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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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Most popular questions from this chapter

In bright light, the pupils of the eyes of a cat narrow to a vertical slit \(0.30 \mathrm{mm}\) across. Suppose that a cat is looking at two mice $18 \mathrm{m}$ away. What is the smallest distance between the mice for which the cat can tell that there are two mice rather than one using light of $560 \mathrm{nm} ?$ Assume the resolution is limited by diffraction only.
A pinhole camera doesn't have a lens; a small circular hole lets light into the camera, which then exposes the film. For the sharpest image, light from a distant point source makes as small a spot on the film as possible. What is the optimum size of the hole for a camera in which the film is $16.0 \mathrm{cm}$ from the pinhole? A hole smaller than the optimum makes a larger spot since it diffracts the light more. A larger hole also makes a larger spot because the spot cannot be smaller than the hole itself (think in terms of geometrical optics). Let the wavelength be \(560 \mathrm{nm}\).
A grating spectrometer is used to resolve wavelengths \(660.0 \mathrm{nm}\) and \(661.4 \mathrm{nm}\) in second order. (a) How many slits per centimeter must the grating have to produce both wavelengths in second order? (The answer is either a maximum or a minimum number of slits per centimeter.) (b) The minimum number of slits required to resolve two closely spaced lines is $N=\lambda /(m \Delta \lambda),\( where \)\lambda$ is the average of the two wavelengths, \(\Delta \lambda\) is the difference between the two wavelengths, and \(m\) is the order. What minimum number of slits must this grating have to resolve the lines in second order?
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.)
About how close to each other are two objects on the Moon that can just barely be resolved by the \(5.08-\mathrm{m}-\) (200-in.)-diameter Mount Palomar reflecting telescope? (Use a wavelength of \(520 \mathrm{nm}\).)
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