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Physics 42200
Waves & Oscillations
Lecture 32 – Geometric Optics
Spring 2013 Semester
Matthew Jones
Thin Lens Equation
First surface:
Second surface:
Add these equations and simplify using
1 1
1
1
1
(Thin lens equation)
1 and
→ 0:
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 (f.f.l.)
Primary principal plane
First principal point
Second focal point (b.f.l.)
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 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
ℎ
ℎ
1
1
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
– Law of reflection:
sin
(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:
nt1α t1 = ni1αi1 − D 1 yi1
yt1=yi1
Equivalent matrix
representation:
yt1 = 0 ⋅ ni1α i1 + yi1
note: paraxial approximation
Reminder:
 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 = nt1α t1 + 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
system matrix of thick lens
For thin lens dl=0
1 - 1/ f 
A = 

1 
0
1
Thick Lens Matrix and Cardinal Points
ni1 (1 − a11 )
V1H1 =
− a12
ni 2 (a22 − 1)
V2 H 2 =
− a12
 D 2d l
1 −
nl
A =
 dl
 n
l

D 1D 2d l 
- D1 - D 2 - +

nl   a11 a12 

= 
Dd
  a21 a22 
1− 1 l

nl

ni1
nt 2
a12 = −
=−
fo
fi
in air
a12 = −
1
f
effective focal length
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
note: R<0
 − 1 − 2n / R 
M = 

1 
 0
 nα r 
 nαi 


 = M
 yr 
 yi 
yr = yi
nα r = −nαi − 2nyi / R
α r = −αi − 2 yi / R