Transcript Ch. 34 - Maxwell's equations

Electromagnetic Waves
Chapter 34, sections 4-9
Energy and pressure
Polarization
Reflection and Refraction
Maxwell’s Equations in a Vacuum
Consider these equations in a vacuum: no charges or currents
 E  dA  
q
0
 B  dA  0
dB 
 E  dl   dt


dE
 B  dl  0I  00 dt



 E  dA  0
 B  dA  0

dB
E  dl  
dt
dE
 B  dl  00 dt
Plane Electromagnetic Waves
Solved by:
Ey
Bz
Works for any wavelength
l=2p/k as long as
E p Bp   /k  c
c  1/ 00
E(x, t) = EP sin (kx-t) jˆ
B(x, t) = BP sin (kx-t) zˆ
c
x
Since =2pf and k=2p/l, this means
lf=c.
l is inversely proportional to f.
The Electromagnetic Spectrum
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.
Energy in Electromagnetic Waves
B
dx
area
A
c
E
propagation
direction
Energy in Electromagnetic Waves
Energy per unit volume in an EM wave:
1
1 2
2
u = uE + uB  (0 E  B )
2
0

B
dx
area
A
c
E
propagation
direction
Energy in Electromagnetic Waves
Energy per unit volume in an EM wave:
1
1 2
2
u = uE + uB  (0 E  B )
2
0
dx
area
A
Thus the energy dU in a box of
area A and length dx is


1
1 2
2
dU  (0 E  B )Adx
2
0
B
c
E
propagation
direction
Energy in Electromagnetic Waves
Energy per unit volume in an EM wave:
1
1 2
2
u = uE + uB  (0 E  B )
2
0
dx
area
A
Thus the energy dU in a box of
area A and length dx is

1
1 2
2
dU  (0 E  B )Adx
2
0
B
c
E
propagation
direction
Let the length dx equal cdt. Then all of this energy flows
through the front face in time dt. Thus energy flows at
dU 1
1 2
 the rate
2
 (0 E  B )Ac
dt 2
0

Energy in Electromagnetic Waves
Rate of energy flow:
dU c
1 2
2
 (0 E  B )A
dt 2
0
B
dx
area
A
c
E
propagation
direction
Energy in Electromagnetic Waves
Rate of energy flow:
dU c
1 2
2
 (0 E  B )A
dt 2
0

B
dx
area
A
We define the intensity S as the rate
of energy flow per unit area:

c
1 2
2
S  (0 E  B )
2
0
c
E
propagation
direction
Energy in Electromagnetic Waves
Rate of energy flow:
dU c
1 2
2
 (0 E  B )A
dt 2
0

B
dx
area
A
We define the intensity S, as the rate
of energy flow per unit area:
c
1 2
2
S  (0 E  B )
2
0
c
E
propagation
direction
Rearranging by substituting E=cB and B=E/c, we get

c
1
1
EB
2
S  (0cEB
EB) 
(00c  1)EB 
2
0c
20
0
The Poynting Vector
B
In general we write:
dx
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 S
energy flow per unit area.
We call this the “Poynting vector”.
Units of S are Jm-2 s-1, or Watts/m2.
area
A
E
propagation
direction
The Poynting Vector
E2
For a plane EM wave the intensity is S 

0 c0
EB

The Poynting Vector
E2
For a plane EM wave the intensity is S 

0 c0
EB
Because the fields depend on position and time, so does the intensity:
1
2
2
S
E 
sin
kx  t 
p
c0

The Poynting Vector
E2
For a plane EM wave the intensity is S 

0 c0
EB
Because the fields depend on position and time, so does the intensity:
1
2
2
S
E 
sin
kx  t 
p
c0
If you sit at a certain position S will change in time. The average is
1
1
1
2
2
2
2

I  Savg 
E p sin kx  t avg 
Ep 
E rms
c0
2c0
c0
_
Sometimes the notation S is used for Savg.
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
P
S
2
4 pr
Source
r
Area  r 2
1
S 2
r

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 / 4pr2
P
1 2
I

E rms
2
4pr
0c
 E rms
P0c
(250W )(4p107 H /m)(3.108 m /s)


2
4pr
4p (1.8m) 2
 E rms  48V /m
E rms
48V /m
B 

 0.16T
8
c
3.10 m /s
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 N
This 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/(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 A=1000N / 9.4x10-6N/m2 =1.0x108 m2 = 100 km2
Polarizatio
n
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.
Ey
Bz
Polarization
direction
x
Polarizatio
n
A polarizer lets through light of only one polarization:
q
E0
E
Transmitted light
has its E in the
direction of the
polarizer’s
transmission axis.
E = E0 cosq
hence S = S0 cos2q
Malus’s Law
If the initial beam has bits with random polarizations,
then S = S0 (cos2qavg= S0/2: half gets through.
OPTICS
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 full
wave 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.
Geometrical Optics
q1 = angle of incidence
q1
Normal to surface
Incident ray
Surface
Angles are measured with respect to the normal to the surface
Reflection
q1
q’1
q1 = q’1
This is called
“specular” 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)
Refraction
q1
q’1
Medium 1
Medium 2
q2
More generally, when light passes
from one transparent medium to
another, part is reflected and part
is transmitted. The reflected ray
obeys q1 = q’1.
Refraction
q1
q’1
q2
Medium 1
More generally, when light passes
from one transparent medium to
another, part is reflected and part
is transmitted. The reflected ray
obeys q1 = q’1.
Medium 2
The transmitted ray obeys
Snell’s 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.
Refraction
q1
q’1
Medium 1
Reflected ray
Incident ray
Medium 2
Refracted ray
q2
q1 = angle of incidence
q’1= angle of reflection
q1 = angle of refraction
Law of Reflection
q1 = q’1
Law of Refraction
n1 sinq1= n2 sinq2
n  index of refraction
ni = c / vi
vi = velocity of light in
medium i
Refraction
l1=v1T
1
q1
q1
2
q2
q2
The period T doesn’t 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.
l2=v2T
The little shaded triangles have the same
hypotenuse: so l1/sinq1= l2/sinq2, or
v1/sinq1=v2/sinq2
Define the index of refraction: n=c/v.
Then Snell’s law is: n1sinq1 = n2sinq2
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: n=1.00
Water: n=1.33
40
air
water
2m
q
d
(1.00)sin40 = (1.33)sinq
sinq=sin40/1.33 so q=28.9o
Then d/2=tan28.9o which gives
d=1.1 m.
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: n=1.00
Water: n=1.33
40
air
water
2m
q
d
(1.00)sin40 = (1.33)sinq
sinq=sin40/1.33 so q=28.9o
Then d/2=tan28.9o which gives
d=1.1 m.
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: n=1.00
40
air
water
2m
q
d
Water: n=1.33
(1.00) sin(40) = (1.33) sinq
Sinq= sin(40)/1.33 so q= 28.9o
Then d/2 = tan(28.9o)
which gives  d=1.1 m.
Turn this around: if you shine a light from the bottom at
this position it will look like it’s coming from further right.
Air-water interface
Air: n1 = 1.00
Water: n2 = 1.33
n1 sinq1 = n2 sinq2
n1/n2 = sinq2 / sinq1
q1
air
water
q2
When the light travels from air to
water (n1 < n2) the ray is bent
towards the normal.
When the light travels from water
to air (n2 > n1) the ray is bent
away from the normal.
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 q2 = (n1 / n2) sin q1.
• Since sin q2 <= 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 q2=p/2, which gives
qc=sin-1(n2/n1).
Total Internal Reflection
n1 > n2
q2
q1
q2
q1
n2
qc
q1
n1
n2sin p/2 = n1sin q1
... sin q1 = sin qc = n2 / n1
Some light is refracted
and some is reflected
Total internal reflection:
no light is refracted
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 qC = n2 / n1  qC = 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(l )
Typically many optical
materials, (glass, quartz)
have decreasing n with
increasing wavelength in the
visible region of spectrum
n
1.55
1.53
1.51
l
400
500
600
Dispersion by a prism:
700 nm
400 nm
700
nm
Example: dispersion at a right angle prism
Find the angle between outgoing red (lr = 700nm) and
violet (lv = 400nm) light [ n400 =1.538, n700 = 1.516, q1 = 40° ].
n1 sinq1 = n2 sinq2
n2 = 1 (air)
q1
q2
red
violet
Red: 1.538 sin(40°) = 1 sinq400  q400 = sin-1(1.538 0.643) = 81.34°
Violet: 1.516 sin(40°) = 1 sinq700  q700 = sin-1(1.516 0.643) = 77.02°
  = 4.32°  angular dispersion of the beam
Reflection and Transmission
at Normal Incidence
Geometrical optics can’t tell how much is reflected and how
much transmitted at an interface. This can be derived from
Maxwell’s 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:
TI
I
RI
2
n2  n1 
4n1 n2

R
, T  1 R 

( n2  n1 )2
 n2  n1 
Notice that when n1=n2 (so that there is not really any
interface), R=0 and T=1.
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.
incident
E parallel
reflected
n1
E perpendicular
n2
transmitted
Reflection and Transmission
at Oblique Incidence
Light with the perpendicular
polarization is reflected more
strongly than light with the
parallel polarization.
R (%)
100
50
perp
10
20
30
parallel
40
50
60
70 80
Angle of incidence
90
Hence if unpolarized light
is incident on a surface, the
reflected beam will be
partially polarized.
Notice that at grazing incidence everything is reflected.
Reflection and Transmission
at Oblique Incidence
qp
100
Polarizing angle, or
“Brewster’s angle”
50
R (%)
perp
10
20
30
parallel
40
50
60
70 80
n2
tan q p 
n1
90
Angle of incidence
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.