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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
Energy in Electromagnetic Waves
area A
dx
c propagation direction
B
E
Energy in Electromagnetic Waves
area A
dx
Energy per unit volume in an EM wave:
u = uE + uB
c propagation direction
€
=1
2(ε0E 2 +
1
μ0
B2)
B
E
Energy in Electromagnetic Waves
area A
dx
Energy per unit volume in an EM wave:
u = uE + uB
Thus the energy dU in a box ofarea A and length dx is
c propagation direction
€
dU =1
2(ε0E 2 +
1
μ0
B2)Adx€
=1
2(ε0E 2 +
1
μ0
B2)
B
E
Energy in Electromagnetic Waves
area A
dx
Energy per unit volume in an EM wave:
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
c propagation direction
€
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)
Energy in Electromagnetic Waves
area A
dx
c propagation direction
B
E
€
dU
dt=
c
2(ε0E 2 +
1
μ0
B2)A
Rate of energy flow:
Energy in Electromagnetic Waves
area A
dx
c propagation direction
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
Rate of energy flow:
Energy in Electromagnetic Waves
area A
dx
c propagation direction
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
Rate of energy flow:
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
For a plane EM wave the intensity is
€
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
For a plane EM wave the intensity is
€
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)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 (%)
Reflection and Transmission at Oblique Incidence
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.
Reflection and Transmission at Oblique Incidence
p
perp parallel tanp =nn1