Lecture 32 Chapter 34 Electromagnetic Waves

Lecture 32
Chapter 34
Electromagnetic Waves
• EM Waves –
Review
– Wavelengths of 108 to 1016 meters (10-1024 Hz)
– Traveling wave of both E
and B fields
– E field is ⊥ B field
– Wave moves in direction ⊥
to both E and B fields
– E and B vary sinusoidally
with same frequency
– At large distances fields
are in phase
r r
E×B
E = E m sin( kx − ω t )
B = Bm sin( kx − ω t )
Review
• EM waves move at the
8
c = 3 × 10 m / s
speed of light, c in free
ω
space (vacuum or air)
v=
• Relate velocity of wave by
k
• Using definition of ω and
2π
wave number k
ω = 2πf k =
λ
• Find velocity of wave is
• Also defined as
Em
c=
Bm
v = c = fλ
c=
1
µ0ε 0
Review
r 1 r r
S=
E×B
• Poynting vector, S – rate of
energy transported per unit
area
• Instantaneous energy flow rate
µ0
S =
1
µ0
• Defined intensity I to be time
averaged value of S
I = S avg
 energy / time 
 power 
=
 =

area

 ave  area  ave
I = S avg =
1
cµ 0
E rms
2
EB
EM Waves (12)
• Problem – Isotropic point light source as power
of 250 W. You are 1.8 meters away. Calculate
the rms values of the E and B fields.
• To find Erms need
1
Ps
2
I =
E rms
I =
2
cµ 0
4π r
• Find intensity I from
Erms = Icµ0 =
Erms
Ps cµ0
2
4πr
(250)(3 × 108 )(1.26 × 10−8 )
=
= 48.1V / m
2
(4π )(1.8)
EM Waves (13)
• Problem – Isotropic point light source as power
of 250 W. You are 1.8 meters away. Calculate
the rms values of the E and B fields.
• To find Brms need
Erms
c=
Brms
Brms
Brms
Erms
=
c
48.1V / m
−7
=
= 1.6 × 10 T
8
3 × 10 m / s
EM Waves (14)
• Look at sizes of Erms and Brms
Erms = 48.1V / m
B rms = 1 . 6 × 10
−7
T
• This is why most instruments measure E
• Does not mean that E component is stronger
than B component in EM wave
– Can’t compare different units
• Average energies are equal for E and B
EM Waves (15)
• The energy density of electric field, uE is equal
to energy density of magnetic field, uB
uE = ε 0 E
1
2
2
E = Bc
uE = ε 0 ( cB ) = ε 0c B
1
2
2
2
1
2
2
1
1
B
2
uE = ε 0
B =
2 µ0ε 0
2 µ0
uE = uB
2
c=
1
µ0ε 0
2
B
uB =
2µ0
EM Waves (16)
• EM waves linear have momentum
momentum as well as energy
• Light shining on object exerts a
pressure – radiation pressure
• Object’s change in momentum is
related to its change in energy
• If object absorbs all radiation from EM
∆
p
wave (total absorption)
• If object reflects all radiation back
original direction (total reflection)
in
∆U
=
c
2 ∆U
∆p =
c
EM Waves (17)
• Just defined intensity, I as power
per unit area A so power is
• Change in energy is amount of
power P in time t
P = IA
∆ U = P ∆ t = IA ∆ t
• Want force of radiation on object
• For total absorption
∆U
∆p =
• Find force is
∆p
F=
∆t
c
∆ p ∆U IA ∆ t IA
=
=
F=
=
∆t c∆t
c∆t
c
EM Waves (18)
• For total reflection back along original path
2 ∆U
∆p =
c
∆p 2 ∆U 2 IA ∆t 2 IA
F=
=
=
=
c∆t
c∆t
c
∆t
• Express in terms of radiation
pr which is force/area
• SI unit is N/m2 called pascal Pa
• Total absorption
I
pr =
c
pressure
F
pr =
• Total reflection
2I
pr =
c
A
EM Waves (19)
• Source emits EM waves
with E field always in same
plane wave is polarized
– Example, television station
• Indicate a wave is polarized
by drawing double arrow
• Plane containing the E field
is called plane of oscillation
EM Waves (20)
• Source emits EM waves
with random planes of
oscillation (E field changes
direction) is unpolarized
– Example, light bulb or Sun
• Resolve E field into
components
• Draw unpolarized light as
superposition of 2 polarized
waves with E fields ⊥ to
each other
EM Waves (21)
• Transform unpolarized
light into polarized by
using a polarizing sheet
• Sheet contains long
molecules embedded
in plastic which was
stretched to align the
molecules in rows
• E field component || to polarizing direction
of sheet is passed (transmitted), but ⊥
component is absorbed
EM Waves (22)
• What is the intensity, I
of the light transmitted
by polarizing sheet?
• For unpolarized light,
separate E field into
components
• Sum of 2 components
1
are equal but only light ||
2
to polarizer is transmitted
• One-half rule: Intensity of unpolarized
wave after a polarizer is half of original
I = I0
EM Waves (23)
• For polarized light, resolve E
into components
• Transmitted || component is
E y = E cos θ
• Use definition of intensity
I =
1
cµ 0
E
2
=
1
cµ 0
E 2 cos 2 θ = I 0 cos 2 θ
• Cosine-squared rule: Intensity of
polarized wave changes as cos2θ
I = I 0 cos θ
2
EM Waves (24)
• Have 2 polarizing sheets
– First one called polarizer
– Second one called analyzer
• Intensity of unpolarized
light going through
polarizer is
I =
1
2
I0
• Light is now polarized and
intensity of light after
analyzer is given by
I = I 0 cos θ
2
EM Waves (25)
• Checkpoint #4 – Unpolarized light hits a polarizer
and then an analyzer. The polarizing direction of
each sheet is indicated by dashed line. Rank
pairs according to fraction of initial intensity which
is passed, greatest first.
EM Waves (26)
• Look at relative orientation of polarization
direction between the 2 sheets.
• What is the intensity if the sheets are…
– Polarized || – all light passes
– Polarized ⊥ to each other – no light passes
– For angles in between – get more light if closer to ||
a,d,b,c