16.8 Power, Intensity, and Decibels

A traveling wave transfers energy from one point to another. The sound wave from a loudspeaker sets your eardrum into motion. Light waves from the sun warm the earth. The power of a wave is the rate, in joules per second, at which the wave transfers energy. As you learned in Chapter 9, power is measured in watts. A loudspeaker might emit 2 W of power, meaning that energy in the form of sound waves is radiated at the rate of 2 joules per second.

A focused light, like that of a projector, is more intense than the diffuse light that goes in all directions. Similarly, a loudspeaker that beams its sound forward into a small area produces a louder sound in that area than a speaker of equal power that radiates the sound in all directions. Quantities such as brightness and loudness depend not only on the rate of energy transfer, or power, but also on the area that receives that power.

FIGURE 16.28 shows a wave impinging on a surface of area a. The surface is perpendicular to the direction in which the wave is traveling. This might be a real, physical surface, such as your eardrum or a photovoltaic cell, but it could equally well be a mathematical surface in space that the wave passes right through. If the wave has power P, we define the intensity I of the wave to be

I = \frac{P}{a} = power - t o - a r e a r a t i o

(16.57)

The SI units of intensity are ${W/m}^{2}$ . Because intensity is a power-to-area ratio, a wave focused into a small area will have a larger intensity than a wave of equal power that is spread out over a large area.

The figure illustrates plane waves of power P impinge on area “a” with intensity I equals P over “a”.

FIGURE 16.28 Full Alternative Text

EXAMPLE 16.9 The intensity of a laser beam

A typical red laser pointer emits 1.0 mW of light power into a 1.0-mm-diameter laser beam. What is the intensity of the laser beam?

MODEL The laser beam is a light wave.

SOLVE The light waves of the laser beam pass through a mathematical surface that is a circle of diameter 1.0 mm. The intensity of the laser beam is

I = \frac{P}{a} = \frac{P}{π r^{2}} = \frac{0.0010 W}{π {(0.00050 m)}^{2}} = 1300 {W/m}^{2}

REVIEW This is roughly the intensity of sunlight at noon on a summer day. The difference between the sun and a small laser is not their intensities, which are about the same, but their powers. The laser has a small power of 1 mW. It can produce a very intense wave only because the area through which the wave passes is very small. The sun, by contrast, radiates a total power $P_{sun} \approx 4 \times 10^{26} W$ . This immense power is spread through all of space, producing an intensity of $1400 W / m^{2}$ at a distance of $1.5 \times 10^{11} m,$ the radius of the earth’s orbit.

If a source of spherical waves radiates uniformly in all directions, then, as FIGURE 16.29 shows, the power at distance r is spread uniformly over the surface of a sphere of radius r. The surface area of a sphere is $a = 4 π r^{2}$ , so the intensity of a uniform spherical wave is

I = \frac{P_{source}}{4 π r^{2}} (intensity of a uniform spherical wave)

(16.58)

The inverse-square dependence of r is really just a statement of energy conservation. The source emits energy at the rate P joules per second. The energy is spread over a larger and larger area as the wave moves outward. Consequently, the energy per unit area must decrease in proportion to the surface area of a sphere.

The figure illustrates a source emitting uniform spherical waves.

FIGURE 16.29 Full Alternative Text

If the intensity at distance $r_{1}$ is $I_{1} = P_{source} / 4 π r_{1}^{2}$ and the intensity at $r_{2}$ is $I_{2} = P_{source} / 4 π r_{2}^{2}$ then you can see that the intensity ratio is

\frac{I_{1}}{I_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}

(16.59)

You can use Equation 16.59 to compare the intensities at two distances from a source without needing to know the power of the source.

NOTE Wave intensities are strongly affected by reflections and absorption. Equations 16.58 and 16.59 apply to situations such as the light from a star or the sound from a firework exploding high in the air. Indoor sound does not obey a simple inverse-square law because of the many reflecting surfaces.

For a sinusoidal wave, each particle in the medium oscillates back and forth in simple harmonic motion. You learned in Chapter 15 that a particle in SHM with amplitude A has energy $E = \frac{1}{2} k A^{2}$ , where k is the spring constant of the medium, not the wave number. It is this oscillatory energy of the medium that is transferred, particle to particle, as the wave moves through the medium.

Because a wave’s intensity is proportional to the rate at which energy is transferred through the medium, and because the oscillatory energy in the medium is proportional to the square of the amplitude, we can infer that

I \propto A^{2}

(16.60)

That is, the intensity of a wave is proportional to the square of its amplitude. If you double the amplitude of a wave, you increase its intensity by a factor of 4.

16.8 Power, Intensity, and Decibels

FIGURE 16.28 Plane waves of power P impinge on area a with intensity $I = P / a$ .

EXAMPLE 16.9 The intensity of a laser beam

FIGURE 16.29 A source emitting uniform spherical waves.