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Wave Phenomena Physics 15c Lecture 18 Interference

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Page 1: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

Wave Phenomena Physics 15c

Lecture 18 Interference

Page 2: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 3: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 4: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 5: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 6: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 7: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 8: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 9: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

⎨⎪⎪

⎩⎪⎪

Page 10: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 11: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 12: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 13: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 14: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 15: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 16: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 17: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 18: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 19: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 20: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 21: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 22: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 23: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 24: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 25: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 26: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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

Page 27: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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…

Page 28: Lecture 18 Interference - Harvard Universityusers.physics.harvard.edu/~morii/phys15c/lectures/Lecture18.pdf · Lecture 18 Interference . What We Did Last Time Studied microscopic

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