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Chapter 32: Problem 25

An intense light source radiates uniformly in all directions. At a distance of 5.0\(\mathrm { m }\) from the source, the radiation pressure on a perfectly absorbing surface is \(9.0 \times 10 ^ { - 6 }\) Pa. What is the total average power output of the source?

Short Answer

Expert verified
The total average power output of the source is approximately 848.23 W.

Step by step solution

01

Understanding Radiation Pressure

Radiation pressure is the pressure exerted by the electromagnetic radiation on a surface. For a perfectly absorbing surface, the radiation pressure \( P \) is related to the intensity \( I \) of the light as \( P = \frac{I}{c} \), where \( c \) is the speed of light \( (3 imes 10^8 \, \text{m/s}) \). We know \( P = 9.0 \times 10^{-6} \, \text{Pa} \).
02

Calculating Intensity

Rearrange the formula \( P = \frac{I}{c} \) to find the intensity: \( I = P \times c \). Substituting \( P = 9.0 \times 10^{-6} \, \text{Pa} \) and \( c = 3 imes 10^8 \, \text{m/s} \), we get \( I = (9.0 \times 10^{-6}) \times (3 \times 10^8) = 2.7 \, \text{W/m}^2 \).
03

Understanding Intensity and Power

Intensity \( I \) is the power \( P \) per unit area \( A \). In a spherical distribution, intensity is related to the total power output by \( I = \frac{P}{A} \), where \( A = 4\pi r^2 \) is the surface area of a sphere with radius \( r \).
04

Calculating the Surface Area at Distance

The distance \( r = 5.0 \, \text{m} \). Substitute into the area formula \( A = 4\pi r^2 \) to get \( A = 4 \pi (5.0)^2 = 100 \pi \, \text{m}^2 \).
05

Calculating the Total Power Output

Using \( I = \frac{P}{A} \), rearrange to find \( P = I \times A \). Substitute \( I = 2.7 \, \text{W/m}^2 \) and \( A = 100 \pi \, \text{m}^2 \): \( P = 2.7 \times 100 \pi = 270 \pi \, \text{W} \approx 848.23 \, \text{W} \).

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Key Concepts

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

Intensity of Light
The intensity of light is a measure of the power per unit area that is incident on a surface. It tells us how much energy is hitting a specific spot at a given time. In physics, intensity is usually expressed in watts per square meter (W/m虏). Intensity can vary based on the light source and the distance from it. The closer you are to a light source, the more intense the light appears because there is more energy concentrated in a smaller area. Intensity plays a crucial role in calculating radiation pressure, especially when considering surfaces that either absorb or reflect light.
Power Output
Power output refers to the total amount of energy a light source emits per second. In our context of light, we often measure power output in watts (W). Power output is crucial for determining other properties like intensity. To find the total power output in problems involving radiation pressure and intensity, one might have to consider the distribution of light over a specific area, as well as the distance from the source. When dealing with spherical light sources, power output helps to determine how much light reaches a particular point at a specific distance away.
Electromagnetic Radiation
Electromagnetic radiation encompasses a range of energy forms, including light. It refers to the waves of the electromagnetic field that carry energy across space. These waves include not just visible light but also radio waves, microwaves, infrared, ultraviolet, X-rays, and gamma rays. The different types of electromagnetic radiation differ in the frequency and wavelength of their waves. When electromagnetic radiation strikes a surface, it can exert a force known as radiation pressure, which is the focus of our problem. Understanding this concept is pivotal when analyzing how light interactions bring about measurable effects like changes in pressure.
Absorbing Surface
An absorbing surface is one that takes in all the electromagnetic radiation that falls on it. In other words, it doesn't reflect or transmit light. This kind of surface is ideal for accurately measuring radiation pressure because none of the incoming light energy is lost. In solving problems related to radiation pressure and intensity, recognizing whether a surface is perfectly absorbing or partially reflecting is essential. In our problem example, calculating radiation pressure involves knowing that we have a perfectly absorbing surface, which simplifies the relationship between intensity and radiation pressure.
Uniform Radiation
Uniform radiation describes a situation where electromagnetic energy is spread out evenly in all directions from a source. This means that at any given distance from the source, the intensity of the radiation is the same, assuming no other intervening medium or object that affects the spread. Uniformity is often assumed in physics problems to simplify calculations, as it allows us to apply the same intensity across a spherical surface around the light source. Understanding uniform radiation is crucial when considering scenarios where radiation pressure affects surfaces equally, a key factor in our exercise.
Spherical Distribution
Spherical distribution refers to the way energy radiates from a point source like a light bulb or the sun. The energy moves outward in a spherical shape. This spherical pattern means that as distance from the source increases, the energy is spread over a larger area. With this, the intensity decreases following the inverse square law. For solving problems related to power output and intensity, understanding spherical distribution helps determine the area over which the light's power is distributed, affecting calculations of how much energy strikes a particular surface area.
Speed of Light
The speed of light, denoted as "c," is a fundamental constant in physics, essential for calculations involving electromagnetic radiation. It is approximately equal to \(3 \times 10^8\) meters per second.In calculations involving radiation pressure, the speed of light helps link the intensity of the radiation to the pressure it exerts when striking a surface. Understanding this relationship forms the basis for calculating other associated variables like power and area.The speed of light is fundamental to the equations used in the original exercise, such as the relationship between intensity, radiation pressure, and electromagnetic radiation traveling through space in spherical distributions.

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Most popular questions from this chapter

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho .\) (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r .\) The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about 3000\(\mathrm { kg } / \mathrm { m } ^ { 3 } .\) Find the particle radius \(R\) such the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10 ^ { 26 }\) W. Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. LHint: Construct the ratio of the two force expressions found in parts (a) and (b).]

A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4\(\mathrm { cm }\) and an electric-field amplitude of \(5.40 \times 10 ^ { - 2 } \mathrm { V } / \mathrm { m }\) at a distance of 250\(\mathrm { m }\) from the phone. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; ( ) the intensity of the wave.

The sun emits energy in the form of electromagnetic waves at a rate of \(3.9 \times 10 ^ { 26 } \mathrm { W }\) . This energy is produced by nuclear reactions deep in the sun's interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius \(r = R = 6.96 \times 10 ^ { 5 } \mathrm { km }\) ) and at \(r = R / 2 ,\) in the sun's interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 32.4 for sunlight just before it enters the earth's atmosphere. (b) The gas pressure at the sun's surface is about \(1.0 \times 10 ^ { 4 }\) Pa; at \(r = R / 2 ,\) the gas pressure is calculated from solar models to be about \(4.7 \times 10 ^ { 13 }\) Pa. Comparing with your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not?

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec { \boldsymbol { E } }\) in the \(+ x\) -direction, \(\vec { \boldsymbol { B } }\) in the \(+ y\) - direction; (b) \(\vec { \boldsymbol { E } }\) in the \(- y\) - direction, \(\vec { \boldsymbol { B } }\) in the \(+ x\) -direction; \(( \mathrm { c } ) \vec { \boldsymbol { E } }\) in the \(+ z\) -direction, \(\vec { \boldsymbol { B } }\) in the \(- x\) -direction; (d) \(\vec { \boldsymbol { E } }\) in the \(+ y\) -direction, \(\vec { \boldsymbol { B } }\) in the \(- z\) -direction.

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0\(\mathrm { MHz }\) in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing-wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing-wave pattern is determined to be in its eighth harmonic, how long is the cavity?
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