Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Depth of field wikipedia, lookup

Fourier optics wikipedia, lookup

Anti-reflective coating wikipedia, lookup

Airy disk wikipedia, lookup

Camera wikipedia, lookup

Aperture wikipedia, lookup

Retroreflector wikipedia, lookup

Ray tracing (graphics) wikipedia, lookup

Optician wikipedia, lookup

Eyepiece wikipedia, lookup

Cataract wikipedia, lookup

Superlens wikipedia, lookup

Nonimaging optics wikipedia, lookup

F-number wikipedia, lookup

Image stabilization wikipedia, lookup

Schneider Kreuznach wikipedia, lookup

Lens (optics) wikipedia, lookup

Harold Hopkins (physicist) wikipedia, lookup

Optical aberration wikipedia, lookup

Transcript
```Geometric Optics of thick lenses and Matrix methods
Mathematical treatment of refraction
for thick lenses shows the existence
of principal planes in the paraxial
approximation which serve as
reference planes for the refraction of
rays entering and leaving the system.
Symmetry and shape of the lens determine the
location of the principal planes.
After an analytical treatment1 of refraction for a thick lens geometry:
 1 1 nl  1d l 
1
 nl  1 


f
R
R
n
R
R
2
l 1 2 
 1
f nl  1d l
f nl  1d l
and V1 H1  h1 ; V2 H 2  h2 ; h1  
; h2  
R2 nl
R1nl
1 1 1
 
(Gaussian form),
so si f
where
Rough approximation for ordinary glass lenses in air: H1H  V1V2 / 3
2
1Complete
derivation can be found in Morgan,
Introduction to Geometrical and Physical Optics.
Also, the Newtonian form holds: xo xi  f
and
yo
Fig. 6.4 Thick-Lens geometry
MT 
yi
x
f
 i 
yo
f
xo
2
Note that as dl  0, this will yield the thin lens result.
Convention: h1, h2 > 0 then H1, H2 is to the right of V1,V2, and
conversely if h1, h2 < 0 then H1, H2 is to the left of V1,V2
Again, H1 and H2 refer to axial points through the principal planes.
Now, consider a compound lens consisting of two thick lenses L1 and
L2, with the usual parameters so1, si1, f1 and so2, si2 and f2, as shown
on the next slide.
si, so are image and object
distances for the combination
 si1  si 2 
si
  
M T     
as a whole and are measured
s
s
s
o
 o1  o 2 
with respect to H1 and H2.
1 1 1
d
fd
  
, H11H1 
, and
f
f1 f 2 f1 f 2
f2
H 22 H 2  
fd
f1
Note that the sign is important and distances > 0 indicate that H1 or H2 are
to the right of H11 or H22.
Equivalent thick lens representation of a compound lens
Fig. 6.5 A compound thick lens.
Note that if the lenses
are thin, the pairs of
points H11, H12 and
H21, H22 coalesce into a
single point and d
becomes the center to
center lens separation.
Consider the example of a compound thin lens below (individual
lenses are thin). f1 = -30 cm, f2 = 20 cm, d = 10 cm.
Then, the effective focal length is
1 1 1
d
  
f
f1 f 2 f1 f 2
Since these are thin lenses the H11H1  O1 H1 
principal planes converge to
single points O1, O2:

f  30 cm
fd (30)(10)

 15 cm
f2
20
fd
(30)(10)
H 22 H 2  O2 H 2  

 10 cm
f1
 30
A compound thin lens
Analytical Ray
Tracing:

3D:
 

2D:
ni kˆi  uˆn  nt kˆt  uˆn  ni sin i  nt sin t
Example of
a computer
program
for ray
tracing.
Consider a ray tracing analysis using the paraxial approximation, sin  
At point P1: ni1 sin i1  nt1 sin t1
 ni1  i1  1   nt1  t1  1 
 ni1i1  nt1t1
and 1  y1 / R1
 ni1  i1  y1 / R1   nt1  t1  y1 / R1 
Fig. 6.7 Ray Geometry
for thick lenses
n n 
 nt1  t1  ni1 i1   t1 i1  y1
 R1 
n n
Let D1  t1 i1
 nt1 t1  ni1 i1  D1 y1
R1
D1 is called the power of a single refracting surface.
For a thin lens, D  D1  D2  1  nl  1 1  1 


f
 R1
R2 
Also, from the geometry
y2  y1  d 21 t1
This is done for
cosmetic reasons.
where d 21  V2V1
 nt1 t1  ni1  i1  D1 yi1
Thus, in matrix form we can write:
and
yt1  0  yi1  y1
nt1 t1  1  D1  ni1 i1 
 y   0
 y 
1
  i1 
 t1  
Development of Matrix Method
Introduce ray vectors
2  1 column matrix:
 n  
rt1   t1 t1  and
 yt1 
 n  
ri1   i1 i1 
 yi1 
and a 2  2 refraction
matrix:
1  D1 


R1  

r

R
r
t1
1 i1
1 
0
Also ni 2  nt1 ,  i 2   t1
so ni 2 i 2  nt1 t1  0
and
y2  yi 2  d 21 t1  yt1
0 nt1 t1 
ni 2 i 2   1
 


 y 
y
d
/
n
1
 i 2   21 t1   t1 
Thus, we can define a 2  2
transfer matrix:
ray at P2 ray at P1



Thus ri 2  T21rt1  T21 R1ri1
0
 1
T21  

d
/
n
1
 21 t1 
Note : T21 R1  T21 R1  1
Continuing with the second interface in the figure (Fig. 6.7)
1  D2 
nt 2  ni 2
with R2  
and D2 

1 
R2
0


so rt 2  R2 T21 R1ri1
Define the system matrix : A  R2 T21 R1


rt 2  R2 ri 2
Note A  1
Let
d 21  d l
and
nt1  nl
Note that the determinant must be 1 and is a check of the system matrix.
After multiplying out the system matrix, its components can be written
explicitly:
 D2 d l  
D1 D2 d l 
   D1  D2 

1 
a 
a
nl  
nl 
A   11 12   

 D1d l 
dl
a21 a22  
1 



n
n


l
l 

Where d21= dl is the lens thickness and the refractive index of the
lens is nt1 = nl.
Note that an examination of the system matrix A gives
The lens is taken to
D1 D2 d l
nl  1
1  nl be in air, as
 a12  D1  D2 
, D1 
, D2 
nl
R1
R2 represented by the
 1 1 nl  1d l 
  a12  nl  1 


R
R
R
R
n
2
1 2 l 
 1
powers D1 and D2.
We observe that this is just the reciprocal of the focal length of a thick
lens such that –a12=1/f , and the lens power is 1/f . More generally, if
the media are different on both sides we would have:
ni1
nt 2
ni1 1  a11 
nt 2 a22  1
a12  

; Finally , V1 H1 
and V2 H 2 
fo
fi
 a12
 a12
Thus, the matrix method involving 2  2 refraction and transfer matrices
enables a determination of fundamental optical system parameters such as
the system focal lengths and position of both principal planes relative to
the lens vertices.
image
ray
object
ray
The first operator T10 transfers
the reference point from the
object (i.e., PO to P1).
The next operator A21 then
carriers the ray through the
lens.
A final transfer operator TI2
brings it to the image plane, PI.
T = transfer matrix
R = refraction matrix
A = system matrix
Example of a complex lens system
analyzed with the Matrix Method:
A71  R7 T76 R6 T65 R5 T54 R4 T43 R3 T32 R2 T21 R1
0
0
 1
 1
0
.
357
0
.
189
; T32  

T21  
1
1.6116 1
 1

1.6116  1

0
 1
1

; R1  
T43   0.081
1.628 
1
0

1.6053 
1


1  1.6116 
1.6053  1


1

1

R2  
 27.57 ; R3  
 3.457 
0

0

1
1




0.848  0.198
 A71  

1.338 0.867 
Fig. 6.10 A Tessar lens
system.
The result of the matrix
method easily allows for
the solution of the basic
lens parameters such as
the focal length and
position of the principal
planes relative to the
vertices of the outer
lenses.
f  5.06
V1 H1  0.77
V7 H 2  0.67
As another example, consider a system of thin lenses in which dl  0. Note
that the power of a thin lens is D1 + D2 = D. Then
1  D1  D2  1  D 1  1 / f 
A


(thin lens )



1
1 
0
 0 1  0
Suppose that two thin lenses are separated by distance d:
H2
H1
O1
O2
d
Then, the system matrix can be written as
1  1 / f 2   1 0 1  1 / f1  1  d / f 2 
A






d
1  d 1 0
1  
0
 1 / f1  d / f1 f 2  1 / f 2 

 d / f1  1

Remember that
ni1
nt 2
a12  

;
fo
fi
ni1 1  a11 
nt 2 a22  1
V1 H1 
and V2 H 2 
 a12
 a12
Where ni1 = nt2 =1 and V1 = O1 and V2 = O2 for a thin lens
Therefore  a12 
1 1 1
d
  
,
f
f1 f 2 f12
O1 H1 
fd
 fd
, O2 H 2 
f2
f1
Note that the locations of the principal planes H1 and H2 strongly depend
on d, which can affect on which side of the lenses the planes are located. It
is worth noting that a lens system composed of N thin lenses can easily be
treated in the same manner for calculating the focal lengths and locations
of the principal planes.
1 2 3 ………N
A similar analysis can be performed for a mirror in the
paraxial approximation. The result is
  1  2n / R 
Mo  

0
1




so rr  M o ri
remembering
 n i   n r 
with ri  
, rr   y 
y
 r 
 i 
Note that for a plane mirror
R   and the system matrix
for a mirror reduces to
  1 0
M|  

 0 1
2
f 
R
Lens Aberrations: Deviations from the corresponding paraxial approximation
Chromatic Aberrations: n() and ray components having different colors have
different effective focal lengths
Monochromatic Aberrations: Spherical aberration, coma, astigmatism
Recall that we used sin    (first order theory in the paraxial approx.)
Including addition terms in sin    -  3/3! leads to the third-order theory
which can explain the monochromatic aberrations.
Remember that for a single refracting spherical interface in the 1st order approx:
n1 n2 n2  n1
 
 D1
so si
R
If the approximation for the OPL (lo + li) are improved, the 3rd order treatment
gives:
2
2

n1 n2 n2  n1
n1  1 1 
n2  1 1  
2
   
   
 
h 
so si
R
 2so  so R  2si  R si  
Where h is the distance above the optical axis as shown in the figure.
Rays striking the surface at a greater distance (marginal rays) are focused
closer to the vertex V than are the paraxial rays and creates spherical
aberration.
Marginal rays are bent too much and
focused in front of paraxial rays.
Distance between the intersection of
marginal rays and the paraxial
focus, Fi, is known as the LSA
(longitudinal spherical aberration).
Note: SA is positive for convex lens
and negative for a concave lens.
TSA (transverse SA) is the
transverse deviation between the
marginal and paraxial rays on a
screen placed at Fi.
If the screen is moved to the
position LC the image blur will
have its smallest diameter, known as
the “circle of least confusion,”
which is the best place to observe
the image.
Rule of thumb: Incident ray will undergo a minimum deviation when i  T.
Remember the dispersing prism:
i
1T
2T
For an object at , the
round side of lens facing
the object will suffer a
minimum amount of SA.
Note that a planarconvex lens can be
approximated as two
prisms.
2T > 1T and the
lower prism results in
a greater deviation.
Fig. 6.16 Spherical Aberration for a planarconvex lens in both orientations.
Similarly if the object and image are to be
nearly equidistant from the lenses (so = si =2f),
an Equi-Convex shaped lens minimizes SA.
Marginal rays give smaller
image  negative coma
f
2f
Marginal rays give larger
image  positive coma
2f
f
Coma (comatic aberration) is associated with the
fact that the principle planes are really curved
surfaces resulting in a different MT for both
marginal and central rays.
Since MT = -si/so , the curved nature of the
principal surface will result in different effective
object and image distances, resulting in different
transverse magnifications. The variation in MT
also depends on the location of the object which
can result in a negative (a) or positive coma (b)
and (c), as demonstrated in the left figure.
The imaging of a point at S can result in a
“comet-like” tail, known as a coma flare and
forms a “comatic” circle on the screen 
(positive coma in this case). This is often
considered the worst out of all the aberrations,
primarily because of its asymmetric
configuration.
Astigmatism:
The Meridional Plane contains the
chief ray which passes through the
center of the aperture and the
optical axis.
The Sagittal Plane contains the
chief ray and is perpendicular to
the meridional plane.
Fermat’s principle shows that
planes containing the tilted rays
will give a shorter focal length,
which depends on the (i) power of
the lens and the (ii) angle of
inclination. The result is that there
is both a meridional focus FT and a
sagittal focus FS.
Tilted rays have a
shorter focal length.
Astigmatism: Note that the cross-section of the beam changes from a
circle (1)  ellipse (2)  line (primary image 3) ellipse (4)  circle
of least confusion (5)  ellipse (6)  line (secondary image 7) .
Focal length difference FS-FT depends on
power D of lens and angle of rays.
Chromatic Aberrations: Since the index depends on the wavelength then
we can expect that the focal length will depend on the wavelength.
1 1 
1
 nl  1  
f
 R1 R2 
f

nl  nl ( )
Circle of least confusion
```
Related documents