Chapter 24

ElectromagneticWaves

24.1 The Nature of Electromagnetic Waves

Electric part of wave:

In each part of the drawing, the red arrow represents E produced at point P by the oscillating charges on the antenna at the indicated time. The black arrows represent E created at earlier times.

For simplicity, only the electric wave traveling to the right is shown here.

How to produce an electromagnetic wave

Two straight wires connected to the terminals of an AC generator can create an electromagnetic wave.

24.1 The Nature of Electromagnetic Waves

Magnetic part of wave:

The current used to generatethe electric wave creates amagnetic field.

Using RHR-2 to find the directionof B at point P for this current direction shows that B isperpendicular to E since E isparallel to I.

24.1 The Nature of Electromagnetic Waves

We just showed how the electromagnetic (E&M) wave is initially generatedby the ac voltage source near the antenna (near field). As the wave moves farther away, it propagates itself by the changing E-field producinga B-field and the changing B-field producing an E-field (radiation field).

E&M wave is transverse and can travel through a vacuum

The speed of an electromagnetic wave in a vacuum is: sm1000.3

8!=c

radiation field wave far from the antenna.

24.1 The Nature of Electromagnetic Waves

A radio wave can be detected with a receiving antenna wirethat is parallel to the electric field:

E generates an oscillating current along the antenna wire.

24.1 The Nature of Electromagnetic Waves

With a receiving antenna in the form of a loop, the magneticfield of a radio wave can be detected,

From Faraday’s law, the changing magnetic flux in the loop will create an oscillating current in it.

24.2 The Electromagnetic Spectrum

Like all waves, electromagnetic waves have a wavelength and frequency, related by:

!fc =

24.2 The Electromagnetic Spectrum

Example 1 The Wavelength of Visible Light

Find the range in wavelengths for visible light in the frequency rangebetween 4.0 x 1014 Hz and 7.9 x 1014 Hz.

nm 750m105.7Hz104.0

sm1000.3 714

8

=!=!

!==

"

f

c#

nm 380m108.3Hz107.9

sm1000.3 714

8

=!=!

!==

"

f

c#

red

violet

24.2 The Electromagnetic Spectrum

AM m 244Hz101230

sm1000.3

3

8

=!

!==

f

v"

FM m 26.3Hz1091.9

sm1000.3

6

8

=!

!==

f

v"

Example 2 The Wavelength of Radio Waves

A station broadcasts AM radio waves whose frequency is 1230 x 103 Hz and FM radio waves whose frequency is 91.9 x 106 Hz. Find the wavelength of each type of wave.

c

c

~ threefootball fields

~ 10 ft

24.6 Polarization

A transverse wave is linearly polarizedwhen its vibrations always occur alongone direction.

A linearly polarized wave on a ropecan pass through a slit that is parallelto the direction of the rope vibrations.

The rope wave cannot pass througha slit that is perpendicular to thevibrations.

POLARIZED ELECTROMAGNETIC WAVES

24.6 Polarization

In polarized light, the electric fieldof the electromagnetic wave fluctuates along a single direction.

Unpolarized light consists of shortbursts of electromagnetic waves emittedby many different atoms. The electricfield directions of these bursts areperpendicular to the direction of wavetravel, as in polarized light, but aredistributed randomly about it.

24.6 Polarization

Polarized light may be produced from unpolarized light with the aid of polarizing material.

The intensity of the transmitted polarized light has 1/2 the intensity of the incident unpolarized light (perpendicular component blocked).

The transmission axis of the material is the direction of polarizationof the light that passes through the material.

24.6 Polarization

MALUS’ LAW !2cosoSS =

intensity beforeanalyzerintensity after

analyzer

Consider two sheets of polarizing material, one to polarize unpolarized light(Polarizer) and one at an angle θ to the polarized light (Analyzer). Theintensity of the light after the analyzer is related to the intensity before theanalyzer by Malus’ Law:

( Intensity ~ E2 )

24.6 Polarization

Example 7 Using Polarizers and Analyzers

What value of θ should be used so the average intensity of the polarizedlight reaching the photocell is one-tenth the average intensity of theunpolarized light?

24.6 Polarization

!25

1 cos=5

1cos =! o4.63=!

Applying Malus’ Law at the Analyzer: S = S0,polar cos2 θ

But S0,polar = (1/2)S0 and we want S = (1/10)S0

So (1/10)S0 = (1/2)S0 cos2 θ

S0 S0,polar S

24.6 Polarization

Conceptual Example 8 How Can a Crossed Polarizer andAnalyzer Transmit Light?

Suppose that a third piece of polarizing material is inserted between the crossed polarizer and analyzer. Does light now reach the photocell?

S0 S1 S2 S3

S1 = S0/2

S2 = S1 cos2 θ

= (S0/2) cos2 θ

S3 = S2 cos2 (90 - θ)

= (S0/2) cos2 θ sin2 θ

Clearly, S3 > 0 as long asθ ≠ 0 or 90o

24.6 Polarization

Polarized light is produced by the scattering of unpolarizedsunlight by molecules in the atmosphere.

Molecules re-radiate sunlight.As you look at larger and larger angles with respect to the incident sunlight, the re-radiatedlight becomes more and morehorizontally polarized.

Polaroid Sun Glasses

Besides sunlight re-radiated from atmospheric molecules, there areother sources of horizontally polarized light that occur in nature, suchas sunlight reflected from horizontal surfaces such as lakes.

Polaroid sun glasses take advantage of this fact by using polarizerswith their axes oriented vertically:

in addition to 1/2 of the unpolarized light which is already blocked bythe polarizers, all of the horizontally polarized light is completely blocked,thus blocking out some of the reflected light which can confuse youduring some outdoor activities (e.g. driving, piloting, fishing…….).

24.6 Polarization

IMAX movie projector Movie viewer using polarized glasses

Another application using polarized glasses: watching 3-D movies.

In a 3-D movie, two separate rolls of film are projected using a projectorwith two lenses, each with its own polarizer. The two polarizers arecrossed. Viewers watch the action on-screen through glasses that have correspondingly crossed polarizers for each eye.

Example. Partially polarized and partially unpolarized light.A light beam passes through a polarizer whose transmission axis makesangle θ with the vertical. The beam is partially polarized andpartially unpolarized, and the average intensity of the incident light,S0, is the sum of the average intensity of the polarized light, S0,polar,and the average intensity of the unpolarized light, S0,unpolar.

As the polarizer is rotated clockwise, the intensity of the transmitted lighthas a minimum value of 2.0 W/m2 when θ = 20.0o and has a maximumvalue of 8.0 W/m2 when the angle is θ = θmax.

Find a) S0,unpolar, and b) S0,polar.

Incident lightS0 = S0,polar + S0,unpolar

Transmitted lightS = Spolar + Sunpolar

Incident lightS0 = S0,polar + S0,unpolar

Transmitted lightS = Spolar + Sunpolar

a) Minimum transmitted intensity S = Smin = 2.0 W/m2 at θ = 20.0o.

S is minimum when Spolar = 0 since Sunpolar is not effected by θ.

Sunpolar = Smin - Spolar = 2.0 - 0 = 2.0 W/m2

Sunpolar = (1/2)S0,unpolar S0,unpolar = 2Sunpolar = 2(2.0) = 4.0 W/m2

b) Maximum transmitted intensity S = Smax = 8.0 W/m2 occurs at θmax.

S is maximum when Spolar = S0,polar (when θ = 20o + 90o = 110o = θmax)

Smax = S0,polar + Sunpolor S0,polar = Smax - Sunpolar = 8.0 - 2.0 = 6.0 W/m2