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Ch 36: Diffraction
Chapter 35, Problem 36

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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1
First, understand that the problem involves single-slit diffraction, where light passing through a narrow slit spreads out and creates a pattern of bright and dark regions on a screen.
Use the formula for the position of the first minimum in a single-slit diffraction pattern, which is given by \( y = L \tan(\theta) \), where \( L \) is the distance from the slit to the screen, and \( \theta \) is the angle at which the first minimum occurs.
Calculate the angle \( \theta \) for the first minimum using the formula \( \theta = \frac{m \lambda}{a} \), where \( m = 1 \) for the first minimum, \( \lambda \) is the wavelength of the light, and \( a \) is the width of the slit.
Substitute the values of \( L \), \( \lambda \), and \( a \) into the equations to find \( \theta \) and then use it to find \( y \), the distance from the center of the central maximum to the first minimum on the screen.
Since \( \theta \) is typically small in diffraction problems, you can approximate \( \tan(\theta) \approx \sin(\theta) \approx \theta \) in radians for simplifying calculations.

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Key Concepts

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

Diffraction

Diffraction is the bending of waves around obstacles and the spreading of waves when they pass through narrow openings. In the context of light, diffraction patterns arise when light encounters a slit, leading to regions of constructive and destructive interference. The width of the slit and the wavelength of the light significantly influence the diffraction pattern observed on a screen.
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Minima in Diffraction Patterns

In a single-slit diffraction pattern, minima occur at specific angles where destructive interference takes place. The position of the first minimum can be calculated using the formula a sin(θ) = mλ, where 'a' is the slit width, 'm' is the order of the minimum (m=1 for the first minimum), and 'λ' is the wavelength of the light. This relationship helps determine the distance from the central maximum to the first minimum on the observation screen.
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Intensity of Light

The intensity of light is defined as the power per unit area and is measured in watts per square meter (W/m²). In diffraction patterns, the intensity varies with position due to the interference of light waves. The intensity at the central maximum is typically the highest, and it decreases as one moves towards the minima, which can be calculated using the intensity distribution formula derived from the diffraction pattern.
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Wave Intensity
Related Practice
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
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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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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Textbook Question
A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation?
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