Transcript Lecture-5-Optics

Consider Refraction at Spherical Surfaces:
Starting point for the development of lens equations
Vast majority of quality lenses that are used today have segments
containing spherical shapes. The aim is to use refraction at surfaces to
simultaneously image a large number of object points which may
emit at different wavelengths.
Point V (Vertex)
sO  SV (object distance)
si  VP (image distance)
i - Angle of incidence
t - Angle of refraction
r - Angle of reflection
The ray SA emitted from point S will strike the surface at A, refract towards the
normal, resulting in the ray AP in the second medium (n2) and strike the point P.
All rays emerging from point S
and striking the surface at the
same angle i will be refracted and
converge at the same point P.
Let’s return to Fermat’s Principal
d
OPL  0 with OPL  n1 lo  n2li
dx
Using the Law of cosines:
lo2  R 2  so  R   2 Rso  R  cos
2


l  R  s  R   2 Rs  R  cos 
 lo  R  so  R   2 Rso  R  cos
and
2
2
2
2
i
i
with cos(   )   cos
i
1/ 2
1/ 2
Therefore, with OPL  n1lo  n2li

 OPL  n1 R  so  R   2 Rso  R  cos

2
2
 n2 R  si  R   2 Rsi  R  cos
2
2


1/ 2
1/ 2
Note: Si, So, R are all positive variables here. Now, we can let
d(OPL)/d = 0 to determine the path of least time.
Then the derivative becomes


1 / 2
d (OPL)
1 2
2
 n1 R  so  R   2 Rso  R  cos
2 Rso  R sin 
d
2
1 / 2
1 2
2
 2 R(si  R) sin    0
 n2 R  si  R   2 Rsi  R  cos
2


We can express this result in terms of the original variables lo and li:

n1 Rso  R sin  n2 Rsi  R sin 

 0 and
lo
li

n1 n2 1  n2 si n1so 

  

lo li
R  li
lo 
However, if the point A on the surface changes, then the new ray will not
intercept the optical axis at point P.
Assume new small vaules of the radial angle  so that cos   1, lo  so, and
li  si.
Again, subscripts o and i
n1 n2 n2  n1
then
 
refer to object and image
so si
R
locations, respectively.
This is known as the first-order theory, and involves a paraxial approximation.
The field of Gaussian Optics utilizes this approach.
Note that we could have also started with Snell’s law: n1sin 1= n2sin2 and
used sin    .
Using spherical (convex) surfaces for imaging and focusing
Object focus
i) Spherical waves from the object focus
refracted into plane waves.
Suppose that a point at fo is imaged at a
point very far away (i.e., si = ).
so  fo = object focal length
n1 n2 n2  n1
then
 
so 
R

n1 n2  n1

fo
R

n1
fo 
R
n2  n1
ii) Plane waves refracted into spherical waves.
Suppose now that plane waves (parallel rays)
are incident from a point emitting light from a
point very far away (i.e., so = ).
when so  

n1 n2 n2  n1
 
 si
R
n2 n2  n1

si
R

n2
f i  si 
R
n2  n1
Diverging rays revealing a virtual image point
using concave spherical surfaces.
Virtual image point
n2
f i  si  
R
n2  n1
Signs of
variables are
important.
R<0
fi < 0
si < 0
Parallel rays impinging on a
concave surface. The refracted
rays diverge and appear to
emanate from the virtual focal
point Fi. The image is therefore
virtual since rays are diverging
from it.
A virtual object point resulting from
converging rays. Rays converging
from the left strike the concave
surface and are refracted such that
they are parallel to the optical axis.
An object is virtual when the rays
converge toward it.
so < 0 here.
when si  

n1
n2
n2  n1


so si  
R
n1 n2  n1

so
R

n1
f o  so  
R 0
n2  n1
The combination of
various surfaces of
thin lenses will
determine the signs
of the corresponding
spherical radii.
S
(a)
(b)
(c)
As the object distance so is
gradually reduced, the
conjugate image point P
gradually changes from real
to virtual.
The point P’ indicates the
position of the virtual
image point that would be
observed if we were
standing in the glass
medium looking towards S.
We will use virtual image points to locate conjugate image points.
In the paraxial approximation:
(A)
nm nl nl  nm
 
so1 si1
R1
n1 n2 n2  n1
from
 
so si
R
The 2nd surface “sees” rays coming towards it from the P’ (virtual
image point) which becomes the 2nd object point for the 2nd surface.
Therefore
so2  si1  d , so2 
Thus, at the 2nd surface:
so2 ,
(B)
si1   si1 , so2   si1  d
nl
nm nm  nl


 si1  d si 2
R2
Add Equations (A) & (B) 
1 1
nm nm
nl d

 nl  nm    
so1 si 2
 R1 R2  si1  d si1
Let d  0 (this is the thin lens approximation) and nm  1:

1 1 
1 1
  nl  1   (c)
so si
 R1 R2 
and is known as the thin-lens equation, or the Lens maker’s formula,
in which so1 = so and si2 = si, V1  V2, and d  0. Also note that
lim si  f i ,
so 
For a thin lens (c)  fi = fo = f and
Convex  f > 0
Concave  f < 0
lim so  f o
si  
1 1 
1
 nl  1  
f
 R1 R2 

Also, known as the Gaussian lens formula
1 1 1
 
so si f
Location of focal lengths for converging and diverging lenses
nl
nlm 
1
nm
nl
nlm 
1
nm
If a lens is immersed in a medium
with
nl
nlm 
nm
 1
1
1 
 nlm  1  
f
 R1 R2 
f
Object
2f
2f
Real
image
f
Convex thin lens
Simplest example showing symmetry in which so = 2f  si = 2f
2
f
Object
f
so
si
Concave, f < 0, image is upright
and virtual, |si| < |f|
1
3
Virtual
image
Note that a ray passing
through the center is drawn as
a straight line.
Ideal behavior of 2 sets of parallel
rays; all sets of parallel rays are
focused on one focal plane.
1 1
1
 
so si
f
si  
f so
so  f


1
1 1
 
si
f so

f
1 

where so   f ,   1 
For case (b) below
si  f , si  0
Tracing a few key rays through a positive and negative lens
Consider the Newtonian form of
the lens equations.
S2
yo
S1
From the geometry of similar triangles:
so yo

,
si
yi
1 1 1
 
f so si

1
1
1


f xo  f xi  f

xo  f  f
1

f xo  f  xi  f
f2
f
 xi  f
xo

f
xo  f   x
xo
i
f
 xo xi  f 2
Newtonian Form:
xo > 0 if the object is to the left of Fo.
xi > 0 if the image is to the right of Fi.
The result is that the object
 and image must be on the
opposite sides of their
respective focal points.
Define Transverse (or Lateral) Magnification:
yi
si
xi  f
f 2 / xo  f
f  f / xo  1xo
MT 
 


xo  f xo
yo
so
xo  f
xo  f
xi
f
 
xo
f
Image forming behavior of a thin positive lens.
f
2f
f
2f
MT > 0  Erect image and MT < 0  Inverted image. All real images for a
thin lens will be inverted.
Simplest example 2f-2f
conjugate imaging gives
MT  
f
f
   1
xo
f
f
2f
f
2f
Define Longitudinal Magnification, ML
dxi
ML 
dxo
f2
and xi 
xo
f2
 M L   2   M T2  0
xo
This implies that a positive dxo corresponds to a negative dxi and vice versa.
In other words, a finger pointing toward the lens is imaged pointing away
from it as shown on the next slide.
The number-2 ray entering the lens parallel
to the central axis limits the image height.
Image orientation for a thin lens:
The transverse magnification (MT) is different
from the longitudinal magnification (ML).
(a) The effect of placing a second lens L2 within the focal length of a
positive lens L1. (b) when L2 is positive, its presence adds convergence
to the bundle of rays. (c) When L2 is negative, it adds divergence to the
bundle of rays.
Two thin lenses separated by a distance
smaller than either focal length.
Note that d < si1, so
that the object for
Lens 2 (L2) is virtual.
Note the additional
convergence caused by L2
so that the final image is
closer to the object. The
addition of ray 4 enables
the final image to be
located graphically.
Note that d > si1, so that the object for Lens 2 (L2) is real.
Fig. 5.30 Two thin lenses separated by a distance greater than the sum
of their focal lengths. Because the intermediate image is real, you
could start with point Pi’ and treat it as if it were a real object point for
L2. Therefore, a ray from Pi’ through Fo2 would arrive at P1.
so1 f1
1
1 1
  , si1 
si1 f1 so1
so1  f1
so 2  d  si1
so 2  0 (virtual)
so 2  0 (real)
For the compound lens system, so1
is the object distance and si2 is the
image distance.
f 2 so1 f1
f 2d 

so 2 f 2
d  si1  f 2
so1  f1
1
1
1
 
, si 2 


si 2 f 2 so 2
so 2  f 2 d  si1  f 2  d  f  so1 f1
2
so1  f1
The total transverse magnification (MT) is
given by
M T  M T 1M T 2
 si1  si 2 
f1si 2




    


 so1  so 2  d so1  f1   so1 f1
For this two lens system, let’s determine the front focal length (ffl) f1 and
the back focal length (bfl) f2.
Let si2   then this gives so2  f2.
so2 = d – si1 = f2  si1 = d – f2 but
 1 
1 1
1
1
   
 
f1 si1 f1 d  f 2
 S o1  si 2

ffl  so1 si 2
f1 d  f 2 

d   f1  f 2 
From the previous slide, we calculated si2. Therefore, if so1   we get,
f 2 d  f 2 f1
f 2 d  f1 
bfl  si 2 

d  f 2  f1 d   f 2  f1 
f 2 f1
for d  0, bfl  ffl 
 f ef
f 2  f1

1
1 1
 
f ef
f1 f 2
fef = “effective focal length”
Suppose that we have in general a system of N lenses whose thicknesses are
small and each lens is placed in contact with its neighbor.
1 2 3 ………N
Then, in the thin lens approximation:
1
1 1 1
1
   ...
f ef
f1 f 2 f 3 f N
Fig. 5.31 A positive and negative thin lens combination for a
system having a large spacing between the lenses. Parallel rays
impinging on the first lens enable the position of the bfl.
Example B
Example A
Example A: Two identical converging (convex) lenses have f1 = f2 = +15 cm
and separated by d = 6 cm. so1 = 10 cm. Find the position and magnification of
the final image.
1
1
1
 
so1 si1 f1
 si1 = -30 cm at (O’) which is virtual and erect
Then so2 = |si1| + d = 30 cm + 6 cm = 36 cm
1
1
1


so 2 si 2 f 2
 si2 = i’ = +26 cm at I’
Thus, the image is real and
inverted.
The magnification is
given by
M T  M T 1M T 2
 si1  si 2    30  26 
   
    
    2.17
 so1  so 2   10  36 
Thus, an object of height yo1 = 1 cm has an image height of yi2 = -2.17cm
Example B: f1= +12 cm, f2 = -32 cm, d = 22 cm
An object is placed 18 cm to the left of the first lens (so1 = 18 cm).
Find the location and magnification of the final image.
1
1
1
 
so1 si1 f1
 si1 = +36 cm in back of the second lens,
and thus creates a virtual object for the
second lens.
Image is real
and Inverted
so2 = -|36 cm – 22 cm| = -14 cm
1
1
1


 si2 = i’ = +25 cm; The magnification is given by
so 2 si 2 f 2
Thus, if yo1 = 1
 si1  si 2   36  25 
     
M T  M T 1M T 2     
  3.57 cm this gives yi2 =
-3.57 cm
 so1  so 2   18   14 