Wave PhenomenaPhysics 15c

Lecture 19Interference

(H&L Chapter 11)

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

2 2plasma 1 pn ω ω= −

20

0p

q nm

ωε

=

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

! Used 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! Things can get more interesting than that…

1 1,ε µ 2 2,ε µ

IEIB

RE RB

TETB1

1 2

2T I

nE En n

=+

1 2

1 2R I

n nE En n−

=+

2

1 2

1 2

n nRn n −

= + 1 2

21 2

4( )

n nTn n

=+

2T T→

Two Boundaries

! Consider transitions through two boundaries! At z = 0,

! At z = d

! We can solve them…

2 2,ε µ 1 1,ε µ

FEFB

BE BB

TETB1 1,ε µ

IEIB

RE RB

z0 d

I R F BE E E E+ = +

1 2

I R F BE E E EZ Z− −

= H

E

2 2ik d ik dF B TE e E e E−+ =

2 2

2 1

ik d ik dF B TE e E e E

Z Z

−−=

Exponentials appear because2( )( , ) i t k z

F FE z t E e ω− −=2( )( , ) i t k z

B BE z t E e ω− +=

Two Boundaries

! Solutions are

! OK, so what did we learn?

2

2 2

2 1 22 2

1 2 1 2

2 ( )( ) ( )

ik d

B Iik d ik d

Z Z Z eE EZ Z e Z Z e−

−=

+ − −

2

2 2

2 1 22 2

1 2 1 2

2 ( )( ) ( )

ik d

F Iik d ik d

Z Z Z eE EZ Z e Z Z e

−

−

+=

+ − −

2 2

2 2

2 21 2

2 21 2 1 2

( )( )( ) ( )

ik d ik d

R Iik d ik d

Z Z e eE EZ Z e Z Z e

−

−

− −=

+ − −

2

2 2

1 22 2

1 2 1 2

4( ) ( )

ik d

T Iik d ik d

Z Z eE EZ Z e Z Z e

−

−=+ − −

z

2 2,ε µ 1 1,ε µ

FEFB

BE BB

TETB1 1,ε µ

IEIB

RE RB

0 d

Reflectivity is (this)2

Transmittivity is (this)2

Reflectivity

! Look at the reflectivity

! Reflectivity oscillatesas a function of k2d

! Reflection vanishes atk2d = mπ !

! What’s happening?

2 2 2 2 21 2 2

2 2 2 2 2 21 2 1 2 2

2( ) sin ( )8 2( ) sin ( )

R

I

E Z Z k dRE Z Z Z Z k d

−= =

+ −

z

2 2,ε µ 1 1,ε µ

FEFB

BE BB

TETB1 1,ε µ

IEIB

RE RB

0 d

2 2

2 2

2 21 2

2 21 2 1 2

( )( )( ) ( )

ik d ik dR

ik d ik dI

E Z Z e eE Z Z e Z Z e

−

−

− −=

+ − −

1 2377 , 250Z Z= Ω = Ω

2k d

R

π 2π

Air " glass " air

Path Lengths

! What’s k2d, anyway?! Wavelength is medium 2 is! If! R hits zero when 2d is integer times the wavelength

! Consider two paths of reflection! Lengths differ by 2d! Path B goes through extra

! Phase changes by 2k2d! Reflections from A and B meet

with the same phase if

! But then, shouldn’t they add up and increase R?

2 2 2k λ π=

2k d mπ= 22d mλ=

2 2,ε µ 1 1,ε µ1 1,ε µ

z0 d

A

B

2k d

R

π 2π

22ik de

22 2k d m π= ×

Hard and Soft Boundaries

! Remember at a simple boundary

! Direction of ER is! Same as EI if i.e. slow " fast transition

! Call it a soft boundary

! Opposite to EI if i.e. fast " slow transition! 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

1 1,ε µ 2 2,ε µ

IEIB

RE RB

TETB1 2

1 2R I

n nE En n−

=+

1 2n n>

1 2n n<

Interference

! Boundaries A/B are eitherhard/soft or soft/hard! Two reflected waves have

opposite polarities andcancel each other

" Destructive interference occurs at

! How about ?! Boundaries give π in phase between paths A and B! Path difference gives another π" Constructive interference

2 2,ε µ 1 1,ε µ1 1,ε µ

z0 d

A

B

22 2k d mπ=

22 (2 1)k d m π= −

2k d

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

0R

d

r

θlight

22 2

0 002

rd R R rR

= − − ≈ 0r R!

2 2air

air0 air 0

22 k r rk dR R

πλ

= =

2

air 0

2 2r mR

π πλ

= 2air 0r m Rλ=

Newton’s Rings

! Expect concentric rings

! Actual pattern less clear– and somewhat colored

! Different wavelengths makerings at different radii

! Ring pattern obscured after thefirst few

r

R

First maximum at

air2k d π=

Back to Reflectivity

! Without interference, Rwould be! 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

2 2 2 21 2 2

2 2 2 2 2 21 2 1 2 2

2( ) sin ( )8 2( ) sin ( )

Z Z k dRZ Z Z Z k d

−=

+ −

2k d

R

π 2π

Air " glass " air

~ 2 4% 8%× =

This is whyreflectivity

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 lenseswith 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: or

2 2,ε µ 3 3,ε µ1 1,ε µ

z0 d

A

B

air 1 2 3 glassn n n n n= < < =

22 (2 1)k d m π= − Odd multiple of π makes A and B cancel

22k d π= 2 4d λ=

Quarter wavelength coating

Coating Material

! d = λ/4 is good but not enough! Need to match the amplitudes of

reflection from A and B forperfect cancellation

! We know for each boundary

! For air " glass! Tune thickness for green (550 nm in vacuum)

2 2,ε µ 3 3,ε µ1 1,ε µ

z0 d

A

B

1 2

1 2

R

I A

E n nE n n

−= +

2 3

2 3

R

I B

n nEE n n

−= +

Makeequal 2 1 3n n n=

1 1.0003n = 3 1.5n = 2 1.22n =

550 113nm4 1.22

d = =×

Middle of visible spectrumHuman eyes most sensitive

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 lightwas in fact made of waves

Planewaves

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 dependingon direction

Two-Body Interference

! E field generated by A and B differ in two ways! Amplitude proportional to and

! This is negligible if! Phase differ by

! Ignore the amplitude difference θA

B

d

1x

2x

11 x 21 x

1 2x x d≈ "

1 2( )k x x−

1( )0

i kx tAE E e ω−= 2( )

0i kx t

BE E e ω−=

1 2

1 2

1 2

0

( ) 20 2

( )

2 cos

ikx ikx i tA B

x xi k tk x x

E E E E e e e

E e

ω

ω

−

+ − −

= + = +

=

Amplitude ( )1 22( )

0 2cos k x xS E −∝

How do I get x1 – x2?

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

1 2x x d≈ "

θA

B

d

1x

2x

sind θ

2 1 sinx x d θ− =

2 21 20 0

20

( ) sincos cos2 2

sincos

k x x kdI I I

dI

θ

π θλ

−= =

=

Intensity vs. Angle

! Approximate solution pretty good! Result matches the guess from red/blue circles

Exact

20

sincos dI I π θλ

=

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 the1/r dependence

! Since θ is small

! Simple (cos)2 pattern is found

d

D

yθD d" D y" 1θ !

sin tan y Dθ θ≈ =

2 20 0

sincos cosd dyI I ID

π θ πλ λ

= =

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

20

sincos dI I π θλ

=