Transcript Acknowledgments

Engineering Optics
Understanding light?
•
•
•
•
•
Reflection and refraction
Geometric optics (l << D): ray tracing, matrix methods
Physical optics (l ~ D): wave equation, diffraction, interference
Polarization
Interaction with resonance transitions
Optical devices?
• Lenses
• Mirrors
• Polarizing optics
• Microscopes and telescopes
• Lasers
•
Fiber optics
• ……………and many more
Engineering Optics
In mechanical engineering, a primary motivation for studying
optics is to learn how to use optical techniques for making
measurements. Optical techniques are widely used in many areas
of the thermal sciences for measuring system temperatures,
velocities, and species on a time- and space-resolved basis. In
many cases these non-intrusive optical devices have significant
advantages over physical probes that perturb the system that is
being studied.
Lasers are finding increasing use as machining and manufacturing
devices. All types of lasers from continuous-wave lasers to lasers
with femtosecond pulse lengths are being used to cut and process
materials.
Huygen’s Wavelet Concept
Collimated Plane
Wave
Spherical Wave
At time 0, wavefront is
defined by line (or curve)
AB. Each point on the
original wavefront emits a
spherical wavelet which
propagates at speed c
away from the origin. At
time t, the new wavefront
is defined such that it is
tangent to the wavelets
from each of the time 0
source points. A ray of
light in geometric optics is
found by drawing a line
from the source point to
the tangent point for each
wavelet.
Geometric Optics: The
Refractive Index
The refractive index n: fundamental property of all optical
systems, a measure of the effective speed of
propagation of light in a medium
c
n
v
The optical path length in a medium is the integral of the
refractive index and a differential geometric length:
ds
OPL 

b
a
n ds
a
b
Fermat’s Principle: Law of Reflection
Fermat’s principle: Light rays will travel from point A to point B in a
medium along a path that minimizes the time of propagation.
Law of reflection:
OPLAB  n
(x3, y3)
  x1    y2  y1 
2
2
  x3    y3  y2 
n
2
2
Fix x1 , y1 , x3 , y3
dOPLAB
0
dy2
qr
(0, y2)
qi
0 
y
n
1
2  y2  y1 
2
x
1
2  y3  y2  1
2
 x3    y3  y2 
 x1    y2  y1 
n  y3  y2 
n  y2  y1 

2
2
2
2
 x3    y3  y2 
 x1    y2  y1 
2
0  n sin qi  n sin q r
(x1, y1)

n
 sin qi  sin q r
2
2
2
Fermat’s Principle: Law of Refraction
OPLAB  ni
Law of refraction:
 x2  x1    y1 
2
2
 x3  x2     y3 
 nt
2
2
Fix x1 , y1 , x3 , y3
(x1, y1)
A
d  OPLAB 
0
dy2
y
qi
x
0
(x2, 0)
ni
nt
ni
1
2  x2  x1 
2
(x3, y3)
1
2  x3  x2  1
2
 x2  x1    y1 
 x3  x2    y3 
ni  x2  x1 
nt  x3  x2 

2
2
2
2
 x2  x1    y1 
 x3  x2    y3 
2
0  ni sin qi  nt sin qt
 ni sin qi  nt sin qt
qt

nt
2
2
2
Imaging by an Optical System
Light rays are emitted in all directions or reflected diffusely from an object
point. Spherical wavefronts diverge from the object point. These light rays
enter the (imaging) optical system and they all pass through the image
point. The spherical wavefronts converge on a real image point. The
optical path length for all rays between the real object and real image is the
same.
Later we will discuss scattering, aberrations (a geometric optics concept),
and diffraction (a physical optics concept) which cause image degradation.
Imaging by Cartesian Surfaces
Consider imaging of
object point O by the
Cartesian surface S.
The optical path
length for any path
from Point O to the
image Point I must be
the same by Fermat’s
principle.
no do  ni di  no so  ni si  constant
no x 2  y 2  z 2  ni
 so  si  x 
 no so  ni si  constant
2
 y2  z2
The Cartesian or perfect
imaging surface is a
paraboloid in three
dimensions. Usually,
though, lenses have
spherical surfaces because
they are much easier to
manufacture.
Reflection at Spherical Surfaces I
Reflection from a spherical
convex surface gives rise to
a virtual image. Rays
appear to emanate from
point I behind the spherical
reflector.
Use paraxial or small-angle
approximation for analysis of optical
systems:
3
5
sin    
cos   1 

3!
2
2!



5!
4
4!



1
Reflection at Spherical Surfaces II
Considering Triangle OPC and
then Triangle OPI we obtain:
q  
2q     
Combining these relations we obtain:
      2
Again using the small angle approximation:
  tan  
h
s
   tan   
h
s
  tan  
h
R
Reflection at Spherical Surfaces III
Now find the image distance s'
in terms of the object distance
s and mirror radius R:
h h
h
  2
s s
R

1 1
2
 
s s
R
At this point the sign convention in
the book is changed and the
imaging equation becomes:
1 1
2
 
s s
R
The following rules must be
followed in using this equation:
1. Assume that light
propagates from left to right.
Image distance s is positive
when point O is to the left of
point V.
2. Image distance s' is positive when I is
to the left of V (real image) and negative
when to the right of V (virtual image).
3. Mirror radius of curvature R is positive
for C to the right of V (convex), negative
for C to left of V (concave).
Reflection at Spherical Surfaces IV
s
R
f 
2
The focal length f of the spherical mirror surface is defined as –R/2, where R is
the radius of curvature of the mirror. In accordance with the sign convention of
the previous page, f > 0 for a concave mirror and f < 0 for a convex mirror.
The imaging equation for the spherical mirror can be rewritten as
1 1
1
 
s s
f
Reflection at Spherical Surfaces V
2
1
O'
3
I'
O
V
I F
Ray 1: Enters from O' through C, leaves along same path
Ray 2: Enters from O' through F, leaves parallel to optical axis
Ray 3: Enters through O' parallel to optical axis, leaves along
line through F and intersection of ray with mirror surface
s  7 cm
s 
R   8 cm
f  R/2 
m
s

s
1
1
1

 
s
f
s
C
Reflection at Spherical Surfaces VI
1
O'
2
3
O
C
I F
V
I'
s   17 cm
s 
R   8 cm
m
f  R/2 
s

s
1
1
1

 
s
f
s
Reflection at Spherical Surfaces VII
Real, Inverted Image
s f

m  
s
0
s
1
1
1

  0
s
f
s
Virtual Image, Not Inverted
s f

m  
s
0
s
1
1
1

  0
s
f
s
Geometrical Optics
Index of refraction for transparent optical materials
• Refraction by spherical surfaces
• The thin lens approximation
• Imaging by thin lenses
• Magnification factors for thin lenses
• Two-lens systems
Optical Materials
Source: Catalog, CVI Laser Optics and Coatings.
Optical Materials
Source: Catalog, CVI Laser Optics and Coatings.
Optical Materials
Source: Catalog, CVI Laser Optics and Coatings.
Optical Materials
Source: Catalog, CVI Laser Optics and Coatings.
Refractive
Index of
Optical
Materials
Source:
Catalog, CVI
Laser Optics
and Coatings.
Refractive Index of Optical Materials
Source:
Catalog, CVI
Laser Optics
and Coatings.
Refraction by Spherical Surfaces
At point P we apply the law of
refraction to obtain
n1 sin q1  n2 sin q2
Using the small angle
approximation we obtain
n1q1  n2 q2
Substituting for the angles q1 and
q2 we obtain
n1      n2    
Neglecting the distance QV and
writing tangents for the angles gives
n2 > n1
 h h
h h 
n1     n2   
 s R
 s R 
Refraction by Spherical Surfaces II
Rearranging the equation we
obtain
n1
n2
n1  n2


s
s
R
Using the same sign convention as
for mirrors we obtain
n1
n2
n2  n1


s
s
R
n2 > n1
Refraction at Spherical Surfaces III
O'
I
q1
O
V
q2
C
I'
s  7 cm
R   8 cm
n1
n
n n
 2  2 1
s
s
R

n1  1.0
n2  4.23
The Thin Lens Equation I
n1
n1
n2
O'
C1
O
V1
C2
V2
For surface 1:
n1
n
n n
 2  2 1
s1
s1
R1
s1
t
s'1
The Thin Lens Equation II
For surface 1:
For surface 2:
n1
n2
n2  n1


s1
s1
R1
n2
n1
n1  n2


s2
s2
R2
Object for surface 2 is virtual, with s2 given by:
s2  t  s1
For a thin lens:
t
s2 , s1
 s2   s1
Substituting this expression we obtain:
n1
n
n
n
n
n
n n
n n
 2  2  1  1  1  2 1  1 2
s1
s1
s1
s2
s1
s2
R1
R2
The Thin Lens Equation III
Simplifying this expression we obtain:
n2  n1   1

1
1
1 






s1
s2
n1  R1 R2 
For the thin lens:
s  s1
s  s2

n2  n1   1

1
1
1 





s
s
n1
R
R
2 
 1
The focal length for the thin lens is found by setting s = ∞:
s

n2  n1   1

1
1
1 





s
f
n1  R1 R2 
The Thin Lens Equation IV
In terms of the focal length f the thin lens equation becomes:
1
1
1


s
s
f
The focal length of a thin lens is >0 for a convex lens and <0 a
concave lens.
Image Formation by Thin Lenses
Convex
Lens
m  
Concave
Lens
s
s
Image Formation by Convex Lens
Convex Lens, focal length = 5 cm:
ho
F
RI
F
1
1 1


s
f
s
m   s s 
hi
f   5 cm
s   9 cm 
s 
Image Formation by Concave Lens
Concave Lens, focal length = -5 cm:
ho
hi
F
1
1 1


s
f
s
m   s s 
F
VI
f   5 cm
s   9 cm 
s 
Image Formation Summary Table
Image Formation Summary Figure
Image Formation: Two-Lens System I
60 cm
1
1
1 s1  f1



s1
f1 s1
s1 f1
1
1
1


s2
f 2 s2
m  m1 m2 
f1   15 cm
f 2   15 cm
s2 
s1   25 cm 
s1 

s2 
Image Formation: Two-Lens System II
7 cm
1
1
1


s1
f1 s1
f1   3.5 cm
s1   5.2 cm 
1
1
1


s2
f 2 s2
f 2   1.8 cm
s2 
m  m1 m2 
s1 

s2 
Matrix Methods
• Development of systematic methods of analyzing
optical systems with numerous elements
• Matrices developed in the paraxial (small angle)
approximation
• Matrices for analyzing the translation, refraction,
and reflection of optical rays
• Matrices for thick and thin lenses
• Matrices for optical systems
• Meaning of the matrix elements for the optical
system matrix
• Focal planes (points), principal planes (points), and
nodal planes (points) for optical systems
• Matrix analysis of optical systems
Translation Matrix
1   0
y1  y0  L tan  0  y0  L  0
y1  1 y0   L   0
1   0  y0  1  0
 y1  1 L   y0 
   0 1    
  0
 1 
1 x1  x0   y0 
0
1   0 

Refraction Matrix
  q  q 
  q   q 
q 
y
R
y
R
y
R
Paraxial Snell ' s Law :
  q
n q  nq 
y
y  n 
y y
n
  q         

R  n 
R  n  
R R
1n 
n

1
y

  
  
Rn

n 
Refraction Matrix
y  1 y   0  
1  n

n
      1  y    
 n 
 R  n  
1
0 


y
  
 y

    1  n  1  n    
          
  n 
R  n
Concave surface : R  0
Convex surface : R  0
Reflection Matrix
  q    q 
y
R
Law of Reflection :
  q 
  q   q 
q  q
y
y
2
 q   y
R
R
R
y   1 y   0  
2
 y  1 
R
  
1
 y 
    2
 
R
0
  y
1   

y
R
q 
y
R
Thick Lens Matrix I
Refraction at first surface :
 1
 y1  
    n  nL
 1
 nL R1
Translation from 1st surface to 2nd surface :
Refraction at second surface :
0
 y0 
 y0 


M
n  
1
 0 
0 

nL 
 y2  1 t   y1 
 y1 


M
2
  0 1  

  1
 2 
1 
 1
y
 3 
    nL  n
 3
 n R2
0
 y2 
 y2 
nL     M  
 
 2 
n   2 
Thick Lens Matrix II
Thick lens matrix :
 1
M   nL  n

 n R2
M  M 3 M 2 M1
0
 1
1
t


n  n
nL  

L
0 1 

n 
 nL R1
0
n 
nL 
Assuming n  n :
 1
M   nL  n

 n R2
 t  n  nL 
0  1 
nL R1
nL  
 n  nL
n  
nL R1

t  n  nL 
1
nL R1
t n

nL 
n

nL 



 n  n  t n  n   n  n
L
L
 L
1 

nL R1 
n R1
 n R2 




nL  n
t  1
nL R2

tn
nL
Thin Lens Matrix
The thin lens matrix is found by setting t = 0:
Thin lens matrix :
1


M   nL  n  1
1

 n  R2 R1 
1 nL  n  1
1 
but

  
f
n  R1 R2 
1
M   1

 f
0

1

0

1

nL
Summary of Matrix Methods
Summary of Matrix Methods
System Ray-Transfer Matrix
 y1 
 
 1
 y2 n  2 


 2n2 
Introduction to
Matrix Methods
in Optics, A.
Gerrard and J.
M. Burch
System Ray-Transfer Matrix
Any paraxial optical system, no matter how complicated, can be
represented by a 2x2 optical matrix. This matrix M is usually denoted
A B
M 

C
D


A useful property of this matrix is that
n0
Det M  AD  BC 
nf
where n0 and nf are the refractive indices of the initial and final media of
the optical system. Usually, the medium will be air on both sides of the
optical system and
Det M  AD  BC 
n0
1
nf
Summary of Matrix Methods
Summary of Matrix Methods
System Ray-Transfer Matrix
The matrix elements of the system matrix can be analyzed to determine
the cardinal points and planes of an optical system.
 y f   A B   y0 
   
  
C
D
 0
 f 

y f  Ay0  B 0
 f  Cy0  D 0
Let’s examine the implications when any of the four elements of the
system matrix is equal to zero.
System Ray-Transfer Matrix
Let’s see what happens when D = 0.
 y f   A B   y0 
   
  
C
0
 0
 f 

y f  Ay0  B 0
 f  Cy0
When D = 0, the input plane for the optical system is the input focal
plane.
Two-Lens System
f1 = +50 mm
Input
Plane
F1
F1
r
T1
f2 = +30 mm
F2
F2
s
q = 100 mm
R1
T2
Output
Plane
T3
R2
 yf 
 y0 
   M  
 0
 f
 1
1
s


M  T3 R2 T2 R1 T1  
  1
0
1


 f 2
 1
1
s


M 
  1
0
1


 f 2
0
 1
1
q


1
1 0 1   
 f1

0
 1
1
q


1
1 0 1   

 f1
r 
 1
1
s


r
1
  1 0 1  
f1 
 f 2
q

0 1 
f1

1  1
   f
1

0
 1 r 
1 0 1

qr
f1 

r

 1 
f1

rq
Two-Lens System
f1 = +50 mm
Input
Plane
F1
F1
r
T1
f2 = +30 mm
F2
F2
s
q = 100 mm
R1
T2

1

1 s  
M  T3 R2 T2 R1 T1   

 0 1   1  1 


 f 2 
Output
Plane
R2
q
f1
q 1

f1  f1
T3




1
qr  r
 r  q 


1


f2 
f1  f1 
rq
qr
f1
 qs s 
 r  q qr
q
qr
r 
 1   r  q 
s

1  
1 
f
f
f
f
f
f
f
f1  
1
2 
1
1
2 1
 2



1
q 1
1
qr  r
  1   

 r  q 
  1
f
f
f
f
f
f


2 
1
1
2 
1 
1
Two-Lens System: Input Focal Plane
f1 = +50 mm
Input
Plane
F1
F1
r
T1
D
r
f2 = +30 mm
F2
F2
s
q = 100 mm
R1
1
qr  r
r

q


  1  0
f2 
f1  f1
  30  50   100  50 
 175 mm
100  50  30
T2

R2
r
Output
Plane
 f 2 f1  q f1
q  f1  f 2
check :
T3
System Ray-Transfer Matrix
Let’s see what happens when A = 0.
 yf  0
   
 f  C
B   y0 
D  0 

y f  B0
 f  Cy0  D0
When A = 0, the output plane for the optical system is the output focal
plane.
Two-Lens System: Output Focal Plane
f1 = +50 mm
Input
Plane
F1
F1
r
T1
f2 = +30 mm
F2
F2
s
q = 100 mm
R1
T2
Output
Plane
R2
q s 
q s
A  1   1     0
f1 f 2 
f1  f1
T3

System Ray-Transfer Matrix
Let’s see what happens when C = 0.
 y f   A B   y0 
   
  
0
D
 0
 f 

y f  Ay0  B0
 f  D0
When C = 0, collimated light at the input plane is collimated light at the
exit plane but the angle with the optical axis is different. This is a
telescopic arrangement, with a magnification of D = f/0.
Telescopic Two-Lens System
f1 = +50 mm
Input
Plane
F1
F1
r
T1
C
f2 = +30 mm
F2
F2
s
q = 80 mm
R1
1
q 1
1

0


f2 
f1  f1
T2

R2
T3
q  f1  f 2
1
qr  r
1 
r

q



1


r  f1  f 2 



f2 
f1  f1
f2 
f  f2
f
 1
1  1
f2
f2
D
Output
Plane
 f1  f 2  r  
f1


r
1
f1
System Ray-Transfer Matrix
Let’s see what happens when B = 0.
 y f   A 0   y0 
   
  
C
D
  0
 f 

y f  Ay0
 f  Cy0  D0
When B = 0, the input and output planes are object and image planes,
respectively, and the transverse magnification of the system m = A.
Two-Lens System: Imaging Planes
f1 = +50 mm
Object
Plane
F1
F1
r
T1
B  rq

f2 = +30 mm
F2
Image
Plane
F2
s
q = 100 mm
R1
T2
 r  q qr
qr
r 
s

1    0
f1
f 2 f1
f1 
 f2
R2
T3
rq
 s
r  q qr
r

1 
f2
f 2 f1
f1
f1 f 2  r  q   f 2 qr
r  f1 f 2  f 2 q   f 1 f 2 q

f1  r  q   q r  f1 f 2  f 2 r r  f1  q  f 2   f1 q  f1 f 2
m  A  1
qs s 
q
 1  
f1
f2 
f1 
qr
f1
Cardinal Points (Planes) of an Optical System
Distances measured to the right of the respective reference plane are
positive, distances measured to the left are negative. As shown:
p<0
f1 < 0
r>0
v>0
q>0
f2 > 0
s<0
w<0
Cardinal Points (Planes) of an Optical System
Thick Lens Analysis
RP1
n0 = 1.0
RP2
Find for the lens:
n0 = 1.0
nL = 1.8
V1
V2
R2 = +45 mm
R1 = +30 mm
t = 50 mm
(a)
(b)
(c)
(d)
Principal Points
Focal Points
Focal Length
Nodal Points
Thick Lens Analysis
In Lecture 4 we found the matrix for a thick lens with the same refractive
index on either side of the lens
Thick lens matrix, assuming n  n :

t  n  nL 
1


nL R1
 A B

M 

C
D
n  n  t  n  nL   n  nL


 L
1 

nL R1 
n R1
 n R2 
50*  0.8 

50*1.0 
1



1.8*30
1.8


 0.8  50*  0.8   0.8 0.8*50 
 1

1 

45
1.8*30
30
1.8*
45




Check : det M  AD  BC 




nL  n
t  1
nL R2

tn
nL
Thick Lens Analysis
A  0.2593
B  27.77 mm
C   0.02206 mm1
D  1.494
1.494
  67.72 mm
0.02206
0.2593
q
  11.75 mm
0.02206
1.494  1
r
  22.39 mm
0.02206
1  0.2593
s
  33.58 mm
0.02206
1.494  1
v
  22.39 mm
0.02206
1  0.2593
w
  33.58 mm
0.02206
p
f1  p  r   45.33 mm
f 2  q  s   45.33 mm
Thick Lens Analysis
RP1
PP1
PP2
RP2
n0 = 1.0
n0 =
1.0
F1
nL = 1.8
V2
V1
H2
H1
F2
R2 = +45
mm
R1 = +30
mm
t = 50 mm
Thick Lens Analysis
f1
f
 2 1
s0
si
In general,
for any
optical
system:
m
n si
n s0
for n  n :  f1  f 2  f
PP1

1
1
1
 
s0 si
f
RP1 PP2
m
si
s0
RP2
F1
H2
H1
R1 =
+30
mm
so = -95 mm
t = 50 mm
F2
R2 =
+45
mm
si = +86.7 mm