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Chapter 35: Problem 4

Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) 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 wave-length for which there will be constructive interference at point \(Q ?\)

Short Answer

Expert verified
(a) 240 m, (b) 120 m.

Step by step solution

01

Understand the Problem

We have two radio antennas, A and B, that emit waves in phase. The distance between A and B is 120 meters, and point Q is located 40 meters to the right of antenna B. We are asked to find the longest wavelength for which there is destructive interference and the longest wavelength for constructive interference at point Q.
02

Calculate Path Difference

The path length from antenna A to point Q is the sum of the distances from A to B and from B to Q, which is 160 meters. The path length from antenna B to point Q is 40 meters. The path difference, therefore, is the difference between these two, which is 160 m - 40 m = 120 m.
03

Condition for Destructive Interference

For destructive interference, the path difference should be odd multiples of half the wavelength: \( \Delta d = (m+0.5)\lambda \), where \( m \) is an integer (0, 1, 2,...). The longest wavelength occurs when \( m = 0 \). Thus, \( 120 = 0.5\lambda \) and solving for \( \lambda \), we get \( \lambda = 240 \ m \).
04

Condition for Constructive Interference

For constructive interference, the path difference should be integer multiples of the wavelength: \( \Delta d = m\lambda \), where \( m \) is an integer (0, 1, 2,...). The longest wavelength occurs when \( m = 1 \). Thus, \( 120 = \lambda \) and solving for \( \lambda \), we get \( \lambda = 120 \ m \).
05

Final Answer

Compile the results. For destructive interference, the longest wavelength is 240 meters, and for constructive interference, the longest wavelength is 120 meters.

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Key Concepts

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

Radio Waves
Radio waves are a type of electromagnetic wave used for transmitting information like radio broadcasts, television signals, and even Wi-Fi. These waves are part of the electromagnetic spectrum, which includes other wave types such as microwaves, infrared waves, and visible light.
Radio waves have long wavelengths compared to other forms of waves, which allows them to travel long distances and penetrate through obstacles like buildings and trees. This makes them ideal for communication over vast areas.
When dealing with radio waves, one of the important aspects to consider is their wavelength, especially in situations involving interference, such as in the problem we are discussing.
Path Difference
The concept of path difference is crucial when analyzing interference patterns between waves. It refers to the difference in distance traveled by two waves arriving at the same point.
In the exercise provided, the path difference is the additional distance radio waves from antenna A travel compared to those from antenna B in reaching point Q. The path lengths are 160 meters for antenna A and 40 meters for antenna B, giving a path difference of 120 meters.
Understanding path difference helps determine whether the waves will interfere constructively or destructively when they meet at a point.
Destructive Interference
Destructive interference occurs when two waves meet and cancel each other out, resulting in a lower amplitude or complete nullification of the waves. This phenomenon happens when the waves are out of phase, specifically when their path difference is an odd multiple of half their wavelength. The formula can be expressed as:
\[ \Delta d = (m+0.5)\lambda \]
In this scenario, for the longest wavelength (and hence lowest frequency), where this cancellation occurs most efficiently at point Q, the smallest possible integer value of \( m \) is 0. This gives a longest wavelength of 240 meters, showcasing the condition needed to achieve destructive interference at the farthest possible wavelength.
Constructive Interference
Constructive interference arises when two waves align perfectly with each other, resulting in a combined wave of greater amplitude. This occurs when the waves are in phase and their path difference is a multiple of their wavelength. The formula for this is:
\[ \Delta d = m\lambda \]
In the exercise, at point Q, the path difference of 120 meters for radio waves is exactly equal to one complete wavelength (using the integer \( m = 1 \)). Thus, the longest wavelength for constructive interference in this setup is 120 meters. This means at this wavelength, both radio waves from antennas A and B integrate perfectly to produce a stronger signal at point Q.

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Most popular questions from this chapter

Points \(A\) and \(B\) are 56.0 \(\mathrm{m}\) apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5 -MHz signal horizontally. These transmitters are in phase with each other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 \(\mathrm{km}\) north of the \(A B\) line and initially placed at point \(C,\) directly opposite the midpoint of \(A B .\) The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain a range so that the intensity of the signal it receives from the transmitter is no less than \(\frac{1}{4}\) of its maximum value. How far from point \(C\) (along an east-west line) can the receiver be moved and always be able to pick up the signal?

Jan first uses a Michelson interferometer with the 606 -nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502 -nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at \(\pm 35.20^{\circ}\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm 19.46^{\circ}\) instead. What is the index of refraction of this liquid?

A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction \(1.52 .\) (a) What minimum thickness is required if light with wavelength 550 \(\mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and 600 Hz. The speed of sound is 340 \(\mathrm{m} / \mathrm{s} .\) You find that where you are standing, you hear minimum- intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 \(\mathrm{cm}\) toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?
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