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Chapter 35: Problem 4

Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 m to the right of antenna \(A\). Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna \(B\). The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q\)? (b) What is the longest wavelength for which there will be constructive interference at point \(Q\)?

Short Answer

Expert verified
The longest wavelength for destructive interference is 240 m, and for constructive interference is 120 m.

Step by step solution

01

Understanding the Problem

We need to determine the conditions for both destructive and constructive interference at the point Q on the line that extends through the two radio antennas, A and B. Given the distances, antenna B is 120 m from antenna A, and point Q is 40 m from antenna B.
02

Calculate the Path Difference

The total distance from antenna A to point Q is 120 m + 40 m = 160 m. The distance from antenna B to point Q is 40 m. The path difference, \( \Delta d \), is the difference between these two distances: \( \Delta d = 160 \text{ m} - 40 \text{ m} = 120 \text{ m} \).
03

Determine Condition for Destructive Interference

For destructive interference, the path difference should be an odd multiple of half the wavelength, i.e., \( \Delta d = \left( n + \frac{1}{2} \right) \lambda \), where \( n \) is an integer (0, 1, 2, ...). The longest wavelength occurs when \( n = 0 \). Setting \( n = 0 \) gives: \( 120 = \frac{1}{2} \lambda \), so \( \lambda = 240 \) m.
04

Determine Condition for Constructive Interference

For constructive interference, the path difference should be a multiple of the wavelength, i.e., \( \Delta d = m \lambda \), where \( m \) is an integer. The longest wavelength occurs when \( m = 1 \). Using \( \Delta d = 120 \) m gives: \( 120 = 1 \cdot \lambda \), so \( \lambda = 120 \) m.

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Key Concepts

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

Destructive Interference
When waves meet, their effects can either cancel each other out or amplify. Destructive interference happens when the peak of one wave meets the trough of another, leading them to cancel.
The result is often a reduction in wave amplitude or, in the case of identical waves, complete cancellation.
For destructive interference to occur, the path difference (\(\Delta d\)) between two waves must be an odd multiple of half the wavelength (\(\lambda\)).
  • The formula for the condition is \[\Delta d = \left(n + \frac{1}{2}\right) \lambda\], where \(n\) is an integer (0, 1, 2, ...).
  • The longest wavelength for destructive interference is when \(n = 0\), simplifying the formula to \(\Delta d = \frac{\lambda}{2}\).
In our problem, \(\Delta d\) is 120 m, leading to \(\lambda = 240\) m when \(n = 0\).This means point \(Q\) experiences destructive interference at this wavelength.
Constructive Interference
Constructive interference occurs when waves meet in such a way that their crests and troughs align.
This alignment leads the wave amplitudes to add together, resulting in a stronger and more pronounced wave.
The condition for constructive interference is met when the path difference (\(\Delta d\)) is a whole number multiple of the wavelength (\(\lambda\)).
  • Expressed mathematically as \[\Delta d = m \lambda\], where \(m\) is an integer (1, 2, 3, ...).
  • The longest wavelength occurs when \(m = 1\), as it represents the fundamental mode of interference.
In our scenario, with \(\Delta d = 120\) m, the longest wavelength for constructive interference is \(\lambda = 120\) m, making point \(Q\) a zone of amplification at this wavelength.
Path Difference
Path difference is critical in understanding wave interference. It refers to the difference in distance traveled by two waves from their sources to a common point.
In terms of interference, this difference determines whether waves will constructively or destructively interfere at that point.
The concept can be explained as follows:
  • For constructive interference, the path difference should be a multiple of the wavelength: \(m \lambda\).
  • For destructive interference, it should be an odd multiple of half the wavelength: \(\left(n+\frac{1}{2}\right) \lambda\).
In the given exercise, the path difference from antennas \(A\) and \(B\) to point \(Q\) is calculated as 120 m (160 m - 40 m).
This path difference guides us in determining the conditions for both constructive and destructive interference.
Wavelength Calculation
Calculating the wavelength is key to understanding interference patterns. Wavelength (\(\lambda\)) can be found using the interference conditions based on the path difference.
The method relies on understanding how the path difference relates to \(\lambda\):
  • For constructive interference, use the formula \(\Delta d = m \lambda\)
  • For destructive interference, use \(\Delta d = \left(n + \frac{1}{2}\right) \lambda\)
By plugging the path difference and the required condition's integer (\(m\) or \(n\)) into these formulas, we can solve for \(\lambda\).
In our problem, the given path difference of 120 m helps determine wavelengths for both types of interference:
  • For constructive interference: \(\lambda = 120\) m (\(m = 1\)
  • For destructive interference: \(\lambda = 240\) m (\(n = 0\)
This comprehensive understanding assists in visualizing how waves interact at different points.

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Most popular questions from this chapter

Eyeglass lenses can be coated on the \(inner\) surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm\)19.0\(^\circ\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(1 \over 10\) the maximum intensity on the screen?

A plate of glass 9.00 cm long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 mm thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 nm. How many interference fringes are observed per centimeter in the reflected light?

Red light with wavelength 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe (\(m\) = 32) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_0\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_0\) = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?
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