electromagnetic (em) waves also can exhibit a doppler effectwoolf/2020_jui/mar29.pdf24.5 the doppler...
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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
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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.
Very much less than
cvc
vff so <<
±= rel
rel if 1
Electromagnetic waves
±=
vvvv
ffs
o
so
1
1
For sound
Not for light
Observed frequency
Source frequency
v
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.
2
v
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.
3
v
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.
4
1 rel
+=
cvff so
frequency “observed” by speeding car
1 rel
+=′
cvff oo frequency observed
by police car so fff −′=∆
( ) sm39 Hz)100.8(2
Hz2100sm100.32 9
8 =×
×=∆
≈sf
fcv
so fcvf −
+= 1 ss f
cv
cvf −
+
+= 11
−
++= 121 2
2
cv
cvf s
+= 2
2
2cv
cvf s c
vffcv
s21 ≈∆→<<
sffcv
2∆
≈→
ss fcv
cvf −
++= 2
2
21
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!!!
221Energy CV=
vacuumin
density Energy 2
21
221
VolumeEnergy
E
Eu
o
o
ε
κε
=
===
( ) ( ) )()(Energy 2212
21 volumeuAdEEd
dA
oo ==
= κεκε
Ad= Volume
5 Electric field carries energy of “density”
Formulation done for a parallel plate capacitor but generally applicable
dAC 0κε
=EdV =
22.8 Self Inductance
Magnetic Energy: (1) Inductance Take an IDEAL solenoid of N turns, cross-sectional area A, length h and carrying current I.
N turns
h
Cross-sectional
area A
We now want to increase the current by ∆I over time interval ∆t.
This will cause the magnetic field generated by the solenoid to change, and then induce a back- EMF that tends to prevent the increase in current (Lenz’s Law!).
+ −
Increasing current
tBNA
tN
∆∆
−=∆∆Φ
−=E
tN
∆∆Φ
−=E
“−” sign indicates “back”-emf
hNInIB 0
0µµ ==
tI
hNNA
hNI
tNA
∆∆
−=
∆
∆−= 001 µµ
E
hANL
tIL
20 , µ
=∆∆
−=E
Where L is defined to be the (self-) inductance of the solenoid
(Henry) H 1As1V =⋅SI Unit of L
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22.8 Self Inductance
(2) The energy stored in an inductor h
ANLtIL
20 , µ
=∆∆
−=EThe work required to increase the current from I by a small ∆I is given by:
( ) ILItItIL
tIqW
∆=∆
∆∆
=
∆−=∆−=∆
)(
)(EE
LII
W=
∆∆ I
W∆
∆
( ) ILIW ∆=∆I∆
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 I
W∆
∆
I∆
I
LII
W=
∆∆
LIIW =∆∆
2
21))((
21
LIILI
MPE
=
=
)Volume(2
12
121
21
2
02
2220
0
22
02
=
=
==
BAhh
IN
Ih
ANLIMPE
µµ
µ
µ
7
24.4 The Energy Carried by Electromagnetic Waves
The total energy density carried by an electro-magnetic wave (which is itself a function of location and time)
[ ] [ ]22 ),(2
1),(21
Volume)( Total),( txBtxEMPEEPEtxu
oo µ
ε +=+
=
8
2
21
Volume Total EEPEu oE ε==
From Ch. 20 electric energy density
From Ch. 22 magnetic energy density
2
21
Volume Total BMPEu
oB µ
==
+
=
λππ
µλππε xftBxftEtxu o
ooo
22sin2
122sin21),( 2222
Averaging over space and time: 2122sin2 =
λππ xft
+
=→ 2
02
0 21
21
21
21 BEu
oo µ
ε
20
20 4
141 BEu
oo µ
ε += But 00
0 Ec
EB ooµε== 20
20 EB ooµε=→ 2
0
20 EB
oo
εµ
=→
202
1 Eu oε=2
021 Bu
oµ=and
EPE = MPE
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)?
9
y
x
z area A
tS
∆==
(Area)energy EM Total
AreaPower
t∆=
(Area)olume)density)(Venergy (EM
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.
tcx ∆=∆ ( ) uctA
AtcuS ⋅=∆
⋅∆⋅=
Intensity: S (unit: W/m2)
ucS ⋅=→
202
1 EcS oε⋅=2
021 BcS
oµ⋅=and
Average intensity of an electromagnetic wave:
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.
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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.
23
2
2
212822
0 mW 10 38.1
CN 720
mNC 1085.8
sm 1000.3
21 )a( ×=
⋅
×
×=== −
rmsoo EcEcS εε
T 1040.2 m/s 1000.3
N/C 720 )b( 68
00
−×=×
==→=c
EBc
EB rmsrms
723.0a.u. 00.1a.u. 723.0 ,
4 and
4 )c(
2
2
2
2
2
==
==→==
−
E
V
E
V
V
E
E
V
E
SUNE
V
SUNV r
rrr
rr
SS
rPS
rPS
ππ
( )28
2
8
2
2
W/m 1063.2 (0.723)
W/m 1038.1/
×=×
==
=
−
EV
EE
E
VV rr
SSrrS
2202
1rms
oo
BcBcSµµ
=⋅=
T 1032.3 m/s 1000.3
)W/m 10m/A)(2.63T 104( 6-8
237
×=×
×⋅×==→
−πµcSB o
rms
24.6 Polarization
In (linearly) polarized light, the electric field oscillates along a single transverse direction.
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POLARIZED ELECTROMAGNETIC WAVES
In unpolarized light, the electric field oscillates along random transverse direction at a given moment in time.
θ E0
E0’=E0cosθ
24.6 Polarization
Polarized light may be produced from unpolarized light with the aid of polarizing material.
13
2002
1 EcS ε=
( ) ( )200
200 cos
21'
21' θεε EcEcS ==
θcos' 00 EE =
Average Intensity of light:
Absorption and retransmission by antennae in polarizer
θθε 22200 coscos
21' SEcS =
=
Malus’ Law For unpolarized incident light Random Input polarization:
21coscos 22 =→ θθ
( ) SSS21cos' 2 == θ
Primed = after polarizer Unprimed = before polarizer
24.6 Polarization
MALUS’ LAW (general case)
θ2cosoSS =
intensity before analyzer
intensity after analyzer 14 The textbook’s
notational convention
E0
E0cosθ
θ = angle between polarization before and after analyzer
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?
15
( ) θ221
101 cosoo SS =
θ251 cos=
51cos =θ
4.63=θ
24.6 Polarization
THE OCCURANCE OF POLARIZED LIGHT IN NATURE
16
Chapter 25
The Reflection of Light: Mirrors
17
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
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).
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At large distances from the source, the wave fronts become less and less curved.
−=
λππψψ xfttx 22sin),( 0
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.
They become plane-waves described by:
25.2 The Reflection of Light
LAW OF REFLECTION
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.
19
2 types of reflecting surfaces: (a) smooth surface
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
B
B
25.4 Spherical Mirrors
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
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.
If the outside surface is silvered or polished, it is a convex mirror.
The principal axis of the mirror is a straight line drawn through the center (C) and the midpoint of the mirror (B).
21
Applications of spherical mirrors
Wide-angle rear-view convex mirror found on many vehicles.
8.2 m diameter (largest single piece) reflector (concave mirror***) of the SUBARU Telescope National Astronomical Observatory of Japan (Mauna Kea Hawaii)
*** actually the SUBARU main mirror is closer to an hyperboloid
Large radius-of-curvature (R) concave mirrors to enlarge nearby object.
e.g. compact mirror
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.
22
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.
In Class Demo The focal length, f , is the distance between the focal point and the middle of the mirror.
Rf 21=
25.4 Spherical Mirrors
The effect is called spherical aberration.
23
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.
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.
Rf 21−=
Since the light rays do not really converge to this point, it is a virtual focus
The SIGN convention is to assign a negative value to virtual quantities
In Class Demo