Recollection of Refraction at a Spherical Interface: For a refraction at a single spherical interface: n

S+

!n!S=

!n " nR

or

R

!n " n#$%

&'(n

S+

R

!n " n#$%

&'(

!n!S= 1

with

!n " nR

being termed the surface power we could write it as: Inserting an object distance of infinity we can infer the value of the second focal length, inserting an infinite image distance we can infer the first focal distance (remembering the sigh convention that f1 is that it is negative for a positive surface). The two relations are:

f2=

!n R

!n " n

f1= "

!n R

!n " n

Inserting these into the relation immediately above: ! f

1

S+f2

"S= 1

(Which matches equation 18-5 for a thick lens between different media). !n " n

R is called the surface power so we could rewrite the above expressions

f2=

!n

Ps

f1= "

n

Ps

For any system of lenses we develop a framework which will allow us to use the same relations as applied for a refraction at a single spherical interface, provided we measure object distance from H1 and image distances from H2.

MATRIX METHOD for THICK LENSES & LENS SYSTEMS We recast the rule for refraction at a single surface by using Snell’s law at the interface between two media n and n’.

Snell’s law at the interface yields:

)sin()sin( 0 !"!" +#=+ fnn In the small angle approximation:

)()( 0 !"!" +#=+ fnn ( ) fnnnn !!" #=+#$

0

R

y=! "" tan

φ

R

α0

y

αf φ

φ

So :

( ) fnnR

ynn !! "=+"#

0

or rewriting:

( )0

!!n

n

Rn

ynnf

"+

""#=

REFRACTION MATRIX: Consider a column vector

�

y

!

"

# $

%

& '

in which y will always represent the height of the ray off the optical axis and α always represents the angle with respect to the optical axis (by convention the upward moving angle has positive sign). A refraction at a spherical interface will change the direction of the ray from α0 to αf as described above, but does not change the height off of the axis. So, we want a matrix that satisfies:

�

A B

C D

!

" #

$

% &

y0

a0

!

" #

$

% & =

y0

n ' ( n ( )y0

( n R+

n

( n )0

!

"

# #

$

%

& &

So obviously :

�

A B

C D

!

" #

$

% & =

1 0

n ' ( n ( )( n R

n

( n

!

"

# #

$

%

& &

A moment’s reflection will show you that the determinant of the refraction matrix is always equal to n/n’.

TRANSLATION MATRIX: A ray moving at angle α0 relative to the optical axis is translated a distance L:

�

y f ! y0

L= tan "

0( ) #"0

A translation takes you from yo to yf without changing the direction, so the matrix is characterized by:

�

A B

C D

!

" #

$

% &

y0

'0

!

" #

$

% & =

yo + '0L

'0

!

" #

$

% &

The appropriate matrix is obviously:

�

A B

C D

!

" #

$

% & =

1 L

0 1

!

" #

$

% &

Obviously the determinant of a translation matrix is equal to 1.

y0

L

α0 yf

We will have the lens system obey a similar relation: n

S+

!n

!S= Peq

f1="n

Peq

f2=

!n

Peq

or: ! f

1

S+f2

"S= 1

but object distance S must be measured relative to the pane H1 and the image distance S’ must be measured from plane H2. That is.:

�

S = do

+ r

! S = di" s

CONSTRUCTING THE THICK LENS SYSTEM MATRIX:

�

We construct the thick lens by taking the refraction matrix for the refraction at the front interface- from index n to nlens across a spherical interface of radius R1. Then we take the translation matrix for a translation by the thickness t of the lens (distance from front vertex to back vertex), followed by the refraction matrix for a spherical interface between index nlens and n’ across a radius of R2.

�

A B

C D

!

" #

$

% & =

1 0

nlens

' ( n ( )( n R2

nlens

( n

!

"

# #

$

%

& &

1 t

0 1

!

" #

$

% &

1 0

n ' nlens( )

nlens

R1

n

nlens

!

"

# #

$

%

& &

Multiplying the two rightmost matrices:

�

A B

C D

!

" #

$

% & =

1 0

nlens

' ( n ( )( n R2

nlens

( n

!

"

# #

$

%

& &

1+t n ' n

lens( )

nlens

R1

tn

nlens

n ' nlens( )

nlens

R1

n

nlens

!

"

# # # #

$

%

& & & &

and thus:

A B

C D

!

"#

$

%& =

1+t n ' n

lens( )nlensR1

tn

nlens

nlens

' (n( )(n R2

+n ' n

lens( )(n R1

+nlens

' (n( ) n ' nlens( )tnlens

(n R1R2

n

(n+tn

nlens

nlens

' (n( )(n R2

!

"

#####

$

%

&&&&&

Any system of lenses could be characterized by a series of refractions and translations by continuing the appropriate matrix multiplications:

�

M = ...R4T3R3T2R2T1R1

Since:

�

detM1*detM

2= det M

1M

2( ) and all the translation matricies have determinant of 1, and all the refraction matricies are equal to the input index over output index, the determinant for the system matrix as a whole is always

�

detM =nin

nout

where nin is the index before the front vertex, and nout is the index after the back vertex. INTERPRETING THE SYSTEM MATRIX An idealization of the system matrix would consider the system to consist of (1) a translation T1 from the system input plane to the principal plane H1, which is a distance of +r, followed by (2) an ideal refraction R1 as if at a spherical interface between the media n and n’ with a surface power of Peq having plane H1 as its input plane and H2 as its output plane followed (3) a translation from H2 to the system output plane, a distance of –s ( the negative sign results from s being measured from the back vertex to H2 while in our case we go the other direction.)

�

TR1T1

=2

1 !s

0 1

"

# $

%

& '

1 0

!Peq

( n

n

( n

"

#

$

$

%

&

'

'

1 r

0 1

"

# $

%

& ' =

A B

C D

"

# $

%

& '

�

1 !s

0 1

"

# $

%

& '

1 r

!Peq

( n

n

( n !

Peqr

( n

"

#

$

$

%

&

'

'

=A B

C D

"

# $

%

& '

�

1+ sPeq

! n r "

sn

! n "

srPeq

! n "Peq

! n

n

! n "

Peqr

! n

#

$

%

%

%

&

'

(

(

(

=A B

C D

#

$ %

&

' (

Solving:

�

A =1+ sPeq

! n

gives:

�

s =A !1Peq

" n

#

$ %

&

' (

=1! A

C

Solving:

�

D =n

! n "

Peqr

! n

gives:

�

r =

D!n

" n !Peq

" n

#

$ %

&

' (

=

D!n

" n

C

Returning to the relation:

�

n

S+

! n

! S = Peq

�

! n

f2

= Peq

Therefore

�

C =!Peq

" n =!1

f2

and since from equation 4-2 we have:

�

f2

= !" n

nf1

�

f1

=n

! n C

Substituting:

�

1

f1

=! n C

n

into the result for C in a thick lens:

�

C =n

lens! " n ( )" n R2

+n ! n

lens( )" n R1

+n

lens! " n ( ) n ! n

lens( )t

nlens

" n R1R2

we can recover equation 18-1:

�

1

f1

=nlens ! " n ( )

nR2

!nlens ! n( )

nR1

+nlens ! " n ( ) nlens ! n( )t

nlensnR1R2

Similarly the thick lens result for A:

A = 1+t n ! n

lens( )nlensR1

can substitute into:

s =1! A

C

s =!t n ! nlens( )nlensR1C

=! nlens ! n( ) f2t

nlensR1

As found in 18-3. If we substitute the expression for the thick lens value for D:

D =n

!n+tn

nlens

nlens

" !n( )!n R2

into:

r =

D !n

"n

C

gives:

r =tn

nlens

nlens ! "n( )"n R2C

=nlens ! "n( ) f1tnlensR2

agreeing with the other equation of 18-3.

MEASURING THE SYSTEM MATRIX: Consider a ray leaving an object point do from the front vertex of the system at height off ho off the axis (equal to the object height), as shown in the figure below: Using translation matrixes to take us from the object point to the front vertex, and from the back vertex to the image point, we transport the ray going all the way through the system from the object point to the image point a distance di beyond the back vertex and at height hi.

ho

do

α0

hi

αf

0

di

�

1 di

0 1

!

" #

$

% &

A B

C D

!

" #

$

% &

1 do

0 1

!

" #

$

% &

ho

'0

!

" #

$

% & =

hi

' f

!

" #

$

% &

Multiplying out the rightmost pair:

�

1 di

0 1

!

" #

$

% &

A Ado + B

C Cdo + D

!

" #

$

% &

ho

'0

!

" #

$

% & =

hi

' f

!

" #

$

% &

and multiplying again:

�

A + Cdi Ado + B + Cdo + D( )diC Cdo + D

!

" #

$

% & ho

'0

!

" #

$

% & =

hi

' f

!

" #

$

% &

Multiplying out the first to get the value for hi :

�

A + Cdi( )ho + Ad

o+ B + Cd

o+ D( )di[ ]!0

= hi

It is required, however, that the image height be independent of the incident angle, since all the rays from the object point should reach the same image point. Thus it is necessary that the content of the square bracket equal zero:

�

Ado

+ B + Cdo

+ D( )di = 0 So

�

A + Cdi( )ho = h

i

m !hi

ho

= A + Cdi( )

The matrix above is now:

�

A + Cdi

0

C Cdo

+ D

!

" #

$

% &

The determinant, as for any system matrix, must be equal to the ratio of the input index over the output index. So:

�

A + Cdi( ) Cdo + D( ) =

nin

nout

So we get:

�

m Cdo

+ D( ) =nin

nout

Cdo

+ D =nin

noutm

And since

�

Ado

+ B = ! Cdo

+ D( )di ,

�

Ado

+ B =!d

inin

mnout

The typical case has nin equal to nout so in those typical cases:

�

Cdo

+ D =1

m

�

Ado

+ B =!d

i

m

along with:

�

m = A + Cdi( )

this gives us three linear relations involving A, B. C. D and measured quantites di, do and m. MEASURING A, B, C, and D: The procedure to measure A,B,C,D is:

(1) Plot 1/m vs do – in a linear fit the slope is C and the intercept D (2) Plot –di/m vs d0 – the slope is A and the intercept is B

(3) Plot m vs di – the slope is C and the intercept is A

(4)

1/m

di

Slope C intercept D