Transcript Real and Virtual Images

ECEN 4616/5616
Optoelectronic Design
Class website with past lectures, various files, and assignments:
http://ecee.colorado.edu/ecen4616/Spring2014/
(The first assignment will be posted here on 1/22)
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If anyone has trouble accessing either of these
sites, please notify me as soon as possible.
Gaussian Optics
Gaussian Optics refers to the use of a linear model to represent idealized optical
lenses, mirrors, etc. We’ve already shown that the imaging and ray tracing
equations are linear in the paraxial limit – how can they be modified to
represent ideal finite-sized lenses? This is not an idle mathematical exercise, as
most optical designs need to start with an ideal system which solves the desired
problem, such as a simple, two-lens zoom camera lens:
Simple zoom lens based on the
combination of powers formula:
K  K1  K2  K1K2d1
The ideal lenses are then replaced
with lens combinations designed
to be as ideal as possible:
Gaussian Imaging
Recall that the focal length of a (positive) lens is defined as the distance past the lens
that an incident plane wave will converge to a point, and that the lens power is
defined as the inverse of the focal length: K  1
f
f
Since the converging wave has a curvature
1
c   K just past the lens, this leads to
f
the conclusion that the lens adds a
curvature of K to an incoming wave:
l
o
K
K
l’
i
This observation leads to the imaging equation:
1
1
K 
l
l
1 1
 K
l l
The imaging equation follows from the definition of focal length and power, hence
already describes an ideal lens. (We showed that real lenses behave ideally in the
paraxial limit, but the imaging equation applies to any lens combination designed to
be approximately linear.)
The paraxial ray refraction equation, u  u  hK , however, was only shown to hold
true in the paraxial approximation where the angles, sines and tangents are all
approximately equal:
u  sin(u)  tan(u)
K
h
u
u’
l’
l
What definition of u, u’ will allow the linear paraxial refraction equation to also
describe finite-sized ideal lenses?
i.e., the problem is to find “angle variables” u, u 
such that:
1
1
K 
l
l
and
u  u  hK
for all h
Solve both equations for K, and equate:
Imaging Equation:
Refraction Equation:
u  u  hK
u u
  K
h h
⇒
⇒

u u 1 1
  
h h l l
Equating the input variables
(unprimed) and the output
(primed) variables
independently, we see that:
,
,
1
1
K 
l
l
1 1
K 
l l
⇒
u   u 
h
h

if u   , and u  
l
l
h h

l l
then the linear ray trace
equation and the imaging
equation will be true for
finite size lenses (h finite).
Hence, for Gaussian ray tracing (ideal lenses), we will use the
tangents of the ray angles in the diffraction equation.
Summary of Gaussian Optics (Ideal Lens) Rules
1) Lenses are represented by plane
surfaces. All calculations about a lens
are done in the lens’ local coordinates,
where the origin is the lens’ intersection
with the z-axis.
2) Lens power, K, is the curvature that a
lens imposes upon an incident wave.
(Hence, in the drawing at right, 1  K  1
l
K
h
u
l
u’
l’
l
3) Rays are characterized by their positions and the
tangents of the angles they make with the z-axis.
(These are generally referred to as the ‘angle variables’)
4) The change in the angle variable of a ray is the
negative product of the lens power and
distance from the axis: u’ – u = -hK. (Hence rays
are bent towards the axis by lenses with positive
power and away from the axis by lenses with
negative power.)
Sign Rules:
1. The (local) origin is at the intersection
of the surface and the z-axis.
2. All distances are measured from the
origin: Right and Up are positive
3. All angles are acute. They are
measured from:
a) From the z-axis to the ray
b) From the surface normal to the
ray
c) CCW is positive, CW is negative
4. Indices of refraction are positive for
rays going left-to-right; negative for
rays going right-to-left. (Normal is leftto-right.)
Summary of Gaussian Optics (continued)
Lens1
n1
h1
u1
l1
Lens2
n1  n2
l2
u’1
u2
l’1
n2
l’2
h2
u’2
d1
Image calculation at a lens:
Ray refraction at a lens:
ni ni
  Ki
l i l i
niui  ni ui  hi Ki
Image transfer between lenses:
Ray transfer between lenses:
l i 1  l i  di
ui 1  ui
hi 1  hi  di ui
Focal Points of a Lens
f’
f
Since lenses can work in either direction, they have two focal points.
f is called the primary focal point and f’ the secondary focal point.
While f and f’ follow the primed variable rule for positive lenses,
they are on opposite sides for negative lenses:
f’
f
Lenses: Real and Virtual Images
A real image can be projected on a screen.
Projection
screen
Real Object
f
f’
Real
Image
Lenses: Real and Virtual Images
A Virtual Image appears to come from somewhere
where there are no actual rays.
Virtual Image
f
Real
Object
f’
Observer
Lenses: Real and Virtual Images
Real
Object
f’
f
Virtual
Image
Lenses: Real and Virtual Images
Real
Object
f’ Virtual
Image
f
Negative lenses only form virtual images
from real objects.
Lenses: Real and Virtual Images
f’
Virtual f
Object
A negative lenses can form a real image from
a virtual object however.
Real Image
Lenses: Real and Virtual Images
A virtual object for one lens can be a real image from the
previous lens in a system.
L1
f’2
Real
Object
L2
Real image from Lens 1 and
virtual object for Lens 2
f’
f1
Real image
from Lens 2
and system.
Optical Devices:
Consider a positive lens with an object less than one
focal length behind it:
What is the effective magnification?
Let the object height = h,
S2
and assume the user places
his/her eye as close to the
S1
lens as feasible. Then, the
angle that the central ray
h
makes is the angular extent
a
of the image for the user:
Virtual
f Object
f’
h
image
a
S1
The Simple Magnifier:
We need to compare this to the maximum angle that the object can be viewed without
using the lens. This is obviously dependent on how close someone can focus their eyes,
so is not a universal constant. For the purpose of these kinds of calculations, however, it
is usual to use -250 mm as the average close viewing distance achievable for most of the
population (the older population may have to use reading glasses!). Hence the
maximum angle the object can subtend to the unaided eye is: b   h
250
The effective magnification is therefore: M 
a
250

b
S1
S2
S1
h
Virtual
image
f
a
Object
f’
We will further assume that the distance, S2, of the virtual image is made to also
equal -250 mm. This will give us slightly more magnification, since the object can
be closer to the lens than for an object at infinity.
1 1 1
 
The imaging equation then gives us:
S1 f S2
The magnification is then: M  
 S1 
S2f
250f

f  S2
250  f
250 250  f

S1
f
Note that, if the object is at the focal distance, then the object is at ∞, and
250
the resulting magnification is:
M
f
In the last lecture, we found
the combination of power
formula by tracing rays through
a pair of surfaces:
n2 ’
u’1=u2
u1 =0
h1
h2
d1
n1
BFL
n1’=n2
f
K2
K1
Refract through first element:
n1u1  n1u1  h1K1
u1  
h1K1
, since u1  0
n1
 u2  
h1K1
, u2  u1, n2  n1
n2
Transfer to second element:
h2  h1  u1d1
hK
 h2  h1  1 1 d1
n2
u’2
Refract at second element:
n2 u2  n2u2  h2K 2
 hK  

hK
 n2   1 1    h1  1 1 d1  K 2
n2
 n2  

KK
 h 
n2   1   h1K1  h1K 2  h1 1 2 d1
n2
 f 
Eliminating h1 and using K 
1
where K is the
f
equivalent power of the combination, we get:
General formula for combination of two surfaces:
n2 K  K1  K 2 
K1K 2
d1
n2
The “Lensmakers” Equation
The ‘thin lens’ is an approximation, as any real lens will have a finite
thickness. In the last lecture (1-17) we saw that, by applying the paraxial
assumption to a surface separating materials of indices n, n’ that the power
of a surface is:
K  c n  n Where c is the curvature (inverse radius) of


the surface, n the index before and n’ the index after the surface.
c1
To adapt the surface power
A “Thick Lens”:
combination formula to a thick lens
c2
n1  n2  n
in air, let:
n1  1
n1  n2  1, n1  n2  n
Substituting for the powers and indices:
d1
n2  1
K1  c1  n  1, K 2  c2 1  n 
We get the formula for the power of a thick lens:
c1c2  n  1
K   n  1 c1  c2  
d
n
2