tue. nov. 11, 2008physics 208, lecture 211 from last time… em waves inductors in circuits i? + -
TRANSCRIPT
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 1
From last time…
EM waves
Inductors in circuits I?
+
-
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 2
•A Transverse wave.
•Electric/magnetic fields perpendicular to propagation direction
•Can travel in empty space
f = v/, v = c = 3 x 108 m/s (186,000 miles/second)
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 3
A microwave oven irradiates food with electromagnetic radiation that has a frequency of about 1010 Hz. The wavelengths of these microwaves are on the order of
A. kilometers
B. meters
C. centimeters
D. micrometers
Quick Quiz
€
=c / f =3 ×108 m /s
1010 /s= 3cm
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 4
Electromagnetic waves
€
Bo = Eo /c
€
rE =
r E o cos kz −ωt( )
r B =
r B o cos kz −ωt( )
€
c =1/ εoμo =1/ 8.85 ×10−12C2 /N ⋅m2( ) 4π ×10−7 N / A2
( )
= 2.9986 ×108 m /s
€
rE ⊥
r B
€
k =2π
λ, ω =
2π
f
z
x
y
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 5
Energy and EM Waves
Energy density in E-field
Energy density in B-field
€
uE = εoE 2 r, t( ) /2
€
uB = B2 r, t( ) /2μo
Total
€
uTot = εoE 2 /2 + B2 /2μo
= εoE 2 /2 + E 2 /2c 2μo = εoE 2 r, t( ) = B2 r, t( ) /μo
€
uTot = εoE 2 = εoEo2 cos2 kz −ωt( ) moves w/ EM wave
at speed c
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 6
Power and intensity in EM waves Energy density uE moves at c
Instantaneous energy transfer = energy passing plane per second. = This is power density W/m2
Time average of this is Intensity =
€
cuTot = cεoE 2 r, t( ) = cB2 r, t( ) /μo
€
cεoEmax2 /2 = cBmax
2 /2μo
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 7
Example: E-field in laser pointer
1 mW laser pointer.
Beam diameter at board ~ 2mm
Intensity =
€
10−3W
π 0.001m( )2 = 318W /m2
How big is max E-field?
€
cεoEmax2 /2 = 318W /m2
Emax =2 318W /m2
( )
3×108 m /s( ) 8.85 ×10−12C2 /N ⋅m2( )
= 489N /C = 489V /m
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 8
Spherical waves Sources often radiate EM wave in all directions
Light bulb The sun Radio/tv transmission tower
Spherical wave, looks like plane wave far away Intensity decreases with distance
Power spread over larger area
€
I =Psource
4π r2
Source power
Spread over thissurface area
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 9
QuestionA radio station transmits 50kW of power from its
antanna. What is the amplitude of the electric field at your radio, 1km away.
€
I =50,000W
4π 1000m( )2 = 4 ×10−3W / m2
€
cεoEmax2 /2 = 4 ×10−3W /m2
Emax =2 4 ×10−3W /m2( )
3 ×108 m /s( ) 8.85 ×10−12C2 /N ⋅m2( )
=1.73N /C =1.73V /m
A. 0.1 V/m
B. 0.5 V/m
C. 1 V/m
D. 1.7 V/m
E. 15 V/m
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 10
The Poynting Vector Rate at which energy flows through a unit area perpendicular
to direction of wave propagation
Instantaneous power per unit area (J/s.m2 = W/m2) is also
Its direction is the direction of propagation of the EM wave
This is time dependent Its magnitude varies in time Its magnitude reaches a maximum at the
same instant as E and B
€
rS =
1
μo
r E ×
r B ≡ Poynting Vector
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 11
Radiation Pressure Saw EM waves carry energy They also have momentum When object absorbs energy U from EM wave:
Momentum p is transferred
Result is a force
Pressure = Force/Area = €
p = U /c ( Will see this later in QM )
€
F = Δp /Δt =U /Δt
c= P /c
€
prad =P / A
c= I /c
Radiation pressure on perfectly absorbing object
Power
Intensity
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 12
Radiation pressure & force
EM wave incident on surface exerts a radiation pressure prad (force/area) proportional to intensity I.
Perfectly absorbing (black) surface:
Perfectly reflecting (mirror) surface:
Resulting force = (radiation pressure) x (area) €
prad = I /c
€
prad = 2I /c
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 13
QuestionA perfectly reflecting square solar sail is 107m X 107m. It has
a mass of 100kg. It starts from rest near the Earth’s orbit, where the sun’s EM radiation has an intensity of 1300 W/m2.
How fast is it moving after 1 hour?
€
prad = 2I /c
Frad = prad A = 2IA /c =2 1300W /m2( ) 1.145 ×104 m2
( )
3 ×108 m /s= 0.1N
a = Frad /m =10−3 m /s2
v = at = 10−3 m /s2( ) 3600s( ) = 3.6m /s
A. 100 m/s
B. 56 m/s
C. 17 m/s
D. 3.6 m/s
E. 0.7 m/s
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 14
Polarization of EM waves Usually indicate the polarization direction by
indicating only the E-field. Can then be indicated with a line:
Unpolarized Plane Polarizedx
yz
€
rE = Eo cos kz −ωt( ) ˆ x r B = Bo cos kz −ωt( ) ˆ y
Superposition of plane polarized waves
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 15
Producing polarized light Polarization by selective absorption: material that transmits
waves whose E-field vibrates in a plain parallel to a certain direction and absorbs all others
This polarizationabsorbed
This polarizationtransmitted transmission axis
Polaroid sheet
Long-chain hydrocarbon molecules Demo on MW and metal grid
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 16
Transmission at an angle
Incident wave is equivalent to superposition
x
y
transmission
€
rE inc =
r E o cos kx −ωt( )
Plane-polarizedincident wave
polarizer
€
E inc cosθ( ) ˆ x + E inc sinθ( ) ˆ y
absorbedtransmitted
Transmitted wave =
€
rE trans = Eo cosθ cos kx −ωt( ) ˆ x
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 17
Detecting polarized light Polarizer
transmits component of E-field parallel to transmission axis absorbs component of E-field perpendicular to transmission axis
Transmitted intensity: I = I0cos2 I0 = intensity of polarized beam on analyzer (Malus’ law)
Allowed componentparallel to analyzer axis
Polaroid sheets
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 18
Malus’ law
Transmitted amplitude is Eocos (component of polarization along polarizer axis)
Transmitted intensity is Iocos2( square of amplitude)
Perpendicular polarizers give zero intensity.
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 19
Polarization by reflectionUnpolarizedIncident light
Reflection polarized with E-field parallel to surface
Refractedlight
Unpolarized light reflected from a surface becomes partially polarized
Degree of polarization depends on angle of incidence n
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 20
Reducing glare
Transmission axis
Reflected sunlight partially polarized.
Horizontal reflective surface ->the E-field vector of reflected light has strong horizontal component.
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Tue. Nov. 11, 2008 Physics 208, Lecture 21 21
Circular and elliptical polarization
Circularly polarized light is a superposition of two waves with orthogonal linear polarizations, and 90˚ out of phase.
The electric field rotates in time with constant magnitude.