24.7

Concepts & Calculations

One of the central ideas of this chapter is that electromagnetic waves carry energy. Two concepts are used to describe this energy—the wave’s intensity and its energy density. The next example reviews these important ideas.

| Concepts & Calculations Example 9 | | | Intensity and Energy Density of a Wave

Figure 24-31 shows the popular dish antenna that receives digital TV signals from a satellite. The average intensity of the electromagnetic wave that carries a particular TV program is , and the circular aperture of the antenna has a radius of . (a) Determine the electromagnetic energy delivered to the dish during a one-hour program. (b) What is the average energy density of the electromagnetic wave?

Figure 24-31 A dish antenna picks up the TV signal from a satellite. The signal extends over the dish’s entire aperture (radius r).

Concept Questions and Answers How is the average power passing through the circular aperture of the antenna related to the average intensity of the TV signal?

> Answer According to Equation 16.8, the average intensity of a wave is equal to the average power that passes perpendicularly through a surface divided by the area A of the surface. Thus, the average power is , where .

How much energy does the antenna receive in a time t?

> Answer Since the average power is the energy per unit time according to Equation 6.10b, the energy received by the antenna is the product of the average power and the time, or Energy .

What is the average energy density, or average energy per unit volume, of the electromagnetic wave?

> Answer The energy delivered to the dish is carried by the electromagnetic wave. The wave passes perpendicularly through an imaginary surface that has an area A that matches the circular area of the dish’s aperture. In a time t, all the energy that passes through the surface is contained, therefore, in a cylinder of length x (see Figure 24-31). Since the energy is being carried by an electromagnetic wave, it travels at the speed of light c. Thus, the length of the cylinder is . The energy density u of the wave, or energy per unit volume, is the energy contained within the cylinder divided by its volume.

Solution

a. | According to Equation 6.10b, the energy received by the antenna during a time t is equal to the average power multiplied by the time, Energy . On the other hand, Equation 16.8 gives the average power as the product of the average intensity and the area, , so that

b. | The energy density is equal to the energy divided by the volume of the cylinder in Figure 24-31. The volume of the cylinder is equal to the cross-sectional area times its length ct.

This calculation is equivalent to using Equation 24.4, which gives the energy density as

We have seen how the intensities of completely polarized or completely unpolarized light beams can change as they pass through a polarizer. But what about light that is partially polarized and partially unpolarized? Can the concepts that we discussed in Section 24.6 be applied to such light? The answer is “yes,” and Example 10 illustrates how.

| Concepts & Calculations Example 10 | | | Partially Polarized and Partially Unpolarized Light

The light beam in Figure 24-32 passes through a polarizer whose transmission axis makes an angle with the vertical. The beam is partially polarized and partially unpolarized, and the average intensity of the incident light is the sum of the average intensity of the polarized light and the average intensity of the unpolarized light; . The intensity of the transmitted light is also the sum of two parts: . As the polarizer is rotated clockwise, the intensity of the transmitted light has a minimum value of when and has a maximum value of when the angle is . (a) What is the intensity of the unpolarized incident light? (b) What is the intensity of the polarized incident light?

Figure 24-32 Light that is partially polarized and partially unpolarized is incident on a sheet of polarizing material.

Concept Questions and Answers How is related to ?

> Answer When unpolarized light strikes a polarizer, one half is absorbed and the other half is transmitted, so that . This is true for any angle of the transmission axis.

How is related to ?

> Answer When polarized light strikes a polarizer, the transmitted intensity is given by Malus’ law (Equation 24.7), , where is the angle between the direction of polarization of the incident light and the transmission axis. Note that is not the same as because the direction of polarization of the incident beam is not given and may not be in the vertical direction.

The minimum transmitted intensity is . Why isn’t it ?

> Answer When the polarized part of the incident beam passes through the polarizer, the intensity of its transmitted portion changes as the angle changes, ranging between some maximum value and , in accord with Malus’ law. However, when the unpolarized part of the beam passes through the polarizer, its transmitted intensity does not change as the angle changes. Thus, the minimum intensity of the transmitted beam is , rather than , because of the unpolarized light.

At what angle is the intensity of the light leaving the polarizing material a maximum?

> Answer When the polarized part of the beam passes through the polarizer, the intensity of its transmitted portion is zero when . For this angle, the direction of polarization of the polarized light is perpendicular to the transmission axis of the polarizer. The intensity rises to a maximum when the polarizer is rotated by relative to this position, at which orientation the direction of polarization is aligned with the transmission axis. Thus, the intensity is a maximum when .

Solution

a. | The average intensity of the transmitted light is (see Figure 24-32) .We are given that the minimum intensity is when , and we know that for this angle. Thus, . The intensity of the unpolarized incident light is twice this amount because the polarizer absorbs one-half the incident light:

b. | When , the intensity of the transmitted beam is at its maximum value of . Writing the transmitted intensity as the sum of two parts, we have

The angle is the angle between the direction of polarization of the incident polarized light and the transmission axis of the polarizer. Since is a maximum, we know that . Solving this equation for yields

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.