Post on 22-Apr-2018
What We Did Last Time
! Studied reflection and refraction! Derived Snell’s law three times
! Using Huygens’ principle! Using Fermat’s principle
! Light “chooses” the fastest path between two points
! Solving boundary-condition problem! Fresnel coefficients " intensities of reflected/refracted light
! Brewster’s angle " Reflection is polarized
2 1
1 2
sinsin
nn
θθ
=
1
2
ZZ
β ≡
2
1
coscos
a θθ
≡2
VR α βα β
−= +
2
4( )
VT αβα β
=+
211
HR αβαβ
−= +
2
4( 1)
HT αβαβ
=+
IE
IB
TETB
RERB1θ
2θ
Reflectivity and Transmittivity
! R + T = 1 for both cases
2VR α β
α β −
= + 2
4( )
VT αβα β
=+
211
HR αβαβ
−= +
2
4( 1)
HT αβαβ
=+
Vertical polarization Horizontal polarization
Brewster’sangle
For air " glass, β = 1.5
VR
VT
HR
HT
Slow " Fast Transition
! What if n1 > n2?
! R = 1, T = 0 beyond the critical angle
For glass " air, β = 1/1.5
Total internal reflection
Total internal reflection
VR
VT
HR
HT
Total Internal Reflection
! Angle of refraction is determined by the difference between wavelengths! Wavelength along the surface
satisfy
! What if we choose θ2 so thatis shorter than λ1?
! Can’t match the wavefrontsno matter what θ1 we use
! No EM waves above water
1λ
2λ
1θ
2θ
sλ1 2
1 2sin sinsλ λλ
θ θ= =
2 2sinsλ λ θ=1λ
2λ2θ
sλ
Boundary Condition
! Examine boundary conditions! No EM Field above water!
!
!
!
! Only solution is! Something went wrong…
θ θIE
IB RERB
1 2E E=! !
1 1 2 2E Eε ε⊥ ⊥=
1 2H H=! !
1 2B B⊥ ⊥=
( ) cos 0I RE E θ+ =
2 ( )sin 0I RE Eε θ− =
0I RH H− =
Nothing0I RE E= =
Oops!No waves anywhere
1 1,ε µ
2 2,ε µ
Imaginary Waves
! There must be EM field above water! Boundary conditions cannot be satisfied without it! But it cannot be usual EM waves either
! Solution: above water
! Wave equation
! Wavelength along the surface is linked to
( ) ( )0 0
i kx i z t z i kx te e eκ ω κ ω+ − − −= =E E E2 2
2 12 2
nc t
∂∇ =
∂EE
22 2 21
2
nkc
κ ω− =
2s kλ π=
2
2
2sin sins
cn
λ πλθ ω θ
= = 2 sinnkc
ω θ=
2 2 22 1sinn n
cωκ θ= −
Imaginary Waves
! EM “waves” above water satisfies wave equation and boundary conditions! It shrinks as e–κz " Not traveling waves
! “Range” of the imaginary waves is
! Comparable to
! Total internal reflection creates “leakage field” that extends a few wavelengths! You can detect it by having two
boundaries close enough to each other
( )0
z i kx te eκ ω− −=E E
2 2 22 1
1sin
cn nκ ω θ
=−
11
2 cn
πλω
=
Brewster’s Angle
! OK, so we know reflected light is polarized, so what?! Is there anything deeper than solving equations?
! Look at the condition α = β again! Assume
! A little trig gives us
2 1 2 2 1
1 2 1 1 2
sinsin
nn
ε µ ε θβε µ ε θ
≡ = = =2
1
coscos
θαθ
≡
1 2µ µ=
2 1
1 2
cos sincos sin
θ θθ θ
=
1 2sin 2 sin 2θ θ= 1 2θ θ= 1 2 2πθ θ+ =or
This isn’t the case This must be
2VR α β
α β −
= +
Goals for This Week
! At Brewster’s angle,! Reflected light and refracted light are
perpendicular to each other! Is this a coincidence?
! Seems farfetched! If not, what is the significance?
! We need to dig into the origin of refraction! What makes matter dielectric?
! We did this for coaxial cable in Lecture #12! How does it cause light to slow down?
1 2 2πθ θ+ =
1θ 1θ
2θ
Creating EM Waves
! Look at the “full” version of Maxwell’s equations
! We need electric charge to create EM waves
! The charge must be moving! Stationary charge create only static E field! Constant current create only static B field
t∂
∇× = −∂BE
0∇ ⋅ =B0
ρε
∇ ⋅ =E
0 00 tµε µ ∂
∇× =∂
+EB J
SI
Charge density Current density
! Let’s move a point charge +q with constant velocity v! H&L Section 10.2 calculates E(r) and B(r)! Poynting vector
! No power is radiated outward
! We shouldn’t be surprised! Think about Relativity! There is a coordinate system in which +q is at rest
! No energy is radiated by EM waves in that frame! Same should be true in any coordinate system
Moving Charge
vq+
EB
Sr⊥S r
Constant velocity " No radiation
inertial frame
E&M Puzzle
! Charge +q creates E at position r! Exactly which direction does E point?
! Parallel to r, right?
! EM field travels at speed c! Field at r is not determined by
where +q is now, but T = r/c ago! In the meantime, +q moved by! So…
! Which point does the tail of E point?! Where +q is now, or where it was at t = −T ?
vq+
E
r
TvvrvTc
= nowt = -T
E&M Puzzle
! Answer: where +q is now! See your E&M textbook for a real explanation
! Sloppy explanation:! Consider two charges +q and −q
moving in parallel! Let –q fall toward +q
vq+
E
r
vq+
vq−
E
F
E
F
“now”
“past”
Hit
Miss
# Watch this in a frame where v = 0" They will collide
# “Now” hypothesis is correct
How Does It Know?
! But how does E at r know where the charge is now?! It can’t – Special Relativity!
! Sounds odd, but that’s the only Relativistically-Correct way
! E doesn’t necessarily point back to the actual location of the charge if it accelerated (or decelerated)! We are on to something now…
vq+
E
rE(r, t) is determined by what q was doing at t – r/c, in such a way that it points back to where q should be now if v has beenconstant
Accelerating Charge
! Consider this picture:! Charge +q is at rest until t = 0! It accelerates for ∆t by a = dv/dt! Keep constant velocity v after ∆t
! Want to know E at large distance r! Easy for t < 0 and t > ∆t! What happens during the
acceleration?q+
v
t = 0 t = ∆t
r
E
Before/After Acceleration
A B C
: 0A t =:B t t= ∆:C t t T= ∆ +
acceleratebetween
now
Radius c(∆t + T) from A# Outside this circle,
acceleration has not happened
# E points outward from A
Radius cT from B# Inside this circle, the
acceleration has finished# E points outward from C
How can we connect them?
During Acceleration
! E field lines must becontinuous! They start and end
only where there ischarge
! So we just connectthem like "
! Acceleration causes“kinks” in E field! Let’s looks at it
more closely
A B C
Radiated Field
! We are interested in E at large distance r! Radial component is
usual Coulomb field! Geometry gives us the ratio
! ET proportional to accelerationA B C0t = t∆ t T∆ +
c(∆t
+ T)
cT
θ
0ETE
0 204
qErπε
=
0
sinTE vTE c t
θ≈
∆v c"
t T∆ "assuming
20
20
sin4
sin4
Tq vTE
r c tqa
c r
θπε
θπε
=∆
=
c t∆sinvT θ
vat
=∆
rTc
=
Radiated Field
! At large r,! i.e., E becomes transverse! This is the EM radiation!
! We like to write it this way:! E is transverse! Proportional to acceleration a
! Direction opposite to a! Stronger radiation at 90°! Decreases with 1/r
00
24qE
rπε= 2
0
sin4Tqa
rE
cθ
πε=
0TE E#
0 sin ( )4
rT c
qE E a tr
µ θπ
= = −
20 0 1/ cµ ε =
Delayed by r/c due to propagation speed
a
Radiated Power
! Poynting vector is
! Intensity of radiation is not isotropic, but! It goes with square of q
! Sign of charge doesn’t matter! It goes with square of a
! Deceleration radiates just the same! It goes with 1/r2
! As it should for spherical waves
0 sin ( )4
rT c
qE a tr
µ θπ
= − 0 sin ( )4
T rT c
qEB a tc c r
µ θπ
= = −
2 220
2 20
sin ( )16
T rT c
qBS E a tc r
µ θµ π
= = −
2sin θ∝a
EM Waves in Matter
! Matter is made of charged particles! Electrons and protons, in particular
! EM waves in matter make them oscillate! They radiate EM waves! EM waves in matter = original + radiated waves! This is why EM waves propagate differently in matter
IncomingRadiated
Brewster’s Angle
! Why is there reflection in the first place?! Because! Electrons in medium 2 is responsible
! Which way do the electrons oscillate?! Parallel to ET
! They radiate EM waves as
! At Brewster’s angle, reflection isparallel to ET " θ = 0 " No reflection!
1θ 1θ
2θ
IE
IB RERB
TE
TB
1 2Z Z≠
TE2 220
2 2
sin ( )16
rc
qS a tc r
µ θπ
= −Angle of radiation
relative to ET
Larmor Formula
! What’s the total power radiated by a moving charge?! Integrate S over the surface
of a sphere at r
! Accelerating charge loses its energy at a rate proportional to acceleration squared! Think about an atom as electrons circling around the nucleus
! Circular motion " acceleration " radiation " loss of energy " electrons fall into the nucleus
! Paradox stimulated early development of QM
Joseph Larmor(1857–1942)
2 22 2 300 0
00
2 2
sin sin8 6q a q aP d Sr
cd d
cπ π πµφ θ θ θ θ
ππµ
= = =∫ ∫ ∫
2 220
2 2
sin ( )16
rc
qS a tc r
µ θπ
= −
Larmorformula
Summary
! Reviewed reflection and refraction! Total internal reflection is more subtle than it looks
! Imaginary waves extend a few λ beyond the surface! Studied how to create EM waves
! Accelerated charge radiates EM waves! Power proportional to (acceleration)2
! Polarization parallel to the acceleration! Why reflection is polarized at Brewster’s angle
! Next: why light travels slower in matter?
0 sin ( )4
rT c
qE a tr
µ θπ
= −
2 20
6q aP
cµ
π=