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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 interfence occurs. (a) Is it constructive or destructive interference, and \((\mathrm{b})\) where are the places located?

Short Answer

Expert verified
(a) Destructive interference; (b) at 2.00 m from each source.

Step by step solution

01

Understanding In-Phase Sources and Path Difference

When two sources are in-phase and produce waves of the same wavelength, constructive interference occurs at points where the path difference between the waves is an integer multiple of the wavelength, i.e., \( n\lambda \), where \( n \) is an integer. Destructive interference occurs when this path difference is a half-integer multiple of the wavelength, i.e., \( \left(n + \frac{1}{2}\right)\lambda \).
02

Determine Constructive Interference Conditions

Constructive interference occurs when the path difference is \( n\lambda \) (i.e., 0, 5.00 m, 10.00 m, etc.). However, since the separation between the sources is 4.00 m, the only possibility for constructive interference is when the path difference is 0, which happens at the midpoint of the line between the sources.
03

Determine Destructive Interference Conditions

Destructive interference occurs at points where the path difference is \( \left(n + \frac{1}{2}\right)\lambda \) (i.e., 2.5 m, 7.5 m, etc.). Considering the separation of 4.00 m, the path difference will be \( \frac{5}{2} \) for the middle region, so at 2.00 m from each source.
04

Check for Two Interference Points

Given the 4.00 m separation and the wavelength of 5.00 m, both points of interference must occur within this distance. Since only the midpoint (2.00 m from each source) fulfills one condition and there are no other points within the distance that allow a complete wave or half-wave to fit exactly, evaluate all possibilities.
05

Identification of Interference Type

For 4.00 m separation between sources with a 5.00 m wavelength, the same type of interference in two places along the line between the sources must be destructive, as constructive interference is only possible at the midpoint (not at two different points). Thus, two destructive interference points appear at places where the path difference is \( 2.5 m \), which are 2.00 m apart (from each source).

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Key Concepts

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

Constructive Interference
Constructive interference happens when two wave sources are perfectly in sync, or in phase, creating a situation where their peaks and troughs align. This results in the waves combining to produce a wave with greater amplitude. Imagine two people jumping on a trampoline at the same time in the same spot. As their jumps add up, they can jump even higher.

The condition for constructive interference is that the path difference—a difference in distance from each wave source to a point—is an integer multiple of the wavelength. In other words, if the difference in distance is exactly a full wave (or two, three, etc.), the waves will amplify each other. In equations, this path difference is expressed as \( n\lambda \), where \( n\) is an integer representing the number of whole waves.

  • Maximized Effect: The effect is strongest at points where the waves meet perfectly.
  • Key Equation: \( n\lambda \).
  • Occurs naturally in sound, light, and other waveforms.
Destructive Interference
Destructive interference occurs when two waves meet and cancel each other out, much like two equally strong breezes blowing directly against each other. This happens when the peaks of one wave line up with the troughs of another, reducing the overall wave energy.

The equation that describes this condition is \(( n + \frac{1}{2})\lambda \). Here, the path difference creates a scenario where one wave is half a wavelength out of sync with the other, leading to cancellation.

  • Result: The waves lessen the impact, often leading to areas of little to no wave activity.
  • Key Equation: \(( n + \frac{1}{2})\lambda \).
  • This type of interference can occur in all types of waves including acoustic and electromagnetic.
Path Difference
The concept of path difference is central to understanding wave interference. It measures how much further one wave has traveled compared to another when reaching a specific point. Imagine two cars starting from the same point but taking different routes to arrive at a destination. The difference in distance they travel is akin to path difference.

For constructive interference, the path difference should be a whole number of wavelengths, allowing waves to align perfectly. For instance, if one wave travels 10 meters and another 15 meters with a wavelength of 5 meters, they align constructively when the path difference is equal to that wavelength.

  • In the context of wave interference, only specific path differences correspond to clear interference patterns.
  • This concept is crucial for predicting the interaction points of in-phase wave sources.
Wave Sources
Wave sources are the origins from which waves are emitted. In experiments or problems involving interference, these sources are often fixed at certain positions. Think about speakers in a stereo setup, where each speaker acts as a wave source, emitting sound waves.

In scenarios like the one presented, the positioning of wave sources directly affects the interference pattern observed along the line between them. If wave sources are in-phase, the observed pattern will reflect scenarios where peaks (constructive) or troughs (destructive) meet. The consistent distance of 4.00 m between the sources in the exercise creates predictable interference patterns linked to the described path difference.

  • Crucial for predicting where constructive or destructive interference occurs.
  • Their position affects the wave's path difference and interference types.
  • Examples: Antennae, loudspeakers, any emitter of waves.
Wavelength
Wavelength, symbolized by the Greek letter lambda (\(\lambda\)), refers to the distance between successive peaks (or troughs) of a wave. In simpler terms, it's like the space between waves on the ocean. Understanding wavelength is essential for grasping interference concepts.

In the problem discussed, the wavelength is 5.00 meters. This specific measurement dictates how waves interfere when they meet other waves.

  • Determines the specific points where constructive and destructive interference occur.
  • Involved in calculations for path differences.
  • Integral for predicting behavior of waves in interference setups.

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Most popular questions from this chapter

In a Young's double-slit experiment, the wavelength of the light used is 520 \(\mathrm{nm}\) (in vacuum), and the separation between the slits is \(1.4 \times 10^{-6} \mathrm{m}\) . Determine the angle that locates \((\mathrm{a})\) the dark fringe for which \(m=0, \quad\) (b) the bright fringe for which \(m=1\) (c) the dark fringe for which \(m=1,\) and \(\quad\) (d) the bright fringe for which \(m=2\)

In a Young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe is \(5.4^{\circ} .\) Find the ratio \(d / \lambda\) of the slit separation \(d\) to the wavelength \(\lambda\) of the light.

A large group of football fans comes to the game with colored cards that spell out the name of their team when held up simultaneously. Most of the cards are colored blue \(\left(\lambda_{\text { vacum }}=480 \mathrm{nm}\right)\) . When displayed, the average distance between neighboring cards is 5.0 \(\mathrm{cm} .\) If the cards are to blur together into solid blocks of color when viewed by a spectator at the other end of the stadium \((160 \mathrm{m}\) away), what must be the maximum diameter (in \(\mathrm{mm}\) ) of the spectator's pupils?

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The distance between adjacent slits of a certain diffraction grating is \(1.250 \times 10^{-5} \mathrm{m}\) . The grating is illuminated by monochromatic light with a wavelength of \(656.0 \mathrm{nm},\) and is then heated so that its temperature increases by 100.0 \(\mathrm{C}^{\circ} .\) Determine the change in the angle of the seventh order principal maximum that occurs as a result of thermal expansion of the grating. The cofficient of linear expansion for the diffaction grating is \(1.30 \times 10^{-4}\left(\mathrm{C}^{9}\right)^{-1} .\) Be sure to include the proper algebraic sign with your answer: \(+\) if the angle increases, - if the angle decreases.

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