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Chapter 35: Problem 1

35.1. Two coherent sources \(A\) and \(B\) of radio waves are 5.00 \(\mathrm{m}\) apart. Each source emits waves with wavelength 6.00 \(\mathrm{m}\) . Consider points along the line between the two sources. At what distances, if any, from \(A\) is the interference (a) constructive and (b) destructive?

Short Answer

Expert verified
(a) Constructive at 2.5 m from A. (b) No destructive points between A and B.

Step by step solution

01

Understand the Problem

We have two coherent sources of radio waves, A and B, 5.00 meters apart with a wavelength of 6.00 meters. We need to find out where along the line between A and B the interference is constructive or destructive.
02

Define Constructive Interference Condition

Constructive interference occurs where the path difference between the waves from A and B is an integer multiple of the wavelength. The condition is \(d_B - d_A = n\lambda\), where \(n\) is an integer (0, 1, 2,...), \(d_A\) is the distance from A, \(d_B\) is the distance from B, and \(\lambda = 6.00\,\mathrm{m}\).
03

Define Destructive Interference Condition

Destructive interference occurs where the path difference is an odd multiple of half the wavelength, i.e., \(d_B - d_A = (n + 0.5)\lambda\).
04

Express Distance Between Points

Since the line is between A and B, we can write \(d_A + d_B = 5.00\, \mathrm{m}\). Thus \(d_B = 5.00\, \mathrm{m} - d_A\).
05

Solve for Constructive Interference

Substitute \(d_B = 5.00\, \mathrm{m} - d_A\) into the constructive condition: \((5.00 - d_A) - d_A = n \times 6.00\). Simplify to get \(5.00 - 2d_A = n \times 6.00\). Solving for \(d_A\), we get \(d_A = \frac{5.00 - 6.00n}{2}\).
06

Solve for Destructive Interference

Substitute \(d_B = 5.00\, \mathrm{m} - d_A\) into the destructive condition: \((5.00 - d_A) - d_A = (n + 0.5)\times 6.00\). Simplify to get \(5.00 - 2d_A = (n + 0.5)\times 6.00\). Solving for \(d_A\), we get \(d_A = \frac{5.00 - 6.00(n + 0.5)}{2}\).
07

Calculate Valid Distances for Constructive Interference

We calculate the values of \(d_A\) using \(d_A = \frac{5.00 - 6.00n}{2}\). For valid real-world solutions, \(d_A\) must be between 0 and 5. Solve for different integer values of \(n\) to find \(n=0\), giving \(d_A = 2.5\,\mathrm{m}\). Higher values of \(n\) result in invalid distances.
08

Calculate Valid Distances for Destructive Interference

Now calculate \(d_A\) using \(d_A = \frac{5.00 - 6.00(n + 0.5)}{2}\). For valid solutions between 0 and 5, solve for different \(n\). Using \(n = 0\), we find \(d_A = -0.75\,\mathrm{m}\) which is invalid. Using \(n = 1\), we find \(d_A = -3.75\,\mathrm{m}\) which is also invalid. It seems no valid solutions for these distances.

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Key Concepts

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

Constructive Interference
In the beautiful world of waves, constructive interference is like a grand concert where multiple voices come together in harmony. When two waves meet this way, they effectively add up, resulting in a higher amplitude. This occurs when the path difference between the two waves is an integer multiple of the wavelength. So, if the difference in their paths is zero, one wavelength, two wavelengths, or so on, they combine perfectly.
  • The formula for constructive interference is given as: \[ d_B - d_A = n\lambda \]
  • Here, \(n\) is an integer (0, 1, 2,...), representing multiples of the wavelength \(\lambda\).
  • In the exercise, with wavelength \(\lambda = 6.00\, \mathrm{m}\), setting \(n = 0\) gives us a constructive point at \(d_A = 2.5\, \mathrm{m}\) from source \(A\).

The key idea is that as long as the path differences are in whole numbers of wavelengths, the waves add up constructively, producing a point of higher intensity on a line of such points. This makes constructive interference an intriguing phenomenon where alignment creates enhanced effects.
Destructive Interference
Destructive interference, on the other hand, is like two opposing forces cancelling each other out. Picture two people pushing equally on a door from opposite sides; the door stays put. In the world of waves, this happens when the path difference between them is an odd multiple of half the wavelength.
  • The condition for destructive interference can be expressed as: \[ d_B - d_A = (n + 0.5)\lambda \]
  • Here, \(n\) is an integer (0, 1, 2,...), and \(0.5\lambda\) signifies half the wavelength, which causes the waves to cancel out partially.
  • Destructive interference would have been possible if valid distance `d_A` were found, but in the exercise, no such valid real-world solution exists between 0 and 5 meters.

It's fascinating because such interference can cause waves to diminish or vanish at specific points. Where constructive interference amplifies, destructive interference neutralizes, showing how waves can interact in complex ways.
Coherent Sources
Coherent sources are the true champions when it comes to predictable wave interference. Imagine a choir where each singer knows their part perfectly, staying in sync with one another. Two wave sources are coherent when they emit waves with a constant phase relationship, meaning the frequency and wavelength are the same.
  • They need to maintain a consistent phase difference to continue the constructive or destructive patterns over time.
  • With radio waves, coherence ensures waves meet predictably, creating stable interference patterns.
  • For our exercise, sources \(A\) and \(B\) are coherent, maintaining a fixed frequency and wavelength, ensuring predictable interference points along the line between them.

Coherent sources are crucial to deriving understandable and usable interference patterns as they maintain a constant phase, making complex wave interactions more straightforward to analyze and predict.

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Most popular questions from this chapter

35.34. Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

35.28. 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.

35.17. Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ? )\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

35.19. In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ}\) , what is the intensity? (b) What is the path difference for 480 -nm light from the two slits at a point where the phase angle is \(60.0^{\circ} ?\)

35.57. Two thin parallel slits are made in an opaque sheet of film. When a monochromatic beam of light is shone through them at normal incidence, the first bright fringes in the transmitted light occur in air at \(\pm 18.0^{\circ}\) with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid, the same bright fringes now occur at \(\pm 12.6^{\circ} .\) Find the index of refraction of the liquid.
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