maxwell’s equations differential forms electromagnetic waves y x z e ⊥ b, and e ⊥ k, b ⊥ k,...
TRANSCRIPT
Maxwell’s Equations
0
d
qAE
0d AB
L
sE
d At
B d
isBL
0d
At
E d00
Differential forms
D
t
BE
0 B
t
DjH
e
At
Ei emm
d000
DEe
0
qAD
d
HB
m
0
At
DisdH
d
0 ,0 iqif
0
d00 At
E
electromagnetic waves
y
x
z
E⊥B, and E⊥k, B⊥k, k is the direction of the wave.
Polarization of electromagnetic waves
])(sin[ v
ztBB m
])(sin[ v
ztEE m
E and B are in phase
EB
C
The speed of electromagnetic wave00
1
c
Energy Transport and the Poynting Vector
The direction of the propagation of the electromagnetic wave is given by: BE
This wave carries energy. This energy transport is defined by the Poynting vector S as:
BES
0
1
EBS0
1
2
0
1E
cS
2
0
or Bc
S
x
y
EB
C
Find the Poynting vector at point Pparallel-plate capacitor of radius R. a conducting current i
Example:
Q O
ruR
rqiBES ˆ
2
142
00
do iR
rrB
2
2
2
20 R
qE
iR
rB
20
2
0 q
E dt
di Ed
0dt
dq E
BS
Q OE
B S
charge increase field increaseenergy increase
(1)Find the E and B at point P(2) Find the Poynting vector at point P
A conducting current i in a wire with radius R and resistivity .
Example:
BES
0
1
2
1
RijE
iRB 02 PR
iB
2
0
i
BE
EBS0
1
32
2
2 R
iS
SL
2
22
Lip RLS
R
2Ri
Example:
BES
0
1
LE
irB 02
B
rRr
iB
2200
E
EBS0
1
rRLS
2
2
SFind the Poynting vector R
On wire E=0, S=0
On resistance
On wire On resistance On
On Rrr
iB
'2'200
'L
E
E
BS
EBS0
1
BES
0
1
''2
2
RLrS
RrLSp
2
2
RSLrp
2
''2
Example:
r
iB
2
0
SdAP
Er
Bz
E
B1r2r
1
210
10 ln2
1 r
rrEdr
r
rEEdr
r
r
1
2ln rrr
E
1
2ln2
12
2
0 rrRr
EBS
?,S
2
1
d2r
r
rrS 2
1 1
2d2
ln2 2
2r
r rr rr
Rr
R
2
R 0122 lErrlE
r
ErE 01
rR
2 0
i
1
2ln10
rrr
E
The Doppler effect for light
uv
vff
0
v
uvff
0
1. Sound wave, observer fixed, source moving away
2. Sound wave, source fixed, observer moving away
3. Light wave, source and observer separating
?f
* o u
22 2
2 2 2
/'
1 /
t ux ct
u c
1
1
0
0
x
t
S
1 /'
1 /
u cf f
u c
Source or observer leaving
Source or observer approaching1 /
'1 /
u cf f
u c
2
2
x cT
t T
2 2
/
1 /
T uT c
u c
2 2
1 /
1 /
u cT
u c
2
1 /'
1 /
u ct T
u c
'T
1
'f 1
f
Chapter 39 Light Waves
Radio waves
Microwaves Infrared radiation
visible region
Ultraviolet
X rays Gamma rays
The sensitivity of the human eye as a function of wavelength
The Wavelength vs Temperature
Thermal radiation—the sun light
Luminescence—cool light source
Glass
ITO
Matel Organic
The speed of light
c
L2L
c2
s/m103 8c
in vacuum
1
v00
1
em
em
c
matterin ])(sin[ v
rtEE m
Propagation of light in matter
The speed of light in a material depends on the the frequency or wavelength.
The phase change
1
00
Reflection and refraction of light waves
Reflection:11
Refraction:
2211 sinsin nn
Index of refraction: nv
c
em
cv
n
c
emn eemn )(n
rainbow In second rainbow pattern is reversed
Reflection and refraction of electromagnetic waves
Reflection and refraction of electromagnetic waves
Total Internal Reflection
021 90sinsin nn c
1
21sinnn
c
In case of ni larger than nR
c
Examples: refraction at water/air interface
Diver sees all of horizonrefracted into a 97°cone.
Diver’s illusion
Optical fiber:1
21c sin
nn
At each contact the glass air interface, if the light hits at greater than the critical angle, it undergoes total internal reflection and stays in the fiber.
Total Internal Reflection only works if noutside < ninside
ninside
noutside
Optical fiber
n2
n1
d
d
d d
n2
n1
Apparent depth:
Apparent Depth
50
actual fish
apparent fish
B
P
a
d’d
θ2
θ1
n2
n1
21 ' tgddtg
2
1'
tg
tg
d
d
2
1
sin
sin
1
2
n
n
θ2
θ1
ACT: RefractionAs we pour more water into bucket, what will
happen to the number of people who can see the ball?
ACT: RefractionAs we pour more water into bucket, what will
happen to the number of people who can see the ball?
Huygens’ Principle:
All points on a wavefront can be considered as point sources for the production of spherical secondary wavelets. After a time t the new position of a wavefront is the surface tangent to these secondary wavelets.
Huygens’ Principle:
All points on a wavefront can be considered as point sources for the production of spherical secondary wavelets. After a time t the new position of a wavefront is the surface tangent to these secondary wavelets.
1sinACtcBC
21 sinsin
2sinACtcAD
21
21
Deriving the law of refraction
1
2
n1
n2
n1 n2 v1 v2
1 cannot equal 2 !
1212 , vvnn
11 sinACtvBC
C
22 sinACtvAD
1
2
2
1
2
1
sin
sin
n
n
v
v
2211 sinsin nn
1
2
Fermat’s Principle:
A
B
2222 )( xdbxaL
0d
d
x
t 0
d
)/d(
x
cL
0)(
2222
xdb
xd
xa
x
sinsin 11 11
A light ray traveling from one fixes point to another fixed point follows a path such that, compared with nearby paths, the time requires is either a minimum or a maximum or remains unchange (that is, stationary).
0d
d
x
L
c
Lt
AB
Pd
x d-x
θ1’1a b
A
B
P
a
b
d
xθ1
θ2
n1
n2
2211 sinsin nn
0d
)(d
/
1
d
d
/
1 22
2
22
1
x
xdb
ncx
xa
nc
0d
d
x
t2
22
1
22 )(
v
xdb
v
xat
0/
1
/
122
222
1
x)(db
d-x
ncxa
x
nc
2
22
1
22
/
)(
/ nc
xdb
nc
xat
Snell’s Law
1 1
1
2
L n1
n2
1
2
2 2 2
The two triangles above each have hypotenuse L
1
1
2
2
sinsin
L2
1
2
1
sin
sin
2211 sinsin nn
fv
fv
/
/
2
1
2
1
1
2
n
n
2
1
v
v
1
2
n
n
Why is the sky blue?• Light from Sun scatters off of air particles––Rayleigh scattering is wavelength-dependent, More scatter for shorter wavelengths (blue end of the visible spectrum)Less scatter for longer wavelengths (red end of the visible spectrum)
Sun lightred and orange
longer wavelengths scatter lessShorter wavelengths scatter more
blueThis is also why sunsets are red
Example1n
2n
3n321 )( nnnA
312 )( nnnB
231 )( nnnC
213 )( nnnD
1
2
2
1
sin
sin
n
n
1
2
1221 , nn
2
3
2323 , nn
1313 , nn 3
2
2
3
sin
sin
n
n
3
1
1
3
sin
sin
n
n
Example
n
1
2
11 ?2
n
sin
sin 1
090
n )90sin(
sin0
2
221
2 sinsin n
220222
2 cos)90(sinsin nn
22
21
2 sinsin n
'22
Example
cosc
Lt
L
Example
cosc
Lt
L
n1sin
sin
n
?'t
1 1cos'
c
nLt t
n
1cos
cos
)sin
arcsin(1 n
t
n
nt
)]sin
(cos[arcsin
cos'
Exercises:P906-908 19, 36, 44ProblemsP910-911 7, 12