Example
ÎAn
observer stands 1.8 m from an isotropic point
source of light with power Ps = 250 W. Calculate
the rms values of the electric and magnetic fields
of the light at the position of the observer.
Strategy: source power -> intensity at distance
r -> E and B fields
Intensity I ?
Ps
average power at r total power
I≡
=
=
2
area at r
4πr
4πr 2
Electric
field Erms ?
I=
1 2
E rms
cµ 0
PHY2049: Chapter 33
1
(continued)
E rms =
Brms
(cµ0
4π )Ps
r
(3 × 10 (4π × 10 ) 4π )250 =
−7
8
=
1. 8
30(250)
= 48 V/m
1. 8
Erms
48
−7
=
=
=
1
.
6
×
10
T
8
c
3 × 10
B field is very weak.
PHY2049: Chapter 33
2
Energy of EM Wave
ÎE/B=c,
yet E and B are
depicted with the same amplitude. Why?
Compare
energy densities uE and uB of the E
and B fields.
1 2
1
2
uB =
B
(J/m3)
u E = ε0 E
2 µ0
2
uE=uB everywhere along the EM wave! Energy
is shared equally by the E and B fields in the
EM wave.
u emw = 2u E = 2u B
PHY2049: Chapter 33
3
(continued)
Rewrite
in terms of E and B:
u emw = 2u E = 2u B
u emw
1 2
1
=
B =
EB (J/m3)
µ0
µ0 c
Energy
transported per unit time through a
plane drawn perpendicular to the propagation
direction of the EM wave (a good definition of
intensity):
1
< u emw > c =<
EB >
(J/m3)(m/s)=(W/m2)
µ0
This is why the intensity has been defined to
be <S> (Eq. 33-23) and the Poynting vector
has the meaning claimed by Eq. 33-20.
PHY2049: Chapter 33
4
Momentum of EM Wave & Radiation
Pressure
ÎEM
waves (e.g. light) carry energy and travel at
velocity c. It is natural to expect that they also
carry momentum, although they have no mass.
Can be shown that they do, but the derivation
is above the level of intro physics.
ÎMomentum
received by an object
Can show (derivation is tough)
∆U
∆p =
c
for total absorption
For
total reflection back along path vs total
absorption (analogy to completely elastic
head-on collision)
∆p ref = 2 ∆p abs Eq. 33-29 is misleading
PHY2049: Chapter 33
5
(continued)
ÎMomentum
received -> Force received ->
pressure
impulse=momentum
∆p
F=
received/given
From mechanics
∆t
∆U
If EM wave is totally absorbed ∆p =
c
Intensity
I≡
Therefore
F=
average power ∆U ∆t
=
area
A
∆p ∆U c IA
=
=
∆t
∆t
c
for total absorption
I
F
pr =
pr ≡
(p for pressure, not
A
c
momentum)
For total reflection back along path
2I
pr =
c
∆pref = 2 ∆pabs
PHY2049: Chapter 33
6
Example
ÎThe
intensity of the solar radiation near the
equator is 1370 W/m2 at noon. What is the
radiation pressure on a completely reflecting
surface that directly faces the sun?
pr = 2I/c = 2(1370)/3x108 = 9.1 x 10-6
Unit?
W/m 2 Ws/m
N
m/s
=
m
2
=
m
2
= Pa
9.1
x 10-6 Pa (Only ~10-10 of atmospheric
pressure)
It is feeble but can be used:
Artist’s view of the Cosmos 1
satellite with the solar sail
unfolded. But the launching
rocket failed.
PHY2049: Chapter 33
7
Radiation Pressure
ÎMore
examples
Radiation pressure, together with gas
pressure, prevents stars from collapsing due
to gravity.
Comet tails—shapes and directions.
PHY2049: Chapter 33
8
Polarization
ÎPolarized
EM waves
E field oscillates along a fixed
direction, as does the B field
Radio waves, EM waves for TV
broadcasting
Laser light
ÎUnpolarized
EM waves
Light from ‘ordinary’ sources
(the Sun, light bulbs, etc.)
ÎPeople--most
mammals for that matter-cannot detect light polarization, but many
insects and cephalopods can, as can some
fish and amphibians.
PHY2049: Chapter 33
9
Polarizing filter
ÎIdeally,
absorbs only E field oscillating along one
direction, unaffecting E field oscillating along the
perpendicular direction, “polarizing” direction.
Initially
polarized light with an E field amplitude E0. Let
θbe the angle between the direction of the E field and
the polarizing direction y of the filter. Going through the
filter, only the component parallel to the polarizing
direction survives.
E = E 0 cos θ
Since
light intensity is proportional to E2 by definition,
after the filter
I = I 0 cos 2 θ
(polarized along y)
If the light is initially unpolarized, after the filter
I = I 0 < cos 2 θ >= I 0 2
(polarized along y)
PHY2049: Chapter 33
10
0
