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