91Ó°ÊÓ

Two radio towers are a distance \(d\) apart as shown in the overhead view. Each antenna by itself would radiate equally in all directions in a horizontal plane. The radio waves have the same frequency and start out in phase. A detector is moved in a circle around the towers at a distance of $100 \mathrm{km}.$ The waves have frequency \(3.0 \mathrm{MHz}\) and the distance between antennas is \(d=0.30 \mathrm{km} .\) (a) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=0^{\circ} ?\) (b) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=90^{\circ} ?\) (c) At how many angles $\left(0 \leq \theta<360^{\circ}\right)$ would you expect to detect a maximum intensity? Explain. (d) Find the angles \((\theta)\) of the maxima in the first quadrant \(\left(0 \leq \theta \leq 90^{\circ}\right) .\) (e) Which (if any) of your answers to parts (a) to (d) would change if the detector were instead only $1 \mathrm{km}$ from the towers? Explain. (Don't calculate new values for the answers.)

Step by step solution

01

(a) Path length difference at \(\theta=0^{\circ}\)

Remember, the path length difference is \(\Delta x = x_2 - x_1\), where \(x_1\) and \(x_2\) are the distances traveled by the two waves from the antenna to the detector. At \(\theta=0^{\circ}\), the detector is in a direct line with both radio towers. The distance traveled by both waves is the same, meaning the path length difference is \(\Delta x = 0\).

02

(b) Path length difference at \(\theta=90^{\circ}\)

When the detector is at \(\theta=90^{\circ}\), we can use the Pythagorean theorem to find the path lengths traveled by two waves: For wave 1: \(x_1 = \sqrt{100^2 - \left(\frac{d}{2}\right)^2}\) For wave 2: \(x_2 = \sqrt{100^2 - \left(-\frac{d}{2}\right)^2}\) Now, we substitute \(d=0.3 \mathrm{km}\) and calculate the path length difference: \(\Delta x = x_2 -x_1 = \sqrt{100^2 - 0.15^2} - \sqrt{100^2 - 0.15^2} = 0.3 \mathrm{km}\)

03

(c) Number of maxima in the range \(0 \leq \theta < 360^{\circ}\)

To detect a maximum intensity, the path difference must be an integer multiple of the wavelength \(\lambda\). We can find the wavelength using the given frequency and the speed of light: \(\lambda = \frac{c}{f} = \frac{3 \times 10^8 \mathrm{m/s}}{3.0 \times 10^6 \mathrm{Hz}}= 100 \mathrm{m} \) As the maximum path length difference is the distance between the radio towers (\(d=0.3 \mathrm{km} = 300 \mathrm{m}\)), the number of maxima can be found by dividing this distance by the wavelength: Number of maxima = \(300 \, \mathrm{m}/ 100 \, \mathrm{m} = 3\)

04

(d) Angles of maxima in the first quadrant

We can find the angles by setting the path length difference, \(\Delta x\), to integer multiples of the wavelength. Let's denote the integer multiples of the wavelength as \(n\lambda\): \(\Delta x = d \sin{\theta} = n\lambda ; \, n = 1,2,3\) Now, substitute the values for \(d\) and \(\lambda\): \(\theta = \sin^{-1}\left(\frac{n\lambda}{d}\right)\) For \(n=1\): \(\theta_1 =\sin^{-1}\left(\frac{100}{300}\right) \approx 19.5^{\circ}\) For \(n=2\): \(\theta_2 =\sin^{-1}\left(\frac{200}{300}\right) \approx 41.8^{\circ}\) For \(n=3\): The value of \(\sin^{-1}\left(\frac{300}{300}\right)\) is greater than \(90^{\circ}\), so we consider only the first two angles in the first quadrant.

05

(e) Changes in answers if the detector were 1 km away

If the detector were only 1 km away, the path length differences would be smaller than before. The wavelength would remain the same because it only depends on the frequency and the speed of the waves. Therefore, there could be fewer maxima since the maximum distance between the towers (\((d=0.30 \mathrm{km})\) would become a more significant portion of the overall distances. So, this change would affect the answers to parts (b), (c), and (d).

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