Electromagnetic Waves

1
Electromagnetic Waves

• Chapter 34, sections 4-9
• Energy and pressure
• Polarization
• Reflection and Refraction

2
Maxwells Equations in a Vacuum
Consider these equations in a vacuum no charges
or currents
3
Plane Electromagnetic Waves
Solved by
Ey
Bz
Works for any wavelength l2p/k as long as
c
x
Since w2pf and k2p/l, this means lfc. l is
inversely proportional to f.
4
The Electromagnetic Spectrum
infra -red
ultra -violet
Radio waves
g-rays
m-wave
x-rays
5
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 density
• uB (1/2?0) 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.

6
Energy in Electromagnetic Waves
B
dx
area A
E
c
propagation direction
7
Energy in Electromagnetic Waves
Energy per unit volume in an EM wave u uE
uB
B
dx
area A
E
c
propagation direction
8
Energy in Electromagnetic Waves
Energy per unit volume in an EM wave u uE
uB Thus the energy dU in a box of area A
and length dx is
B
dx
area A
E
c
propagation direction
9
Energy in Electromagnetic Waves
Energy per unit volume in an EM wave u uE
uB Thus the energy dU in a box of area 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 at the rate
B
dx
area A
E
c
propagation direction
10
Energy in Electromagnetic Waves
B
Rate of energy flow
dx
area A
E
c
propagation direction
11
Energy in Electromagnetic Waves
B
Rate of energy flow
dx
area A
E
We define the intensity S as the rate of energy
flow per unit area
c
propagation direction
12
Energy in Electromagnetic Waves
B
Rate of energy flow
dx
area A
E
We define the intensity S, as the rate of energy
flow per unit area
c
propagation direction
Rearranging by substituting EcB and BE/c, we get
13
The Poynting Vector
B
In general we write S (1/?0) E x B S
is a vector that points in the direction of
propagation of the wave and represents the rate
of energy flow per unit area. We call this the
Poynting vector. Units of S are Jm-2 s-1, or
Watts/m2.
dx
area A
E
propagation direction
14
The Poynting Vector
For a plane EM wave the intensity is
15
The Poynting Vector
For a plane EM wave the intensity is
Because the fields depend on position and time,
so does the intensity
16
The Poynting Vector
For a plane EM wave the intensity is
Because the fields depend on position and time,
so does the intensity
If you sit at a certain position S will change in
time. The average is
_
Sometimes the notation S is used for Savg.
17
Poynting vector for spherical waves
A point source of light, or any EM radiation,
spreads out as a spherical wave
Power, P, flowing through sphere is same for
any radius.
Source
18
ExampleAn 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 /
4pr2
19
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.
20
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
?
21
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 Earths orbit? Make the sail
reflective.
Can ignore gravity. Need Fma(1000kg)(1
m/s2)1000 N This comes from pressure FPA, so
AF/P. Here P is the radiation pressure of
sunlight Suns power 4 x 1026 W, so
Spower/(4pr2) gives S (4 x 1026 W) /
(4p(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 A1000N / 9.4x10-6N/m2
1.0x108 m2 100 km2
22
Polarization
The direction of polarization of a wave is the
direction of the electric field. Most light is
randomly polarized, which means it contains a
mixture of waves of different polarizations.
Polarization direction
23
Polarization
A polarizer lets through light of only one
polarization
E
Transmitted light has its E in the direction of
the polarizers transmission axis.
E0
q
E E0 cosq hence S S0 cos2q Maluss
Law If the initial beam has bits with random
polarizations, then S S0 (cos2q)avg S0/2 half
gets through.
24
OPTICS
25
Geometrical Optics

• Optics is the study of the behavior of light (not
  necessarily visible light).
• This behavior can be described by Maxwells
  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 inphenomena
  such as interference and diffraction.

26
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.
27
Reflection and Transmission
Some materials reflect light. For example, metals
reflect light because an incident oscillating
light beam causes the metals 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.
28
Geometrical Optics
q1 angle of incidence
Normal to surface
Incident ray
Surface
Angles are measured with respect to the normal to
the surface
29
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)

q1
q1
q1 q1
This is called specular reflection
30
Refraction
More generally, when light passes from one
transparent medium to another, part is reflected
and part is transmitted. The reflected ray obeys
q1 q1.
31
Refraction
More generally, when light passes from one
transparent medium to another, part is reflected
and part is transmitted. The reflected ray obeys
q1 q1.
The transmitted ray obeys Snells Law of
Refraction It stays in the plane, and the angles
are related by n1sinq1 n2sinq2
Here n is the index of refraction of a medium.
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Refraction
q1 angle of incidence ?1 angle of
reflection q1 angle of refraction
Law of Reflection q1 ?1 Law of Refraction n1
sin?1 n2 sin?2
n ? index of refraction ni c / vi vi velocity
of light in medium i
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Refraction
l1v1T
The period T doesnt change, but the speed of
light can be different. in different materials.
Then the wavelengths l1 and l2 are unequal.
This also gives rise to refraction.
q1
1
q1
2
q2
l2v2T
The little shaded triangles have the same
hypotenuse so l1/sinq1 l2/sinq2, or
v1/sinq1v2/sinq2
q2
Define the index of refraction nc/v. Then
Snells law is n1sinq1 n2sinq2
34
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 the bottom?
Air n1.00 Water n1.33 (1.00)sin40
(1.33)sinq sinqsin40/1.33 so q28.9o Then
d/2tan28.9o which gives d1.1 m.
40
air
water
2m
q
d
35
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 the bottom?
Air n1.00 Water n1.33 (1.00)sin40
(1.33)sinq sinqsin40/1.33 so q28.9o Then
d/2tan28.9o which gives d1.1 m.
40
air
water
2m
q
d
36
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 the bottom?
Air n1.00 Water n1.33 (1.00) sin(40)
(1.33) sinq Sinq sin(40)/1.33 so q
28.9o Then d/2 tan(28.9o) which gives ? d1.1
m.
40
air
water
2m
q
d
Turn this around if you shine a light from the
bottom at this position it will look like its
coming from further right.
37
Air-water interface
Air n1 1.00 Water n2 1.33
n1 sin?1 n2 sin?2 n1/n2 sin?2 / sin?1
When the light travels from air to water (n1 lt
n2) the ray is bent towards the normal. When
the light travels from water to air (n2 gt n1) the
ray is bent away from the normal.
This is valid for any pair of materials with n1 lt
n2
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Total Internal Reflection

• Suppose the light goes from medium 1 to 2 and
  that n2ltn1 (for example, from water to air).
• Snells law gives sin q2 (n1 / n2) sin q1.
• Since sin q2 lt 1 there must be a maximum value
  of q1.
• At angles bigger than this critical angle, the
  beam is totally reflected.
• The critical angle is when q2p/2, which
  gives qcsin-1(n2/n1).

39
Total Internal Reflection
n1 gt n2
q2
n2
q2
q1
qc
q1
q1
n1
n2sin p/2 n1sin q1 ... sin q1 sin qc n2 /
n1
Total internal reflection no light is refracted
Some light is refracted and some is reflected
40
Example Fiber Optics
An optical fiber consists of a core with index n1
surrounded by a cladding with index n2, with
n1gtn2. 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 qC n2 / n1 ? qC sin-1(n2
/ n1) sin-1(1.6/1.7) 70o. (Need to graze at
lt 20o)
41
Dispersion
The index of refraction depends on frequency or
wavelength n n(l )
Typically many optical materials, (glass,
quartz) have decreasing n with increasing
wavelength in the visible region of spectrum
Dispersion by a prism
700 nm 400 nm
42
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 .
?2
n1 sin?1 n2 sin?2
n2 1 (air)
Red 1.538 sin(40) 1 sin?400 ? ?400
sin-1(1.538 0.643) 81.34 Violet 1.516
sin(40) 1 sin?700 ? ?700 sin-1(1.516 0.643)
77.02 ? ? 4.32 ? angular dispersion of
the beam
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Reflection and Transmission at Normal Incidence
Geometrical optics cant tell how much is
reflected and how much transmitted at an
interface. This can be derived from Maxwells
equations. These are described in terms of
the reflection and transmission coefficients R
and T, which are, respectively, the fraction of
incident intensity reflected and transmitted. For
the case of normal incidence, one finds
Notice that when n1n2 (so that there is not
really any interface), R0 and T1.
44
Reflection and Transmission at Oblique Incidence
In this case R and T depend on the angle of
incidence in a complicated way and on the
polarization of the incident beam. We relate
polarization to the plane of the three rays.
E parallel
incident
reflected
n1
E perpendicular
n2
transmitted
45
Reflection and Transmission at Oblique Incidence
Light with the perpendicular polarization is
reflected more strongly than light with
the parallel polarization. Hence if unpolarized
light is incident on a surface, the reflected
beam will be partially polarized.
R ()
100
50
perp
parallel
10 20 30 40 50 60 70
80 90
Angle of incidence
Notice that at grazing incidence everything is
reflected.
46
Reflection and Transmission at Oblique Incidence
qp
100
Polarizing angle, or Brewsters angle
50
R ()
perp
parallel
10 20 30 40 50 60 70
80 90
Angle of incidence
Brewsters angle of incidence is the angle at
which light polarized in the plane is not
reflected but transmitted 100All the reflected
light has perpendicular polarization.