimum is known, use the condition \( d \sin \theta = m \lambda \) to calculate the width of the slit \( d \). Here, \( m = 1 \) for the first minimum. Rearrange the equation to solve for \( d \): \( d = \frac{\lambda}{\sin \theta} \).
To find the other angles where no waves hit the shore, use the same condition \( d \sin \theta = m \lambda \) for higher-order minima (\( m = 2, 3, 4, \dots \)). Solve for \( \sin \theta \) for each \( m \), and then calculate the corresponding angles \( \theta \). These angles represent the directions where destructive interference occurs, and no waves reach the shore.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Wave Interference

Wave interference occurs when two or more waves overlap and combine to form a new wave pattern. This can result in constructive interference, where wave amplitudes add together, or destructive interference, where they cancel each other out. In the context of the problem, the absence of waves at certain angles indicates destructive interference due to the barrier and the hole, creating regions where the wave amplitudes effectively cancel.
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Diffraction

Diffraction is the bending of waves around obstacles or through openings, which allows waves to spread out after passing through a hole. In this scenario, the water waves passing through the hole in the barrier will diffract, creating a pattern of wave fronts that can lead to areas of constructive and destructive interference along the shore. Understanding diffraction is crucial for determining where waves will and will not reach the shore.
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Wave Speed and Frequency

The speed of a wave is determined by its frequency and wavelength, described by the equation v = fλ, where v is wave speed, f is frequency, and λ is wavelength. In this problem, the wave speed is given as 15.0 cm/s, and the frequency can be calculated from the number of wave crests observed per minute. This relationship is essential for analyzing how the waves interact with the barrier and the resulting patterns observed at the shore.
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Related Practice
Textbook Question

Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point O in Fig. 36.5a . The microphone is then moved in a direction perpendicular to the line from the center of the slit to point O. At what distances from O will the intensity detected by the microphone be zero?

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Textbook Question

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at ±61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. How wide is the hole in the barrier?

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Textbook Question

Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin u can be? What does this tell you is the largest that m can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Textbook Question

A slit 0.240 mm wide is illuminated by parallel light rays of wavelength 540 nm. The diffraction pattern is observed on a screen that is 3.00 m from the slit. The intensity at the center of the central maximum (θ = 0°) is 6.00 x 10-6 W/m2. What is the distance on the screen from the center of the central maximum to the first minimum?

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Textbook Question

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at ±90.0°, so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at θ = 45.0° to the intensity at θ = 0?

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Textbook Question

A slit 0.240 mm wide is illuminated by parallel light rays of wavelength 540 nm. The diffraction pattern is observed on a screen that is 3.00 m from the slit. The intensity at the center of the central maximum (θ = 0°) is 6.00 x 10-6 W/m2. What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

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