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Chapter 34: Problem 56

Many times, radio antennas occur in pairs. The effect is that they will then produce constructive interference in one direction while producing destructive interference in another direction - a directional antenna-so that their emissions don't overlap with nearby stations. How far apart at a minimum should a local radio station, operating at \(88.1 \mathrm{MHz},\) place its pair of antennae operating in phase such that no emission occurs along a line \(45.0^{\circ}\) from the line joining the antennae?

Short Answer

Expert verified
Answer: The minimum distance between the two radio antennas is approximately 2.41 m.

Step by step solution

01

Calculate the wavelength of the radio wave

First, we need to find the wavelength (λ) of the radio wave. To calculate the wavelength, we use the formula: \(λ = \dfrac{c}{f}\) Where c is the speed of light in a vacuum (approximately \(3*10^8 m/s\)) and f is the frequency of the radio wave. In this case, the frequency is \(88.1 \mathrm{MHz}\): \(λ = \dfrac{3*10^8 m/s}{88.1 * 10^6 Hz}\)
02

Solve for the wavelength

Now, we will calculate the wavelength of the radio wave: \(λ = \dfrac{3*10^8 m/s}{88.1 * 10^6 Hz} = 3.41 m\) The wavelength of the radio wave is 3.41 m.
03

Apply the destructive interference condition

The destructive interference occurs when the path difference between the two radio waves is equal to an odd multiple of half wavelength (n(λ/2)), where n is an odd integer: Path difference = \(n * \cfrac{λ}{2}\) In this problem, we want to find the minimum distance, which means n = 1. Therefore, the path difference between the two radio waves should be: Path difference = \(\cfrac{λ}{2} = \cfrac{3.41 m}{2} = 1.705 m\)
04

Apply trigonometry to find the minimum distance between the antennae

We can use the formula for sine function in a right triangle: \(\sin (θ) = \cfrac{\textrm{opposite side}}{\textrm{hypotenuse}}\) Since we know the angle (45 degrees) and the path difference (opposite side) is 1.705 m, we can use the sine function to find the minimum distance between the antennae (hypotenuse): \(\sin(45^{\circ}) = \cfrac{1.705 m}{\textrm{hypotenuse}}\) Now, solve for the hypotenuse: Hypotenuse = \(\cfrac{1.705 m}{\sin(45^{\circ})}\)
05

Calculate the minimum distance between the antennae

Finally, we compute the minimum distance between the antennae: Hypotenuse = \(\cfrac{1.705 m}{\sin(45^{\circ})} = 2.41 m\) The minimum distance between the two antennas to produce destructive interference at an angle of 45 degrees from the line joining the antennae is approximately 2.41 m.

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Most popular questions from this chapter

The Michelson interferometer is used in a class of commercially available optical instruments called wavelength meters. In a wavelength meter, the interferometer is illuminated simultaneously with the parallel beam of a reference laser of known wavelength and that of an unknown laser. The movable mirror of the interferometer is then displaced by a distance \(\Delta d,\) and the number of fringes produced by each laser and passing by a reference point (a photo detector) is counted. In a given wavelength meter, a red He-Ne laser \(\left(\lambda_{\mathrm{Red}}=632.8 \mathrm{nm}\right)\) is used as a reference laser. When the movable mirror of the interferometer is displaced by a distance \(\Delta d\), a number \(\Delta N_{\text {Red }}=6.000 \cdot 10^{4}\) red fringes and \(\Delta N_{\text {unknown }}=7.780 \cdot 10^{4}\) fringes pass by the reference photodiode. a) Calculate the wavelength of the unknown laser. b) Calculate the displacement, \(\Delta d\), of the movable mirror.

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A \(5.000-\mathrm{cm}\) -wide diffraction grating with 200 grooves is used to resolve two closely spaced lines (a doublet) in a spectrum. The doublet consists of two wavelengths, \(\lambda_{\mathrm{a}}=\) \(629.8 \mathrm{nm}\) and \(\lambda_{\mathrm{b}}=630.2 \mathrm{nm} .\) The light illuminates the entire grating at normal incidence. Calculate to four significant digits the angles \(\theta_{1 \mathrm{a}}\) and \(\theta_{1 \mathrm{~b}}\) with respect to the normal at which the first-order diffracted beams for the two wavelengths, \(\lambda_{\mathrm{a}}\) and \(\lambda_{\mathrm{b}}\), respectively, will be reflected from the grating. Note that this is not \(0^{\circ}\) What order of diffraction is required to resolve these two lines using this grating?
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