Physics 42200
Waves & Oscillations
Spring 2013 SemesterMatthew Jones
Lecture 32 – Geometric Optics
Thin Lens Equation
Add these equations and simplify using �� � 1 and � → 0:
1
�1
��� �� 1
1
�� 1
��
(Thin lens equation)
First surface:��
����
������ ��
��Second surface:
��
��� ���
������ ��
��
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
First focal point (f.f.l.)
Second focal point (b.f.l.)
Primary principal plane
Secondary principal plane
First principal point
Second principal point
Thick Lens: definitions
Nodal points
If media on both sides
has the same n, then:
N1=H1 and N2=H2
Fo Fi H1 H2 N1 N2 - cardinal points
Thick Lens: Principal Planes
Principal planes can lie outside the lens:
Thick Lens: equations
fss io
111 =+
Note: in air (n=1)
2fxx io = effective
focal length:( ) ( )
−+−−=2121
1111
1
RRn
dn
RRn
f l
lll
( )2
1
1
Rn
dnfh
l
ll −−= ( )1
2
1
Rn
dnfh
l
ll −−=Principal planes:
o
i
o
i
o
iT x
f
f
x
s
s
y
yM −=−=−=≡Magnification:
Thick Lens Calculations
1. Calculate focal length1
�� � 1
1
�� 1
��
� 1 �
�����
2. Calculate positions of principal planes
ℎ� � � � 1 �
���
ℎ� � � � 1 �
���
3. Calculate object distance, �, measured from principal plane
4. Calculate image distance:1
�1
���1
�
5. Calculate magnification, �� � ��/�
Thick Lens: exampleFind 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
cm 30.22cm22.0cm30 =+=os
so si
30 cm
fss io
111 =+
( ) ( )cm
1
40205.1
15.0
40
1
20
15.0
1111
1
2121
⋅⋅⋅+
−−=
−+−−=RRn
dn
RRn
f l
lll
cm 8.26=f
cm22.0cm5.140
15.08.261 =
⋅−⋅⋅−=h
f
cm44.0cm5.120
15.08.262 −=
⋅⋅⋅−=h
f
cm8.26
11
cm22.30
1 =+is
cm 238=is
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):
2121
111
ff
d
fff−+=
1222
2111
ffdHH
ffdHH
=
=
Example: page 246
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
note: paraxial approximation
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.
111111 iiitt ynn D−= αα
1 1 1 10t i i iy n yα= ⋅ +
Refraction:
yt1=yi1
=
1
111
1
11
10
1
i
ii
t
tt
y
n
y
n αα D-Equivalent matrix
representation:
Reminder:
++
≡
DyC
ByA
yDC
BA
ααα
≡ rt1 - output ray
≡ ri1 - input ray
≡ R1 - refraction matrix111 it rRr =
Matrix: Transfer Through Space
11122 0 tttii ynn ⋅+= αα
11212 tti ydy +⋅= α
=
1
11
1212
22
1
01
t
tt
ti
ii
y
n
ndy
n ααEquivalent matrix
presentation:
≡ ri2 - output ray
≡ rt1 - input ray
≡ T21 - transfer matrix
1212 ti rTr =
Transfer:
yt1
yi2
112122 it rRTRr =
System Matrix
yt1yi1
yi2 yi2
ri1 rt1 ri2 rt2
R1 T21
ri3 rt3
R3T32
111 it rRr = 11211212 iti rRTrTr ==
Thick lens ray transfer:
1212 RTR=ASystem matrix:12 it rr A=
Can treat any system with single system matrix
1212 RTR=A
=10
1 D-R
=1
01
ndTReminder:
++++
≡
DdCbDcCa
BdAbBcAa
dc
ba
DC
BA
=10
1
1
01
10
1 12 D-D-
ll ndA
yt1yi1
yi2 yi2
nl
dThick Lens Matrix
−
+−=
l
l
l
l
l
l
l
l
n
d
n
dn
d
n
d
1
2121
2
1
1
D
DD-D-D-
D
A
system matrix of thick lens
For thin lens dl=0
=10
/11 f-A
1
�� �� 1
1
�� 1
��
Thick Lens Matrix and Cardinal Points
=
−
+−=
2221
1211
1
2121
2
1
1
aa
aa
n
d
n
dn
d
n
d
l
l
l
l
l
l
l
l
D
DD-D-D-
D
A
i
t
o
i
f
n
f
na 21
12 −=−= in air
fa
112 −= effective focal length
( )12
11111
1
a
anHV i
−−=
( )12
22222
1
a
anHV i
−−=
Matrix Treatment: example
rI
rO
OlI rTTr 21A=
=
O
OO
OOIII
II
y
n
ndaa
aa
ndy
n αα1
01
1
01
12221
1211
2
(Detailed example with thick lenses and numbers: page 250)
−−=
10
/21 RnM
Mirror Matrix
=
i
i
r
r
y
n
y
n ααM
ir yy =
Rnynn iir /2−−= αα
note: R<0
Ryiir /2−−= αα