/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 A steep cliff west of Lydia's ho... [FREE SOLUTION] | 91影视

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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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Most popular questions from this chapter

When a double slit is illuminated with light of wavelength \(510 \mathrm{nm},\) the interference maxima on a screen \(2.4 \mathrm{m}\) away gradually decrease in intensity on either side of the 2.40 -cm-wide central maximum and reach a minimum in a spot where the fifth-order maximum is expected. (a) What is the width of the slits? (b) How far apart are the slits?
In a double-slit interference experiment, the wavelength is \(475 \mathrm{nm}\), the slit separation is \(0.120 \mathrm{mm},\) and the screen is $36.8 \mathrm{cm}$ away from the slits. What is the linear distance between adjacent maxima on the screen? [Hint: Assume the small-angle approximation is justified and then check the validity of your assumption once you know the value of the separation between adjacent maxima.] (tutorial: double slit 1 ).
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?
A lens \((n=1.52)\) is coated with a magnesium fluoride film \((n=1.38) .\) (a) If the coating is to cause destructive interference in reflected light for \(\lambda=560 \mathrm{nm}\) (the peak of the solar spectrum), what should its minimum thickness be? (b) At what two wavelengths closest to 560 nm does the coating cause constructive interference in reflected light? (c) Is any visible light reflected? Explain.
The photosensitive cells (rods and cones) in the retina are most densely packed in the fovea- the part of the retina used to see straight ahead. In the fovea, the cells are all cones spaced about \(1 \mu \mathrm{m}\) apart. Would our vision have much better resolution if they were closer together? To answer this question, assume two light sources are just far enough apart to be resolvable according to Rayleigh's criterion. Assume an average pupil diameter of \(5 \mathrm{mm}\) and an eye diameter of \(25 \mathrm{mm}\). Also assume that the index of refraction of the vitreous fluid in the eye is \(1 ;\) in other words, treat the pupil as a circular aperture with air on both sides. What is the spacing of the cones if the centers of the diffraction maxima fall on two nonadjacent cones with a single intervening cone? (There must be an intervening dark cone in order to resolve the two sources; if two adjacent cones are stimulated, the brain assumes a single source.)
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