Young
s Experiment

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 )= E
0 cos( w t + 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

Fig. 35-13
  
E
0 sin w t

E
0 sin
 w f  
E

2

E
0 cos
  
2 E
0 cos 1
2 f
?
E
2 
E
2
4 cos
0
2 1
2 f
I E
2
 
2
I E
0 0
4 cos
2
1
2 f
I 4 I
0 cos
2
1
2 f
Eq. 35-22

 phase difference
 
  path length difference
2
 



 phase difference f 
2


 d

 sin

 
2



 path length difference


Eq. 35-23
35-
3

Intensity in Double-Slit Interference
E
2
E
1
E
1

E
0 sin w t and E
2

E
0 sin
 
I

4 I
0 cos
2 1
2 f f 
2

 d sin
 maxima when: 1
2 f  m

for m

0,1, 2, 2 m
 
2

 d sin

 d sin
  m

for m

0,1, 2, (maxima) minima when: 1
2 f   m

1
2
   d sin
   m

1
2
 
for m

0,1,2, (minima)
35-
4

I avg

2 I
0
Intensity in Double-Slit Interference
Fig. 35-12
35-
5

Ex.11-2 35-2 wavelength
600 nm n
2
=1.5 and m = 1 → m = 0

n
1 n
1
> n
2 n
2 n
1 n
1
< n
2 n
2
Fig. 35-16
Reflection
Off lower index
Off higher index
Reflection Phase Shift
0
0.5 wavelength
35-
8

Fig. 35-17

2
0
Three effects can contribute to the phase difference between r
1 and r
2
.
1. Differences in reflection conditions
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

½ wavelength phase difference to difference in reflection of r
1 and r
2
2 L
 odd number
2
2 L
 

wavelength = odd number
2
  n 2
(in-phase waves)
  n 2
(out-of-phase waves)
  n 2
 n
2
2 L
  m

1
2

 n
2
2 L
 m
 n
2
for m

0,1, 2, (maxima-- bright film in air)
for m

0,1, 2, (minima-- dark film in air)
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

爾
蝴
蝶
大
藍
魔

Ex.11-3 35-3 Brighted reflected light from a water film thickness 320 nm n
2
=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

L 2 d
1

2 d
2
(interferometer)
Fig. 35-23

L m

L M
1
)
35-
17

Determining Material thickness L
N m
=
2

L m
=
2 Ln

(number of wavelengths
in slab of material)
N a
2

L
= (number of wavelengths
in same thickness of air)
N m
N a
=
2 Ln


2

L
=
2

L  
for paths with and without
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
φ
1 the phase difference with air ； 2 ： vacuum f f
1
 
2
2 L
L
2
N

 n

2


O
Q
4
 b g


1 L
4
 b g


1 L

2 N
 N fringes n 1
N
2

L
1 c c


9
2 m m h h

.
35-
20

11-3 Diffraction and the Wave Theory of Light
Diffraction Pattern from a single narrow slit.
Side or secondary maxima
Light
Fresnel Bright Spot.
Light
Bright spot
Central maximum
These patterns cannot be explained using geometrical optics (Ch. 34)!
36-
21

 Newton
 corpuscle
 Poisson


Fresnel wave

slit a sin st
(1 minima) a
   nd sin 2 (2 minima)




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


2 a sin
 
2
 a sin
  m

, for m

1, 2, 3
(second minimum)
(minima-dark fringes)
36-
27

Ex.11-6 36-1 Slit width

Intensity in Single-Slit Diffraction,
Qualitatively

 phase difference




2

 
 

 path length difference



2



 


  x sin
 
N=18

= 0
 small
Fig. 36-7
1st min.
1st side max.
36-
29

Intensity and path length difference
Fig. 36-9 sin 1
2 f 
E

2 R f 
E m
R
E


1
2
E m f sin 1
2 f
I
I m

E

2
2

E m
I

I m sin



2 f

2
 

 a sin
 
36-
30

Intensity in Single-Slit Diffraction, Quantitatively
Fig. 36-8
Here we will show that the intensity at the screen due to a single slit is:
I where



I m
1
2 f sin



 a

2
(36-5)
In Eq. 36-5, minima occur when:
  m

, for m

1, 2, 3
If we put this into Eq. 36-6 we find: m
 

 a
 m

1, 2, 3
  m

, for m

1, 2, 3
(minima-dark fringes) 36-
31

Ex.11-7 36-2
  m

1
2

, m

1, 2, 3,

Diffraction by a Circular Aperture
Distant point source, e,g., star d
 lens
 
 sin 1.22
(1st min.- circ. aperture) d a
Image is not a point, as expected from geometrical optics! Diffraction is responsible for this image pattern a
Light Light
 
 
 sin 1.22
(1st min.- single slit) a
3633

Resolvability
Rayleigh’s Criterion: two point sources are barely resolvable if their angular separation θ
R 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.

Fig. 36-
11 sin

1  1.22
 d

R
small


1.22
(Rayleigh's criterion) d
3634

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
λ cm
= 550 nm

The telescopes on some commercial and military surveillance satellites
Resolution of 85 cm and 10 cm respectively
D
L
 
R

 d

= 550 × 10
–
9 m.
(a) L = 400 × 10 3 m ， D = 0.85 m → d = 0.32 m.
(b) D = 0.10 m → d = 2.7 m.
3638

Diffraction by a Double Slit
Single slit a
~ 
Two vanishingly narrow slits a <<

I
Two Single slits a ~


I m
 cos 2

 sin


2
(double slit)




 d

 a
 sin
 sin

3639

Ex.11-10 36-5 a sin d sin
  m
2

for m

0,1, 2,
μ d = 32 m a = 4.050 m
μ
λ
= 405 nm

Diffraction Gratings
Fig. 36-18 Fig. 36-19 Fig. 36-20 d sin
  m

for m

0,1, 2 (maxima-lines)
3641

Width of Lines
Fig. 36-21
Nd sin
  hw
 
, sin
  hw
   hw
  hw


Nd
(half width of central line)
Fig. 36-22
  hw

Nd
 cos

3642

Grating Spectroscope
Separates different wavelengths (colors) of light into distinct diffraction lines
Fig. 36-24
Fig. 36-23
3643

Optically Variable Graphics
Fig. 36-27
3645

全像術

Gratings: Dispersion
D

 
 
(dispersion defined)
D
 d m cos

Angular position of maxima
(dispersion of a grating) (36-30) d sin
  m
 d
 cos
  md
 Differential of first equation
(what change in angle does a change in wavelength produce?)
For small angles d
    d
 cos

and d
   

 
 
 d
 m cos
 
3649

Gratings: Resolving Power
R

 avg
 
(resolving power defined)
R

Nm (resolving power of a grating) (36-32)
Rayleigh's criterion for halfwidth to resolve two lines
Substituting for
  in calculation on previous slide
  hw

Nd
 cos

  hw
  


N m
R






Nm

Dispersion and Resolving Power Compared
3651

X-Ray Diffraction
X-rays are electromagnetic radiation with wavelength ~1 Å
= 10 -10 m (visible light ~5.5x10
-7 m)
X-ray generation
Fig. 36-29
X-ray wavelengths to short to be resolved by a standard optical grating
  sin

1 m
 d
 sin

1
3000 nm


0.0019

3652

Diffraction of x-rays by crystal d ~ 0.1 nm
→ three-dimensional diffraction grating
  m

for m

0,1, 2 (Bragg's law)
Fig. 36-30
3653

X-Ray Diffraction, cont’d
5 d

5
4 a
0
2
or d
 a
0
20

0.2236
a
0
Fig. 36-31

Structural Coloring by Diffraction