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Chapter 25: Problem 2

A steep cliff west of Lydia's home reflects a \(1020-\mathrm{kHz}\) radio signal from a station that is \(74 \mathrm{km}\) due east of her home. If there is destructive interference, what is the minimum distance of the cliff from her home? Assume there is a \(180^{\circ}\) phase shift when the wave reflects from the cliff.

Short Answer

Expert verified
Answer: The minimum distance of the cliff from Lydia's home is approximately 5.364 km.

Step by step solution

01

Determine the wavelength of the radio wave

Given the frequency of the radio wave as \(1020 \ \mathrm{kHz}\), convert it to Hz by multiplying it by \(1000\) and then use the speed of light (\(c = 3 \times 10^8 \ \mathrm{m/s}\)) to find the wavelength, using the formula \(\mathrm{wavelength} = \frac{c}{\mathrm{frequency}}\). \(1020 \ \mathrm{kHz} = 1020 \times 1000 \ \mathrm{Hz} = 1.02 \times 10^6 \ \mathrm{Hz}\) Wavelength, denoted as \(\lambda\), is given by: \(\lambda = \frac{c}{f} = \frac{3 \times 10^8 \ \mathrm{m/s}}{1.02 \times 10^6 \ \mathrm{Hz}} = 294.12 \ \mathrm{m}\)
02

Determine the minimum path difference for destructive interference

As both the direct and reflected waves have a \(180^{\circ}\) phase shift, the minimum path difference required for destructive interference must be an odd multiple of half of the wavelength: \(\mathrm{min \ path \ difference} = (2n + 1) \times \frac{\lambda}{2}\), where n=0,1,2,... To find the minimum, let's set \(n = 0\): \(\mathrm{min \ path \ difference} = (2 \times 0 + 1) \times \frac{294.12 \ \mathrm{m}}{2} = 147.06 \ \mathrm{m}\)
03

Use the given information to find the cliff's distance

Let's denote the distance between Lydia's house and the cliff as \(x\) and the distance between the transmitter and Lydia's house as \(74 \ \mathrm{km}\). Since the distance from the station to Lydia's house and the reflected distance to the cliff are horizontal, we can use the Pythagorean theorem: \((x + 74 \times 10^3)^2 - x^2 = (147.06)^2\) Solving for x: \(x^2 + 148 \times 10^3 x + 74^2 \times 10^6 - x^2 = (147.06)^2\) \(148 \times 10^3 x = (147.06)^2\) \(x = \frac{(147.06)^2}{148 \times 10^3} =5363.97\ \mathrm{m}\) So the minimum distance of the cliff from Lydia's home is approximately \(5364 \ \mathrm{m}\) or \(5.364 \ \mathrm{km}\).

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