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Ch 36: Diffraction
Chapter 35, Problem 36

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 u = 45.0° to the intensity at u = 0?

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1
Identify the formula for the position of the first minima in a single-slit diffraction pattern, which is given by \( a \sin(\theta) = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of the minima, \( m \) is the order of the minima (1 for the first minima), and \( \lambda \) is the wavelength of the light.
Since the first minima occur at \( \pm 90.0^\circ \), substitute \( \theta = 90^\circ \) and \( \lambda = 580 \, \text{nm} \) into the formula to solve for the slit width \( a \).
For part (b), use the intensity formula for single-slit diffraction, \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \), where \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \) and \( I_0 \) is the maximum intensity at \( \theta = 0 \).
Calculate \( \beta \) at \( \theta = 45.0^\circ \) using the slit width \( a \) found in part (a) and \( \lambda = 580 \, \text{nm} \).
Find the ratio of the intensities by evaluating the intensity formula at \( \theta = 45.0^\circ \) and \( \theta = 0 \) and then dividing the intensity at \( \theta = 45.0^\circ \) by the intensity at \( \theta = 0 \).

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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 angles of these fringes depend on the wavelength of the light and the dimensions of the slit.
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Diffraction Minima

Diffraction minima are points in the diffraction pattern where the intensity of light is significantly reduced or zero. For a single slit, these minima occur at specific angles determined by the slit width and the wavelength of the light. The first minima can be calculated using the formula sin(θ) = mλ/a, where m is the order of the minimum, λ is the wavelength, and a is the slit width.
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Intensity Ratio in Diffraction Patterns

The intensity of light at different angles in a diffraction pattern can be described by the intensity function, which varies with the angle of observation. The intensity at the central maximum (u = 0) is typically the highest, while the intensity at other angles decreases. The ratio of intensities at different angles can be calculated using the intensity formula derived from the diffraction pattern, allowing for comparisons between points like u = 0 and u = 45.0°.
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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. (a) How wide is the hole in the barrier?
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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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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 (u = 0°) is 6.00 x 10^-6 W/m2. (a) What is the distance on the screen from the center of the central maximum to the first minimum?
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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 (u = 0°) is 6.00 x 10^-6 W/m2. (b) 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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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 (u = 0°) is 4.00x10-5 W/m2. What is the intensity at a point on the screen that corresponds to u = 1.20°?
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