24.5 The Doppler Effect and Electromagnetic Waves
Electromagnetic (EM) waves also can exhibit a Doppler effect:
1. Increase in observed frequency for source and observer
approaching one another
2. Decrease in observed frequency for source and observer receding
from one another
But Doppler Shift of EM waves differs from that of sound for two reasons:
a) Sound waves travels in a medium, whereas electromagnetic waves can
travel in a vacuum (and in some transparent medium)
b) For sound, it is the motion relative to the medium that is important. For
electromagnetic waves, only the relative motion of the source and
observer is important.
 vo
1±
v
fo = f s 
 1  vs

v

For sound
Not for light






Electromagnetic waves
 vrel 
f o = f s 1 ±

c 

Observed
frequency
Source
frequency
if vrel << c
Very much less than
1
24.5 The Doppler Effect and Electromagnetic Waves
Example Radar Guns and Speed Traps.
The radar gun of a police car emits an
electromagnetic wave with a frequency of 8.0x109Hz. The approach is essentially head on.
The wave from the gun reflects from the speeding car and returns to the police car, where onboard equipment measures its frequency to be greater than the emitted wave by 2100 Hz.
Find the speed of the car with respect to the highway. The police car is parked.
v
2
24.5 The Doppler Effect and Electromagnetic Waves
Example Radar Guns and Speed Traps.
The radar gun of a police car emits an
electromagnetic wave with a frequency of 8.0x109Hz. The approach is essentially head on.
The wave from the gun reflects from the speeding car and returns to the police car, where onboard equipment measures its frequency to be greater than the emitted wave by 2100 Hz.
Find the speed of the car with respect to the highway. The police car is parked.
v
3
24.5 The Doppler Effect and Electromagnetic Waves
Example Radar Guns and Speed Traps
The radar gun of a police car emits an electromagnetic wave with a frequency of 8.0x109Hz.
The approach is essentially head on. The wave from the gun reflects from the speeding car
and returns to the police car, where on-board equipment measures its frequency to be greater
than the emitted wave by 2100 Hz. Find the speed of the car with respect to the highway.
The police car is parked.
v
frequency “observed”
by speeding car
 v 
f o = f s 1 + rel 
c 

∆f = f o′ − f s

= f o 1 +

v
 − fs
c
v 
v
 
=  f s 1 + 1 +  − f s
c 
c
 

v v2  
= f s 1 + 2 + 2  − 1
c c  

v≈c
 v 
f o′ = f o 1 + rel 
c 

 v v2 
= f s  2 + 2 
 c c 
frequency observed
by police car

v v2 
= f s 1 + 2 + 2  − f s
c c 

2v
v
<< 1 → ∆f ≈ f s
c
c
2100 Hz
∆f
= (3.0 × 108 m s )
= 39 m s
2 fs
2(8.0 × 10 9 Hz)
→v≈c
∆f
2 fs
4
19.5 Capacitors and Dielectrics
Electrical Energy stored in a capacitor
can be thought of in two different ways:
(1) Electrical potential energy between the excess +q and –q charges spread evenly
separated by distance d, RELATIVE to the situation where the two plates are
neutral.
(2) Potential energy stored in the electric field (which exists only between the
plates, and is zero outside!!!
Energy = 12 CV 2
V = Ed
Volume = Ad
C=
κε 0 A
d
 κε o A 
2
2
Energy = 
(Ed ) = (12 κε o E )( Ad ) = u( volume)
 d 
1
2
u = Energy density =
Energy
Volume
= 12 κε o E 2
= 12 ε o E 2 in vacuum
Electric field carries energy of “density”
Formulation
done for a
parallel plate
capacitor but
generally
applicable
5
22.8 Self Inductance
Magnetic Energy: (1) Inductance
Take an IDEAL solenoid of N turns, crosssectional area A, length h and carrying current I.
We now want to increase the current by ∆I over
time interval ∆t.
N turns
Crosssectional
area A
This will cause the magnetic field generated by
the solenoid to change, and then induce a backEMF that tends to prevent the increase in current
(Lenz’s Law!).
h
Increasing
current
E = −N
∆Φ
∆B
= − NA
∆t
∆t
E = − NA
−
+
E = −N
∆Φ
∆t
“−” sign indicates “back”-emf
B = µ 0 nI =
µ0 NI
h
1  µ 0 NI 
 µ 0 N  ∆I
∆
=
−
NA



∆t  h 
 h  ∆t
µ0 N 2 A
∆I
E = −L
, L=
∆t
h
SI Unit of L
1V ⋅ s A = 1 H (Henry)
Where L is defined to be the (self-)
inductance of the solenoid
6
22.8 Self Inductance
2
µ
N
A
∆
I
0
(2) The energy stored in an inductor E = − L
, L=
∆t
h
The work required to increase the current
from I by a small ∆I is given by:
∆W = −E∆q = −E ( I∆t )
 ∆I 
=  L ( I∆t ) = (LI )∆I
 ∆t 
∆W
= LI
∆I
Total magnetic potential energy
(MPE) stored is then equal to the
total work required to ramp up the
current from 0 to I, which is given by
the triangular area under the line
∆W ∆I = LI
∆W
∆I
∆W = (LI )∆I
∆I
I
∆W
= LI
∆I
∆W
∆I
1 2 1 µ0 N 2 A 2
MPE = LI =
I
2
2 h
2
 1 2
1  µ0 N 2 I 2 



=
=
Ah
B ( Volume)
2


2µ0  h

 2µ0

MPE =
1
1
∆2I( LI )( I ) = 2 LI
2
7
I
24.4 The Energy Carried by Electromagnetic Waves
From Ch. 20
electric energy
density
Total EPE 1
= εoE 2
uE =
Volume
2
u( x, t ) =
The total energy density
carried by an electromagnetic wave (which is
itself a function of location
and time)
u ( x, t ) =
Averaging over space and time:
1
1
2
2
u = ε o E0 +
B0
4
4 µo
EPE = MPE
From Ch. 22
magnetic energy
density
Total MPE
1 2
=
B
Volume
2 µo
Total ( EPE + MPE ) 1
1
2
[B( x, t )]2
= ε o [E ( x, t )] +
Volume
2
2 µo
1
2πx 
1
2πx 

2
2
+

sin
2
B
ft
ε o Eo 2 sin 2  2πft 
π



2
λ  2 µo o
λ 


2πx  1

sin  2πft 
=
λ

 2
2
uB =
→u =
1 1 2
1 1 2
ε o  E0  +
 B0 
2 2
 2 µo  2

E0
2
2
= ε o µo E0 → B0 = ε o µo E0
But B0 =
c
1
2
u = ε o E0
2
and
u=
→
B0
2
µo
1
2 µo
= ε o E0
B0
2
2
8
24.4 The Energy Carried by Electromagnetic Waves
A related quantity: Intensity of EM waves (How bright is the light hitting you?)
Analogous to sound intensity (how loud is the sound wave arriving at you)?
Intensity: S (unit: W/m2)
S=
=
y
x
z
area
A
Power Total EM energy
=
(Area) ∆t
Area
(EM energy density)(Volume)
(Area) ∆t
We compute the intensity by finding the
energy contained the gray box (of length
c∆t, and area A) passing through the
yellow frame (area A) in time ∆t.
∆x = c∆t
S=
u ⋅ (c∆t ⋅ A)
= c⋅u
A∆t
→ S = c⋅u
Average intensity of
an electromagnetic
wave:
1
2
S = c ⋅ ε o E0
2
and
S = c⋅
1
2 µo
B0
2
9
Example: Sunlight enters the top of the earth's atmosphere with an electric field whose RMS
value is Erms = 720 N/C. Find
(a) the average intensity of this electromagnetic wave and
(b) the rms value of the sunlight's magnetic field At the Earth, orbiting the Sun at 1.00 a.u.
(c) Repeat (a) and (b) for Venus, which orbits the Sun at a radius of 0.723 a.u.
*** a.u. = astronomical unit.
10
Example: Sunlight enters the top of the earth's atmosphere with an electric field whose RMS
value is Erms = 720 N/C. Find
(a) the average intensity of this electromagnetic wave and
(b) the rms value of the sunlight's magnetic field At the Earth, orbiting the Sun at 1.00 a.u.
(c) Repeat (a) and (b) for Venus, which orbits the Sun at a radius of 0.723 a.u.
*** a.u. = astronomical unit.
C 2 
N
1

2
2
8 m 
−12
3 W

720
1
.
38
10
(a ) S = cε o E0 = cε o E rms =  3.00 × 10
=
×
 8.85 × 10


s 
N ⋅ m 2 
C
m2
2

E
E
720 N/C
( b) B0 = 0 → Brms = rms =
= 2.40 × 10 −6 T
8
c
3.00 × 10 m/s
c
2
−2
PSUN
PSUN
SV rE2  rV 
rV 0.723 a.u.


=
→
=
=
(c) SV =
and
,
=
= 0.723
S
E
4π rV2
4π rE2
1.00 a.u.
S E rV2  rE 
rE
r
SV =  V
 rE
S = c⋅
−2

SE
1.38 × 108 W/m
8
2
 S E =
=
=
2
.
63
×
10
W/m
(0.723) 2
(rV / rE )2

1
2 µo
B0 =
→ Brms =
c
2
µo
µo S
c
Brms
=
2
( 4π × 10 −7 T ⋅ m/A)(2.63 × 10 3 W/m 2 )
= 3.32 × 10 -6 T
8
3.00 × 10 m/s
11
24.6 Polarization
POLARIZED ELECTROMAGNETIC WAVES
In (linearly) polarized light, the electric field
oscillates along a single transverse direction.
In unpolarized light, the electric field
oscillates along random transverse
direction at a given moment in time.
12
24.6 Polarization
Polarized light may be produced from
unpolarized light with the aid of
polarizing material.
Average Intensity
of light:
S=
1
cε 0 E02
2
E0 ' = E0 cos θ
Absorption and
retransmission by
antennae in polarizer Primed = after polarizer
Unprimed = before polarizer
1
1
2
2
cε 0 (E0 ' ) = cε 0 (E0 cosθ )
2
2
1
2
S ' =  cε 0 E 0  cos 2 θ = S cos 2 θ
2

S'=
E0’=E0cosθ
E0
θ
Malus’ Law
For unpolarized incident light
Random Input polarization:
cos 2θ → cos 2θ =
(
)
S ' = S cos 2θ =
1
S
2
1
2
13
24.6 Polarization
MALUS’ LAW (general case)
E0
θ = angle between
polarization before and
after analyzer
E0cosθ
S = S o cos 2 θ
intensity after
analyzer
The textbook’s
notational convention
intensity before
analyzer
14
24.6 Polarization
Example Using Polarizers and Analyzers
What value of θ should be used so the average intensity of the polarized light
reaching the photocell is one-tenth the average intensity of the unpolarized light?
1
10
S o = (12 S o )cos 2 θ
1
5
= cos 2 θ
cos θ =
1
5
θ = 63.4
15
24.6 Polarization
THE OCCURANCE OF POLARIZED LIGHT IN NATURE
16
Chapter 25
The Reflection of
Light: Mirrors
We will start to treat light in terms of “rays” -technical for the poetic “beam” of light.
This works as long as the “width” of our rays are
significantly larger than the wavelength of the light
17
25.1 Wave Fronts and Rays
A hemispherical view of a sound wave emitted
by a small pulsating sphere (i.e. a point-like
source).
The rays are perpendicular to the wave
fronts (locations where the wave is at the
same phase angle, e.g. at the maxima or
compressions.
At large distances from the source, the
wave fronts become less and less curved.
They become plane-waves described by:


ψ ( x, t ) = ψ 0 sin 2πft −
2πx 

λ 
18
25.2 The Reflection of Light
LAW OF REFLECTION
2 types of reflecting surfaces:
(a) smooth surface
The incident ray, the reflected ray, and
the normal to the surface all lie in the
same plane, and the angle of incidence
equals the angle of reflection.
Specular reflection: the reflected
rays are parallel to each other.
(b) rough surface
Diffuse
reflection
Plane mirror
Law of reflection still applies locally, where the normal is the
direction perpendicular to the tangent plane
19
25.4 Spherical Mirrors
Mirrors cut from segments of a
sphere (Spherical mirrors) are
used in many optical instruments.
If the inside surface of the
spherical mirror is silvered or
polished, it is a concave mirror.
B
If the outside surface is silvered or
polished, it is a convex mirror.
B
The principal axis of the
mirror is a straight line drawn
through the center (C) and the
midpoint of the mirror (B).
Just as it does for a plane mirror,
the Law of reflection still applies
locally, where the normal is the
direction perpendicular to the
tangent plane
20
Applications of
spherical mirrors
8.2
m diameter
(largest single
Large
radius-of-curvature
(R) piece)
reflector (concave
concave
mirrors tomirror***)
enlarge of the
SUBARU
Telescope
nearby
object.
National Astronomical Observatory of
e.g. compact
Japan
(Mauna mirror
Kea Hawaii)
Wide-angle rear-view convex
mirror found on many vehicles.
*** actually the SUBARU main
mirror is closer to an hyperboloid
21
25.4 Spherical Mirrors
Light rays near and parallel
to the principal axis are
reflected from the concave
mirror and converge at the
focal point.
In Class Demo
The focal length, f , is the
distance between the focal
point and the middle of the
mirror.
The focal point of a spherical
concave mirror is
(approximately) halfway
between the center of
curvature of the mirror (C) and
the middle of the mirror at B.
f = 12 R
22
25.4 Spherical Mirrors
Rays that lie close to the principal
axis are called paraxial rays.
Rays that are far from the principal
axis do not converge to a single point.
The effect is called spherical
aberration.
When paraxial light rays that are parallel to the
principal axis strike a convex mirror, the rays appear
to originate from the focal point behind the mirror.
Since the light rays do not really
converge to this point, it is a
virtual focus
f = − 12 R
In Class Demo
The SIGN convention is to assign a
negative value to virtual quantities
23