Download Chapter 10: Diffraction-1

Chapter 10 Diffraction
March 9, 11 Fraunhofer diffraction: The single slit
10.1 Preliminary considerations
Diffraction: The deviation of light from propagation in a straight line.
There is no essential physical distinction between interference and diffraction.
Huygens-Fresnel Principle: Every unobstructed point of a wave front serves as a source of
spherical wavelets. The amplitude of the optical field at any point beyond is the
superposition of all these wavelets, taking into account their amplitudes and phases.
Fraunhofer (far field) diffraction: Both the incoming and outgoing waves approach
being planar. a2/lR<<1, where R is the smaller of the two distances from the source to the
aperture and from the aperture to the observation point. a is the size of the aperture. The
diffraction pattern does not change when moving the observation plane further away.
Fresnel (near field) diffraction: Plane of observation is close to the aperture. General
case of diffraction. The diffraction pattern changes when the observation plane moves.
P
S
a
R1
R2
1
Mathematical criteria for Fraunhofer diffraction:
The phase for the rays meeting at the observation
point is a linear function of the aperture variables. S
P
y'
Waves from a point source:
Harmonic spherical wave:
E ( r, t ) 
y' sinq
A
A
cos( kr  t ), or E ( r, t )  ei ( kr t )
r
r
y
A is the source strength.
P (x,y)
D/2
Coherent line source:
r
dy'
x
exp( ikr)
E ( x, y )  εL 
dy '
D / 2
r
D/2
-D/2
eL is the source strength per unit length.
This equation changes a diffraction problem into an integration (interference) problem.
2
y
10.2 Fraunhofer diffraction
10.2.1 The single slit
P (x,y)
D/2
The slit is along the z-axis and has a width of D.
y'
q
r ( y ' )  R 2  y '2 2 Ry ' sin q
cos2 q 2
 R  y ' sin q 
y' 
2R
r
R
x
-D/2
exp( ikr)
In the phase, r is approximated by R-y' sinq, if D2/Rl <<1.
E ( x, y )  e L 
dy '
D / 2
Fraunhofer diffraction condition.
r
D / 2 exp[ ik ( R  y ' sin q )]
 eL 
dy '
D / 2
R
In the amplitude, r is approximated by R.
e D
sin[( kD / 2) sin q ]
 L exp( ikR)
R
(kD / 2) sin q
e D sin 
 L
exp( ikR),
  (kD / 2) sin q
R 
D/2
 sin  

 I (q )  I (0)



2
The overall phase is the same as a point source
at the center of the slit.
Integrate over z gives the same function.
3
 sin  

I (q )  I (0)
  
2
y

kD
sin q
2
P (x,y)
D/2
y'
dI
2 sin  (  cos   sin  )
 I (0)
0
d
3
r
q
R
x
-D/2
minima
sin   0,    ,  2 ,  3 ...

  tan  ,   0, 1.43 , 2.46 , 3.47 ,... maxima
 sin 
I (q )  I (0)
 
I/I(0)=
0.047 0.016
(©WIU OptoLab)



2

Width of the central peak
2
2l
  2  q 

kD / 2 D
Widths of the side peaks
    q 

kD / 2

l
D
4
Phasor model of single slit Fraunhofer diffraction: rolling paper
5
Read: Ch10: 1-2
Homework: Ch10: 3,7,8,9
Due: March 27
6
March 23 Double slit and many slits
10.2.2 The double slit
E ( x, z ) 
eL
R

b/2
b / 2
eL
P (x,z)
exp[ ik ( R  z ' sin q )]dz '
a b / 2
exp[ ik ( R  z ' sin q )]dz '
R a b / 2
e b sin 
exp( ikR)1  exp( ika sin q )
 L
R 

z
b
a
q
x
 be 
Let I 0   L  , intensity at the axis when there is only one slit. (We omit the 1/2  factor.)
 R 
  (ka / 2) sin q
  (kb / 2) sin q
2
2
 sin  
 cos 2 
I (q )  4 I 0 
  
The result is a rapidly varying double-slit interference pattern (cos2) modulated by
a slowly varying single-slit diffraction pattern (sin2/ 2).
7
2
 sin  
 cos 2 
I (q )  4 I 0 
  
 sin(  b sin q / l ) 
2   a sin q 
I (q )  I (0) 
cos


l
  b sin q / l 


2
Single-slit diffraction Two-slit interference
Fringes
Envelope
Question:
Which interference maximum coincides
with the first diffraction minimum?
b sin q  l 
a
m
a sin q  ml 
b
“Half-fringe” (split fringe) may occur there.
Our author counts a half-fringe as 0.5
fringe.
half-fringe
(©WIU OptoLab)
8
10.2.3 Diffraction by many slits
P (x,z)
z
b
R
a
q
x
C  e L / R, F ( z ' )  exp[ ik ( R  z ' sin q )]
E ( x, z )  C 
b/2
b / 2
 bC
 bC
sin 

sin 

F ( z ' )dz 'C 
a b / 2
a b / 2
F ( z ' )dz '  C 
2 a b / 2
2 a b / 2
F ( z ' )dz '...  C 
( N 1) a  b / 2
( N 1) a b / 2
F ( z ' )dz '
exp( ikR)1  exp( i 2 )  exp( i 4 )  ...  exp[ i 2( N  1) ]
exp( ikR)
2
1  exp( i 2 N )
1  exp( i 2 )
 sin    sin N 
 
I (q )  I 0 


sin





  (ka / 2) sin q
  (kb / 2) sin q
2
9
2
 sin    sin N 
 
I (q )  I 0 


sin





  (ka / 2) sin q
  (kb / 2) sin q
2
Principle maxima:   0,   ,  2 , ...
Minima (totally N-1):

2
3
( N  1)
  ,
,
, ... 
N
N
N
N
Subsidiary maxima (totally N-2):
 
3
5
(2 N  3)
,
, ..., 
2N
2N
2N
 sin 

 
2
  sin N 
 

  sin  
a  4b
N 6
2

10
Phasor model of three-slit interference: rotating sticks
11
Read: Ch10: 2
Homework: Ch10: 10,11,12
Due: April 3
12
March 25 Rectangular aperture
10.2.4 The rectangular aperture
Coherent aperture:
E eA
Y
y
exp( ikr)
dS

r
Aperture
dS=dydz
P(Y,Z)
r
R
R  X 2 Y 2  Z2
r  X 2  (Y  y ) 2  ( Z  z ) 2
 R 1  ( y 2  z 2 ) / R 2  2(Yy  Zz ) / R 2
x
z
X
Z
 R 1  2(Yy  Zz ) / R 2

 R 1  (Yy  Zz ) / R 2
E (Y , Z ) 

e A exp( ikR)
R
Fraunhofer diffraction condition
 exp ik (Yy  Zz) / RdS
Aperture
13
Y
y
Rectangular aperture:
dS=dydz
P(Y,Z)
r
R
b
z
E (Y , Z ) 

e A exp( ikR)
e A exp( ikR)
R
R
a/2
 
x
a
Z
 exp  ik (Yy  Zz) / RdS
Aperture
b/2
 a / 2 b / 2
exp  ik (Yy  Zz ) / R  dydz
ka Z
ka


'


(

sin q a ),
abe A exp( ikR)  sin  '  sin  '  
2 R
2

, 


R
  '   '    '  kb  Y (  kb sin q ).
b

2 R
2
 sin  ' 
I (Y , Z )  I (0)


'


2
 sin  ' 



'


2
14
 sin  ' 
I (Y , Z )  I (0)


'


Y minimum:
Z minimum:
2
 sin  ' 



'


2
 '  kaZ / 2 R,

 '  kbY / 2 R.
2R
kb
2R
 '  kaZ / 2 R   ,  2 , ... Z   m
ka
 '  kbY / 2 R   ,  2 , ... Y   m
15
Read: Ch10: 2
Homework: Ch10: 14,17
Due: April 3
16
March 27 Circular aperture
10.2.5 The circular aperture
Importance in optical instrumentation: The image of
a distant point source is not a point, but a diffraction
pattern because of the limited size of the lenses.
Yy  Zz  q sin    sin   q cos    cos 
 q cos(   )
E (Y , Z ) 
e A exp( ikR)
e A exp( ikR)
R
2
Aperture
0 0 exp  ik q / R cos(   )dd
R
  0 e exp( ikR ) 2 a
 A
0 0 exp  ik q / R cos  dd
R
a
e exp( ikR)
 A
2  J 0 ( kq / R )d
0
R
e exp( ikR)
J ( kaq / R )
 A
2a 2 1
R
kaq / R

4e A2 A2  J 1 ( kaq / R ) 
E 
R 2  kaq / R 
2
2
P(Y,Z)


q
a
q
R

x
z
 exp  ik (Yy  Zz) / RdS
a
Y
y
Z
Bessel functions:.
1 2
exp( iu cos v)dv
2 0
i  m 2
J m (u ) 
exp[ i (mv  u cos v)]dv
2 0
d m
[u J m (u )]  u m J m 1 (u )
du
J (u ) 1
lim 1

u 0
u
2
J 0 (u ) 
2
 2 J (kaq / R) 
 2 J1 (ka sin q ) 
I (q )  I (0)  1

I
(
0
)

 ka sin q 
 kaq / R 
17
2
 2 J (ka sin q ) 
I (q )  I (0)  1
 ka sin q 
2
J0(u)
J1(u)
u
I (q ) / I (0)
q1
0.018
Radius of Airy disk:
kasin q
3.83
J1 (kaq / R)  0  kaq1 / R  3.83
q1  1.22
Rl
fl
,  1.22
for a lens
2a
D
P
D
f
18
Read: Ch10: 2
Homework: Ch10: 25,28
Due: April 3
19
March 30 Resolution of imaging systems
10.2.6.0 Equivalence between the far field and the focal plane diffraction pattern
Two coherent point sources:
P
y
R
a
q
a sinq
E1 (q  y / R )  C1 exp(ikR )  C2 exp[ik ( R  a sin q )]


E2 (q  y '/ f )   C1 exp(ikL)   C2 exp[ik ( L  a sin q )]
L
a
q
a sinq
q
f
P'
y'
 E (q ) 2
2
2
  exp ik ( L  R )    constant
 2
 E1 (q )

 (does not depend on q )

y y' y' f
q   ,

R f
y R


• This applies to any number of arbitrarily distributed point sources in space.
• Far field and focal plane produce the same diffraction pattern, but with different sizes.
• R is replaced by f in the focal plane pattern.
A lens pulls a far-field diffraction pattern to its focal plane, reduces the size by f/R.
20
10.2.6 Resolution of imaging systems
Image size of a circular aperture:
P
D
q1  1.22
q 
f
fl
D
q1
l
 1.22
f
D
Rayleigh’s criterion for bare resolution:
The center of one Airy disk falls on the first
minimum of the other Airy disk.
We can actually do a little better.
Image size of a far point source:
P
D
q 
q1
l
 1.22
f
Dlens
f
Angular limit of resolution:
 min
 1.22
l
D
Overlap of two incoherent point sources:
far away
P2
S1

S2
f
D
q
P1
21
Angular limit of resolution:
 min
Our eyes:  min  550 nm  1'
2 mm
Pupil diameter
 1.22
l
D
About 1/3000 rad
Focal length
150mm
Human cone photoreceptor cells
Spot distance on the retina: 20 mm/3000=6.7mm
Space between human photoreceptor cells on the
retina: 5-7mm.
Pixel size of a CCD camera: ~7.5 mm.
Wavelength dependence: CD DVD
Question: Comparing the circular with
the square aperture, why does the
square aperture produce a smaller
diffraction pattern? (l/D vs. 1.22 l/D)
22
Read: Ch10: 2
No homework
23
April 1, 3 Gratings
Diffraction grating: An optical device with
regularly spaced array of diffracting elements.
Transmission gratings and reflection gratings.
Grating equation: a(sin q m  sin qi )  ml
2
a
a
qm
qm
qi
qi
1
m=0
-1
-2
Blazed grating: Enhancing the energy of a
certain order of diffraction.
Blaze angle: g
Specular reflection: qr  qi  2g  qm
a
g
qi
specular
qr
q0
0th
24
Grating spectroscopy:
Angular width for a spectral line:
N-slit interference
Between two minima, (N-1)/N to (N+1)/N .
2
2 
 sin N 


   
2l

sin

N


  q 
Na cos q m
2 
  (ka / 2)(sin q  sin q i )  q 
ka cos q 

 sin  

I (q )  I 0 



2
q
l
Angular width of a spectral line due to instrumental
broadening. Inversely proportional to Na.
d qm
Angular dispersion:
a(sin q m  sin q i )  ml 
dq m
m

dl a cos q m
dl
25
Limit of resolution:
Barely resolved two close wavelengths: l min
2l
Na cos q m



dq m
m
ml  
Angular separation from dispersion

 q separation 
dl a cos q m
a cos q m 



Rayleigh' s criterion for l min : q width  2q separation

l
Na(sin q m  sin qi )

 mN 
l min
l
Angular w idth of each wavelengt h ( q ) width 
Resolving power:
l
 mN
l min
Question: Why does the resolving power
increase with increasing order number and with
increasing number of illuminated slits?
qwidth
qseparation
l
26
Free spectral range:

a(sin q m  sin q i )  m(l  l )  (m  1)l  llfsr
fsr 
m =3
ll
m
m
fsr
m =2
m =1
l ll
sinqm
In higher order diffraction the spectrum is more spread in angle. This results in a
higher resolving power but a narrower free spectral range.
27
Read: Ch10: 1-2
Homework: Ch10: 32,33,34,39,41
Due: April 10
28