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 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 (f.f.l.)
Primary principal plane
First principal point
Second focal point (b.f.l.)
Secondary principal plane
Second principal point
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
Note: in air (n=1)
1 1 1 xo xi = f 2 1 1 (nl − 1)d l
= (nl − 1) −
+ = effective 1
+
so si f focal length:
f R R n R R
1 2 l 1 2
f (nl − 1)d l f (nl − 1)d l
Principal planes: h1 = − h2 = −
nl R2 nl R1
yi si xi f
Magnification: MT ≡ =− =− =−
yo so f xo
Thick Lens Calculations
1. Calculate focal length
1 1 1 1
1
2. Calculate positions of principal planes
1
ℎ
1
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
+ = f f
so si f
30 cm
so si
1 1 (nl − 1)d l 1 0.5 ⋅ 1 1
= (nl − 1) −
1 1
+ = 0.5 − + cm
f R
1 R 2 n R R
l 1 2 20 − 40 1.5 ⋅ 20 ⋅ 40
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 + =
26.8 ⋅ 0.5 ⋅ 1 30.22cm si 26.8cm
h2 = − cm = −0.44cm
20 ⋅ 1.5 si = 238 cm
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): 1 1 1
= + −
d H11H1 = fd f 2
f f1 f 2 f1 f 2
H 22 H 2 = fd f1
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
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 yt1 = 0 ⋅ ni1α i1 + yi1 A B α Aα + By
≡
C D y
Cα + Dy
Equivalent matrix nt1α t1 1 - D 1 ni1αi1
representation: =
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 ni 2αi 2 1 0 nt1α t1
presentation: =
yi 2 d 21 nt1 1 yt1
≡ rt1 - input ray
≡ T21 - transfer matrix
ri 2 = T21rt1
≡ ri2 - output ray
System Matrix
yi2 yi2
yi1 y
t1
ri1 rt1 ri2 rt2 ri3 rt3
R1 T21 T32 R3
rt1 = R1ri1 ri 2 = T21rt1 = T21R1ri1
Thick lens ray transfer: rt 2 = R 2T21R1ri1
System matrix: A = R 2T21R1 rt 2 = A ri1
Can treat any system with single system matrix
d
Thick Lens Matrix
nl A = R 2T21R1
1 - D
yi1 y
yi2 yi2
R =
0 1
t1
1 0
Reminder: T =
A B a b Aa + Bc Ab + Bd d n 1
≡
C D c d Ca + Dc Cb + Dd
1 1 1
1
1 - D 2 1 0 1 - D 1
A =
0 1 d l nl 1 0 1 system matrix of thick lens
D 2d l D 1D 2d l
1 − - D1 - D 2 - + For thin lens dl=0
nl nl
A = 1 - 1/ f
dl D 1d l A =
n 1−
l nl 0 1
Thick Lens Matrix and Cardinal Points
ni1 (1 − a11 )
V1H1 =
− a12
ni 2 (a22 − 1)
V2 H 2 =
− a12
D 2d l D 1D 2d l
1 − - D1 - D 2 - +
nl nl a11 a12
A = =
dl Dd a21 a22
n 1− 1 l
l nl
ni1 nt 2 in air 1
a12 = − =− a12 = − effective focal length
fo fi f
Matrix Treatment: example
rI
rO
rI = T1A l T2rO
nIα I 1 0 a11 a12 1 0 nOαO
=
y I d I 2 nI 1 a21 a22 d1O nO 1 yO
(Detailed example with thick lenses and numbers: page 250)
Mirror Matrix
note: R<0
− 1 − 2n / R
M =
0 1
nα r nαi
= M
yr yi
yr = yi
nα r = −nαi − 2nyi / R
α r = −αi − 2 yi / R
