Transcript PowerPoint Presentation - Lecture 4 Propagation of light

Part 2: Phase structure function, spatial
coherence and r0
Definitions - Structure Function and
Correlation Function
• Structure function: Mean square difference
D (r )   (x )   (x  r )
2



dx

(x
)


(x

r
)

• Covariance function: Spatial correlation of a
random variable with itself
B (r )   (x  r ) (x ) 


dx  (x  r ) (x )

2

Relation between structure function and
covariance function
D (r )  2B (0)  B (r )
To derive this relationship, expand the product in the
definition of D ( r ) and assume homogeneity to take
the averages
Definitions - Spatial Coherence Function
For light wave  exp[i(x)], phase is(x)  kz  t
• Spatial coherence function of field is defined as
r
r * r r
C (r )  ( x) ( x  r )
Covariance for complex fn’s
C (r) measures how “related” the field  is at one
position x to its values at neighboring positions x + r .
Do not confuse  the complex field with its phase 
•
Now evaluate spatial coherence
function C (r)
• For a Gaussian random variable  with zero mean,
•

exp i  exp   2 / 2

• So
r
r
r r
C (r )  expi[ ( x)   ( x  r )]
r
r r 2
r


 exp   ( x)   ( x  r ) / 2  exp  D (r ) / 2 


• So finding spatial coherence function C (r) amounts to evaluating
the structure function for phase D ( r ) !
Next solve for D ( r ) in terms of the
turbulence strength CN2
• We want to evaluate
C (r )  exp D (r )/2
• Remember 
that
r
r

D (r )  2  B (0)  B (r ) 
Solve for D ( r ) in terms of the
turbulence strength CN2, continued
• But
r
 ( x)  k
h  h

v
dz  n( x, z)
for a wave propagating
h
vertically (in z direction) from height h to height h + h.
Here n(x, z) is the index of refraction.
• Hence
B (r )  k
2
h h
h h
h
h
 dz  dz n(x, z)n(x  r, z)
Solve for D ( r ) in terms of the
turbulence strength CN2, continued
z  z   z
• Change variables:
• Then
B (r )  k 2
 k2
h h
h h z 
h
h z 
h h
h h z 
 dz  dz
 dz  dz B
N

(r ,z)
h z 
h
B (r )  k 2h
n(x, z)n(x  r , z z)
h h z 

h z

2
dzB
(r
,z)

k
h  dzBN (r,z)
 N
Solve for D ( r ) in terms of the
turbulence strength CN2, continued
• Now we can evaluate D ( r )

D (r )  2 B (0)  B (r ) 2k 2h  dz BN (0,z)  BN (r ,z)


D (r )  2k 2h  dz  BN (0,0)  BN (r ,z)  BN (0,0)  BN (0,z) 


D (r )  k 2h  dz DN (r ,z)  DN (0,z)

Solve for D ( r ) in terms of the
turbulence strength CN2, completed
• But
r
r
DN (r )  CN2 r
2/3
r
D (r )  k 2 hCN2

C
2
N
r

2
z
 r 2  z2
dz
 

2 1/ 3

1/ 3
so
 z2 / 3 


2 (1/ 2)(1/ 6)r 5 / 3  2.914r5 / 3
5
(2 / 3) 
•

r
2 5/3 2
2 5/3
2
D (r )  2.914 k r CN h  2.914 k r  dhCN (h)
0
Finally we can evaluate the spatial
coherence function C (r)

 1

r
r
2 5/3
2
C (r )  exp  D (r ) / 2   exp    2.914 k r  dhCN (h) 
 
 2 
0
For a slant path you can add
factor ( sec  )5/3 to account for
dependence on zenith angle 
Concept Question: Note the scaling of the coherence
function with separation, wavelength, turbulence
strength. Think of a physical reason for each.
Given the spatial coherence function, calculate
effect on telescope resolution
• Define optical transfer functions of telescope,
atmosphere
• Define r0 as the telescope diameter where the two
optical transfer functions are equal
• Calculate expression for r0
Define optical transfer function (OTF)
• Imaging in the presence of imperfect optics (or
aberrations in atmosphere): in intensity units
Image = Object  Point Spread Function
convolved with
I = O  PSF   dx O( x - r ) PSF ( x )
• Take Fourier Transform: F ( I ) = F (O ) F ( PSF )
• Optical Transfer Function is Fourier Transform of PSF:
OTF = F ( PSF )
Examples of PSF’s and their
Optical Transfer Functions
Seeing limited OTF
Intensity
Seeing limited PSF
Intensity
l/D
l / r0

Diffraction limited PSF
l/D
l / r0

r0 / l
-1
D/l
Diffraction limited OTF
r0 / l
D/l
-1
Now describe optical transfer function of the
telescope in the presence of turbulence
• OTF for the whole imaging system (telescope plus atmosphere)
S(f)=B(f)T(f)
Here B ( f ) is the optical transfer fn. of the atmosphere and T ( f)
is the optical transfer fn. of the telescope (units of f are cycles
per meter).
f is often normalized to cycles per diffraction-limit angle (l / D).
• Measure the resolving power of the imaging system by
R =  df S ( f )
=
 df B ( f )  T ( f )
Derivation of r0
• R
of a perfect telescope with a purely circular aperture of (small)
diameter d is
R
=
 df T ( f ) = ( p / 4 ) ( d / l )
2
(uses solution for diffraction from a circular aperture)
• Define a circular aperture r0 such that the R of the telescope (without
any turbulence) is equal to the R of the atmosphere alone:
 df B ( f ) =  df T ( f )  ( p / 4 ) ( r0 / l )2
Derivation of r0 , continued
• Now we have to evaluate the contribution of the
atmosphere’s OTF:
 df B ( f )
• B ( f ) = C ( l f ) (to go from cycles per meter to cycles
per wavelength)

 1

r
2 5/3
2
C (r )  exp    2.914 k r  dhCN (h) 
 
 2 
0
B( f )  C (lf )  exp Kf
5/ 3
Derivation of r0 , continued
• Now we need to do the integral in order to solve for r0 :
( p / 4 ) ( r0 / l )2 =
 df B ( f ) =  df exp (- K f 5/3)
(6p / 5) (6/5) K-6/5
• Now solve for K:
K = 3.44 (r0 / l )-5/3
B ( f ) = exp - 3.44 ( lf / r0 )5/3 = exp - 3.44 (  / r0 )5/3
Replace by r
Derivation of r0 , concluded
r 5/ 3 1
2 5 /3
2

3.44

2.914k
r
sec

dhC


 
N (h)
2
r0 

r0  0.423k sec 
2
 dhC (h)
2
N
5/ 3

D (r)  6.88(r / ro )
3 / 5
Scaling of r0
• r0 is size of subaperture, sets scale of all AO correction
-3/5
3 / 5


2
2
r0  0.423k sec   CN (z)dz


0
H
 l
6 /5
sec  
3 / 5
 2

 CN (z)dz


• r0 gets smaller when turbulence is strong (CN2 large)
• r0 gets bigger at longer wavelengths: AO is easier in the IR than with visible
light
• r0 gets smaller quickly as telescope looks toward the horizon (larger zenith
angles  )
Typical values of r0
• Usually r0 is given at a 0.5 micron wavelength for
reference purposes. It’s up to you to scale it by l-1.2 to
evaluate r0 at your favorite wavelength.
• At excellent sites such as Paranal, r0 at 0.5 micron is 10
- 30 cm. But there is a big range from night to night, and
at times also within a night.
• r0 changes its value with a typical time constant of 5-10
minutes
Phase PSD, another important parameter
• Using the Kolmogorov turbulence hypothesis, the
atmospheric phase PSD can be derived and is
0.023 11/ 3
( k )  5 / 3 k
r0
• This expression can be used to compute the amount of
phase error over an uncorrected pupil
D
  1.03 
 r0 
2
5/3
Units: Radians of phase / (D / r0)5/6
Tip-tilt is single biggest contributor
Focus,
astigmatism, coma
also big
Reference: Noll76
High-order terms
go on and on….