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13 反射、折射、干涉、繞射
Sections
1. 反射 (reflection)與折射
(refraction)
2. 干涉 (interference)
3. 繞射 (diffraction)
13-1 Reflection and Refraction (反射
與折射)
1  1
n2 sin  2  n1 sin 1
(reflectio n)
(refraction)
折射率
․The Index of Refraction
․The Stealth Aircraft F-117A
Chromatic Dispersion – prisms and
gratings
Rainbows
13-2 Interference – (干涉)
What produces the blue-green of a Morpho’s wing?
How do colorshifting inks shift colors?
airbrush
Huygens’ principle
All points on a wavefront
serve as point sources of
spherical secondary wavelets.
After a time t, the new
position of the wavefront will
be that of a surface tangent
to these secondary wavelets.
Fig. 352
Law of Refraction from Huygens’
principle
Fig. 35-3
35-10
1
1 v1
tec  thg  


v1 v2
2 v2
sin 1 
sin  2 
sin 1 1 v1


sin  2 2 v2
c
n1 
v1
1
hc
2
hc
2
(for triangle hce)
(for triangle hcg)
c
Index of Refraction: n 
v
c
and n2 
v2
sin 1 c n1 n2


sin  2 c n2 n1
Law of Refraction: n1 sin 1
 n2 sin  2
35-11
Phase Difference, Wavelength and
Index of Refraction
Wavelength and Index of
Refraction
n v
v

  n    n 
 c
c
n
v
cn c
fn 

  f
n  n 
The frequency of light in a medium
is the same as it is in vacuum
35-13
Phase Difference
Fig. 35-4
Since wavelengths in n1 and n2
are different, the two beams may
no longer be in phase
L
L
Ln1
Number of wavelengths in n1: N1 


n1  n1

L
L
Ln2
Number of wavelengths in n2 : N 2 


n 2  n2

Assuming n2  n1: N 2  N1 
Ln2


Ln2


L

 n2  n1 
N2  N1  1/2 wavelength  destructive interference
35-14
Ex.13-1 35-1
wavelength 550.0 nm
n2=1.600 and
L = 2.600 m
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
35-17
Intensity and phase
E  t   E0 sin w t  E0 sin w t  f   ?
E  2  E0 cos    2 E0 cos 12 f
E 2  4 E02 cos2 12 f
Phasor diagram
I E2
 2  4cos2 12 f  I  4 I 0 cos2 12 f
I 0 E0
Eq. 35-22
   f
Fig. 35-13
 phase   path length 
 difference   difference 



2

 phase  2  path length 
 difference     difference 




f
2

 d sin  
Eq. 35-23
35-18
Intensity in Double-Slit Interference
E2
E1
E1  E0 sin wt and E2  E0 sin wt  f 
I  4 I 0 cos
2 1
2
f
f
2 d

maxima when: 12 f  m for m  0,1, 2,
 d sin   m for m  0,1, 2,
sin 
 f  2m 
2 d

sin 
(maxima)
minima when: 12 f   m  12    d sin    m  12   for m  0,1,2,
(minima)
35- 19
Intensity in Double-Slit Interference
I avg  2 I 0
Fig. 35-12
35- 20
Ex.13-2 35-2
wavelength 600 nm
n2=1.5 and
m=1→m=0
Interference from 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- 23
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- 24
Equations for Thin-Film
Interference
½ wavelength phase difference to difference in reflection of r1 and r2
odd number
odd number
2L 
 wavelength =
 n 2 (in-phase waves)
2
2
2 L  integer  wavelength = integer  n 2 (out-of-phase waves)
n 2 

n2
2L   m 
2L  m

n2
1
2


n2
for m  0,1,2,
for m  0,1,2,
(maxima-- bright film in air)
(minima-- dark film in air)
35- 25
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- 26
大
藍
魔
爾
蝴
蝶
雙狹縫干涉之強度
Ex.13-3 35-3 Brightest 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.13-4 35-4 anti-reflection
coating
Ex.13-5 35-5 thin air wedge
Michelson Interferometer
L  2d1  2d 2
(interferometer)
Lm  2 L (slab of material of thickness
L placed in front of M 1 )
Fig. 35-23
35- 32
Determining Material thickness L
2 L 2 Ln
Nm =
=
(number of wavelengths
m

in slab of material)
Na =
2L

(number of wavelengths
in same thickness of air)
N m -N a =
2 Ln

2L 2L

=
 n-1 (difference in wavelengths


for paths with and without
thin slab)
35- 33
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- 34
Solution to Problem 35-81
φ1 the phase difference with air ； 2 ：vacuum
f1  f 2
n  1g
L
2n 2 O4b
L
 2 LM  P

N  Q 
b g
N fringes
c
c
h 100030
.
.
h
4 n  1 L
 2 N

60 500  109 m
N
n  1
 1
2L
2 5.0  102 m
35- 35
13-3 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- 36
The Fresnel Bright Spot (1819)
Newton
 corpuscle
 Poisson
 Fresnel
 wave

Diffraction by a single slit
a sin   
(1st minima)
a sin   2
(2nd minima)
單
狹
縫
繞
射
之
強
度
雙狹縫與單狹縫
 Double-slit
diffraction (with
interference)
 Single-slit diffraction
Diffraction by a Single Slit:
Locating the first minimum
a

sin    a sin   
2
2
(first minimum)
36- 41
Diffraction by a Single Slit:
Locating the Minima
a

sin    a sin   2
4
2
a sin   m, for m  1, 2,3
(second minimum)
(minima-dark fringes)
36- 42
Ex.13-6 36-1 Slit width
Intensity in Single-Slit Diffraction,
Qualitatively
 phase   2


 difference   
N=18
=0
  path length 
  difference 


 2 
f  
  x sin  
  
 small
Fig. 36-7
1st min.
1st side
max.
36- 44
Intensity and path length difference
E
sin f 
2R
Em
f
R
1
2
E 
Em
1
2
f
sin 12 f
I   E2
 sin  
 2  I    I m 

Im
Em



Fig. 36-9
2
 2 
f     a sin  
  
36- 45
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 
(36-5)

  
1
a
where   f 
sin  (36-6)
2

2
In Eq. 36-5, minima occur when:
  m ,
for m  1, 2,3
If we put this into Eq. 36-6 we find:
Fig. 36-8
a
m 
sin  , for m  1, 2,3

or a sin   m , for m  1, 2,3
(minima-dark fringes)
36- 46
Ex.13-7 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

d

(1st min.- circ. aperture)
Image is not a point, as
expected from geometrical
optics! Diffraction is
responsible for this image
pattern
a
Light
Light

sin  

a
(1st min.- single slit)
a

36- 48
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.
 
Fig. 361 
 R  sin 1.22 
11
d

R
small

1.22

d
(Rayleigh's criterion)
36-49
13-3 Diffraction – (繞射)
Why do the colors in a pointillism
painting change with viewing distance?
Ex.13-8 36-3 pointillism
D = 2.0 mm
d = 1.5 mm
(diameter of
the pupil)
Ex.13-9 36-4
d = 32 mm
f = 24 cm
λ= 550
nm
(a) angular
separation
(b) separation
in the focal
plane
The telescopes on some commercial
and military surveillance satellites
Resolution of 85 cm and 10 cm respectively
D

  R  122
.
L
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- 53
Diffraction by a Double Slit
Single slit a~
Two vanishingly narrow slits a<<
Two Single slits a~
 sin  
I    I m  cos   

  
2
2
d

sin 

(double slit)
a

sin 

36- 54
Ex.13-10 36-5
d = 19.44 μm
a = 4.050 μm
λ= 405
a sin   
d sin   m2 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
(maxima-lines)
36- 56
Width of Lines
Fig. 36-21
Nd sin  hw   ,
 hw 

Nd
sin  hw   hw
(half width of central line)
Fig. 36-22
 hw 

Nd cos
(half width of line at  )
36- 57
Grating Spectroscope
Separates different wavelengths (colors)
of light into distinct diffraction lines
Fig. 36-24
Fig. 36-23
36- 58
Compact Disc
Optically Variable Graphics
Fig. 36-27
36- 60
全像術
Viewing a holograph
A Holograph
Gratings: Dispersion

D

(dispersion defined)
D
m
(dispersion of a grating) (36-30)
d cos 
Angular position of maxima
d sin   m
Differential of first equation
(what change in angle
does a change in
wavelength produce?)
d  cos   d  md 
For small angles
d   and d   
d  cos     m

m

 d  cos  
36- 64
Gratings: Resolving Power
avg
R

(resolving power defined)
R  Nm (resolving power of a grating) (36-32)

Rayleigh's criterion for halfwidth to resolve two lines
 hw 
Substituting for  in calculation
on previous slide
 hw  
d  cos     m


N
Nd cos 
 m

R
 Nm

36- 65
Dispersion and Resolving Power Compared
36- 66
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 too short to be
resolved by a standard optical grating
Fig. 36-29
m
1 1 0.1 nm 
  sin
 sin
 0.0019
d
3000 nm
1
36- 67
Diffraction of x-rays by crystal
d ~ 0.1 nm
→ three-dimensional diffraction grating
2d sin   m for m  0,1, 2
(Bragg's law)
Fig. 36-30
36- 68
X-Ray Diffraction, cont’d
5d 
5
4
2
0
a
a0
or d 
 0.2236a0
20
Fig. 36-31
36- 69
Structural Coloring by Diffraction