Geometrical Optics
Geometrical light rays
Ray matrices and ray vectors
Matrices for various optical
components
The Lens Maker’s Formula
Imaging and the Lens Law
Mapping angle to position
1
Leerdoelen
In dit college leer je:
• Wat de benaderingen zijn van geometrische optica
• Wat de belangrijkste karakteristieken van lenzen zijn
• Hoe je de stralengang door optische systemen met o.a. lenzen
kunt analyseren
• Een matrix-methode voor geometrische optica
• Toepassing: microscopie, fotografie en bijbehorende
eigenschappen
• Hecht: 5.1 t/m 5.4; 5.7; 6.1 en 6.2
2
Is geometrical optics the whole story?
No. We neglect the wave
nature of light.
~0
Also, our ray pictures seem to
imply that, if we could just
remove all aberrations, we
could focus a beam to a point
and obtain infinitely good
spatial resolution.
Not true. The smallest
possible focal spot is the on
the order of the wavelength l.
Same for the best spatial
resolution of an image. This is
due to diffraction, which is
not included in geometrical
optics.
≈l
3
Arguments in favor of geometrical optics
Remember week 1 in which we discussed three reasons to believe that
the ray picture of light is useful:
1. The straight propagation of light that we see entering through a
window.
2. The camera obscura in which a small aperture produces an image
that is upside down.
3. The very narrow beams that are produced by a laser.
4
Relation between object, image and focus
1 1 1
 
f s 0 si
yo
yi
f
f
s1
s0
Magnification:
yi  si
MT 

yo
so
5
Lensmaker’s formula
1 1 
1
1

 nl  1  
so1 si 2
 R1 R2 
1 1 
1
 nl  1  
f
 R1 R2 
1 1 1
 
f s 0 si
8
Types of lenses
Lens nomenclature:
Which type of lens to use (and how to orient it) depends on the
aberrations (next lecture) and application.
9
Sign convention for R
10
Tracing a few rays (Fig 5.20 Hecht)
• Rays through the center of the lens go straight
• Parallel rays from the object pass through Fi
• Rays through Fo propagate parallel to the axis behind the lens
11
Image formation by a thin lens
1 1 1
 
f s 0 si
yi  si
MT 

yo
so
•
•
•
•
The image is formed at the point where these rays cross
A real image is formed if s0 > f
The magnification of the image is given by Mt
For a single positive lens, the magnification will be negative
12
Virtual
Images
A virtual image occurs when the outgoing rays
from a point on the object never actually intersect
at a point but can be traced backwards to one.
Negative-f lenses have virtual images, and positive-f lenses do
also if the object is less than one focal length away.
Virtual
image
Virtual
image
Object infinitely
far away
Object
f<0
Simply looking at a flat mirror yields a virtual image.
f>0
13
Combination of 2 lenses spaced < F1, F2
14
Ray Optics
axis
We'll define light rays as directions in space, corresponding,
roughly, to k-vectors of light waves.
We won’t worry about the phase. We also ignore reflections.
Each optical system will have an axis, and all light rays will be
assumed to propagate at small angles to it. This is called the
Paraxial Approximation: sin q  tan q  q
15
The Ray
Vector
xin , qin
xout, qout
A light ray can be defined by two coordinates:
its position, x
q
its slope, q
x
Optical axis
These parameters define a ray vector,
which will change with distance and as
the ray propagates through optics.
 x
q 
 
16
Ray Matrices
For many optical components, we can define 2 x 2 ray matrices.
An element’s effect on a ray is found by multiplying this matrix with
the ray vector.
Ray matrices
can describe
simple and complex systems.
Optical system ↔ 2 x 2 Ray matrix
 xin 
q 
 in 
A
C

B
D 
 xout 
q 
 out 
These matrices are often (uncreatively) called ABCD Matrices.
17
Physical meaning of two
of the matrix elements
𝐴=
𝑥𝑜𝑢𝑡
𝑥𝑖𝑛
𝑖𝑓 𝜃𝑖𝑛 = 0
𝐷=
𝜃𝑜𝑢𝑡
𝜃𝑖𝑛
𝑖𝑓 𝑥𝑖𝑛 = 0
spatial
magnification
 xout   A B   xin 
q   C D  q 
  in 
 out  
angular
magnification
18
For cascaded elements, we simply
multiply ray matrices.
 xin 
q 
 in 
O2
O1

 xout 
q   O3 O2
 out 


 O1

O3
 xout 
q 
 out 
 xin   
 xin 
q     O3 O2 O1 q 
 in   
 in 
Notice that the order looks opposite to what it should be,
but it makes sense when you think about it.
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Ray matrix for free space or a medium
If xin and qin are the position and slope upon entering, let xout and
qout be the position and slope after propagating from z = 0 to z.
xout qout
Xin, qin
xout = xin + z qin
q out = qin
Rewriting these expressions
in matrix notation:
z
z=0
Ospace
1 z 
= 

0
1


é x
ê out
êq
ë out
ù
é 1 z ùé xin ù
ú = ê
ú
úê
ú
êq ú
0
1
ë
û
û
ë in û
20
Ray Matrix for an Interface
qout
At the interface, clearly:
qin
xout = xin.
n1
Now calculate qout.
Snell's Law says:
xin
xout
n2
n1 sin(qin) = n2 sin(qout)
which becomes for small angles: n1 qin = n2 qout
 qout = [n1 / n2] qin
é 1
0
Ointerface = ê
êë 0 n1 / n2
ù
ú
úû
21
Ray matrix for a curved interface [radius R ]
At the interface, again:
xout = xin.
To calculate qout, we must
calculate q1 and q2.
If qs is the surface slope at
the height xin, then
R
qs
q2
q1
qin
xin
n1
z=0
qout
qs
z
qs = xin /R
n2
z
q1 = qin+ qs and q2 = qout+ qs
q1 = qin+ xin / R and q2 = qout+ xin / R
Snell's Law: n1
q1 = n2 q2
 n1 (qin  xin / R)  n2 (qout  xin / R)
 qout  (n1 / n2 )(qin  xin / R)  xin / R
é
1
0
 qout  (n1 / n2 )qin  (n1 / n2  1) xin / R Ocurved = ê (n / n -1) / R n / n
interface ê
1
2
ë 1 2
22
ù
ú
úû
A thin lens is just two curved interfaces.
We’ll neglect the glass in between (it’s
a really thin lens!), and we’ll take n1 = 1.
Ocurved
interface
1
0 



(
n
/
n

1)
/
R
n
/
n
 1 2
1
2
R1
n1=1
R2
n2≠1
n1=1
1
0 
1
0 

Othin lens  Ocurved Ocurved  
 (1/ n  1) / R 1/ n 
(
n

1)
/
R
n
interface 2
interface 1

2

1

1
0  
1
0





(
n

1)
/
R

n
(1/
n

1)
/
R
n
(1/
n
)
(
n

1)
/
R

(1

n
)
/
R
1

2
1
 
2
1

1
0



(
n

1)(1/
R

1/
R
)
1

2
1

This can be written:
where: 1/ f  (n  1)(1/ R1  1/ R2 )
 1
 1/ f

0
1 
The Lens-Maker’s Formula
23
Next: justify that f = focal length
Ray matrix for a lens
1/ f  (n  1)(1/ R1  1/ R2 )
Olens
é 1
= ê
êë -1/f
0
1
ù
ú
úû
The quantity, f, is the focal length of the lens. It’s the single most
important parameter of a lens. It can be positive or negative.
R1 > 0
R2 < 0
f>0
If f > 0, the lens deflects
rays toward the axis.
R1 < 0
R2 > 0
f<0
If f < 0, the lens deflects
rays away from the axis.
24
A lens focuses parallel rays to a point
one focal length away.
A lens followed by propagation by one focal length:
é x
ê out
êq
ë out
ù
ú=
ú
û
é 0
é x ù
ù
0 ê in ú
ú
= ê
1 úûêë 0 úû
êë -1/f
é 1 f ùé 1
ê
úê
êë 0 1 úûêë -1/f
f
f
f ùé xin
úê
1 úûêë 0
For all rays
xout = 0!
ù é
ù
0
ú=ê
ú
úû êë -xin / f úû
Assume all input
rays have qin = 0
At the focal plane, all rays
converge to the z axis (xout = 0)
independent of input position.
[Parallel rays at a different angle
focus at a different xout.]
Notice we have now proven the Lensmakers’ Law.
25
Concave Spherical Mirror
A concave mirror
A radio telescope
Like a lens, a curved mirror will focus a beam. Its focal length is R/2,
with R the radius of curvature.
26
Ray Matrix for a Curved Mirror
Consider a mirror with radius of curvature, R, with its optic axis
perpendicular to the mirror:
q1  qin  q s
q s  xin / R
R
qout
qs
qin
qout  q1  q s  (qin  q s )  q s
 qin  2 xin / R
q1
q1
xin = xout
z
0
 1
 Omirror = 


2
/
R
1


Like a lens, a curved mirror will focus a beam. Its focal length is R/2.
Note that a flat mirror has R = ∞ and hence an identity ray matrix.
27
Consecutive lenses
Suppose we have two lenses
right next to each other (with
no space in between).
 1
Otot = 
-1/f 2
f1
0  1
0
1

= 



1  -1/f1 1 
-1/f1  1/ f 2
f2
0
1 
1/f tot = 1/f1 + 1/f 2
So two consecutive lenses act as a single lens whose focal length is
computed by the inverse sum.
As a result, we define a measure of inverse lens focal length, the
diopter.
1 diopter = 1 m-1
28
A system images an object when B = 0.
When B = 0, all rays from a
point xin arrive at a point xout,
independent of their angle.
xout = A xin
é x
ê out
êq
ë out
ù é
é x ù é
ù
Axin
ú = ê A 0 úê in ú = ê
ú ë C D ûê q ú ê C x + D q
in
û
ë in û ë in
When B = 0, A is the magnification.
29
ù
ú
ú
û
The Lens Law
From the object to
the image, we have:
1) A distance do
2) A lens of focal length f
3) A distance di
0  1 d o 
1 d i   1
O
  1/ f 1  0 1 
0
1




do 
1 di   1

  1/ f 1  d / f 
0
1


o

1  di / f d o  di  d o di / f 



1/
f
1

d
/
f
o


B  d o  di  d o di / f 
do di 1/ d o  1/ di  1/ f  
0 if
1 1 1
 
d o di f
This is the Lens Law.
30
Imaging
Magnification
If the imaging condition,
1 1 1
 
d o di f
is satisfied, then:
1  di / f
O
 1/ f

1  do / f 
1 1
A  1  di / f  1  di   
 d o di 
di

M 
do
0 
1/ M 
1 1
D  1  do / f  1  do   
d o di 

do
 
 1/ M
31
di
0
So:
 M
O
 1/ f
Microscopes
Image
plane #1
Objective
M1
Eyepiece
Image
plane #2
M2
Microscopes basically consist of a set of lenses, and is used to:
1. In a microscope, the object is really close and we wish to magnify it.
2. Lens aberrations should be minimized (next lecture).
A microscope is effectively a telescope in reverse. The goal is to
magnify the object, not its angular diameter.
Eyepiece
The microscope objective yields a magnification
of from 5 to 100, corresponding to focal lengths
of 40 mm to 2 mm.
The eyepiece yields a magnification of 2 to 10,
corresponding to various focal lengths and
distances.
Objective
32
Microscope
Terminology
33
F-number
The F-number, “f / #”, of a lens is the ratio of its focal length and its
diameter.
f/# = f/d
Confusing!!
For example, a lens with a 25 mm aperture and a 50 mm focal length
has an f-number of 2, which is usually designated as f/2
f
d1
f
f
d2
f/# =1
f
f/# =2
Small f-number lenses collect more light but are harder to engineer.
34
Depth of Field
Only one plane is imaged (i.e., is in focus) at a time. But we’d like
objects near this plane to at least be almost in focus. The range of
distances in acceptable focus is called the depth of field.
It depends on how much of the lens is used, that is, the aperture.
Out-of-focus
plane
Image
Object
f
Aperture
Size of blur in
out-of-focus
plane
Focal
plane
The smaller the aperture, the greater the depth of field.
35
Depth of field example
A large depth of field
isn’t always desirable.
f/32 (very small aperture;
large depth of field)
f/32 means f/D = 32, the focal length of
the lens is 32 times larger than the
aperture diameter
f/5 (relatively large aperture;
small depth of field)
A small depth of field is also
desirable for portraits.
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Tot slot
Wat hebben we gezien:
• Geometrische optica
• Berekeningen met lenzen
• ABCD-matrix voor optische systemen
• Matrix-berekeningen in geometrische optica
• Inleiding in microscopie
37