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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
Chapter 36, Problem 7a

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?

Verified step by step guidance
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Step 1: Start by identifying the given quantities. The wave speed is 15.0 cm/s, the frequency of the waves can be calculated from the observation that 75.0 wave crests pass by each minute (convert this to waves per second), and the distance from the barrier to the shore is 3.20 m. Additionally, the region where no waves reach the shore is ±61.3 cm from the point directly opposite the hole.
Step 2: Calculate the wavelength of the waves using the relationship between wave speed, frequency, and wavelength: v = fλ, where v is the wave speed, f is the frequency, and λ is the wavelength. Rearrange to solve for λ: λ = v/f.
Step 3: Use the concept of diffraction to analyze the problem. The hole in the barrier acts as a single slit, and the waves diffract through it. The angular position of the first minimum in the diffraction pattern is given by the equation for single-slit diffraction: a sin(θ) = mλ, where a is the width of the slit (hole), θ is the angle to the first minimum, m is the order of the minimum (for the first minimum, m = 1), and λ is the wavelength.
Step 4: Determine the angle θ using the geometry of the problem. The distance from the point directly opposite the hole to the edge of the region where no waves reach the shore is 61.3 cm, and the distance from the barrier to the shore is 3.20 m. Use the tangent function: tan(θ) = x/L, where x is the distance to the edge of the region (61.3 cm) and L is the distance from the barrier to the shore (3.20 m). Since θ is small, you can approximate sin(θ) ≈ tan(θ).
Step 5: Substitute the values for λ, θ, and m into the diffraction equation a sin(θ) = mλ to solve for the width of the hole a. Rearrange the equation to find a = λ/sin(θ).

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Key Concepts

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

Wave Speed and Frequency

Wave speed is the distance traveled by a wave per unit time, while frequency is the number of wave crests passing a point per second. In this scenario, the wave speed is given as 15.0 cm/s, and the frequency can be calculated from the observation of 75.0 crests passing per minute, which translates to 1.25 crests per second. Understanding the relationship between wave speed, frequency, and wavelength is crucial for analyzing wave behavior.
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Intro to Waves and Wave Speed

Diffraction of Waves

Diffraction refers to the bending and spreading of waves when they encounter an obstacle or pass through an opening. In this problem, the waves passing through the hole in the barrier will diffract, creating a pattern of wave fronts that spread out. The width of the hole affects the extent of diffraction, which is essential for determining how far the waves can reach the shore.
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Interference Patterns

Interference patterns occur when two or more waves overlap, leading to regions of constructive and destructive interference. In this case, the observation that no waves reach the shore at ±61.3 cm from the point directly opposite the hole suggests a pattern of destructive interference. Understanding how waves interact with each other and the barrier is key to solving for the width of the hole.
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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. At what other angles do you find no waves hitting the shore?

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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

Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens, the distance from the central maximum to the first minimum is 8.65 mm. What is the width of the slit?

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