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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 3
Chapter 36, Problem 3

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?

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Step 1: Begin by understanding the diffraction condition for dark fringes in a single-slit experiment. The condition for destructive interference is given by the equation: a⁢sin⁢u=m⁢λ, where a is the slit width, u is the angle of the dark fringe, m is the order of the fringe (an integer), and λ is the wavelength of the light.
Step 2: To determine the maximum number of dark fringes, consider the physical constraint that sin⁢u cannot exceed 1. Rearrange the equation to find the maximum value of m: m=aλ. Substitute the given values for a (0.0666 mm = 6.66 × 10⁻⁵ m) and λ (585 nm = 5.85 × 10⁻⁷ m) to calculate the maximum integer value of m. This will give the total number of dark fringes on one side of the central bright spot.
Step 3: Multiply the maximum value of m by 2 to account for both sides of the central bright spot. Add 1 to include the central bright spot itself in the count. This will give the total number of dark fringes.
Step 4: For part (b), use the diffraction condition a⁢sin⁢u=m⁢λ to find the angle u for the most distant dark fringe. Substitute the maximum value of m into the equation and solve for sin⁢u. Then use the inverse sine function to calculate the angle u.
Step 5: Summarize the results conceptually: The total number of dark fringes is determined by the maximum value of m, and the angle of the most distant dark fringe is calculated using the diffraction condition. Ensure that the physical constraints (e.g., sin⁢u ≤ 1) are satisfied.

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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, resulting in a new wave pattern. In the context of light, this can lead to constructive interference (bright fringes) or destructive interference (dark fringes). The condition for destructive interference in a single slit is given by the equation a sin(θ) = mλ, where 'a' is the slit width, 'm' is the order of the dark fringe, and 'λ' is the wavelength of light.
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Sinusoidal Function Limits

The sine function, sin(θ), has a maximum value of 1, which occurs at θ = 90 degrees. This limit is crucial when determining the maximum order 'm' of dark fringes in a diffraction pattern. If sin(θ) exceeds 1, it indicates that no further dark fringes can be formed, thus providing a boundary for the possible values of 'm' in the interference equation.
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Dark Fringe Position

The position of dark fringes in a diffraction pattern can be calculated using the formula a sin(θ) = mλ. The angle θ for the most distant dark fringe corresponds to the maximum value of 'm' that satisfies the sine function's limit. This angle indicates where the first minimum occurs on either side of the central bright fringe, providing insight into the distribution of light and dark regions on the observation screen.
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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

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

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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