# Document

```Young’s Experiment
Coherence
Two sources to produce an interference that is stable over
time, if their light has a phase relationship that does not
change with time: E(t)=E0cos(wt+f)
Coherent sources: Phase f must be well defined and
constant. When waves from coherent sources meet, stable
interference can occur － laser light (produced by
cooperative behavior of atoms)
Incoherent sources: f jitters randomly in time, no stable
interference occurs － sunlight
35- 2
Intensity and phase
E  t   E 0 sin w t  E 0 sin  w t  f   ?
E  2  E 0 cos 
E  4 E 0 cos
2
I
I0
2

E
2
E
2
0
  2 E 0 cos 12 f
2 1
2
f
 4 cos
2 1
2
f  I  4 I 0 cos
2 1
2
f
Eq. 35-22
 phase   path length 

 

difference
difference

 


2

  f
Fig. 35-13
 phase  2   path length 




difference
difference





f 
2

 d sin  
Eq. 35-23
35- 3
Intensity in Double-Slit Interference
E2
E1
E 1  E 0 sin w t
I  4 I 0 cos
m axim a w hen:
1
2
2 1
2
f
1
2
2 d

f  m  for m  0,1, 2,
 d sin   m  for m  0,1, 2,
m inim a w hen:
f 
f  m 
1
2

an d
E 2  E 0 sin  w t  f 
sin 
 f  2 m 
2 d

sin 
(m axim a)
 d sin    m 
1
2

for m  0,1, 2,
(m inim a)
35- 4
Intensity in Double-Slit Interference
I avg  2 I 0
Fig. 35-12
35- 5
Ex.11-2 35-2
wavelength 600 nm
n2=1.5 and
m=1→m=0
Interference form Thin Films
Reflection Phase Shifts
n1
n1
n1 > n2
n1 < n2
n2
n2
Reflection
Reflection Phase Shift
Off lower index
0
Off higher index
0.5 wavelength
Fig. 35-16
35- 8
Phase Difference in Thin-Film
Interference
Three effects can contribute to the phase
difference between r1 and r2.
1. Differences in reflection conditions

2
Fig. 35-17
0
2. Difference in path length traveled.
3. Differences in the media in which the waves
travel. One must use the wavelength in each
medium ( / n), to calculate the phase.
35- 9
Equations for Thin-Film
Interference
½ wavelength phase difference to difference in reflection of r1 and r2
2L 
odd num ber
 w avelength =
odd num ber
2
2
  n 2 (in-phase w aves)
2 L  integer  wavelength = integer   n 2 (out-of-phase waves)
n 2 

n2
2L  m 
2L  m

1
2


for m  0,1, 2,
(m axim a-- bright film in a ir)
n2
for m  0,1, 2,
(m inim a-- dark film in air)
n2
35- 10
Color Shifting by Paper Currencies,paints
and Morpho Butterflies
weak mirror
soap film
better mirror
looking directly down ： red or red-yellow
tilting ：green
35- 11

Ex.11-3 35-3 Brighted reflected
light from a water film
thickness 320 nm
n2=1.33
m = 0, 1700 nm, infrared
m = 1, 567 nm, yellow-green
m = 2, 340 nm, ultraviolet
Ex.11-4 35-4 anti-reflection
coating
Ex.11-5 35-5 thin air wedge
Michelson Interferometer
L  2d1  2d 2
 Lm  2 L
(in terferom eter)
(slab of m aterial of thickness
L placed in front of M 1 )
Fig. 35-23
35- 17
Determining Material thickness L
Nm=
2L
m
=
2 Ln

(num ber of w avelengths
in slab of m a terial)
Na=
2L

(num ber of w avelengths
in sam e thickness of air)
N m -N a =
2 Ln


2L

=
2L

 n-1 
(difference in w avelengths
for paths w ith and w ithout
thin slab)
35- 18
Problem 35-81
In Fig. 35-49, an airtight chamber of
length d = 5.0 cm is placed in one of the
arms of a Michelson interferometer. (The
glass window on each end of the chamber
has negligible thickness.) Light of
wavelength λ = 500 nm is used.
Evacuating the air from the chamber
causes a shift of 60 bright fringes. From
these data and to six significant figures,
find the index of refraction of air at
atmospheric pressure.
35- 19
Solution to Problem 35-81
φ1 the phase difference with air ； 2 ：vacuum
f1  f2
n  1g
L
2 n 2  O 4  b
L
 2 LM 

P
N  Q 
b g  2 N
4 n  1 L
N fringes

n  1
N
2L
 1
c
2c
5.0  10
60 500  10
h 1.00030 .
mh
9
2
m
35- 20
11-3 Diffraction
and the Wave Theory of Light
Diffraction Pattern from a single narrow slit.
Side or secondary
maxima
Light
Central
maximum
Fresnel Bright Spot.
Light
Bright
spot
These patterns
cannot be explained
using geometrical
optics (Ch. 34)!
36- 21
The Fresnel Bright Spot (1819)
Newton
 corpuscle
 Poisson
 Fresnel
 wave

Diffraction by a single slit
a sin   
st
(1 minima)
a sin   2 
(2
nd
minima)

 Double-slit
diffraction (with
interference)
 Single-slit diffraction
Diffraction by a Single Slit:
Locating the first minimum
a
2
sin  

2
 a sin   
(first minimum)
36- 26
Diffraction by a Single Slit:
Locating the Minima
a
4
sin  

 a sin   2 
(second minimum)
2
a sin   m  , for m  1, 2, 3
(minima-dark fringes)
36- 27
Ex.11-6 36-1 Slit width
Intensity in Single-Slit Diffraction,
Qualitatively
 phase   2   path length 




difference
difference



 

N=18
=0
 2 
f  
   x sin 
  
 small
Fig. 36-7
1st min.

1st side
max.
36- 29
Intensity and path length difference
sin f 
1
2
E
2R
E 
I 
Im
Fig. 36-9

f 
Em
2

E
2
Em
R
Em
1
2
f
sin 12 f
 sin  
 I    I m 




 2 
f 
  a sin 
  
2

36- 30
Intensity in Single-Slit Diffraction, Quantitatively
Here we will show that the intensity at the screen due to
a single slit is:
 sin  
I    I m 




w here  
1
f 
2
a

2
(36-5)
sin 
(36-6)
In Eq. 36-5, minima occur when:
  m ,
for m  1, 2, 3
If we put this into Eq. 36-6 we find:
a
m 
sin  ,
for m  1, 2, 3

Fig. 36-8
or
a sin   m  ,
for m  1, 2, 3
(m inim a-dark fringes)
36- 31
Ex.11-7 36-2
1

   m    , m  1, 2, 3,
2

Diffraction by a Circular Aperture
Distant point
source, e,g., star
d
lens
sin   1.22


(1st m in.- circ. aperture)
Image is not a point, as
expected from geometrical
optics! Diffraction is
responsible for this image
pattern
d
a
Light
Light

sin   1.22

a
(1st m in.- single slit)
a

36- 33
Resolvability
Rayleigh’s Criterion: two point sources are barely
resolvable if their angular separation θR results in the
central maximum of the diffraction pattern of one
source’s image is centered on the first minimum of the
diffraction pattern of the other source’s image.
   sm all

Fig. 361 
 R  sin  1.22   1.22
11
d 
d

R
(R ayleigh's criterion)
36-34
11-4.9 Diffraction – (繞射)
Why do the colors in a pointillism
painting change with viewing distance?
Ex.11-8 36-3 pointillism
D = 2.0 mm
d = 1.5 mm
Ex.11-9 36-4
d = 32 mm
f = 24 cm
λ= 550
nm
The telescopes on some commercial
and military surveillance satellites
Resolution of 85 cm and 10 cm respectively
D
L
  R  1.22

d
 = 550 × 10–9 m.
(a) L = 400 × 103 m ， D = 0.85 m → d = 0.32 m.
(b) D = 0.10 m → d = 2.7 m.
36- 38
Diffraction by a Double Slit
Single slit a~
Two vanishingly narrow slits a<<
Two Single slits a~
I 

 sin  
 I m  cos   

  
2
 
2
(double slit)
 
d

a

sin 
sin 
36- 39
Ex.11-10 36-5
d = 32 μm
a = 4.050 μm
λ= 405
a sin   
d sin   m 2  for m  0,1, 2,
nm
Diffraction Gratings
Fig. 36-18
Fig. 36-19
d sin   m  for m  0,1, 2
Fig. 36-20
(m axim a-lines)
36- 41
Width of Lines
Fig. 36-21
N d sin   hw   ,
  hw 

sin   hw    hw
(half w idth of central line)
Nd
Fig. 36-22
  hw 

N d cos 
(half w idth of line at  )
36- 42
Grating Spectroscope
Separates different wavelengths (colors)
of light into distinct diffraction lines
Fig. 36-24
Fig. 36-23
36- 43
Compact Disc
Optically Variable Graphics
Fig. 36-27
36- 45

Viewing a holograph
A Holograph
Gratings: Dispersion
D 


(dispersion defined)
D 
m
d cos 
(dispersion of a grating) (36-30)
Angular position of maxima
d sin   m 
Differential of first equation
(what change in angle
does a change in
wavelength produce?)
d  co s   d   m d 
For small angles
d     an d
d   
d  co s      m  



m
d  cos 

36- 49
Gratings: Resolving Power
R
 avg

(resolving pow er defined)
R  N m (resolving pow er of a grating) (36-32)

Rayleigh's criterion for halfwidth to resolve two lines
  hw 
Substituting for  in calculation
on previous slide
  hw   


N d cos 
 m
N
R 


 Nm
36-50
Dispersion and Resolving Power Compared
36- 51
X-Ray Diffraction
X-rays are electromagnetic radiation with wavelength ~1 Å
= 10-10 m (visible light ~5.5x10-7 m)
X-ray generation
X-ray wavelengths to short to be
resolved by a standard optical grating
Fig. 36-29
  sin
1
m
d
 sin
1
1   0.1 nm 
 0.0019 
3000 nm
36- 52
Diffraction of x-rays by crystal
d ~ 0.1 nm
→ three-dimensional diffraction grating
2 d sin   m  for m  0,1, 2
(B ragg's law )
Fig. 36-30
36- 53
X-Ray Diffraction, cont’d
5d 
5
4
a
2
0
or d 
a0
20
 0.2236 a 0
Fig. 36-31
36-54
Structural Coloring by Diffraction
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