tue. nov. 11, 2008physics 208, lecture 211 from last time… em waves inductors in circuits i? + -
TRANSCRIPT
Tue. Nov. 11, 2008 Physics 208, Lecture 21 1
From last time…
EM waves
Inductors in circuits I?
+
-
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)
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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
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.
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.