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Ch. 34 - The Wave Nature of Light: Interference and Polarization
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 33, Problem 78

Consider two antennas radiating 6.8-MHz radio waves in phase with each other. They are located at points S₁ and S₂, separated by a distance d = 175 m, Fig. 34–50. Determine the points on the positive y-axis where the signals from the two sources will be out of phase (crests of one meet troughs of the other).

Verified step by step guidance
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Step 1: Understand the problem. The two antennas radiate 6.8-MHz radio waves in phase, and we need to find points on the positive y-axis where the signals are out of phase. This occurs when the path difference between the waves from S₁ and S₂ is an odd multiple of half the wavelength.
Step 2: Calculate the wavelength of the radio waves using the formula λ = c / f, where c is the speed of light (approximately 3 × 10⁸ m/s) and f is the frequency (6.8 MHz). Substitute the values to find λ.
Step 3: Determine the condition for destructive interference. Destructive interference occurs when the path difference ΔL = |L₁ - L₂| is equal to (n + 0.5)λ, where n is an integer (0, 1, 2, ...). Here, L₁ and L₂ are the distances from S₁ and S₂ to a point on the positive y-axis.
Step 4: Express the distances L₁ and L₂ geometrically. For a point on the positive y-axis at a distance y from the origin, L₁ = √(y²) = y and L₂ = √(y² + d²), where d = 175 m is the separation between the antennas.
Step 5: Solve for y by setting the path difference ΔL = |L₁ - L₂| equal to (n + 0.5)λ. Substitute L₁ = y, L₂ = √(y² + d²), and λ from Step 2 into the equation. Rearrange and solve for y for different values of n to find the points on the positive y-axis where destructive interference occurs.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Wave Interference

Wave interference occurs when two or more waves overlap and combine to form a new wave pattern. This can result in constructive interference, where wave crests align, or destructive interference, where a crest meets a trough, leading to cancellation. Understanding this concept is crucial for analyzing how the signals from the two antennas interact at various points in space.
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Phase Difference

Phase difference refers to the difference in the phase of two waves at a given point in time. For waves to be out of phase, the phase difference must be an odd multiple of π (180 degrees), meaning that the crest of one wave coincides with the trough of another. This concept is essential for determining the specific points along the y-axis where the signals from the antennas will interfere destructively.
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Path Length Difference

Path length difference is the difference in distance traveled by two waves from their sources to a common point. In this scenario, the path length difference between the waves from antennas S₁ and S₂ must equal half the wavelength for destructive interference to occur. Calculating this difference helps identify the locations on the y-axis where the signals will be out of phase.
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Related Practice
Textbook Question

Two narrow slits 0.070 mm apart are illuminated by a very bright 488-nm light source forming an interference pattern on a screen 4.0 m away. Calculate (a) the distance between the m = 0 and m = 1 lines in the pattern and (b) the distance between the m = 100 and m = 101 lines.

Textbook Question

A highly reflective mirror can be made for a particular wavelength at normal incidence by using two thin layers of transparent materials of indices of refraction n₁ and n₂ ( 1 < n₁ < n₂ ) on the surface of the glass (n > n₂). What should be the minimum thicknesses d₁ and d₂ in Fig. 34–49 in terms of the incident wavelength λ, to maximize reflection?

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Textbook Question

Suppose the mirrors in a Michelson interferometer are perfectly aligned and the path lengths to mirrors M₁ and M₂ are identical. With these initial conditions, an observer sees a bright maximum at the center of the viewing area. Now one of the mirrors is moved a distance x. Determine a formula for the intensity at the center of the viewing area as a function of x, the distance the movable mirror is moved from the initial position.

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Textbook Question

Two polarizers are oriented at 55° to each other and plane-polarized light is incident on them. If only 25% of the light gets through both of them, what was the initial polarization direction of the incident light?

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Textbook Question

Light of wavelength 690 nm passes through two narrow slits 0.66 mm apart. The screen is 1.75 m away. A second source of unknown wavelength produces its second-order fringe 1.23 mm closer to the central maximum than the 690-nm light. What is the wavelength of the unknown light?

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Textbook Question

Unpolarized light falls on two polarizer sheets whose axes are at right angles. (a) What fraction of the incident light intensity is transmitted? (b) What fraction is transmitted if a third polarizer is placed between the first two so that its axis makes a 58° angle with the axis of the first polarizer? (c) What if the third polarizer is in front of the other two?

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