/*! 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 51 A \(65-k W\) radio station broad... [FREE SOLUTION] | 91影视

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Chapter 25: Problem 51

A \(65-k W\) radio station broadcasts its signal uniformly in all directions. (a) What is the average intensity of its signal at a distance of \(250 \mathrm{m}\) from the antenna? (b) What is the average intensity of its signal at a distance of \(2500 \mathrm{m}\) from the antenna?

Short Answer

Expert verified
(a) 8.3 W/m虏 at 250 m, (b) 0.083 W/m虏 at 2500 m.

Step by step solution

01

Understand the Concept of Intensity

The intensity of a wave is defined as the power per unit area, which can be expressed as \( I = \frac{P}{A} \). For a radio station emitting uniformly in all directions, this means over a spherical surface area \( A = 4\pi r^2 \) where \( r \) is the distance from the source.
02

Calculate Total Surface Area for 250 m Distance

At a distance of 250 meters, the surface area over which the power is distributed is calculated by \[ A = 4\pi (250)^2. \]
03

Compute Intensity at 250 m

Use the formula \( I = \frac{P}{A} \). Given \( P = 65,000 \text{ W} \), the average intensity at 250 m is \[ I = \frac{65,000}{4\pi (250)^2}. \]
04

Calculate Total Surface Area for 2500 m Distance

For a distance of 2500 meters, calculate the total surface area as \[ A = 4\pi (2500)^2. \]
05

Compute Intensity at 2500 m

Similarly, use \( I = \frac{P}{A} \) to find the intensity at 2500 m: \[ I = \frac{65,000}{4\pi (2500)^2}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Distribution
When a radio station emits a signal, the power of that signal is distributed over a certain area. This is similar to spreading butter on a toast; the larger the toast, the thinner the butter if the amount of butter stays the same. The key to understanding power distribution in radio signals is to see how this power is shared over a spherical area. The concept can be expressed as
  • Power Density: This tells us how much power is spread over a unit area.
  • Intensity: This is the power density at a given point, essentially the concentration of power.
The formula used is: \[ I = \frac{P}{A} \] Here, \( I \) stands for intensity, \( P \) is the power, and \( A \) is the area over which the power is spread. This is an essential concept in understanding how the signal weakens as we move away from the source.
Spherical Surface Area
In many radio signal problems, the signal spreads out in a spherical shape from the source. This is because signals often radiate equally in all directions, just like how light spreads from a lamp. To compute the intensity, we need to calculate the surface area of this imaginary sphere at a given distance from the source.
  • Spherical Geometry: Understanding circles and spheres is crucial; circles are to flat surfaces what spheres are to space.
  • Surface Area Formula: The formula for the surface area of a sphere is: \( A = 4\pi r^2 \).
Where \( r \) is the radius or the distance from the source to the point where we're measuring intensity. This area grows with the square of the distance (\( r^2 \)), meaning that as you move further away, the area gets much larger, and thus, the intensity decreases.
Distance from Source
The distance from the source is a critical factor in determining the intensity of a radio signal. As you move further from the transmitter, the total surface area over which the signal is distributed increases, which in turn reduces the intensity. Here are some important points to consider:
  • Exponential Increase in Area: The spherical area increases exponentially because of the square in \( r^2 \). This means small increases in distance create large increases in area.
  • Direct Impact on Intensity: The formula for intensity \( I = \frac{P}{4\pi r^2} \) shows that intensity is inversely proportional to the square of the distance. Doubling the distance from the source results in a reduction of intensity by a factor of four.
Being mindful of this relationship helps to predict and analyze how strong a signal will be at various distances. Understanding these principles is fundamental in fields like telecommunications and broadcasting where signal coverage needs to be maximized efficiently.

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Most popular questions from this chapter

Unpolarized light with intensity \(I_{0}\) falls on a polarizing filter whose transmission axis is vertical. The axis of a second polarizing filter makes an angle of \(\theta\) with the vertical. Plot a graph that shows the intensity of light transmitted by the second filter (expressed as a fraction of \(l_{0}\) ) as a function of \(\theta\). Your graph should cover the range \(\theta=0^{\circ}\) to \(\theta=360^{\circ}\).

A cell phone transmits at a frequency of \(1.75 \times 10^{8} \mathrm{Hz}\). What is the wavelength of the electromagnetic wave used by this phone?

Experiments show that the ground spider Drassodes cupreus uses one of its several pairs of eyes as a polarization detector, In fact, the two eyes in this pair have polarization directions that are at right angles to one another. Suppose linearly polarized light with an intensity of \(825 \mathrm{W} / \mathrm{m}^{2}\) shines from the sky onto the spider, and that the intensity transmitted by one of the polarizing eyes is \(232 \mathrm{W} / \mathrm{m}^{2}\) (a) For this eye, what is the angle between the polarization direction of the eye and the polarization direction of the incident light? (b) What is the intensity transmitted by the other polarizing eye?

The sugar concentration in a solution (e.g. in a urine specimen) can be measured conveniently by using the optical activity of sugar and other asymmetric molecules. In general, an optically active molecule, like sugar, will rotate the plane of polarization through an angle that is proportional to the thickness of the sample and to the concentration of the molecule. To measure the concentration of a given solution, a sample of known thickness is placed between two polarizing filters that are at right angles to each other, as shown in Figure \(25-34\). The intensity of light transmitted through the two filters can be compared with a calibration chart to determine the concentration. (a) What percentage of the incident (unpolarized) light will pass through the first filter? (b) If no sample is present, what percentage of the initial light will pass through the second filter? (c) When a particular sample is placed between the two filters, the intensity of light emerging from the second filter is \(40.0 \%\) of the incident intensity. Through what angle did the sample rotate the plane of polarization? (d) A second sample has half the sugar concentration of the first sample. Find the intensity of light emerging from the second filter in this case.

A typical home may require a total of \(2.00 \times 10^{3} \mathrm{kWh}\) of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of \(1.00 \times 10^{3} \mathrm{W} / \mathrm{m}^{2} .\) Assuming that sunlight is available \(8.0 \mathrm{h}\) per day, 25 d per month (accounting for cloudy days), and that you have a way to store energy from your collector when the Sun isn't shining, determine the smallest collector size that will provide the needed energy, given a conversion efficiency of \(25 \%\)
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