Lecture Note No. 5

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Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



1

Engineering

Optics and Optical Techniques

Lecture Note No. 5


Ray Tracing for Thick Lens (Secs. 6.1, 6.2)

Aberrations (Sec.
6.3)

Reading Assignment: Secs. 6.4, 6.5


FOCAL LENGTH OF THICK LENS























After
a grea
t deal of algebraic manipulation
, the focal length measured from the
principal plane
, i.e.,
1
h
ffl
f



or
2
h
bfl
f



is given by:









i
o
l
l
l
l
s
s
R
R
n
d
n
R
R
n
f
1
1
1
1
1
1
1
2
1
2
1























l
l
l
l
l
l
n
R
d
n
f
h
n
R
d
n
f
h
1
2
2
1
1
1








(see E
-
o
-
C Prob. 6.18)





o
i
o
i
o
i
T
x
f
f
x
s
s
y
y
M











Note:
i
o
s
s
f
,
,

are measured w.r.t. the Principal Plane

Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



2

Example:

For a lens with
R
1

= 20 cm,
R
2

=
-
40 cm,
d
l

= 1 cm,
n
l

= 1.5, and
an object located
30 cm

from
V
1
, calculate
f
,
h
l
,
h
2
,
s
i
, by using both the

“thin” lens equati
on and the
“thick” lens
equation.
Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



3

RAY TRACING





















For the
surface “1”
, Snell’s Law
(
1
1
1
1
sin
sin
t
t
i
i
n
n



)
applied for
paraxial

(



sin
)
region:







1
1
1
1
1
1
1
1
1
1










t
t
i
i
t
t
i
i
n
n
n
n



Substituting
1
1
1
1
/
sin
R
y





gives the
refraction equation
:


1
1
1
1
1
1
i
i
i
t
t
y
D
n
n






where

1
1
1
1
R
n
n
D
i
t



(Power of the refraction surface “1”)






Note:
0

D

as


R

and
0


n








D

as
0

R





The
transfer equation

is:




1
1
0
i
t
y
y




Now the matrix representation of the two equations is:























1
1
1
1
1
1
1
1
0
1
i
i
i
t
t
t
y
n
D
y
n




or

r
t1

(transmitted ray vector from “1”)

=
R
1

(refraction matrix of the surface “1”) x
r
i1

(in
cident ray vector to “1”)

… (A)

Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



4

For the
inside lens “1 to 2”
,





1
1
21
2
1
1
2
2
0
t
t
i
t
t
i
i
y
d
y
n
n
































1
1
1
1
21
2
2
2
1
/
0
1
t
t
t
t
i
i
i
y
n
n
d
y
n





or

r
i2

(incident ray vector to “2”)

=
T
21

(transfer matrix across “1
-
2”) x
r
t1

(transmitted ray vector from “1”)

… (B)


For thin lens

approximation (
d
21

= 0),
T
21

becomes a unit matrix






1
0
0
1
.




For the
surface “2,”




r
t2

(transmitted ray vector to “2”)

=
R
2

(refraction matrix of the surface “2”) x
r
i2

(incident ray vector from “2”)

… (C)



and






















1
0
1
2
2
2
2
2
R
n
n
D
R
i
t


By combining

(A), (B) and (C) with the use of
l
i
t
t
i
l
n
n
n
,
.
n
n
,
d
~
d




2
1
2
1
21
0
1
, the
whole lens system is expressed as:




r
t2

=
R
2

T
21

R
1
r
i1

=
Ar
i1


Where the
lens
system matrix
A

(
D
1
,
D
2
,
d
l
,
n
l
)=

















l
l
l
l
l
l
l
l
n
d
D
n
d
n
d
D
D
D
D
n
d
D
1
2
1
2
1
2
1
1
,


det
A

= 1

Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



5

Example

(Ref. To Fig. 5.31 on p. 170):















0
.
1
,
5
.
1
,
2
,
2
43
21




m
l
n
n
cm
d
cm
d







d
32

= 10 cm

f
1

=
-
30 cm

f
2

= 20 cm

Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



6

MIRRORS

















i
r
i
i
i
i
R
y





2







*
R
: + if
c

is right of
V
,

-

if
c

is left of
V




Combining these gives


R
/
ny
n
n
R
/
y
i
i
r
i
i
r
2
2
















And

i
r
y
y



The matrix expression is given as:






























i
i
r
r
y
n
R
n
y
n


1
0
2
1


Or equivalently,

r
f

=
M

r
i


r
f

(reflected ray vector) =
M

(mirror matrix)
r
i
(incident ray vector)




For a
flat mirror

(


R
,
n

= 1.0),






M

=







1
0
0
1
,

r
f

=






r
r
y

,


r
i

=






i
i
y






Therefore,
i
r





and
i
r
y
y


Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



7


Aberrations:

Departures from the idealized conditions of Gaussian or ray optics
.



Becomes more
substantial in the n
on
-
paraxial region
.



MONOCHROMATIC ABERRATIONS:


The lens formula, Eq. (5.8), is given as


R
n
n
s
n
s
n
i
o
1
2
2
1




which is valid for
paraxial region,
0


.




Considering refraction at a spherical surface shown in F
ig. 5.6 and Eq. (5.5) on p. 154, and using

!
2
1
cos
2





and
!
sin
3
3






for small

, Eq. (5.8) is extended to

































2
2
2
1
2
1
2
2
1
1
1
2
1
1
2
i
i
o
o
i
o
s
R
s
n
R
s
s
n
h
R
n
n
s
n
s
n



where the height
h

is the distance of ray measured w.r.t. the optical axi
s.



The focal length
f



i
s
s
o



lim

decreases with increasing
h
:
Spherical aberrations




Heterodyning of beams:










(single
-
refraction)



(double refraction)



Coma …
The dependence of
M
T

on
h
, the marginal rays are blurred

(
Fig. 6.21
).


Astigmatism …
Blurred collimation point of oblique rays again because of the focal length
difference (
Figs. 6.26 and 6.27
).


Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



8

CHROMATIC ABERRATIONS

[
n

increases with

]
:

Since



l
n

increases with decre
asing

, the thin lens equation,













2
1
1
1
1
1
R
R
n
f
l
, shows
that
the focal length decreases with decreasing

. Thus, the focal length of blue ray is smaller
than the focal length of red ray, i.e.,
R
B
f
f


and
R
B
n
n


(
Figs. 6.36 and 6.37
).



Achromatic Doublet Lens to Reduce CA:

High
-
quality camera optics (p.220) or high
-
quality binoculars (p.226)



Thin lens equation:




1
1
1
1
1
1
1
2
1























l
m
l
i
o
n
R
R
n
n
f
s
s



where the composite curvature is
defined as










2
1
1
1
R
R


and



1
1
1



l
n
f

















2
2
1
1
2
1
1
1
1
1
1








n
n
f
f
f




where










12
11
1
1
1
R
R


and











22
21
2
1
1
R
R




For Red:





2
2
1
1
2
1
1
1
1
1
1








R
R
R
R
R
n
n
f
f
f



For Blue:





2
2
1
1
2
1
1
1
1
1
1








B
B
B
B
B
n
n
f
f
f


Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



9



Let
B
R
f
f
1
1


(Achromatic condition):

R
B
R
B
n
n
n
n
1
1
2
2
2
1








For Yellow:





2
2
2
1
1
1
1
1
1
1






Y
Y
Y
Y
n
f
&
n
f




And





Y
Y
Y
Y
f
n
f
n
1
1
2
2
2
1
1
1









Combining the two “blue
-
boxed” equations gives:











1
1
1
2
1
1
1
2
2
2
1
2
1
1










V
V
n
/
n
n
n
/
n
n
f
f
Y
R
B
Y
R
B
Y
Y


or

0
2
2
1
1


Y
Y
f
V
f
V




where
2
1
V
,
V
: Abbe numbers











R
B
Y
n
n
n
1



and
1
2
1
1


V
,
V

: Dispersive Power











1
Y
R
B
n
n
n
.





Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



10

Fraunhofer Lines

(Table 6.1):


… Reference spectral lines generated by different substances such as H, Na, He, …



C





Red

D

or
d





Yellow

b

or
c




Green

F





Blue

F
,
g

or
K




Violet



Abbe Number
V

for
d

(Yellow) is defined as:






C
F
d
d
n
n
n
V



1

(Fig. 6
-
39),
and
0
2
2
1
1


d
d
d
d
f
V
f
V








d
d
d
f
f
f
2
1
1
1
1






Combining the two “green
-
boxed
” equations gives:






d
d
d
d
d
d
d
d
d
d
V
V
f
V
f
V
V
f
V
f
1
2
2
2
2
1
1
1
1
1







Engineering Optics and Optical Techniques

-

Spring 2007

Lecture Note No. 5, prepared by Prof
. Kenneth D. Kihm



11

[Example]

Design an achromatic lens of
d
f
= 0.5 m choosing BK1 material [
C
n

= 1.50763,
d
n

= 1.51009,
F
n

= 1.51566] for lens 1
(equi
-

or double
-
convex) and F2 material [
C
n

=
1.61503,
d
n

= 1.62004,
F
n

= 1.63208] for lens 2 (concave).













d
V
1

= 63.46,
d
V
2

= 36.37




*
C
F
d
d
n
n
n
V



1




d
f
1

= 0.21344,
d
f
2

=
-
0.37242



*




d
d
d
d
d
d
d
d
d
d
V
V
f
V
f
V
V
f
V
f
1
2
2
2
2
1
1
1
1
1




































11
1
12
11
1
1
1
1
2
1
1
1
1
1
1
R
n
R
R
n
n
f
d
d
d
d

;




12
11
R
R

0.2177 m


































22
12
2
22
21
2
2
2
2
1
1
1
1
1
1
1
1
R
R
n
R
R
n
n
f
d
d
d
d

;



21
R

=
-
0.2177 m &

22
R
-
3.819 m



Homework Assignment #5

E
-
o
-
C. Problems 6.9,

6.13, 6.15, 6.17, 6.22, 6.23,
and 6.24

Due by Tuesday of
February 2
0,

2006