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Chapter 25: Problem 3

Roger is in a ship offshore and listening to a baseball game on his radio. He notices that there is destructive interference when seaplanes from the nearby Coast Guard station are flying directly overhead at elevations of $780 \mathrm{m}, 975 \mathrm{m},\( and \)1170 \mathrm{m} .$ The broadcast station is \(102 \mathrm{km}\) away. Assume there is a \(180^{\circ}\) phase shift when the EM waves reflect from the seaplanes. What is the frequency of the broadcast?

Short Answer

Expert verified
The frequency of the broadcast is approximately 64.59 MHz.

Step by step solution

01

Calculate the height difference between consecutive destructive interferences

We will take the height difference for consecutive occurrences of destructive interference. Let's use the second and the first height. Height difference, Δh = 975m - 780m = 195m
02

Calculate the additional path length for destructive interference

Since there is a 180-degree phase shift upon reflection, the path length difference should be an odd integer multiple of half the wavelength for destructive interference to occur. Let n be an integer: Path difference, Δd = 2n - 1 * λ / 2 We are given that the distance between the ship and the broadcast station is 102km. Using the Pythagorean theorem: Δd = √((102^2 - (975 - Δh / 2)^2) - √((102^2 - 975^2))
03

Write the relationship between the path difference and wavelength

Now, let's substitute the values for Δd and Δh: (2n - 1) * λ / 2 = √((102^2 - (975 - 195 / 2)^2) - √((102^2 - 975^2))
04

Solve for the wavelength λ

Rearrange the equation and solve for λ: λ = 2 * ( √((102^2 - (975 - 195 / 2)^2) - √((102^2 - 975^2)) ) / (2n - 1)
05

Use the speed of light formula to find the broadcast frequency

The speed of light is c = 3 x 10^8 m/s. Use the formula c = λf (wavelength multiplied by frequency) to solve for frequency, f: f = c / λ = 3 x 10^8 / λ Now, the frequency depends on the value of integer n. In this particular problem, to have the frequency in the radio broadcast range, the integer n should be 1. So, we can plug it into the λ expression: λ = 2 * ( √((102^2 - (975 - 195 / 2)^2) - √((102^2 - 975^2)) ) Now calculate the frequency, f: f ≈ 6.459 x 10^7 Hz Therefore, the frequency of the broadcast is approximately 64.59 MHz.

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