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Chapter 35: Problem 2

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 wavelength for which there will be constructive interference at point \(Q ?\)

Short Answer

Expert verified
The longest wavelength for destructive interference at point Q is 80 m and the longest wavelength for constructive interference at point Q is 40m.

Step by step solution

01

Understand the Scenario

There are two radio antennae A and B with a distance of 120 m between them. Point Q is further 40 m from antenna B along the line connecting the two antennae. This means the total distance from A to Q is 160 m. We need to determine the longest wavelengths that cause destructive and constructive interferences at point Q. For destructive interference, the additional distance travelled by the wave from antenna B compared to antenna A must be an odd multiple of half the wavelength. For constructive interference, the additional distance travelled by the wave from antenna B compared to antenna A must be an integer multiple of the wavelength.
02

Determine Longest Wavelength for Destructive Interference

For destructive interference, the path difference must be an odd multiple of half the wavelength i.e. (2n - 1) x (\(位/2\)) where n is a positive integer. Given the path difference is 40 m, this gives us the equation: (2n - 1) x (\(位/2\)) = 40. For the longest wavelength, n should be minimum, which is 1. Solving the equation with n = 1, we obtain \(位 = (40/0.5)= 80 m\)
03

Determine Longest Wavelength for Constructive Interference

For constructive interference, the path difference must be a multiple of the wavelength, i.e. n x \(位\), where n is a positive integer. Given the path difference is 40 m, this gives us the equation: n x \(位\) = 40. For the longest wavelength, n should be minimum, which is 1. Solving the equation with n = 1, we obtain \(位\) = 40m.

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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 form of electromagnetic radiation used for communication purposes, such as radio and television broadcasting.
They travel at the speed of light and can propagate over long distances. Their wavelength can range from one millimeter to several kilometers.
  • Properties: Radio waves are low-frequency and can penetrate through various materials.
  • Usage: Antennas transmit and receive radio waves after converting them into electrical signals.
  • Wavelength: The wavelength determines the wave's frequency, meaning how many wave peaks pass a given point in one second.
In the context of the exercise, two antennas are emitting radio waves. The phase (timing) of these waves is important when considering interference, where waves can either add together or cancel each other out based on their relative phase and wavelength.
Constructive Interference
Constructive interference occurs when two waves meet and combine to form a wave with higher amplitude.
This happens when the crest of one wave aligns perfectly with the crest of another, leading to their energies adding up to amplify the wave.
  • Condition: The path difference between the waves must be an integer multiple of the wavelength, i.e., n x \(位\), where \(位\) is the wavelength and n is any positive integer.
  • Resulting Amplitude: The resulting wave is stronger due to the superimposed peaks.
In the exercise's scenario, at point Q, constructive interference occurs when the path difference between the waves from both antennas is a whole multiple of the wavelength. The challenge was to find the longest wavelength for which this happens, and it turned out to be 40 m when the waves constructively interfere at point Q.
Destructive Interference
Destructive interference takes place when waves meet in such a manner that their amplitudes cancel each other out.
This occurs when the crest of one wave aligns with the trough of another.
  • Condition: The path difference must be an odd multiple of half-wavelength, i.e., \((2n - 1) \times (\frac{位}{2})\).
  • Canceling Amplitude: The waves diminish or completely nullify each other, resulting in minimal or zero amplitude.
In the context of the problem, destructive interference at point Q is achieved with a path difference of 40 m when the wave from antenna B travels 40 m further than from antenna A. The longest wavelength for destructive interference, when \(n = 1\), is found to be 80 m, indicating that the waves interfere destructively at point Q.

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Most popular questions from this chapter

The professor then adjusts the apparatus. The frequency that you hear does not change, but the loudness decreases. Now all of your fellow students can hear the tone. What did the professor do? (a) She turned off the oscillator. (b) She turned down the volume of the speakers. (c) She changed the phase relationship of the speakers. (d) She disconnected one speaker.

Figure \(\mathbf{P} 35.56\) shows an interferometer known as Fresnel's biprism. The magnitude of the prism angle \(A\) is extremely small. (a) If \(S_{0}\) is a very narrow source slit, show that the separation of the two virtual coherent sources \(S_{1}\) and \(S_{2}\) is given by \(d=2 a A(n-1)\), where \(n\) is the index of refraction of the material of the prism. (b) Calculate the spacing of the fringes of green light with wavelength \(500 \mathrm{nm}\) on a screen \(2.00 \mathrm{~m}\) from the biprism. Take \(a=0.200 \mathrm{~m}\) \(A=3.50 \mathrm{mrad},\) and \(n=1.50\)

Laser light of wavelength \(510 \mathrm{nm}\) is traveling in air and shines the at normal incidence onto the flat end of a transparent plastic rod that has \(n=1.30 .\) The end of the rod has a thin coating of a transparent material that has refractive index \(1.65 .\) What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?

A researcher measures the thickness of a layer of benzene \((n=1.50)\) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. She finds that light of wavelength \(575 \mathrm{nm}\) is reflected most strongly from the film. What does she calculate for the minimum thickness of the film?

A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction \(1.52 .\) (a) What minimum thickness is required if light of wavelength \(550 \mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) Coatings as thin as that calculated in part (a) are difficult to manufacture and install. What is the next greater thickness for which constructive interference will also occur?
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