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Wave Phenomena Physics 15c
Lecture 18 Interference
What We Did Last Time Studied microscopic origin of refraction in matter Started from EM radiation due to accelerated charges Uniform distribution of such charges make EM waves appear to slow
down Three levels of describing EM waves in matter
Point charges in vacuum Current density in vacuum Permittivity (or dielectric constant) of matter
Analyzed EM waves in plasma
Plasma reflective for ω < ωp
nplasma = 1−ωp
2 ω 2
ωp =
q2n0
ε0m
Insulators Unlike in plasma or in metal, electrons in insulators are bound to the molecules Binding force is similar to a spring Rest of the molecule M is much heavier Ignore the movement
Equation of motion Forced oscillation – We know the solution
Current density is
M qE ksx
m
mx = qE0e− iω t − ksx
x = x0e− iω t
J = qn0v =
−iωq2n0
m(ω02 −ω 2)
E0e− iω t
−mω 2x0 = qE0 − ksx0
x0 =
qE0
ks − mω 2 =qE0
m(ω02 −ω 2)
ω0 =
ks
m
Dispersion Relation Wave Equation is
Dispersion relation is
It’s dispersive again
Phase velocity:
Index of refraction:
∇2E =
1c2
∂2E∂t 2 + µ0
∂J∂t
−k 2E = −ω 2
c2 E −µ0q
2n0
mω 2
ω02 −ω 2 E
k 2 =
ω 2
c2 1+q2n0
ε0m(ω02 −ω 2)
⎛
⎝⎜
⎞
⎠⎟ =
ω 2
c2 1+ ρω0
2
ω02 −ω 2
⎛
⎝⎜
⎞
⎠⎟
ρ ≡
q2n0
ε0ks
cp =ωk=
c
1+ ρω02 (ω0
2 −ω 2)
n = 1+ ρ
ω02
ω02 −ω 2
Example: Air Air is a mixture of N2, O2, H2O, Ar, etc Many resonances exist in UV and shorter wavelengths Things are simpler in visible light, where
We don’t really know n0 and ks, but we do know n = 1.0003 at STP ρ = 0.0006 ρ is proportional to density n0
Using the ideal gas formula
cp =c
1+ ρω02 (ω0
2 −ω 2)≈
c1+ ρ n ≈ 1+ ρ ≈1+ 1
2 ρ
ρ ≡
q2n0
ε0ks
PV = nmolRT
n = 1+ 0.0003 273K
TP
1atmIndex of refraction of air for low-freq. EM waves
ω ω0
Mirage In a hot day, air temperature is higher near the ground
Light travels faster near the ground
Refraction makes light bend upward You may see the ground
reflect light as if there is a patch of water
T
hot
cool slow
fast n = 1+ 0.0003 273K
TP
1atm
Goals For Today Interference When waves overlap with each other, they strengthen or cancel
each other
Thin-film interference Reflectivity of light through a thin transparent film
Newton’s rings Anti-reflective coatings
Two-body interference Young’s experiment – Wave nature of light
Thin-Film Interference We studied reflection at boundary between insulators Continuity of E and H gave us
Using n instead of Z assuming µ1 = µ2
What if we had 2 (or more) boundaries in series? For a pane of glass window, I said T → T2
Things can get more interesting than that…
ε1,µ1 ε2,µ2
EI HI
ER HR
ET HT ET =
2n1
n1 + n2
EI
ER =
n1 − n2
n1 + n2
EI
R =
n1 − n2
n1 + n2
⎛
⎝⎜⎞
⎠⎟
2
T =
4n1n2
(n1 + n2)2
Two Boundaries Consider transitions through two boundaries At z = 0,
At z = d
Note the exponentials
Solve for ER, ET, EF, EB
EI + ER = EF + EB
EI − ER
Z1
=EF − EB
Z2
H
E
EFeik2d + EBe− ik2d = ET
EFeik2d − EBe− ik2d
Z2
=ET
Z1
ε2,µ2 ε1,µ1 ε1,µ1
z 0 d
EI HI
ER HR
EF HF
EB HB
ET HT
E(z,t) =
EIei (k1z−ω t ) + ERei (−k1z−ω t ) for z ≤ 0
EFei (k2z−ω t ) + EBei (−k2z−ω t ) for 0 < z ≤ d
ETei (k1(z−d )−ω t ) for z > d
⎧
⎨⎪⎪
⎩⎪⎪
Two Boundaries Solutions are
OK, so what did we learn?
ER =(Z1
2 − Z22)(eik2d − e− ik2d )
(Z1 + Z2)2e− ik2d − (Z1 − Z2)2eik2d EI
ET =4Z1Z2e
− ik2d
(Z1 + Z2)2e− ik2d − (Z1 − Z2)2eik2d EI
EF =2Z2(Z1 + Z2)e− ik2d
(Z1 + Z2)2e− ik2d − (Z1 − Z2)2eik2d EI
EB =2Z2(Z1 − Z2)eik2d
(Z1 + Z2)2e− ik2d − (Z1 − Z2)2eik2d EI
Reflectivity is (this)2
Transmittivity is (this)2
ε2,µ2 ε1,µ1 ε1,µ1
z 0 d
EI HI
ER HR
EF HF
EB HB
ET HT
Reflectivity Look at the reflectivity
Reflectivity oscillates as a function of k2d
Reflection vanishes at k2d = mπ (m = 1,2,3,…)
What’s happening?
R =
ER
EI
2
=2(Z1
2 − Z22)2 sin2(k2d)
8Z12Z2
2 + 2(Z12 − Z2
2)2 sin2(k2d)
ER
EI
=(Z1
2 − Z22)(eik2d − e− ik2d )
(Z1 + Z2)2e− ik2d − (Z1 − Z2)2eik2d
Z1 = 377Ω, Z2 = 250Ω
k2d
R
π 2π
Air glass air
ε2,µ2 ε1,µ1 ε1,µ1
z 0 d
EI HI
ER HR
EF HF
EB HB
ET HT
Path Lengths What’s k2d = mπ ? Wavelength in medium 2 is
k2d = mπ means
R = 0 when 2d is a multiple of the wavelength
Consider two paths of reflection Lengths differ by 2d Path B goes through an extra
Phase changes by 2k2d Reflections from A and B meet with
the same phase if 2k2d = m×2π
But then, shouldn’t they add up and increase R?
λ2 =
2πk2
d =
mπk2
= mλ2
2
ε2,µ2 ε1,µ1 ε1,µ1
z 0 d
A
B e
2ik2d
k2d
R
π 2π
Hard and Soft Boundaries Remember at a simple boundary
Direction of ER is Same as EI if n1 > n2, i.e. slow fast
Call it a soft boundary Opposite to EI if n1 < n2, i.e. fast slow
Call it a hard boundary Reflection on a hard boundary flips the polarity of E Or, the phase of the reflected waves is off by π This is important when you are considering interference
ε1,µ1 ε2,µ2
EI HI
ER HR
ET HT ER =
n1 − n2
n1 + n2
EI
ET =
2n1
n1 + n2
EI
Interference Boundaries A/B are either hard/soft or soft/hard Two reflected waves have
opposite polarities to start with
At 2k2d = 2mπ, the paths A and B meet with opposite phase and cancel out Destructive interference
How about 2k2d = (2m − 1)π ? Boundaries cause a π in phase
difference Path difference gives another π Constructive interference
ε2,µ2 ε1,µ1 ε1,µ1
z 0 d
A
B
k2d
R
π 2π
Newton’s Rings Place a convex lens on a glass plane Suppose R0 is large θ is small All surfaces are normal to the light
Reflection depends on 2kaird
Reflection vanishes at
R0
d
r
light
d = R0 − R0
2 − r 2 ≈r 2
2R0
for r R0
2kaird =
kairr2
R0
=2πr 2
λairR0
2πr 2
λairR0
= 2mπ r2 = mλairR0
Newton’s Rings Expect concentric rings
Actual pattern less clear — and somewhat colored
Different wavelengths make rings at different radii
Ring pattern obscured after the first few
r
R
First maximum at 2kaird = π
Back to Reflectivity
Without interference, R would be 2×4% = 8% Just adding the intensities of
reflections (4% each) from two boundaries
Correct solution was obtained by: Adding the amplitudes of E, and then Calculating the intensity = (amplitude)2
R =
2(Z12 − Z2
2)2 sin2(k2d)8Z1
2Z22 + 2(Z1
2 − Z22)2 sin2(k2d)
k2d
R
π 2π
Air glass air
This is why reflectivity doesn’t add up
Add amplitudes, NOT intensities
Anti-Reflective Coating Destructive interference can be used to eliminate reflection on lenses Surface reflection of 4% each is
unacceptable for lenses with many elements
23-element TV camera lens would lose 85% of the light
Idea: put a thin coating of material with intermediate Z Choose the thickness just right so that the reflection from two
boundaries cancel each other
Quarter Wavelength Coating Now we have three materials
Both boundaries are “hard” No difference between reflection
at A and B To eliminate reflection, path difference 2d must satisfy
Larger m won’t work well for white light
Best solution: 2k2d = π, or
nair = n1 < n2 < n3 = nglass
2k2d = (2m −1)π Odd multiple of π makes A and B cancel
d = λ2 4
Quarter wavelength coating
ε2,µ2 ε3,µ3 ε1,µ1
z 0 d
A
B
Coating Material d = λ/4 is good but not enough Need to match the amplitudes of
reflection from A and B for perfect cancellation
We know for each boundary
To make them equal, we need
For air-to-glass, n1 = 1.0003, n3 = 1.5 n2 = 1.22 Tune thickness for green (550 nm in vacuum)
ER
EI
⎛
⎝⎜⎞
⎠⎟ A
=n1 − n2
n1 + n2
ER
EI
⎛
⎝⎜⎞
⎠⎟ B
=n2 − n3
n2 + n3
n2 = n1n3
d =
5504 ×1.22
= 113nm Middle of visible spectrum Human eyes most sensitive
ε2,µ2 ε3,µ3 ε1,µ1
z 0 d
A
B
Two-Body Interference Imagine two charged particles oscillating together Each radiate EM waves What’s the total radiation?
Very similar situation can be built by intercepting plane waves with a wall with two holes Waves going through the holes
spread spherically Young’s double-slit experiment
(1801–1803) demonstrated light was in fact made of waves
Plane waves
Geometrical Solution What do we expect to see? Draw circular waves in red/blue around two
sources Represent peaks/valleys
Look for crossings Red/red or blue/blue add up Constructive interference
Red/blue cancel each other Destructive interference
Wave amplitude varies depending on direction
Two-Body Interference E field generated by A and B differ in two ways Amplitude proportional to 1/x1 and 1/x2
This is negligible if Phase differ by k(x1 − x2)
Ignore the amplitude difference
Intensity goes with Need x1 − x2
θ A
B
d
x1
x2
x1 ≈ x2 d
EA = E0ei (kx1−ω t )
EB = E0ei (kx2 −ω t )
E = EA + EB = E0(eikx1 + eikx2 )e− iω t
= 2E0 cosk(x1 − x2)
2e
i kx1+ x2
2−ω t
⎛
⎝⎜
⎞
⎠⎟
S ∝ E0 cos k (x1− x2 )
2( )2
Path Difference Suppose Two paths are almost parallel From geometry
Intensity I can be written as
NB: still ignoring the 1/r dependence of amplitude Exact solution much uglier than this
x1 ≈ x2 d
θ A
B
d
x1
x2
d sinθ
x2 − x1 = d sinθ
I = I0 cos2 k(x1 − x2)2
= I0 cos2 kd sinθ2
= I0 cos2 πd sinθλ
Intensity vs. Angle
Approximate solution pretty good Result matches the guess from red/blue circles
Exact
I = I0 cos2 πd sinθ
λ
I used r = 5d for this plot
Difference invisible for larger r
Young’s Experiment In Young’s experiment, a screen was set up to observe the interference pattern
It’s safe to ignore the 1/r dependence
Since θ is small
Simple (cos)2 pattern is found
d
D
yθ D d D y θ 1
sinθ ≈ tanθ = y D
I = I0 cos2 πd sinθ
λ= I0 cos2 πdy
λD
y
I
Young’s Experiment We know light is electromagnetic waves It wasn’t so obvious in the 18th century Corpuscular model dominant in Newton’s era
Wavelength of light is too short Few wave-like phenomena can be observed in human-sized
experiment (or daily experience) Young used interference to expand the wave-ness of light to a
visible size
Story flips again in late 19th century Experiments (photoelectric effect, black body radiation) show
particle-ness of light More about this later…
Summary Studied interference >2 waves overlap Amplitudes add up Intensity = (amplitude)2 does not add up
Thin-film interference Reflectivity of thin film depends on thickness,
plus hard/soft-ness of the two boundaries Newton’s rings 1/4-wavelength anti-reflective coatings
Two-body interference Intensity depends on angle Young’s experiment demonstrated wave-ness of light
I = I0 cos2 πd sinθ
λ
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