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Ch 36: Diffraction
Chapter 35, Problem 36

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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Identify the given values: wavelength (λ) = 585 nm and slit width (a) = 0.0666 mm. Convert all units to meters for consistency.
Understand the concept of diffraction through a single slit and the condition for dark fringes, which is given by the equation a \sin(\theta) = m\lambda, where m is the order of the dark fringe (m = ±1, ±2, ±3, ...).
Determine the maximum value of m. Since the sine function has a maximum value of 1, set up the inequality a \sin(\theta) = m\lambda \leq a. Solve for m using m \leq \frac{a}{\lambda}.
Calculate the total number of dark fringes. Since m can be positive or negative, and includes values from -m to +m, the total number of dark fringes is 2m + 1, accounting for both sides of the central maximum.
For part (b), find the angle for the most distant dark fringe using \theta = \arcsin(\frac{m\lambda}{a}) where m is the maximum integer value calculated previously.

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Key Concepts

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

Diffraction and Interference

Diffraction refers to the bending of waves around obstacles and the spreading of waves when they pass through narrow openings. In the context of light, when it passes through a slit, it creates an interference pattern of bright and dark fringes on a screen due to the constructive and destructive interference of light waves. This phenomenon is crucial for understanding how light behaves when encountering slits and is foundational for solving problems related to fringe patterns.
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Dark Fringes and Order of Interference

Dark fringes in a diffraction pattern occur at specific angles where the path difference between light waves from different parts of the slit leads to destructive interference. The order of interference, denoted by 'm', represents the number of wavelengths by which the path difference differs from an integer multiple of the wavelength. The maximum value of 'm' is determined by the geometry of the setup and the wavelength of light, which is essential for calculating the number of dark fringes.
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Number of Dark Fringes on a Screen

Sine Function and Its Limits

The sine function, sin(θ), is a mathematical function that relates the angle θ to the ratio of the opposite side to the hypotenuse in a right triangle. For diffraction problems, the maximum value of sin(θ) is 1, which occurs at θ = 90 degrees. This limit is critical in determining the maximum order of dark fringes (m) that can exist, as it sets a boundary for the angles at which dark fringes can be observed in the interference pattern.
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Related Practice
Textbook Question
Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum 1u = 0°2 is I0. What is the distance on the screen from the center of the central maximum (b) to the point where the intensity has fallen to I0>2?
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Textbook Question
If the planes of a crystal are 3.50 Å (1 Å = 10-10 m = 1 Ångstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0°, and in what part of the electromagnetic spectrum do these waves lie?
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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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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. (a) How wide is the hole in the barrier?
293
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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. (b) At what other angles do you find no waves hitting the shore?
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