Electromagnetic Waves Chapter 34, sections 4-9 Energy and pressure Polarization Reflection and...

$: Electromagnetic Waves Chapter 34, sections 4-9 Energy and pressure Polarization Reflection and Refraction.$
Electromagnetic Waves

Chapter 34, sections 4-9

Energy and pressure

Polarization

Reflection and Refraction

Consider these equations in a vacuum: no charges or currents

€

E∫ • dA =q

ε0

€

B∫ • dA = 0

€

E∫ • dl = −dΦB

dt

€

B∫ • dl = μ0I + μ0ε0

dΦE

dt

€

E∫ • dA = 0

€

B∫ • dA = 0

€

E∫ • dl = −dΦB

dt

€

B∫ • dl = μ0ε0

dΦE

dt

Maxwell’s Equations in a Vacuum

Plane Electromagnetic Waves

x

Ey

Bz

E(x, t) = EP sin (kx-t)

B(x, t) = BP sin (kx-t) z

j

c

Solved by:

Works for any wavelength =2/k as long as

€

E p Bp = ω /k = c

c =1/ ε0μ0 Since =2f and k=2, this means f=c.

is inversely proportional to f.

The Electromagnetic Spectrum

Radio waves

-wave

infra-red -rays

x-rays

ultra-violet

Energy in Electromagnetic Waves

• Electric and magnetic fields contain energy, the potential energy stored in the field:

uE= (1/2)0 E2 electric field energy densityuB= (1/20) B2 magnetic field energy density

• The energy is put into the oscillating fields by the sources that generate them.

• This energy then propagates to locations far away, at the velocity of light.

B

E


area A

dx

c propagation direction

B

E


area A

dx

Energy per unit volume in an EM wave:

u = uE + uB


€

=1

2(ε0E 2 +

1

μ0

B2)

B

E


area A

dx


u = uE + uB

Thus the energy dU in a box ofarea A and length dx is


€

dU =1

2(ε0E 2 +

1

μ0

B2)Adx€

=1

2(ε0E 2 +

1

μ0

B2)

B

E


area A

dx


u = uE + uB

Thus the energy dU in a box ofarea A and length dx is

Let the length dx equal cdt. Then all of this energy flows through the front face in time dt. Thus energy flows atthe rate


€

dU =1

2(ε0E 2 +

1

μ0

B2)Adx

€

dU

dt=

1

2(ε0E 2 +

1

μ0

B2)Ac

€

=1

2(ε0E 2 +

1

μ0

B2)


area A

dx


B

E

€

dU

dt=

c

2(ε0E 2 +

1

μ0

B2)A

Rate of energy flow:


area A

dx


We define the intensity S as the rateof energy flow per unit area:

€

S =c

2(ε0E 2 +

1

μ0

B2)

B

E

€

dU

dt=

c

2(ε0E 2 +

1

μ0

B2)A



area A

dx


We define the intensity S, as the rateof energy flow per unit area:

€

S =c

2(ε0E 2 +

1

μ0

B2)

Rearranging by substituting E=cB and B=E/c, we get

€

S =c

2(ε0cEB +

1

μ0cEB) =

1

2μ0

(ε0μ0c2 +1)EB =

EB

μ0

B

E

€

dU

dt=

c

2(ε0E 2 +

1

μ0

B2)A


The Poynting Vector

area A

dx

B

E

propagation direction

In general we write:

S = (1/0) E x B

S is a vector that points in thedirection of propagation of thewave and represents the rate ofenergy flow per unit area. We call this the “Poynting vector”.

Units of S are Jm-2 s-1, or Watts/m2.

rS

The Poynting Vector

For a plane EM wave the intensity is

€

S =EB

μ0

=E 2

cμ0

The Poynting Vector


€

S =EB

μ0

=E 2

cμ0

Because the fields depend on position and time, so does the intensity:

€

S =1

cμ0

E p2 sin2 kx −ωt( )

The Poynting Vector


€

S =EB

μ0

=E 2

cμ0

Because the fields depend on position and time, so does the intensity:

€

S =1

cμ0

E p2 sin2 kx −ωt( )

If you sit at a certain position S will change in time. The average is

€

I = Savg =1

cμ0

E p2 sin2 kx −ωt( )[ ]

avg=

1

2cμ0

E p2 =

1

cμ0

E rms2

Sometimes the notation S is used for Savg._

Poynting vector for spherical waves

Sourcer

A point source of light, or any EM radiation, spreads out as a spherical wave:

Source

Power, P, flowingthrough sphereis same for anyradius.

€

Area∝ r2€

S =P

4πr2

€

S ∝1

r2

Example:An observer is 1.8 m from a point light source whose average power P= 250 W. Calculate the rms fields in the position of the observer.

Intensity of light at a distance r is S= P / 4r2

€

I =P

4πr2=

1

μ0cE rms

2

∴ E rms =Pμ0c

4πr2=

(250W )(4π10−7 H /m)(3.108 m /s)

4π (1.8m)2

∴ E rms = 48V /m

∴ B =E rms

c=

48V /m

3.108 m /s= 0.16μT

Wave Momentum and Radiation Pressure

It is somewhat surprising to discover that EM radiation possesses momentum as well as energy. The momentum and energy of a wave are related by p = U / c.

Wave Momentum and Radiation Pressure

It is somewhat surprising to discover that EM radiation possesses momentum as well as energy. The momentum and energy of a wave are related by p = U / c.

If light carries momentum then it follows that a beam of light falling on an object exerts a pressure:

Force = dp/dt = (dU/dt)/c Pressure (radiation) = Force / unit area

P = (dU/dt) / (A c) = S / c

Radiation Pressure

€

Prad =S

c

Example: Serious proposals have been made to “sail” spacecraft to the outer solar system using the pressure of sunlight. How much sail area must a 1000 kg spacecraft have if its acceleration is to be 1 m/s2 at the Earth’s orbit? Make the sail reflective.

Can ignore gravity. Need F=ma=(1000kg)(1 m/s2)=1000 NThis comes from pressure: F=PA, so A=F/P.Here P is the radiation pressure of sunlight:Sun’s power = 4 x 1026 W, so S=power/(4r2) gives S = (4 x 1026 W) / (4(1.5x1011m)2 )= 1.4kW/m2.Thus the pressure due to this light, reflected, is: P = 2S/c = 2(1400W/m2) / 3x108m/s = 9.4x10-6N/m2

Hence A=1000N / 9.4x10-6N/m2 =1.0x108 m2 = 100 km2

Polarization

The direction of polarization of a wave is the direction ofthe electric field. Most light is randomly polarized, which

means it contains a mixture of waves of different polarizations.

x

Ey

Bz Polarization direction

Polarization

A polarizer lets through light of only one polarization:

E0

E

E = E0 cos

hence S = S0 cos2 Malus’s Law

If the initial beam has bits with random polarizations, then S = S0 (cos2avg= S0/2: half gets through.

Transmitted lighthas its E in thedirection of thepolarizer’s transmission axis.

OPTICSOPTICS

Geometrical Optics

• Optics is the study of the behavior of light (not necessarily visible light).

• This behavior can be described by Maxwell’s equations.

• However, when the objects with which light interacts are larger that its wavelength,the light travels in straight lines called rays, and its wave nature can be ignored.

• This is the realm of geometrical optics. • The wave properties of light show up in

phenomena such as interference and diffraction.

Geometrical Optics

Light can be described using geometrical optics, as long as the objects with which it interacts are much larger than the wavelength of the light.

This can be described using geometrical optics

This requires the use of fullwave optics (Maxwell’s equations)

Reflection and Transmission

Some materials reflect light. For example, metals reflect light because an incident oscillating light beam causes the metal’s nearly free electrons to oscillate, setting up another (reflected) electromagnetic wave.

Opaque materials absorb light (by, say, moving electrons into higher atomic orbitals).

Transparent materials are usually insulators whose electrons are bound to atoms, and which would require more energy to move to higher orbitals than in materials which are opaque.

1

1 = angle of incidence

Geometrical Optics

Surface

Normal to surface

Incident ray

Angles are measured with respect to the normal to the surface

Reflection

The Law of Reflection:

Light reflected from a surface stays in the plane formed by the incident ray and the surface normal; and the angle of reflection equals the angle of incidence (measured to the normal)

1 ’1

1 = ’1

This is called “specular” reflection

Refraction

More generally, when light passes from one transparent medium to another, part is reflected and part is transmitted. The reflected ray obeys 1 = ’

1.

1 ’1

2

Medium 1

Medium 2

Refraction

1 ’1

2

Medium 1

Medium 2

More generally, when light passes from one transparent medium to another, part is reflected and part is transmitted. The reflected ray obeys 1 = ’

1.

The transmitted ray obeys

Snell’s Law of Refraction:

It stays in the plane, and the angles are related by

n1sin1 = n2sin2

Here n is the “index of refraction” of a medium.

Refraction

n index of refractionni = c / vi

vi = velocity of light in medium i

Incident ray

1 ’1

2

Medium 1

Medium 2

Reflected ray

Refracted ray

1 = angle of incidence

’1= angle of reflection

1 = angle of refraction

Law of Reflection1 = ’1

Law of Refractionn1 sin1= n2 sin2

Refraction

The little shaded triangles have the same hypotenuse: so 1/sin1= 2/sin2, or

v1/sin1=v2/sin2

1=v1T

2=v2T

1

2

1

2

1

2

Define the index of refraction: n=c/v.Then Snell’s law is: n1sin1 = n2sin2

The period T doesn’t change, but the speed of light can be different. in different materials. Then the wavelengths 1 and 2 are unequal. This also gives rise to refraction.

Example: air-water interface

If you shine a light at an incident angle of 40o onto the surface of a pool 2m deep, where does the beam hit thebottom?

air

water

40

2m

Air: n=1.00 Water: n=1.33

(1.00)sin40 = (1.33)sinsin=sin40/1.33 so =28.9o

Then d/2=tan28.9o which givesd=1.1 m.

d

Example: air-water interface

If you shine a light at an incident angle of 40o onto the surface of a pool 2m deep, where does the beam hit thebottom?

air

water

40

2m

Air: n=1.00 Water: n=1.33

(1.00) sin(40) = (1.33) sinSin = sin(40)/1.33 so = 28.9o

Then d/2 = tan(28.9o) which gives d=1.1 m.

d

Turn this around: if you shine a light from the bottom atthis position it will look like it’s coming from further right.

Air-water interface

air

water

1

2

Air: n1 = 1.00 Water: n2 = 1.33

When the light travels from air towater (n1 < n2) the ray is bent towards the normal.

When the light travels from waterto air (n2 > n1) the ray is bent away from the normal.

n1 sin1 = n2 sin2 n1/n2 = sin2 / sin1

This is valid for any pair of materials with n1 < n2

Total Internal Reflection

• Suppose the light goes from medium 1 to 2 and that n2<n1 (for example, from water to air).

• Snell’s law gives sin 2 = (n1 / n2) sin 1.

• Since sin 2 <= 1 there must be a maximum value of 1.

• At angles bigger than this “critical angle”, the beam is totally reflected.

• The critical angle is when 2=/2, which givesc=sin-1(n2/n1).

c

2 2

11

1

n2

n1

Some light is refracted and some is reflected

Total internal reflection:no light is refracted

Total Internal Reflection

n2sin = n1sin 1

... sin 1 = sin c = n2 / n1

n1 > n2

Example: Fiber Optics

An optical fiber consists of a core with index n1 surrounded by a cladding with index n2, with n1>n2. Light can be confined by total internal reflection, even if the fiber is bent and twisted.

Exercise: For n1 = 1.7 and n2 = 1.6 find the minimum angle of incidence for guiding in the fiber.

Answer: sin C = n2 / n1 C = sin-1(n2 / n1) = sin-1(1.6/1.7) = 70o.

(Need to graze at < 20o)

Dispersion

The index of refraction depends on frequency or wavelength: n = n( )

Typically many opticalmaterials, (glass, quartz)have decreasing n with increasing wavelength in thevisible region of spectrum

Dispersion by a prism:700 nm400 nm

1.55

1.53

1.51

400 500 600 700 nm

n

Example: dispersion at a right angle prism

Find the angle between outgoing red (r = 700nm) and violet (v = 400nm) light [ n400 =1.538, n700 = 1.516, 1 = 40° ].

1 red

violet

2

Red: 1.538 sin(40°) = 1 sin400 400 = sin-1(1.538 0.643) = 81.34°

Violet: 1.516 sin(40°) = 1 sin700 700 = sin-1(1.516 0.643) = 77.02°

= 4.32° angular dispersion of the beam

n1 sin1 = n2 sin2

n2 = 1 (air)

Reflection and Transmission at Normal Incidence

Geometrical optics can’t tell how much is reflected and howmuch transmitted at an interface. This can be derived fromMaxwell’s equations. These are described in terms of thereflection and transmission coefficients R and T, which are,respectively, the fraction of incident intensity reflected andtransmitted. For the case of normal incidence, one finds:

Notice that when n1=n2 (so that there is not really anyinterface), R=0 and T=1.

I RI

TI

R n nn n

T R n nn n

= −+

⎛⎝⎜

⎞⎠⎟ = − =

+ 1 1

1

1 1 4,( )

Reflection and Transmission at Oblique Incidence

In this case R and T depend on the angle of incidence ina complicated way – and on the polarization of the incidentbeam. We relate polarization to the plane of the three rays.

E parallel

reflected

incident

transmitted

E perpendicular

n1

n2

100

50

10 20 30 40 50 60 70 80 90

Angle of incidence

R (%)


perp parallel

Light with the perpendicularpolarization is reflected morestrongly than light with theparallel polarization.

Hence if unpolarized light is incident on a surface, thereflected beam will be partially polarized.

Notice that at grazing incidence everything is reflected.

100

50

10 20 30 40 50 60 70 80 90

Angle of incidence

R (%)

Polarizing angle, or“Brewster’s angle”

Brewster’s angle of incidence is the angle at which light polarized in the plane is not reflected but transmitted 100%All the reflected light has perpendicular polarization.


p

perp parallel tanp =nn1

Electromagnetic Waves Chapter 34, sections 4-9 Energy and pressure Polarization Reflection and...

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Transcript of Electromagnetic Waves Chapter 34, sections 4-9 Energy and pressure Polarization Reflection and...