# 16.11 Energy in waves: intensity

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• Calculate the intensity and the power of rays and waves.

All waves carry energy. The energy of some waves can be directly observed. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls.

Loud sounds pulverize nerve cells in the inner ear, causing permanent hearing loss. Ultrasound is used for deep-heat treatment of muscle strains. A laser beam can burn away a malignancy. Water waves chew up beaches.

The amount of energy in a wave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements. Loud sounds have higher pressure amplitudes and come from larger-amplitude source vibrations than soft sounds. Large ocean breakers churn up the shore more than small ones. More quantitatively, a wave is a displacement that is resisted by a restoring force. The larger the displacement $x$ , the larger the force $F=\text{kx}$ needed to create it. Because work $W$ is related to force multiplied by distance ( $\text{Fx}$ ) and energy is put into the wave by the work done to create it, the energy in a wave is related to amplitude. In fact, a wave’s energy is directly proportional to its amplitude squared because

$W\propto \text{Fx}={\text{kx}}^{2}\text{.}$

The energy effects of a wave depend on time as well as amplitude. For example, the longer deep-heat ultrasound is applied, the more energy it transfers. Waves can also be concentrated or spread out. Sunlight, for example, can be focused to burn wood. Earthquakes spread out, so they do less damage the farther they get from the source. In both cases, changing the area the waves cover has important effects. All these pertinent factors are included in the definition of intensity     $I$ as power per unit area:

$I=\frac{P}{A}$

where $P$ is the power carried by the wave through area $A$ . The definition of intensity is valid for any energy in transit, including that carried by waves. The SI unit for intensity is watts per square meter ( $\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ ). For example, infrared and visible energy from the Sun impinge on Earth at an intensity of $\text{1300}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ just above the atmosphere. There are other intensity-related units in use, too. The most common is the decibel. For example, a 90 decibel sound level corresponds to an intensity of ${\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ . (This quantity is not much power per unit area considering that 90 decibels is a relatively high sound level. Decibels will be discussed in some detail in a later chapter.

## Calculating intensity and power: how much energy is in a ray of sunlight?

The average intensity of sunlight on Earth’s surface is about $7\text{00}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$ .

(a) Calculate the amount of energy that falls on a solar collector having an area of $0\text{.}\text{500}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ in $4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{h}$ .

(b) What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own?

Strategy a

Because power is energy per unit time or $P=\frac{E}{t}$ , the definition of intensity can be written as $I=\frac{P}{A}=\frac{E/t}{A}$ , and this equation can be solved for E with the given information.

Solution a

1. Begin with the equation that states the definition of intensity:
$I=\frac{P}{A}.$
2. Replace $P$ with its equivalent $E/t$ :
$I=\frac{E/t}{A}.$
3. Solve for $E$ :
$E=\text{IAt}.$
4. Substitute known values into the equation:
$E=\left(\text{700}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}\right)\left(0\text{.}\text{500}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left[\left(4\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{h}\right)\left(\text{3600}\phantom{\rule{0.25em}{0ex}}\text{s/h}\right)\right].$
5. Calculate to find $E$ and convert units:
$5\text{.}\text{04}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J},$

Discussion a

The energy falling on the solar collector in 4 h in part is enough to be useful—for example, for heating a significant amount of water.

Strategy b

Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel.

Solution b

1. Take the ratio of intensities, which yields:
$\frac{I\prime }{I}=\frac{P\prime /A\prime }{P/A}=\frac{A}{A\prime }\phantom{\rule{0.25em}{0ex}}\left(\text{The powers cancel because}\phantom{\rule{0.25em}{0ex}}P\prime =P\right)\text{.}$
2. Identify the knowns:
$A=\text{200}A\prime ,$
$\frac{I\prime }{I}=\text{200}.$
3. Substitute known quantities:
$I\prime =\text{200}I=\text{200}\left(\text{700}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}\right).$
4. Calculate to find $I\prime$ :
$I\prime =\text{1.40}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}.$

Discussion b

Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire.

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