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Coherent sources \(A\) and \(B\) emit electromagnetic waves with wavelength 2.00 \(\mathrm{cm} .\) Point \(P\) is 4.86 \(\mathrm{m}\) from \(A\) and 5.24 \(\mathrm{m}\) from B. What is the phase difference at \(P\) between these two waves?

Short Answer

Expert verified
The phase difference at point P is 0 radians.

Step by step solution

01

Calculate the Path Difference

The path difference between the two sources at point \(P\) is given by the distance from \(B\) to \(P\) minus the distance from \(A\) to \(P\). Therefore, the path difference \(\Delta d\) is \(5.24 \, \text{m} - 4.86 \, \text{m} = 0.38 \, \text{m}\).
02

Relate Path Difference to Wavelength

The relationship between the path difference \(\Delta d\) and the wavelength \(\lambda\) is found by dividing the path difference by the wavelength. So, \( \frac{\Delta d}{\lambda} = \frac{0.38 \, \text{m}}{0.02 \, \text{m}} = 19\). This means the path difference is 19 wavelengths.
03

Calculate the Phase Difference

Now that we know the path difference in terms of the number of wavelengths, we can calculate the phase difference using the formula \( \Delta \phi = 2\pi \times \frac{\Delta d}{\lambda}\). Therefore, \(\Delta \phi = 2\pi \times 19 = 38\pi\, \text{radians}\). Since phase difference is usually expressed within a range of \(0\) to \(2\pi\), we need to use modulo operation: \(38\pi \mod 2\pi = 0\).

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Key Concepts

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

Coherent Sources
Coherent sources are crucial in understanding wave interference. They are sources that emit waves with a constant phase difference and the same frequency. This means the waves are synchronized over time, maintaining a consistent phase relationship.

In practical terms, coherent sources produce wave patterns that can either constructively or destructively interfere. Constructive interference occurs when the phase difference is a multiple of \(2\pi\), leading to amplified waves. Destructive interference happens at odd multiples of \(\pi\), causing the waves to cancel each other out.
  • Synchronized wave emission
  • Same frequency
  • Constant phase difference
These properties are essential in experiments and technologies involving wave interference, such as laser technology and holography.
Path Difference
Path difference refers to the difference in lengths between the paths taken by two waves from their sources to a specific point. The calculation of path difference is foundational in determining how waves will interfere at a given location.

In the exercise, the path difference \(\Delta d\) was found by subtracting the distance from source \(A\) to point \(P\) from the distance from source \(B\) to point \(P\). The calculation of path difference is given by:
\[\Delta d = 5.24 \, \text{m} - 4.86 \, \text{m} = 0.38 \, \text{m}\]
  • Relates directly to constructive or destructive interference
  • Measured in meters
  • Essential for calculating phase difference
Understanding path difference allows us to predict the interaction of waves, which is critical in fields like radio wave communication and acoustics.
Phase Difference
Phase difference is a measure of how "in-step" or "out-of-step" two waves are, relative to each other. It's typically expressed in radians or degrees. In the context of wave interference, the phase difference determines whether waves will constructively or destructively interfere.

For the exercise, using the determined path difference and wavelength, the phase difference \(\Delta \phi\) is calculated by:
\[\Delta \phi = 2\pi \times \frac{\Delta d}{\lambda}\ = 2\pi \times 19 = 38\pi\, \text{radians}\]
Phase difference influences wave behavior greatly:
  • Constructive interference if \((\Delta \phi\) is multiple of \(2\pi\))
  • Destructive interference if \((\Delta \phi\) is odd multiple of \(\pi\))
By calculating phase difference, engineers and scientists can design systems that rely on precise wave interactions, such as antenna arrays and noise-canceling headphones.
Electromagnetic Waves
Electromagnetic waves are waves that propagate through the electromagnetic field, which encompasses electric and magnetic components. These waves travel at the speed of light in a vacuum and are fundamental to our understanding of classical and quantum physics.

Electromagnetic waves include visible light, radio waves, microwaves, and more. Each type of wave is characterized by its wavelength and frequency. In our exercise, we dealt with electromagnetic waves having a wavelength of 2.00 cm.
  • Consists of oscillating electric and magnetic fields
  • Propagates through vacuum at speed of light \((c = 3 \times 10^8\, \text{m/s})\)
  • Encompasses a broad spectrum from gamma rays to radio waves
Understanding electromagnetic waves is crucial for many applications such as wireless communication, medical imaging (like MRI), and optical devices.

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Most popular questions from this chapter

Young's experiment is performed with light from excited helium atoms \((\lambda=502 \mathrm{nm}) .\) Fringes are measured carefully on a screen 1.20 \(\mathrm{m}\) away from the double slit, and the center of the 20 \(\mathrm{th}\) fringe (not counting the central bright fringe) is found to be 10.6 \(\mathrm{mm}\) from the center of the central bright fringe. What is the separation of the two slits?

What is the thinnest film of a coating with \(n=1.42\) on glass \((n=1.52)\) for which destructive interference of the red component \((650 \mathrm{nm})\) of an incident white light beam in air can take place by reflection?

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 \(\mathrm{nm} ?\) The index of refraction of the film is \(1.33,\) and there is air on both sides of the film.

Nonglare Glass. When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use \(\mathrm{TiO}_{2}\) , which has an index of refraction of \(2.62,\) as the coating, what is the minimum film thickness that will cancel light of wavelength 505 \(\mathrm{nm} ?\) (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

White light reflects at normal incidence from the top and bottom surfaces of a glass plate \((n=1.52) .\) There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 \(\mathrm{nm}\) . What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 \(\mathrm{nm}\) ?

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