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Chapter 32: Problem 25

An intense light source radiates uniformly in all directions. At a distance of 5.0 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 power output of the source is approximately \( 8.5 \times 10^5 \) W.

Step by step solution

01

Understanding Radiation Pressure

Radiation pressure is the force exerted by electromagnetic radiation, such as light, on a surface. For a perfectly absorbing surface, the radiation pressure \( P \) can be described by the equation \( P = \frac{I}{c} \), where \( I \) is the intensity of the light and \( c \) is the speed of light in a vacuum, approximately \( 3.0 \times 10^8 \) m/s.
02

Calculate Intensity from Radiation Pressure

Given that the radiation pressure \( P \) is \( 9.0 \times 10^{-6} \) Pa, we can rearrange the equation \( P = \frac{I}{c} \) to find the intensity \( I \) as follows: \( I = P \cdot c = 9.0 \times 10^{-6} \times 3.0 \times 10^8 = 2.7 \times 10^{3} \) W/m虏.
03

Understanding Power and Intensity

Intensity \( I \) is defined as the power \( P_{source} \) per unit area. For a point source radiating uniformly in all directions, the area at a distance \( r \) from the source is the surface area of a sphere, calculated by \( A = 4\pi r^2 \).
04

Calculate the Surface Area at 5.0 m

The surface area of a sphere with a radius \( r = 5.0 \) m is \( A = 4\pi (5.0)^2 = 100\pi \) m虏.
05

Calculate the Total Power Output of the Source

Using the relationship between intensity and power, \( I = \frac{P_{source}}{A} \), where \( A = 100\pi \) m虏 and \( I = 2.7 \times 10^3 \) W/m虏, we solve for \( P_{source} \):\[P_{source} = I \cdot A = 2.7 \times 10^{3} \times 100\pi = 2.7 \times 10^{5}\pi \]which approximately equals \( 8.5 \times 10^5 \) 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 term "intensity of light" refers to the amount of energy that light delivers per unit area, per unit time. Think of it as how "strong" or "bright" the light is when it hits a surface. The mathematical expression used to define intensity, \( I \), is the power, \( P_{source} \), that a light source emits, divided by the area over which the light spreads. It's measured in watts per square meter (W/m虏). For a perfectly absorbing surface, the relationship between intensity and radiation pressure is given by, \( P = \frac{I}{c} \), where \( c \) is the speed of light. This relationship shows that intensity directly affects the force applied by light on a surface. This force per unit area is what we measure as radiation pressure.
Electromagnetic Radiation
Electromagnetic radiation is a form of energy that travels through space as waves. It includes a spectrum of different types, such as light, radio waves, X-rays, and microwaves, to name a few. Light, in particular, is one type of electromagnetic radiation that is essential for understanding radiation pressure and intensity.
Light travels at approximately \( 3.0 \times 10^8 \) meters per second, also known as the speed of light. This speed is a constant when light moves through a vacuum. What makes electromagnetic radiation unique is its ability to exert pressure on surfaces. This pressure, although small, is measurable and plays a critical role in various scientific fields, including astrophysics and solar energy research.
Power Output
Power output refers to the total energy produced by a source per unit time. It's measured in watts (W), where one watt equals one joule of energy per second. In the context of light sources, power output is a measure of how much light energy is emitted by the source in total.
In our exercise, we determined the power output by relating it to intensity and the area over which the light is spread. Using the formula for intensity, \( I = \frac{P_{source}}{A} \), and modifying it to solve for \( P_{source} \), we found that the total power output of the source is approximately \( 8.5 \times 10^5 \) watts. Understanding power output is crucial for applications in designing lighting systems or assessing energy needs for different devices and environments.
Spherical Surface Area
When dealing with light sources that emit in all directions, it is essential to think in three-dimensional terms. A point source of light radiates uniformly in all directions, forming a spherical wavefront as it moves outward.
To calculate the surface area of this sphere at a certain distance, we use the formula \( A = 4\pi r^2 \), where \( r \) is the radius, or distance from the source. In our specific case, with \( r = 5.0 \) meters, the area is \( 100\pi \) m虏. This calculation allows us to understand how light spreads out, influencing both intensity and radiation pressure. Spherical surface area is an important concept because it affects how light interacts with objects, such as focusing on lenses or the amount of light reaching distant planets.

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Most popular questions from this chapter

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x, t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max} e^{-k_C x} cos(k_Cx - \omega t)\), where \(k_C = \sqrt{(\omega \mu/2\rho}\). Verify this by substituting E_y(x, t) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (\(Hint\): The field does work to move charges within the conductor. The current of these moving charges causes \(i^2R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of 1/\(e\) in a distance \(1/k_C = \sqrt{2\rho/\omega\mu}\), and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); permeability \(\mu = \mu_0\)). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

The electric field of a sinusoidal electromagnetic wave obeys the equation \(E = (375 V/m)\) cos [(1.99 \(\times\) 107 rad/m)x + (5.97 \(\times\) 10\(^{15}\) rad/s)\(t\)]. (a) What is the speed of the wave? (b) What are the amplitudes of the electric and magnetic fields of this wave? (c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?

A sinusoidal electromagnetic wave is propagating in vacuum in the +\(z\)-direction. If at a particular instant and at a certain point in space the electric field is in the +\(x\)-direction and has magnitude 4.00 V/m, what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?

If the eye receives an average intensity greater than 1.0 \(\times\) 10\(^2\) W/m\(^2\), damage to the retina can occur. This quantity is called the \(damage\) \(threshold\) of the retina. (a) What is the largest average power (in mW) that a laser beam 1.5 mm in diameter can have and still be considered safe to view head-on? (b) What are the maximum values of the electric and magnetic fields for the beam in part (a)? (c) How much energy would the beam in part (a) deliver per second to the retina? (d) Express the damage threshold in W/cm\(^2\).

Medical x rays are taken with electromagnetic waves having a wavelength of around 0.10 nm in air. What are the frequency, period, and wave number of such waves?
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