/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Two in-phase sources of waves ar... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 3

Two in-phase sources of waves are separated by a distance of \(4.00 \mathrm{m}\). These sources produce identical waves that have a wavelength of \(5.00 \mathrm{m}\). On the line between them, there are two places at which the same type of interference occurs. (a) Is it constructive or destructive interference, and (b) where are the places located?

Short Answer

Expert verified
(a) Destructive interference occurs; (b) places located at 1.75 m and 2.25 m.

Step by step solution

01

Understanding the Context

We have two in-phase wave sources separated by a distance of 4.00 m, emitting identical waves with a wavelength of 5.00 m. Our task is to determine the type of interference occurring at two specific points between these sources and the locations of these points.
02

Interference Type Identification

Since the sources are in-phase, constructive interference occurs when the path difference is an integer multiple of the wavelength, and destructive interference occurs when the path difference is an odd multiple of half the wavelength. We'll determine which type aligns with a 4.00 m separation.
03

Calculating Phase Difference

We calculate the path difference at any point between the sources. For constructive interference, \( d \sin \theta = m \lambda \), where \( m \) is an integer, and for destructive interference, \( d \sin \theta = (m+0.5) \lambda \). Since \( \theta \) is 0 on the line, the path difference is 4.00 m.
04

Applying Condition for Interference

Using the path difference of 4.00 m and the wavelength of 5.00 m, we apply the conditions: - Constructive: \( m \lambda = 4.00 \ ext{m} \) - Destructive: \( (m + 0.5) \lambda = 4.00 \ ext{m} \) Only destructive interference fits as \( 4.00 = 0.5 \times 5 \), giving close to an integer.
05

Solving for Locations of Interference

For destructive interference, the interference points occur where the difference in distances from the sources is an odd multiple of half the wavelength (here: 2.5 m). Since points must be symmetrically placed around the midpoint of 2.00 m (halfway between), solutions are offset by 2.5 m. Locations are 1.75 m and 2.25 m.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constructive Interference
Constructive interference occurs when two or more waves meet in such a way that their amplitudes add together, resulting in a larger wave. This happens when the path difference between the waves is a whole number multiple of the wavelength.
  • The waves are "in phase," meaning they peak and trough at the same times.
  • Mathematically, the condition is expressed as: \( \Delta L = m\lambda \), where \( \Delta L \) is the path difference, \( \lambda \) is the wavelength, and \( m \) is any integer (0, 1, 2,...).
Think of it as two people jumping on a trampoline at the exact same rhythm; their combined jumps make them go higher together. In practical terms, this can be seen in phenomena like the "beats" you hear when two musical notes slightly differ in frequency. The waves from each note combine constructively at times to create louder sounds.
Destructive Interference
Destructive interference is the opposite of constructive interference. It happens when two waves meet and cancel each other out, resulting in a reduced or no wave amplitude.
  • This type of interference occurs when the waves are "out of phase," with crests aligning with troughs of the other wave.
  • The condition for destructive interference is \( \Delta L = (m+0.5)\lambda \), where \( m \) is an integer.
Imagine two people jumping on a trampoline, but one jumps just as the other lands. Instead of going higher, their energy cancels out. In real-world applications, noise-canceling headphones use destructive interference. They emit sound waves that are out of phase with unwanted noise, effectively reducing it.
Path Difference
Path difference is a crucial concept in understanding wave interference. It refers to the difference in distance traveled by two waves before they meet at a point.
  • When this difference is an exact multiple of the wavelength, you get constructive interference.
  • When the path difference is an odd multiple of half the wavelength, it results in destructive interference.
  • In many experiments, observing the path difference allows predictions about where interference patterns will form.
For instance, if two wave sources are 4 meters apart, and the wavelength is 5 meters, understanding the path difference helps specify where waves overlap constructively or destructively. In the given exercise, the path difference calculated helped identify where destructive interference happens, leading to the cancellation at specific points along the line between the sources.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength \(=640.0 \mathrm{nm}\) in vacuum). Assuming that the visible spectrum extends from 380 to \(750 \mathrm{nm},\) for which visible wavelength(s) in vacuum will the film appear bright due to constructive interference?

Orange light \(\left(\lambda_{\text {vacuum }}=611 \mathrm{nm}\right)\) shines on a soap film \((n=1.33)\) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?

In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18000 times farther from the slit than the slit is wide. What is the ratio \(\lambda / W,\) where \(\lambda\) is the wavelength of the light shining through the slit and \(W\) is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that \(\sin \theta \approx \tan \theta\).

Point A is the midpoint of one of the sides of a square. On the side opposite this spot, two in-phase loudspeakers are located at adjacent corners, as shown in the figure. Standing at point A you hear a loud sound because of constructive interference between the identical sound waves coming from the speakers. As you walk along the side of the square toward either empty corner, the loudness diminishes gradually to nothing and then increases again until you hear a maximally loud sound at the corner. If the length of each side of the square is \(4.6 \mathrm{m},\) find the wavelength of the sound waves.

A sheet that is made of plastic \((n=1.60)\) covers one slit of a double slit (see the drawing). When the double slit is illuminated by monochromatic light \(\left(\lambda_{\text {vacuum }}=586 \mathrm{nm}\right),\) the center of the screen appears dark rather than bright. What is the minimum thickness of the plastic?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Magnetism

Read Explanation

Nuclear Physics

Read Explanation

Electromagnetism

Read Explanation

Physics of Motion

Read Explanation

Mechanics and Materials

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.