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Ch 36: Diffraction
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 36: DiffractionProblem 13
Chapter 35, Problem 13

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?

Verified step by step guidance
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Step 1: Understand the problem and identify the key concepts. This problem involves Fraunhofer diffraction through a single slit. The first part (a) requires finding the slit width using the condition for the first diffraction minima, and the second part (b) involves calculating the intensity ratio at specific angles using the single-slit diffraction intensity formula.
Step 2: For part (a), use the condition for the first diffraction minima in single-slit diffraction: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of the minima, \( m \) is the order of the minima (\( m = \pm 1 \) for the first minima), and \( \lambda \) is the wavelength of the light. Rearrange the formula to solve for \( a \): \( a = \frac{m \lambda}{\sin \theta} \). Substitute \( m = 1 \), \( \lambda = 580 \; \text{nm} = 580 \times 10^{-9} \; \text{m} \), and \( \theta = 90.0^\circ \).
Step 3: For part (b), use the single-slit diffraction intensity formula: \( I(u) = I_0 \left( \frac{\sin u}{u} \right)^2 \), where \( u = \frac{\pi a \sin \theta}{\lambda} \), \( I_0 \) is the maximum intensity at \( \theta = 0 \), and \( \theta \) is the angle of observation. First, calculate \( u \) for \( \theta = 45.0^\circ \) and \( \theta = 0 \). For \( \theta = 0 \), \( u = 0 \), and for \( \theta = 45.0^\circ \), substitute \( a \) (from part (a)), \( \lambda \), and \( \sin 45.0^\circ \) into the formula for \( u \).
Step 4: Compute the ratio of intensities \( \frac{I(45.0^\circ)}{I(0)} \). Since \( I(0) = I_0 \), the ratio simplifies to \( \left( \frac{\sin u}{u} \right)^2 \) for \( u \) at \( \theta = 45.0^\circ \). Substitute the value of \( u \) calculated in Step 3 into this expression to find the ratio.
Step 5: Summarize the results. The slit width \( a \) is determined from part (a) using the diffraction minima condition, and the intensity ratio is calculated in part (b) using the single-slit diffraction intensity formula. Ensure all units are consistent and verify the calculations for accuracy.

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Key Concepts

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

Fraunhofer Diffraction

Fraunhofer diffraction occurs when light waves pass through a slit and are observed at a distance where the wavefronts can be considered parallel. This type of diffraction is characterized by the formation of a pattern of bright and dark fringes on a screen, which can be analyzed using mathematical equations. The angle of diffraction and the slit width are crucial in determining the positions of these minima and maxima.
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Diffraction Minima

Diffraction minima are points in the diffraction pattern where the intensity of light is zero due to destructive interference. For a single slit, the positions of these minima can be calculated using the formula a sin(θ) = mλ, where 'a' is the slit width, 'θ' is the angle of the minima, 'm' is the order of the minima, and 'λ' is the wavelength of the light. Understanding these minima is essential for determining the slit width and analyzing the intensity distribution.
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Intensity Ratio in Diffraction Patterns

The intensity of light in a diffraction pattern varies with angle and can be described by the intensity function derived from the amplitude of the wave. The ratio of intensities at different angles, such as u = 45.0° and u = 0°, can be calculated using the intensity formula I(θ) = I0 (sin(β)/β)², where β is related to the angle and slit width. This ratio provides insight into how the diffraction pattern changes with angle and is important for understanding light behavior in diffraction.
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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

Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 cm from the slits. The bright bands are 1.00 cm apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

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Textbook Question

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 5.00 x 10-4 W/m2, what is the intensity at a point on the screen that is 0.900 mm from the center of the central maximum?

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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

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (θ = 0°) is 4.00x10-5 W/m2. What is the intensity at a point on the screen that corresponds to θ = 1.20°?