Download Waves & Oscillations Physics 42200 Spring 2015 Semester Lecture 28 – Geometric Optics

Physics 42200
Waves & Oscillations
Lecture 28 – Geometric Optics
Spring 2015 Semester
Matthew Jones
Thin Lens Equation
First surface:
Second surface:
Add these equations and simplify using 1 and → 0:
1 1
1
1
1
(Thin lens equation)
Thick Lenses
• Eliminate the intermediate image distance, • Focal points:
– Rays passing through the focal point are refracted parallel
to the optical axis by both surfaces of the lens
– Rays parallel to the optical axis are refracted through the
focal point
– For a thin lens, we can draw the point where refraction
occurs in a common plane
– For a thick lens, refraction for the two types of rays can
occur at different planes
Thick Lens: definitions
First focal point (front focal length)
Primary principal plane
First principal point
Second focal point (back focal length)
Secondary principal plane
Second principal point
Nodal points
Fo Fi H1 H2 N1 N2 - cardinal points
If media on both sides
has the same n, then:
N1=H1 and N2=H2
Thick Lens: Principal Planes
Principal planes can lie outside the lens:
Thick Lenses and Principal Planes
• For a single refracting surface, we measured and with respect to the vertex (ie, the
surface of the lens)
• For a thick lens, we need to define and with respect to the principal planes.
• We need to calculate where they are, but it
makes the algebra simpler.
• We are not going to derive the following
formula…
Thick Lens: equations
Note: in air (n=1)
1 1 1 xo xi = f 2
+ =
so si f
effective
focal length:
Principal planes:
Magnification:
1
1
1 (nl − 1)d l 
= (nl − 1) −
+

f
R
R
n
R
R
2
l 1 2 
 1
f (nl − 1)d l
f (nl − 1)d l
h1 = −
h2 = −
nl R2
nl R1
yi
si
xi
f
MT ≡
=− =− =−
yo
so
f
xo
Thick Lens Calculations
1. Calculate focal length
1
−1 1
1
= −1
−
+
2. Calculate positions of principal planes
−1 −1 ℎ = −
ℎ = −
3. Calculate object distance, , measured from principal plane
4. Calculate image distance:
1 1 1
+ =
5. Calculate magnification, = − /
Thick Lens: example
Find the image distance for an object positioned 30 cm from the
vertex of a double convex lens having radii 20 cm and 40 cm, a
thickness of 1 cm and nl=1.5
1 1 1
+ =
so si
f
f
30 cm
so
f
si
1
1
1 (nl − 1)d l 
1
0.5 ⋅1  1
1
= (nl − 1) −
+
−
 = 0.5 −
 cm
f
R
R
n
R
R
20
−
40
1
.
5
⋅
20
⋅
40

2
l 1 2 
 1
f = 26.8 cm
so = 30cm + 0.22cm = 30.22 cm
26.8 ⋅ 0.5 ⋅ 1
h1 = −
cm = 0.22cm
1
1
1
+ =
− 40 ⋅ 1.5
30.22cm si 26.8cm
26.8 ⋅ 0.5 ⋅ 1
h2 = −
cm = −0.44cm
si = 238 cm
20 ⋅ 1.5
Compound Thick Lens
Can use two principal points (planes) and effective focal length f
to describe propagation of rays through any compound system
Note: any ray passing through the first principal plane will emerge
at the same height at the second principal plane
For 2 lenses (above):
Example: page 246
1 1 1
d
= + −
f
f1 f 2 f1 f 2
H11H1 = fd f 2
H 22 H 2 = fd f1
Ray Tracing
• Even the thick lens equation makes approximations and
assumptions
– Spherical lens surfaces
– Paraxial approximation
– Alignment with optical axis
• The only physical concepts we applied were
– Snell’s law: sin = sin – Law of reflection: = (in the case of mirrors)
• Can we do better? Can we solve for the paths of the rays
exactly?
– Sure, no problem! But it is a lot of work.
– Computers are good at doing lots of work (without complaining)
Ray Tracing
• We will still make the assumptions of
– Paraxial rays
– Lenses aligned along optical axis
• We will make no assumptions about the lens
thickness or positions.
• Geometry:
Ray Tracing
• At a given point along the optical axis, each ray can
be uniquely represented by two numbers:
– Distance from optical axis, – Angle with respect to optical axis, • If the ray does not encounter an optical element its
distance from the optical axis changes according to
the transfer equation:
= + – This assumes the paraxial approximation sin ≈ Ray Tracing
• At a given point along the optical axis, each ray can
be uniquely represented by two numbers:
– Distance from optical axis, – Angle with respect to optical axis, • When the ray encounters a surface of a material with
a different index of refraction, its angle will change
according to the refraction equation:
= − − =
– Also assumes the paraxial approximation
Ray Tracing
• Geometry used for the refraction equation:
sin = ≈ = + = + /
= + = + /
= +
= +
− = −
Matrix Treatment: Refraction
At any point of space need 2 parameters to fully specify ray:
distance from axis (y) and inclination angle (α) with respect to
the optical axis. Optical element changes these ray parameters.
Refraction:
note: paraxial approximation
nt1α t1 = ni1αi1 − D 1 yi1
Reminder:
yt1=yi1
Equivalent matrix
representation:
yt1 = 0 ⋅ ni1α i1 + yi1
 A B  α   Aα + By 

  ≡ 

C
D
y
C
α
+
Dy

  

 nt1α t1   1 - D 1  ni1αi1 

 = 


 y t 1   0 1   yi 1 
≡ ri1 - input ray
rt1 = R1ri1
≡ R1 - refraction matrix
≡ rt1 - output ray
Matrix: Transfer Through Space
Transfer:
ni 2αi 2 = nt1αt1 + 0 ⋅ yt1
yi2
yi 2 = d 21 ⋅ α t1 + yt1
yt1
Equivalent matrix
presentation:
0  nt1α t1 
 ni 2αi 2   1

 = 


 yi 2   d 21 nt1 1  yt1 
≡ rt1 - input ray
ri 2 = T21rt1
≡ T21 - transfer matrix
≡ ri2 - output ray
System Matrix
yi2
yi1 y
t1
ri1
rt1
R1
ri2
T21
rt1 = R1ri1
Thick lens ray transfer:
yi2
rt2
ri3
T32
rt3
R3
ri 2 = T21rt1 = T21R1ri1
rt 2 = R 2T21R1ri1
System matrix:
A = R 2T21R1
Can treat any system with single system matrix
rt 2 = A ri1
Thick Lens Matrix
d
A = R 2T21R1
nl
yi2
yi1 y
t1
yi2
Reminder:
 A B  a b   Aa + Bc Ab + Bd 


 ≡ 

C
D
c
d
Ca
+
Dc
Cb
+
Dd


 

 1 - D 2  1
A = 

 0 1  d l nl
 D 2d l
1 −
nl
A =
 dl
 n
l

0  1 - D 1 


1  0 1 
D 1D 2d l 
- D1 - D 2 - +

nl 
D 1d l

1−

nl

1 - D 
R = 

0 1 
 1 0
T = 

 d n 1
1
1
1
1
system matrix of thick lens
For thin lens dl=0
1 - 1/ f 
A = 

1 
0
Matrix Treatment: example
rI
rO
rI = T1A l T2rO
 nIα I   1

 = 
 y I   d I 2 nI
0  a11 a12  1


1  a21 a22  d1O nO
0  nOαO 


1  yO 
(Detailed example with thick lenses and numbers: page 250)
Mirror Matrix
Sign convention: R > 0
 − 1 2n / R 

M = 
1 
0
 nα r 
 nαi 


 = M
 yr 
 yi 
yr = yi
nα r = −nα i + 2nyi / R
α r = −α i + 2 yi / R
Tray Tracing Example
• Transfer matrix (distance in medium ):
1
0 = ⁄ 1
• Refraction matrix (spherical surface)
1 − =
0 1
− =
• This example:
! = "# $ "% " !
Ray Tracing Example
"
!
• Initial ray:
"%
%
"$
!
! =
0
• Final ray should cross the optical axis at a
distance from the second vertex.
• Multiply the matrices, solve for …
Ray Tracing Example
• Use Mathematica…
Ray Tracing Example
• Use Mathematica…
Ray Tracing Example
• Use Mathematica…
– Object position was at the focal point of the first
refracting surface:
=
−1
• It works!