Chapter

Electromagnetic Waves

Blinn College - Physics 1402 - Terry Honan

. - The Nature of Electromagnetic Radiation

The Discoveries

With Faraday’s Law, we saw that a changing magnetic field can produce an EMF, an electromotive force. It turns out that this EMF can be written in terms an induced electric field; that field pushes the charges around a circuit causing the EMF. James Clerk Maxwell then showed that in addition to changing magnetic fields producing electric fields, changing electric fields can produce magnetic fields. This makes electromagnetic radiation possible. Sinusoidally varying electric field induce sinusoidally varying magnetic fields, which in turn, produce sinusoidally varying electric fields. This arrangement then propagates (in a vacuum) at the speed of light.

In 1864, Maxwell showed mathematically that the equations describing electromagnetism give rise to wave solutions and the speed of those waves can be described in terms of our electromagnetic constants (Subscript[ε, 0]) and (Subscript[μ, 0]).

c=(1/(Sqrt[(Subscript[μ, 0]) (Subscript[ε, 0])])).

Maxwell then realized that the speed of these waves was the previously measured value of the speed of light in a vacuum. In addition to explaining light, Maxwell predicted the rest of the electromagnetic spectrum.

After Maxwell in 1887, Heinrich Hertz was able to show how to produce and receive these electromagnetic waves in a laboratory. Guglielmo Marconi was able to find a practical application of Maxwell’s and Hertz’s discoveries, using them for communication. In 1901 Marconi sent Morse code as a radio signal across the Atlantic. This became the basis of modern wireless communications.

The Propagating Fields

In an electromagnetic wave the electric and magnetic field are perpendicular and both are perpendicular to the direction of wave propagation. If we choose the direction of the electric fields to be in the y-direction and the magnetic fields to be in the z-direction then the wave propagates at the speed of light in the x-direction.

There is a right-hand rule to get the direction of wave propagation: Put your thumb in the direction of (OverVector[E]) and your fingers in the direction of (OverVector[B]) then your palm is in the direction of wave propagation. Note that if you look at any x position the electric field varies as a sinusoidal function, like what we studied in the previous chapter.

Example . - Directions of Fields

Solution

Producing and Receiving Electromagnetic Waves

Accelerating electric charges are the ultimate source of electromagnetic waves. To produce waves consider an AC voltage source connected across two separated conductors, shown as gray bars in the diagram below. As the polarity of the source varies the charges on the conductors varies and this creates a sinusoidally varying electric field. These fields then propagate as a wave.

This, essentially, is a broadcasting antenna.

A receiving antenna is basically the same thing in reverse. An incoming electromagnetic wave creates a very small sinusoidal voltage across an antenna and that small voltage can be amplified into a signal. Radio waves hit an antenna at all frequencies. A simple resonance circuit selects for the correct frequency. The LC Resonance frequency is f=1/(2 π(Sqrt[L C])). Although modern radio tuners use complexe solid state circuitry, the knob to adjust frequency on an old radio tuner is simply a variable capacitor that tunes for the correct f in an LC circuit.

. - The Speed of Light and the Electromagnetic Spectrum

The Speed of Light

The speed of light in a vacuum is to three digits

c=3.00×(10^8) m/s.

Because of its most fundamental nature we can choose our units so that c have the exact value

c=2.99792458×(10^8) m/s.

The definition of a second is given in terms of our most accurate method of measuring time, with atomic clocks, and the above definition of c then gives a definition of the meter.

The first documented attempt to measure the speed of light was due to Galileo. He and an assistant both had lanterns they could cover and they were separated by a large distance. When the distant assistant saw Galileo cover his lantern, he covered his. Galileo then looked for a time delay. Of course, light is much too fast for that, but that is only obvious to us because we know light is very fast. A similar set-up could measure the speed of sound. By viewing light as something physical that traveled at a finite speed, Galileo showed great insight.

Historically, the first reasonable estimate of the speed of light was made by Roemer.  By plotting the period of Io, one of Jupiter's moons, Roemer was able to explain the lack of periodicity in Io's orbit in terms of the time difference for light from Io to reach the Earth as the relative distance between Jupiter and Earth varies as both are in different positon relative to the sun in their orbits.

An accurate measure of c was first found by Fizeau. By shining a light through a rapidly rotating toothed wheel and reflecting it off a distant mirror, he could essentially redo Galileo’s experiment with very short time intervals and large distances.

The speed of light is a fundamental speed limit.  It is impossible to send any information faster than c.

Example . - Time of Travel for Light

(Subscript[R, E,S])=1.50×(10^11)(QuantityUnit)

Solution

t=(d/c)=((Subscript[R, E,S])/c)=500(QuantityUnit)=8.33(QuantityUnit)

(Subscript[R, E,M])=3.84×(10^8)(QuantityUnit)

Solution

t=(d/c)=((Subscript[R, E,M])/c)=1.28(QuantityUnit)

Solution

t=(10^(-9))(QuantityUnit)  ⟹  d=c t=0.3(QuantityUnit)≃1 foot

Solution

1(QuantityUnit)=c×(1(QuantityUnit))=3.00×(10^8) (m/((QuantityUnit)))×365.24(QuantityUnit)×((24(QuantityUnit))/day)×((3600(QuantityUnit))/((QuantityUnit)))=9.47×(10^15)(QuantityUnit)

The Electromagnetic Spectrum

The Full Spectrum

Electromagnetic radiation can be described by its wavelength and frequency.  Since frequency and wavelength are related by

f λ=c

for radiation in a vacuum, it follows that all electromagnetic radiation can be written along a line of increasing frequency and decreasing wavelength.

Long wavelength waves are radio waves.  Wavelengths shorter than around 0.3 meter are usually labeled microwaves. Smaller than about 1 mm start the infrared (IR) waves. At 700 nm we get into the visible spectrum which is a narrow band of wavelengths down to 400 nm.  Below 400 nm to around 3 nm are the ultraviolet (UV) waves.  Wavelengths smaller than that are called X-rays and the small wavelength limit, beyond around 0.003 nm, are called gamma rays.

The Visible Spectrum and Primary Colors

The visible spectrum consists of the narrow band of wavelengths between 400 and 700 nm.  In order of decreasing wavelength (increasing frequency) we have ROYGBIV: Red, Orange, Yellow, Green, Blue, Indigo and Violet. There is approximately a factor of two of wavelengths (and frequencies) we can see; this is crucial to our perception of light. With sound we can hear many octaves, where going up an octave is doubling a frequency. As you go through the “do-re-mi” scale from the low “do” to the high “do”, you jump up an octave, doubling the frequency. With light we can see the equivalent of only one octave of light. Our brain matches the ends of the visible spectrum into a circle; violet appears as a reddish blue.  If we perceived sound similarly, the low “do” and high “do” would sound the same.

Notice that the visible spectrum is not quite a factor of two of wavelengths and frequencies. There is a missing color, magenta. Although we may perceive magenta as a combinations of other colors, there is no pure color, meaning color of a single wavelength, corresponding to magenta.

Matching the ends of the visible spectrum together makes it possible to represent colors as combinations of three primary colors. There are two notions of primary colors: additive mixing and subtractive mixing.  Additive mixing is used with computer monitors and television screens.  We begin with black and add colors.  The additive primary colors are Red, Green and Blue.  Combining all three we can get white as shown.  Subtractive mixing is where we begin with white and remove colors.  This is used when mixing paints or for color printers.  Here the primary colors are Cyan (a blue-green color), Magenta (a reddish violet) and Yellow.  Removing all three subtractive colors gives black.

Example . - Frequencies and Wavelengths

Solution

c=3.00×(10^8)(((QuantityUnit))/((QuantityUnit)))  and  f=2450(QuantityUnit)=2450×(10^6)(QuantityUnit)   ⟹   λ=(c/f)=0.122(QuantityUnit)=12.2(QuantityUnit)

Solution

f=90.9×(10^6)(QuantityUnit)   ⟹   λ=(c/f)=3.30(QuantityUnit)   and   f=1705×(10^3)(QuantityUnit)   ⟹   λ=(c/f)=176(QuantityUnit)

Solution

λ=0.020×(10^(-9))(QuantityUnit)   ⟹   f=(c/λ)=1.5×(10^19)(QuantityUnit)

The Doppler Effect

We have all observed the Doppler effect; when a train is blowing its horn or a police car blows its siren while moving toward you, the frequency is increased and when moving away it is decreased. This is is called the Doppler effect and it occurs for all waves. When the relative motion between you and some wave source is toward each other, either the source is moving toward you or you toward the source, the observed frequency of the wave increases. When the the relative motion is away, the frequency decreases.

For electromagnetic waves, as long as the relative speed u is mush less that the speed of light c (which we write v<<c) we get

f'=f (1±(u/c))  or  Δ f=f'-f=±f(u/c)   for  u<<c

where f is the original frequency, f'is the shifted frequency and Δ f=f'-f is the frequency shift. We will see a corrected formula for this in the Relativity chapter that will apply as the relative speed is larger, approaching c.

When a wave reflects off a moving target, there is a frequency shift Δ f between the source and the target and another Δ f shift in the returning waves giving a double shift.

Δ (Subscript[f, tot])=2 Δ f=±2f(u/c)   (reflection from a moving target)  for  u<<c

Example . - Doppler Effect for a Jet

Solution

f=475×(10^6)(QuantityUnit) ,  u=325(((QuantityUnit))/((QuantityUnit)))  and  c=3.00×(10^8)(((QuantityUnit))/((QuantityUnit)))  ⟹  Δ f=f'-f=±f(u/c)=±515(QuantityUnit)

Example . - A Police Radar Gun

Solution

f=10.5×(10^9)(QuantityUnit) ,  u=90(((QuantityUnit))/((QuantityUnit)))×((1609(QuantityUnit))/((QuantityUnit)))×(((QuantityUnit))/(3600(QuantityUnit)))=40.225(((QuantityUnit))/((QuantityUnit)))  ⟹   Δ (Subscript[f, tot])=2 Δ f=±2f(u/c)=±2820(QuantityUnit)

. - Energy and Momentum in Electromagnetic Radiation

Energy and Intensity

Energy Density

The energy density u in an electromagnetic field can be written as a sum over electric and magnetic contributions

u=(Subscript[u, E])+(Subscript[u, B])=(1/2)(Subscript[ε, 0]) (E^2)+(1/(2 (Subscript[μ, 0]))) (B^2)

We can find the average energy density (Subscript[u, ave]) using the fact that the fields are sinusoidal and that the average of (sin^2) is (1/2).

(Subscript[u, ave])=(Subscript[u, E,ave])+(Subscript[u, B,ave])=(1/2)(Subscript[ε, 0]) (1/2)  (Subscript[E, max]^2)+(1/(2 (Subscript[μ, 0])))(1/2)  (Subscript[B, max]^2)

Because of the symmetry between electric and magnetic fields in electromagnetism, the electric and magnetic contributions are equal,  (Subscript[u, E,ave])=(Subscript[u, B,ave]).

It follows that

(Subscript[u, ave])=((Subscript[ε, 0])/2) (Subscript[E, max]^2)=((Subscript[B, max]^2)/(2 (Subscript[μ, 0])))=(Subscript[ε, 0]) (Subscript[E, rms]^2)=((Subscript[B, rms]^2)/(Subscript[μ, 0]))

where we used the relationship between rms and peak quantities for any sinusoidal function: (Subscript[E, rms])=(Subscript[E, max])/(Sqrt[2]) and (Subscript[B, rms])=(Subscript[B, max])/(Sqrt[2])

c=(1/(Sqrt[(Subscript[μ, 0]) (Subscript[ε, 0])])) =((Subscript[E, max])/(Subscript[B, max]))=((Subscript[E, rms])/(Subscript[B, rms]))=(E/B)

Intensity

For light, intensity is a measure of its brightness. We will define intensity for all electromagnetic radiation. Intensity I is a measure of the power (energy/time) per area.  If A is the area of a surface normal to the radiation, Δt is a time, U is the energy passing through the surface and 𝒫 is the power (U/Δt) through the surface, then these are related by

I=(𝒫/A)=(U/(A Δt)).

Consider all the radiation in the right cylinder (with any shape cross-section, even rectangular) and length c Δt.  Since all the radiation, and thus all the energy, is flowing at speed c, it follows that all the energy in the cylinder passes A in Δt.  The volume of the cylinder is A c Δt, so

U=(Subscript[u, ave]) A c Δt  and  I=(𝒫/A)=(U/(A Δt))=(Subscript[u, ave])  c

This gives several equivalent expressions for the intensity

I=(Subscript[u, ave])  c=(((Subscript[ε, 0])c)/2) (Subscript[E, max]^2)=(Subscript[ε, 0])c (Subscript[E, rms]^2)

Example . - The Intensity of Sunlight

Solution

(Subscript[ε, 0])=8.85×(10^(-12))(((QuantityUnit))/((QuantityUnit)·(((QuantityUnit))^2))) , c=3.00×(10^8)(QuantityUnit)/(QuantityUnit)   and   I=1370(((QuantityUnit))/(((QuantityUnit))^2))

I=(Subscript[ε, 0])c (Subscript[E, rms]^2)   ⟹   (Subscript[E, rms])=(Sqrt[(I/((Subscript[ε, 0])c ))])=718(((QuantityUnit))/((QuantityUnit)))

((Subscript[E, rms])/(Subscript[B, rms]))=c   ⟹   (Subscript[B, rms])=((Subscript[E, rms])/c)=2.39×(10^(-6))(QuantityUnit)

Solution

r=(Subscript[R, ES])=1.50×(10^11)(QuantityUnit)

I=(𝒫/A)=(𝒫/(4π (r^2)))=(𝒫/(4π (Subscript[R, ES]^2)))    ⟹  𝒫=4 π (Subscript[R, ES]^2)I=3.87×(10^26)(QuantityUnit)

Example . - Laser Light

Solution

A=π (r^2)=π (((((0.0025(QuantityUnit))/2)))^2)=4.9087×(10^(-6))(((QuantityUnit))^2)

𝒫=35mW  and  I=(𝒫/A)  =7130(((QuantityUnit))/(((QuantityUnit))^2))

I=(Subscript[ε, 0])c (Subscript[E, rms]^2)   ⟹   (Subscript[E, rms])=(Sqrt[(I/((Subscript[ε, 0])c ))])=1640(((QuantityUnit))/((QuantityUnit)))

Momentum and Pressure

Momentum Carried by Radiation

Electromagnetic radiation carries energy.  Since there is moving energy there is also momentum.  To see this consider Einstein's famous formula E=m (c^2).  This was shown to be generally true in 1905, but in the case of electromagnetism an analogous expression had been derived previously.  We are using U for energy, so let us write this as U=m (c^2).  Since momentum is p=m v and the radiation is moving at c we can write p=m c.  Combining these expressions we get

p=(U/c).

This is the momentum carried by electromagnetic radiation.

Some comments should be made on the mass referred to above.  The mass in E=m (c^2) is known as the relativistic mass.  This is to be distinguished from the rest mass.  The tabulated values of the masses of particles are their rest masses.  In relativity, a particle with a rest mass can never be accelerated to the speed of light, but it can reach a speed arbitrarily close to that of light.  A particle of light is known as a photon; this is called a massless particle meaning that it has no rest mass.  Massless particles must always move at c.

Momentum from Radiation Normally Incident on a Surface

If the radiation is normally incident on a surface we can derive simple expressions for the momentum given to the surface.  First consider the case of a surface that is a perfect absorber.  All of the momentum of the radiation is given to the surface, giving

p=(U/c)   (perfect absorber).

If the surface is a perfect reflector then the change in the momentum of the radiation is twice the value of the incident radiation.  Recall that momentum is a vector and here we are subtracting two vectors in the opposite direction.  Since momentum must be conserved, the change in the momentum of the radiation is equal (in magnitude) to the momentum given to the surface.

p=2(U/c)    (perfect reflector)

Pressure and Force on a Surface from Normally Incident Radiation

Newton's second law (Subscript[(OverVector[F]), net])=(Δ/(Δ t))(OverVector[p]) relates force to momentum.  The force can be related to the momentum in the case of normally incident radiation by the expression

F=(p/(Δt)).

Pressure is defined as force per area.

Pressure=(F/A).

We can then write the pressure Pressure in terms of the momentum p.

Pressure=(p/(A Δt))

Using the definition of intensity

I=(𝒫/A)=(U/(A Δt)).

We can turn the momentum expressions, which involve the energy U into expressions for the pressure involving the intensity I by dividing both sides of the momentum expressions by A Δt.

Pressure=(I/c)   (perfect absorber)

Pressure=2(I/c)    (perfect reflector)

In the cases of both pressure on a surface and momentum given to a surface, the effect is very small. to get both pressure and momentum we divide by the speed of light and this will typically give a small value. Both effects are difficult to notice and it is hard to find a practical application for them.

Example . - Pushing with a Laser

Solution

𝒫=200mW  ⟹  U=𝒫 t  =𝒫×1 (QuantityUnit)=𝒫×365.24 days×((24(QuantityUnit))/day)×((3600(QuantityUnit))/((QuantityUnit)))=262970(QuantityUnit)

p=2(U/c)=0.00175(QuantityUnit)·(QuantityUnit)/(QuantityUnit)

Example . - Radiation Force on the Earth

Solution

Pressure=(I/c) =4.5667×(10^(-6))(QuantityUnit)

(Subscript[R, E])=6.37×(10^6)(QuantityUnit)   ⟹  F=Pressure×A=Pressure×π (Subscript[R, E]^2)=5.82×(10^8)(QuantityUnit)

. - Polarization

Recall from section 25.1 that the electric field and magnetic field were perpendicular and both were perpendicular to the direction of propagation. There we took x to be the direction of propagation and the electric field was in the y-direction and the magnetic field was in the z-direction. However, there is a plane of possible direction perpendicular to the direction of propagation, which we will keep as x. Rotating the fields in this plane gives different polarizations. We choose the direction of polarization to be the direction of the electric field.

Polarized Light through a Filter

A polarizing filter allows only the component of the electric field along the axis of the filter. The wave that exits the filter is then polarized along the axis of the filter with a smaller amplitude. The amplitude is the peak electric field.

A=(Subscript[E, max])

If (Subscript[A, 0]) is the amplitude before the filter and A is the amplitude after the filter then

A=(Subscript[A, 0])cos θ

The intensity is proportional to the square of the amplitude I∝(A^2), since I=(1/2)(Subscript[ε, 0])c (Subscript[E, max]^2). Using this we can write the intensity of light after the filter to the intensity before.

A=(Subscript[A, 0]) cos θ  and  I∝(A^2)  ⟹  I=(Subscript[I, 0]) (cos^2)θ

Summarizing, when polarized light with intensity (Subscript[I, 0]) passes through a polarizing filter then it leaves with an intensity of I given by.

I=(Subscript[I, 0]) (cos^2)θ

where θ is the angle between the angle of polarization of the light and the polarizing axis of the filter. This relation is known as Malus's law.

Unpolarized Light through a Filter

Viewing normal ambient light through a polarizing filter usually shows no effect when the filter is rotated.  This is because the light is unpolarized; this means that it is a random mixture of all polarizations.  Given this random mixture, we average over all polarizations. Since the average value of (cos^2) is 1/2 we get.

I=(1/2) (Subscript[I, 0])

relating the intensities before and after the filter. The light then leaves the filter polarized along the axis of the filter.

Polarization by Reflection and Scattering

Reflected light tends to be polarized.  When light reflects of a surface at some angle (not normally incident) there is one possible polarization direction that is parallel to the surface and the reflected light tends to be polarized in this parallel direction. For instance, light reflecting off a horizontal surface tends to be horizontally polarized.  Polarizing sunglasses have filters with vertical axes to remove this reflected light or glare. By saying “tends to be polarized” I mean that the light has more of that polarization than the other.

We see the sky because light is scattered. You can view this as light bouncing off air molecules. Light from the sun scatters off an air molecule toward you. The light rays from the sun to the air molecules and from the air molecules to you for a plane and the light tends to be polarized perpendicular to that plane. When you look at the air with the sun behind you, then the light from the sun to the molecules define a vertical plane, and the scattered light then tends to be horizontally polarized. While discussing scattering by air, it should be mentioned that blue light scatters more than the other colors and that is why the sky is blue.

Example . - Light through Two Polarizing Filters

Solution

I=(Subscript[I, 0]) (cos^2)θ

(Subscript[I, 1])=(Subscript[I, 0])(cos^2)(Subscript[θ, 1])=1500(QuantityUnit)/(((QuantityUnit))^2)

(Subscript[I, 2])=(Subscript[I, 1])(cos^2)((Subscript[θ, 2])-(Subscript[θ, 1]))=375(QuantityUnit)/(((QuantityUnit))^2)

Solution

I=(1/2)(Subscript[I, 0])

(Subscript[I, 1])=(1/2)(Subscript[I, 0])=1000(QuantityUnit)/(((QuantityUnit))^2)

(Subscript[I, 2])=(Subscript[I, 1])(cos^2)((Subscript[θ, 2])-(Subscript[θ, 1]))=250(QuantityUnit)/(((QuantityUnit))^2)