Learning Objectives

By the end of this section, you will be able to do the following:

Describe the behavior of electromagnetic radiation
Solve quantitative problems involving the behavior of electromagnetic radiation

Section Key Terms
illuminance	interference	lumens
luminous flux	lux	polarized light

Types of Electromagnetic Wave Behavior

In a vacuum, all electromagnetic radiation travels at the same incredible speed of 3.00 × 10⁸ m/s, which is equal to 671 million miles per hour. This is one of the fundamental physical constants. It is referred to as the speed of light and is given the symbol c. The space between celestial bodies is a near vacuum, so the light we see from the Sun, stars, and other planets has traveled here at the speed of light. Keep in mind that all EM radiation travels at this speed. All the different wavelengths of radiation that leave the Sun make the trip to Earth in the same amount of time. That trip takes 8.3 minutes. Light from the nearest star, besides the Sun, takes 4.2 years to reach Earth, and light from the nearest galaxy—a dwarf galaxy that orbits the Milky Way—travels 25,000 years on its way to Earth. You can see why we call very long distances astronomical.

When light travels through a physical medium, its speed is always less than the speed of light. For example, light travels in water at three-fourths the value of c. In air, light has a speed that is just slightly slower than in empty space: 99.97 percent of c. Diamond slows light down to just 41 percent of c. When light changes speeds at a boundary between media, it also changes direction. The greater the difference in speeds, the more the path of light bends. In other chapters, we look at this bending, called refraction, in greater detail. We introduce refraction here to help explain a phenomenon called thin-film interference.

Have you ever wondered about the rainbow colors you often see on soap bubbles, oil slicks, and compact discs? This occurs when light is both refracted by and reflected from a very thin film. The diagram shows the path of light through such a thin film. The symbols n₁, n₂, and n₃ indicate that light travels at different speeds in each of the three materials. Learn more about this topic in the chapter on diffraction and interference.

Figure 15.11 shows the result of thin film interference on the surface of soap bubbles. Because ray 2 travels a greater distance, the two rays become out of phase. That is, the crests of the two emerging waves are no longer moving together. This causes interference, which reinforces the intensity of the wavelengths of light that create the bands of color. The color bands are separated because each color has a different wavelength. Also, the thickness of the film is not uniform, and different thicknesses cause colors of different wavelengths to interfere in different places. Note that the film must be very, very thin—somewhere in the vicinity of the wavelengths of visible light.

This diagram shows three areas separated by two parallel, horizontal lines. The distance between the two lines is labeled 't'. The top area, labeled n subscript 1, has a white background. The middle area, labeled n subscript 2, has a hazy bluish background. The bottom area, labeled n subscript 3, has a hazy gray background. An arrow labeled 'Incident light' starts from the top left of the white area, points in a southeasterly direction, and ends on the upper horizontal line. At that point, two arr

Figure 15.11 Light striking a thin film is partially reflected (ray 1) and partially refracted at the top surface. The refracted ray is partially reflected at the bottom surface and emerges as ray 2. These rays will interfere in a way that depends on the thickness of the film and the indices of refraction of the various media.

Misconception Alert

Do not confuse polar molecules with polarized light. If a molecule is polar, it refers to a separation of negative and positive electric charges. Polarized light is light whose electric field component vibrates in a specific plane.

You have probably experienced how polarized sunglasses reduce glare from the surface of water or snow. The effect is caused by the wave nature of light. Looking back at Figure 15.3, we see that the electric field moves in only one direction perpendicular to the direction of propagation. Light from most sources vibrates in all directions perpendicular to propagation. Light with an electric field that vibrates in only one direction is called polarized. A diagram of polarized light would look like Figure 15.3.

Polarized glasses are an example of a polarizing filter. These glasses absorb most of the horizontal light waves and transmit the vertical waves. This cuts down glare, which is caused by horizontal waves. Figure 15.12 shows how waves traveling along a rope can be used as a model of how a polarizing filter works. The oscillations in one rope are in a vertical plane and are said to be vertically polarized. Those in the other rope are in a horizontal plane and are horizontally polarized. If a vertical slit is placed on the first rope, the waves pass through. However, a vertical slit blocks the horizontally polarized waves. For EM waves, the direction of the electric field oscillation is analogous to the disturbances on the ropes.

This figure has two panels arranged side by side. The left panel, labeled '(a)', shows a hand holding one end of a wave moving in a vertical plane and passing through a vertically oriented slit. Next to the hand is a double-headed vertical arrow labeled 'Direction of polarization'. The right panel, labeled '(b)', shows a hand holding one end of a wave moving in a horizontal plane and passing through a horizontally oriented slit. Next to the hand is a double-headed horizontal arrow labeled 'D

Figure 15.12 The transverse oscillations in one rope are in a vertical plane, and those in the other rope are in a horizontal plane. The first is said to be vertically polarized, and the other is said to be horizontally polarized. Vertical slits pass vertically polarized waves and block horizontally polarized waves.

Light can also be polarized by reflection. Most of the light reflected from water, glass, or any highly reflective surface is polarized horizontally. Figure 15.13 shows the effect of a polarizing lens on light reflected from the surface of water.

This figure shows two photographs arranged side by side. The left photograph, labeled '(a)', shows a river bed that looks hazy because of interfering gray and white patches. The right photograph, labeled '(b)', shows the same area of the river bed with almost no interfering gray patches, so that the pebbles on the bed are clearly visible.

Figure 15.13 These two photographs of a river show the effect of a polarizing filter in reducing glare in light reflected from the surface of water. Part (b) of this figure was taken with a polarizing filter and part (a) was taken without. As a result, the reflection of clouds and sky observed in part (a) is not observed in part (b). Polarizing sunglasses are particularly useful on snow and water.

Watch Physics

Polarization of Light, Linear and Circular

This video explains the polarization of light in great detail. Before viewing the video, look back at the drawing of an electromagnetic wave from the previous section. Try to visualize the two-dimensional drawing in three dimensions.

Click to view content

Grasp Check

How do polarized glasses reduce glare reflected from the ocean?

They block horizontally polarized and vertically polarized light.
They are transparent to horizontally polarized and vertically polarized light.
They block horizontally polarized rays and are transparent to vertically polarized rays.
They are transparent to horizontally polarized light and block vertically polarized light.

Snap Lab

Polarized Glasses

Safety Warning

EYE SAFETY—Looking at the Sun directly can cause permanent eye damage. Avoid looking directly at the Sun.

Materials

two pairs of polarized sunglasses

OR
two lenses from one pair of polarized sunglasses

Procedure

Look through both or either polarized lens and record your observations.
Hold the lenses, one in front of the other. Hold one lens stationary while you slowly rotate the other lens. Record your observations, including the relative angles of the lenses when you make each observation.
Find a reflective surface on which the Sun is shining. It could be water, glass, a mirror, or any other similar smooth surface. The results will be more dramatic if the sunlight strikes the surface at a sharp angle.
Observe the appearance of the surface with your naked eye and through one of the polarized lenses.
Observe any changes as you slowly rotate the lens, and note the angles at which you see changes.

Grasp Check

If you buy sunglasses in a store, how can you be sure that they are polarized?

When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to refraction of light.
When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to reflection of light.
When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to the polarization of light.
When one pair of sunglasses is placed in front of another and rotated in the plane of the body, the light passing through the sunglasses will be blocked at two positions due to the bending of light waves.

Quantitative Treatment of Electromagnetic Waves

We can use the speed of light, c, to carry out several simple but interesting calculations. If we know the distance to a celestial object, we can calculate how long it takes its light to reach us. Of course, we can also make the reverse calculation if we know the time it takes for the light to travel to us. For an object at a very great distance from Earth, it takes many years for its light to reach us. This means that we are looking at the object as it existed in the distant past. The object may, in fact, no longer exist. Very large distances in the universe are measured in light years. One light year is the distance that light travels in one year, which is $9.46 \times 10^{12}$ kilometers or $5.88 \times 10^{12}$ miles (…and 10¹² is a trillion!).

A useful equation involving c is

15.1

c = f λ

where f is frequency in Hz, and $λ$ is wavelength in meters.

Worked Example

Frequency and Wavelength Calculation

For example, you can calculate the frequency of yellow light with a wavelength of $6.00 \times 10^{- 7}$ m.

STRATEGY

Rearrange the equation to solve for frequency.

15.2

f = \frac{c}{λ}

Solution

Substitute the values for the speed of light and wavelength into the equation.

15.3

f = \frac{3.00 \times 10^{8} m/s}{6.00 \times 10^{- 7} m} = 5.00 \times 10^{14} s^{- 1} = 5.00 \times 10^{14} Hz

Discussion

Manipulating exponents of 10 in a fraction can be tricky. Be sure you keep track of the + and – exponents correctly. Checking back to the diagram of the electromagnetic spectrum in the previous section shows that 10¹⁴ is a reasonable order of magnitude for the frequency of yellow light.

The frequency of a wave is proportional to the energy the wave carries. The actual proportionality constant will be discussed in a later chapter. Since frequency is inversely proportional to wavelength, we also know that wavelength is inversely proportional to energy. Keep these relationships in mind as general rules.

The rate at which light is radiated from a source is called luminous flux, P, and it is measured in lumens (lm). Energy-saving light bulbs, which provide more luminous flux for a given use of electricity, are now available. One of these bulbs is called a compact fluorescent lamp; another is an LED (light-emitting diode) bulb. If you wanted to replace an old incandescent bulb with an energy saving bulb, you would want the new bulb to have the same brightness as the old one. To compare bulbs accurately, you would need to compare the lumens each one puts out. Comparing wattage—that is, the electric power used—would be misleading. Both wattage and lumens are stated on the packaging.

The luminous flux of a bulb might be 2,000 lm. That accounts for all the light radiated in all directions. However, what we really need to know is how much light falls on an object, such as a book, at a specific distance. The number of lumens per square meter is called illuminance, and is given in units of lux (lx). Picture a light bulb in the middle of a sphere with a 1-m radius. The total surface of the sphere equals 4πr² m². The illuminance then is given by

15.4

illuminance = \frac{P}{4 π r^{2}} .

What happens if the radius of the sphere is increased 2 m? The illuminance is now only one-fourth as great, because the r² term in the denominator is 4 instead of 1. Figure 15.14 shows how illuminance decreases with the inverse square of the distance.

This diagram shows a small sphere on the left, labeled 'Source of Electromagnetic Radiation', and a series of arrows emerging from it, pointing rightward. Three curved surfaces are separated from the sphere by distances labeled as 'r', '2r', and '3r'. One of these surfaces is labeled 'Part of the surface of a sphere with radius r'. The arrows penetrate these three surfaces and move farther and farther apart as they move away from the sphere.

Figure 15.14 The diagram shows why the illuminance varies inversely with the square of the distance from a source of light.

Worked Example

Calculating Illuminance

A woman puts a new bulb in a floor lamp beside an easy chair. If the luminous flux of the bulb is rated at 2,000 lm, what is the illuminance on a book held 2.00 m from the bulb?

STRATEGY

Choose the equation and list the knowns.

Equation: $illuminance = \frac{P}{4 π r^{2}}$

P = 2,000 lm

π = 3.14

r = 2.00 m

Solution

Substitute the known values into the equation.

$\begin{matrix} illuminance & = & \frac{P}{4 π r^{2}} \\ = & \frac{2,000 lm}{4 (3.14) {(2.00)}^{2} m^{2}} \\ = & 39.8 {lm/m}^{2} \\ = & 39.8 lx \end{matrix}$

Discussion

Try some other distances to illustrate how greatly light fades with distance from its source. For example, at 3 m the illuminance is only 17.7 lux. Parents often scold children for reading in light that is too dim. Instead of shouting, “You’ll ruin your eyes!” it might be better to explain the inverse square law of illuminance to the child.

Practice Problems

Red light has a wavelength of 7.0 × 10−7 m and a frequency of 4.3 × 1014 Hz. Use these values to calculate the speed of light in a vacuum.

3 × 1020 m/s
3 × 1015 m/s
3 × 1014 m/s
3 × 108 m/s

A light bulb has a luminous flux of 942 lumens. What is the illuminance on a surface

3.00 m

from the bulb when it is lit?

$33.32 lx$
$26.15 lx$
$2.77 lx$
$8.33 lx$

Check Your Understanding

Exercise 6

Give an example of a place where light travels at the speed of 3.00 × 10⁸ m/s.

outer space
water
Earth’s atmosphere
quartz glass

Exercise 7

Explain in terms of distances and the speed of light why it is currently very unlikely that humans will visit planets that circle stars other than our Sun.

The spacecrafts used for travel are very heavy and thus very slow.
Spacecrafts do not have a constant source of energy to run them.
If a spacecraft could attain a maximum speed equal to that of light, it would still be too slow to cover astronomical distances.
Spacecrafts can attain a maximum speed equal to that of light, but it is difficult to locate planets around stars.

15.2 The Behavior of Electromagnetic Radiation

Learning Objectives