Gerrard, A._ Burch, J. M. - Introduction to Matrix Methods in Optics-Dover Publications (1975)

~~~~

Entrance pupil

!

II
II

RP II I RP2

'
1I\IT0015m

IVI

I Ir"'0 om

p=+10D

0·1 m

Figure A

.Stop Exit pupil here

RP,I ~I

I I

I Eyer x RP
6
0±;5m
II

H~;25mI II IRPRPs I
4
I

I

I VI I

II I
~~II :::II'II 0·01 m
D I

I II I
I

A.8

280

L-matrices

Observer's eye 1 Focal distance Focal distance Objective (a trivial
posi.tion, d case)
Eye- ens of eye-lens of objective lens
M = M, [1 0L, =
M. M, M, M, = L, 1].,.10

_[00'1]

L, -
-10 1

L,. = [--100'10l'll]

M,L,,, M! L, = [-0'1 O'll]
o -10
(O'll-lOdl] = M
L6 = [-0o'1 (O'll-lOdl]
-10
-10

Table of data required for calculating aperture
effects in a lens system

Referenae pZane RPI RP3 RP4 RPe

Stop radius J. 0·015 0·00325 0·0025 0·002
'l- 1 0 - 0·1 - 0·1

(L11) i

(L 12 ) i 0 0·1 o-i.i (o·n-lod)

A. 5. 2 Determination of aperture stop

Since the object is at infinity, we identify the iris
by seeking a minimum value for

Ymax(i) = IJi/(Lll)i l

281

The values obtained can be added as an additional row
to the above table:

Reference p"lane RP1 RP3 RPI+ RP6

J.
1-
10 - 0 1 51 00 0'025 0'02
Ymax{i) = (Lll )i

The smallest value of Ymax{i) is shown boxed, and this

indicates that the iris is provided at RP1 by the lens
rim of the objective. Since RP1 is at one end of the

system, the iris is itself the entrance pupil. For

location of the exit pupil, it is clear that the Q-
matrix Q1 is the same as M, since L1 is just a unit
matrix. We thus obtain E2 = - (M12 M22) = (- d + 0-011).
But, since RP6 is itself located a distance d to the

right of the eye-lens, this means that the exit pupil

or Ramsden disc is 11 rom to the right of the eye-lens.

Its radius is IJ1/M22/ = 0'0015. We can see already
that if the observer positions himself 'correctly', so
that d = 0-011, his own 4 rom pupil diameter will be
sufficient to transmit all the light from the 3 rom

diameter exit pupil. But the lateral tolerance on his
head position is only ± 0-5 rom.

A.5.J Determination of fie"ld stops

To determine which reference plane provides the
field stop, we now consider the maximum available field
angle V for a principal ray through the centre of the

entrance pupil. Tabulating the data in the form of an
additional row, we find:

Reference p"lane RP1 RP3 RPI+ RP6

J. 00 0-002
= - -1--
Vmax (1•-) (iris) 0-0325 10- 0 2271 o-rr-ios
(L12) i
(oo i f

d = 0-011)

Ignoring for the moment the effects of the observer's
eye position at RP6' we see that it is the rim of the
eye-lens at RPI+ rather than the focal plane diaphragm
at RP3 that must be considered the field stop.

282

Location of the entrance window is at a distance

F1 = - (L 12 fLu) 4 = + 1'1 metres to the left of the

objective at RP1. Its radius is !J4/(L11)41 = 0·025
metres (that is 5 em diameter). Since the field stop
is the rim of the last lens in the system, it is itself
the exit window, and no calculation is needed.

(Notice that, when a lens rim provides a stop, it
makes no difference which of two reference planes we
use in calculating its obscuration effect. The inter-
vening lens matrix is such that it leaves the L11 and
L12 elements unchanged.)

A.5.4 DetePmination of fieZds of the instrument

Since the entrance window unfortunately does not
coincide with the object plane at infinity, we shall
encounter some effects of vignetting. The angular
field radius corresponding to the cone of principal
rays is obtained, as in section A.4.l, by projection
of the entrance window onto the object plane, and its
value is

f(E1 - F1) (L11) 0·0025
1'1 x 0.1 = 0·0227 radians

(the angle-value already boxed under RP4 in the cal-

culation of Vmax). (Here, f = 4.)

To determine the effects of vignetting, we shall
construct a (~,n) diagram. For each kth stop, the

boundary lines for an object field at infinity are

given by the equations

Since s = 1 for this instrument - the iris is at the

front - we can immediately put (L11)s = 1 and

(L12)S = 0, and obtain the simple equation

~[- (L12)k]+ nJs(Lll)k = ± J k

One of the boundary lines is that joining the point

~ = 0, n = Jk/Js(Lll)k to the point ~ = - Jk/(L12)k,
n = o. The other line runs parallel to it at the

same distance on the other side of the origin.
It will be noticed that, apart from division by a

normalising factor, Ymax(s) = J s ' we have already cal-

283

culated for each stop all the data necessary to plot
the intercepts. The ~-intercepts are listed in section

A.5.3, and, on division by J s = 0·015 and reinserting

signs, the data in section A.5.2 yield the n-intercepts:

Reference plane RPl RP3 RPIt RP6

n-intercepts 1 00 -1·667 -1·333

using these calculated values for the intercepts, we
now construct (~,n) diagrams as shown in Figure A.9.
In the central diagram, for which the observer's eye

is assumed to be in its correct position at the exit
pupil, the main field-limiting action of the eyepiece

is represented by a parallelogram. Only the extreme
tips of this parallelogram are truncated by the lines
~ = ± 0·0325 which represent the focal plane diaphragm.
For visual observation, this Keplerian arrangement is
extremely unsatisfactory, since only a small central
portion of the field remains free from vignetting. It
is for this reason that nearly all astronomical eye-
pieces incorporate a field lens which helps to image
the iris onto a near-central region of the eye-lens.
(For a telescope to be used backwards as a beam-
expander for a low-power laser, the situation is quite
different; the above arrangement might well be used,
supplemented by an extremely small focal plane dia-
phragm to act as a spatial filter.)

Either from the diagram, or by calculating ~-values
for the relevant intersection points, we obtain the
following angular radii for the object field at infin-
ity:

Radius Diameter

(in radians) (in degrees)

Field of 0·0091 1·04
full illumination

Field for 0·0227 2·60
principal rays

Total field 0·0325 3·72

284

Since the system functions as a xlO telescope, the
angles subtended by fields observed in the image space
will be ten times larger. But for rays travelling at
0·2 radians to the axis, the paraxial approximation
will become imperfect: unless the eyepiece has been
designed to be orthoscopic, we may expect discrepancies
of 1 or 2 per cent between actual and nominal values.

A.5.5 Effect of obse~ve~'s eye position

In the left-hand diagram of Figure A.9, an additional
pair of boundary lines has been added to indicate the
effect of moving the observer's eye 5 mm too close to
the eyepiece. (For d = 0·006, the first ~-intercept,
0·002/(0·11 - 10d) becomes 0 004.) In this case, no
additional obscuration is caused, since the situation
is still dominated by the inadequate diameter of the
eye-lens

Focal plane
stop

Ins

-1 -1 -1

Eye at 6mm Eye at 11mm Eye at16mm
from eye-lens from eye-lens from eye-lens

(.p•.,,) Diagrams for x10 telescope with single eye-lens

Figure A.9

285

In the right-hand diagram, however, we see that
moving the head 5 mm away from the instrument (so that
d = 0'016 and the first ¢-intercept is - 0'04) begins
to introduce additional obscuration on the right-hand
side of the diagram, the shape of which is now slightly
more sYmmetrical. It is clear, however, that the focal
plane diaphragm will cease to be visible from this eye
position, the total field now being reduced to 0'0281
radians radius, or 3'220 diameter.

In a well-designed hand-held telescope or binocular,
the entrance window is at infinity, and the resultant
(¢,n) diagram has an ideal rectangular shape, But,
unless the head is centred carefully with respect to
the exit pupil, both laterally and longitudinally, the
performance of the instrument can still be spoiled by·
vignetting, and also by unnecessary eye aberrations.
Caveat observator:

286

Appendix B
Matrix Representation of
Centring and Squaring Errors

B.l THE USE OF AUGMENTED 3 x 3 MATRICES
In an ideal centred optical system, we are able to

use a 2 x 2 ray-transfer matrix to represent the homo-
geneous equations

in the form

In an actual system, however, the position of the
optical axis in RP2 may be displaced slightly from its
assumed position by a small distance ~Yl' for example
because a plane-parallel window has been tilted slightly
so that it is no longer square to the z-axis. In many
cases, also, the exact direction in which the optical
axis is pointing deviates slightly from its assumed
direction by a small angle ~Vl. This effect can occur
if a 'plane-parallel' plate is slightly wedge-shaped
or if one of the lenses has not been properly centred.

Under these circumstances, constant terms ~Yl and
~Vl have to be added to the above two equations, with
the result that they are no longer homogeneous. We
can, if we wish, represent the transformation by the
equation:

287

But the output ray vector is no longer obtained by
simple matrix multiplication of the input ray vector.
There is, however, a trick which enables us to retain
this feature of matrix multiplication: this is the
use of a 3 x 3 'augmented matrix'. To see how such a
matrix is formed, we write down our two ray-transfer
equations and then set down a third 'dummy' equation,
which at first sight seems completely trivial:

1 0(1) + 0(1) + 1
This can be rewritten

It is easily verified that, if [; :}s uni.moduLar ,
then so is the above 3 x 3 'augmented matrix'. Because
two of the elements in the added row are zero, and the
third is + 1, their determinants must be the same.
B.2 MULTIPLICATION OF AUGMENTED MA~RICES

Consider now an optical system for which the multi-
plication of the normal 2 x 2 matrix chain gives the
overall matrix:

M MnMn-l.'. Mr . . . M3M2Ml
where M = [Ar Br ] is the unimodular matrix repres-

r c- Dr

enting transfer from the rth intermediate reference
plane to the (r+l)th. Let us suppose that for each

288

matrix Mr there are additional small misalignment terms
~Yr,~Vr that need to be considered.

We shall therefore need to consider a corresponding
chain of augmented 3 x 3 matrices, which we shall rep-
resent by the curly symbols AI, A2, etc. We can write

A = An An-I' • • Ar . . • A 3 uH2 .Jt1
where

Br

o

Consider now the multiplication of two such matrices:

['A~1 B2 Bl

'Y'l[' 'Y'JC2 D2 ~V2 C1 Dl t,V 1

00 100 1

['A'+ B,C, A2 Bl+ B 2D1 A2~Yl + B 2t,V1+
C2 Bl+ D2Dl
C2A 1+ D2Cl 'Y'JC2~Yl+ D2~Vl+ /W2
0
0 1

If we now examine the elements of this product matrix,
we find that there is a square of four elements which
are exactly the same as would be obtained by calculating
the 2 x 2 product

M2Ml = A 2 B2][AI
[ C2 D2 Cl

Of the other elements, those in the third row are of
little interest; but the top two elements of the third
column can be regarded as new effective values of ~Y
and ~V for the two matrices taken together. Writing

289
them as ~Y21 and ~V21 and extracting them from the
Af2~ product matrix that we have just calculated, we
find by inspection that

Proceeding in the same way to form the product
Af 3 Af2 Af1 = j{3 ( Af2 Jlt1), we find that

and, similarly,

If we use the symbol Qr' as in appendix A, to denote
the product of a matrix chain beginning with Mn and

finishing with ~1r, then the effective values of ~y and

~V for the complete matrix chain M = MnMn-1 • . • M2M1

can be represented by the following expression:

effective

Alternatively, since Qr+1Lr+l = M and Qr+l = -1 ,

MLr +l
we can express this result in terms of L-matrices, and

290

[:]

effective

An example of the calculation of the various L-matrices
has been discussed in appendix A.

The overall matrix of the system can now be written
down as

where A, B, C and D are the elements of the usual over-

all ray-transfer matrix M, and t>y and 6V are the overall
misalignment terms, calculated as above.

B.3 EFFECT OF MISALIGNMENT OF AN OPTICAL RESONATOR
If we assume that the overall augmented matrix ~ that

we have calculated represents the effect of a single
round trip in an optical resonator, it is evident that

a ray [:] ,:avelling down the axis in RPI generates in

[O:JRP, a ray which has a different path. TO find a

new effective optical axis in this situation, we seek

l::Jan input ray that repeats itself exactly - in

effect an eigenvector of the matrix.lt. Obviously, we
must have

291

for this to be achieved.
Multiplying out, we obtain

Ayo + BVo + l1y yo

Cyo + DVo + I1V Vo

and
1 =1

a trivial result again. Rearranging the first two
equations, we find

(1 - A)yo - BVo = l1y

(1 -- Cyo + D)Vo = I1V

Hence I1Y I-B - A)

1I(1-u«:I1V (1 - D) and Vo = -C Y/
( 1 - A) I1
A) -B 1
II1V
-C (1 - D) 1 -C
-B

(1 - D)

The determinant in the denominator has the value 1.

(1 - A) (1 - D) - BC = (2 - A - D), since (AD - BC)

Provided that (2 - A - D) does not vanish, we then

obtain the solution

yo = (1 - D) l1y + BI1 V and Vo Cl1y + (1 - A) 11V
(2 - A - D) (2 - A - D)

Alternatively, this solution can be expressed in the
matrix form

IYo] Y]
Lv = (2 - 1 D) (1 - ~ 1 ) [I1
A- I1V
o

See section III.4 for a discussion of some of the
implications of this result for a simple two-mirror
resonator.

292

Appendix C

Statistical Derivation of
the Stokes Parameters

In this appendix we consider the problem of specifying
the state of polarization for a light beam which con-
tains a large number of independent oscillating com-
ponents.

For a given single fuLLy poLarized disturbance, the

Stokes parameters were defined in section IV.3 in terms
of the amplitudes and phases of the transverse electric
field components

Ex = Hei<jJ
and

Ey = Kei 1jJ

where Hand K are real and positive amplitudes and the
phase difference (1jJ - <jJ> is given by~. In terms of

H, K and ~, the defining equations were

=and V 2HKsin~

The first parameter I represents the intensity, or
irradiance; it will be recalled that the usual method
of calculating this from the electric field vector is
to form the bracket product

293

If we replace each of the above matrices by its com-
plex conjugate transpose, we obtain a new product which
is not a scalar but a 2 x 2 Hermitian matrix, sometimes
referred to as a 'coherency matrix'.

¢] [ -i¢Hei . [HKe-1-lji] = 2 HKe-itJ.]

[ ',1, He • K2

Ke1-'I' HKe1-tJ.

since (lji - ¢) = tJ.. It will be seen immediately that

each of the elements in the above matrix can be re-
expressed in terms of the four real Stokes parameters.
If we denote the coherency matrix by Z, we can write

II+ Q u- iV]

Z ~Lu+ tv I-Q

For any 2 x 2 matrix to be Hermitian, the two diagonal
elements must necessarily be real, and one off-diagonal
element must be the complex conjugate of the other. In
our choice of elements for a Z-matrix, therefore, we
have just four degrees of freedom, and there is one and

only one Z-matrix which corresponds to the four real
elements of a given Stokes column.

Let us now proceed to calculate a Z-matrix

for the case where the transverse electric field com-
ponents arise from the superposition of a large number

N of individual disturbances. We shall assume that

::::':1[::::::]b::::,:::,:'b:':::S:::::d::,:'::::,ll

factor exp(iwt) , which is taken to be the same for all
N disturbances.

(There are perhaps two points to be noted here.
First, the fact that each rth disturbance will tend to
radiate in a slightly different direction, and at a

294

slightly different optical frequency, can be accommo-

dated by taking both ~r and Wr to be rapidly varying

functions of time and of position in the xy-plane.

Second, the student should note that in many textbooks

the time-dependence in a complex exponential expression

for an electromagnetic wave is written with the oppos-
ite sign exp(-iwt).)

Using the principle that for any electromagnetic

field the contributions can be added ZinearZy for each

component we obtain for the resultant electric field

vector the Maxwell column.

N•

LHre'Z-¢r

l'

N.

LKre'Z-wr

l'

The corresponding Z-matrix is then given by

~HN1'e'Z-.~l'

t Kl'ei 1J!r
l"

N •N •
~Hre'Z-~r~Hse-'Z-~S

l' s

N •N •
LKre'Z-W1'~Hse-'Z-~S

rs

In order to prcceed further, we now need to make three
assumptions.

(a) We assume that our main purpose in calculating the
Z-matrix is to obtain values for four Stokes para-
meters - real numbers which can be compared with
those obtained by direct photoelectric measurement.
Ih making such measurements, however, we shall need

to collect the light energy carried by a finite
area of the beam over a finite period of observa-

295

tion, and the values so obtained will correspond to
a process of time-averaging and space-averaging
applied to the elements of the above Z-matrix. The
mean value of Z obtained by such averaging will be
denoted by <Z>.

(bl We shall also assume that for both x-vibrations and

y-vibrations all N oscillating disturbances are

'mutually incoherent'. What this statement means
is that, for any pair of oscillators r and a, the
averaged intensity observed when both are operating

simultaneously is the same as the sum of their
individual intensities. As far as the x-vibrations

are concerned, when both oscillators operate to-
gether they generate an intensity

Ira (Hr e-i~r + Ha e -i~a l (Hr ei~r + H ei~a,

a

Hr2 + Ha2 + 2Hr Ha cos (~r - <Pa'

Ir + I + 2Hr Ha cos(<Pr - ~a'
a

If the two disturbances are mutually incoherent,

this indicates that the averaged value of the

'interference term' or 'mutual intensity' must be

vanishingly small, that is <2HrHacos(~r - ~al> = o.

Since both H-values are positive, and in some cases

can be considered constant over the beam area and

the observation time that is being sampled, the

vanishing of this term implies that the phase dif-

ference (~r - ~al is passing through many complete
cycles of variation, and is equally likely, at any

given instant, to have any value between 0 and 2TI.

Under these circumstances, we can say that all
averaged terms such as <cos (~r - ~al >, <exp[i (~r - ~a)J >
or <exp[-i (~r - ~al] > are vanishingly small.

By similar reasoning applied to the y-vibrations,

we find that <exp[i(ljir - ljia)] > = 0 and

<exp[-i(ljir - ljialJ> = O. There is, of course, an
important exception to this rule, in that, when
r = a, all of these expressions, instead of van-

ishing, are equal to unity.

(cl Finally, we shall assume that for each individual

rth oscillator the phase difference (ljir - ~r) = 6r
is constant. In other words, although ~r itself

296

varies through many cycles during an observation,
exactly the same sequence of variations is followed

::r::~ a:h~:t:r:::::r:~:eM:::e::e::::::b[~rr~- J
Kr e1.-t:.r

and a rapidly varying phase factor exp(i~r •

It follows that the averaged expressions
<exp [±i (1jJr - ~s)J > can be regarded as vanishingly
small if r ~ s, and as equal to exp(± it:.) for the

special case where r = s.

Armed wi th these three assumptions, we can now apply

the averaging process to the Z-matrix, and obtain

NN

L~r.JL"s.HJ r Ks<exp[i (~r -1jJe )] >
<Z> N N

""KLr.JL.s.J r Ks <exp[i (1jJr -1jJe )] >

Because, as we have seen, each of the averaged phase

factors vanishes except when r = S, we can eliminate the

double summation and obtain

N N N
L:Zr
L:H~ LHr Kr exp(-it:.r )
r r
<Z> r
N
N
LK~
L:HrKrexP(ib.r)
r r

We thus obtain the important result that, when the
disturbances being added are mutually incoherent, the
averaged Z-matrix for the compound disturbance is just
the sum of the Z-matrices for the individual disturb-
ances. And, similarly, each of the four resultant
Stokes parameters is equal to the sum of the individual
parameters

u

It follows that, if we know the Stokes parameters

297

for each individual disturbance, or if we can predict
by statistical reasoning their most probable values,
then the parameters for the complete beam can be ob-
tained by straightforward summation. Alternatively,
we can start by determining the (averaged) Z-matrix

elements and then use the formulae

I = (Zll + Z22), Q (Zll - Z22), U
------'=--......;:,.;;...

We shall conclude this appendix by listing, for
several types of light, the Z-matrix, the Stokes
column and, in most cases, the Maxwell column by which
each type may be represented. In listing the columns,
however, we shall normalize them so that they each
represent a beam of unit intensity.

Type (1) UnpoZarized Zight

In this case, the Hr-values and Kr-values will all be

~sitive, and they will have the same mean-square value

H = KT. On the other hand the phase differences ~r

will have no preferred value, so that, on summation
NN

over many disturbances, ~exp(i~r) and ~exp(-i~r)

rr
will be very small in comparison with"N.

The Z-matrix thus becomes

ftH~
lf HKexp(i~ )
~r r r

where =2NH2 2NK2

I = N(H2 + l{2)

When normalized to unit intensity, the corresponding

298 There is, of course,

[~Stokes column Is evi.derrt.Ly ]

no Maxwell column corresponding to this type of light.

Type (2a) Light polar-iced paral-l-el: to the tc-aicie

For this case, every K is zero, so the Z-matrix

becomes

[:~ _[1 :]Z •
: } NH2
0

If type (2a) light is produced by passing type (1)

through an (ideal) linear polarizer, the intensity

remaining is only one-half of the original - namely

Nlf!." instead of (NH2 + NK2).

When normalized to unit intensity, however, the
mStokes column for this :sigh[o:]i S
and the oorres-

ponding Maxwell column ~

Type ( zb) Light p'iane-polax-i zed paral-Lel: to the y-axis
For this case, every H is zero, so the Z-matrix be-

comes

Type (2a) Light pZane-poZarized with the vibration-

p Zane at an anqLe e to the x-axie

This light can be obtained by passing type (1) light

through a polarizer with its pass-plane at angle e to

the x-axis. If we describe the transverse vibration

299

components emerging from the polarizer in terms of

(x' ,y') coordinates, such that the x'-axis coincides

with the pass-plane, we shall have all the Kp-values

equal to zero. If we denote each surviving Hr-value

by Ar , then in terms of the original (x,y) coordinates

we shall have transverse components

Hr Ar cosO and Kr = Ar sinO

\'lith f::,r equal to zero. Our Z-matrix then becomes

N n

4=A~cos20 ~A~cos8sinO

1"

ZN N
~A~cos8sin8 ~A~sin28

rr

[ cos'e Sinecose]
NA2 sin28

sin8cos8

~~~;:]The correspondinr n~rmalized Stokes column and Maxwelland
column are then

Type (2d) Light polarized at angle -8 with respect to
the e-aeie

The results for this case are obtained by substit-

::ai:uting - 8 for 8 throughout section (2c) above. Only
(n~d
2:]un:on[s c::lchanged in sign, and we
l-~~~28
-sin8J

Type (3) Light in circular state of polarization
We start with light of type (2c), with 8 = n/4, or

450 , so that cos28 = 0 and sin28 = 1. If we now

insert a quarter-wave plate with its fast axis along

the x-direction, the y-vibrations of the light will be
retarded in phase by n /2 with respect to the x-

300

vibrations. For each pth disturbance, therefore, we

shall have left-handed circularly polarized light for

which I1p = - 7T/2. Because both cose and sine are 1/12,

the Z-matrix becomes

tA~exp(i;)
p

N

L:A~
p

The corresponding Stokes column and Maxwell column are

therefor{~] and ), [~]

If we rotate the quarter-wave plate through 900 , so

that it advances the y-vibrations by 7T /2, the phase

difference I1p will become + 7T /2 and we shall generate

Pight-handed circularly polarized light, for which the

[~]normalized Stokes column and Maxwell column are

and), [:]

Type (4) ElZiptically polaPized light with axes of the
ellipse in the x- and y-dipections

As with type (3) light, we assume that type (2c) light

is sent through a quarter-wave plate with its fast

axis parallel to the x-axis. In this case, however,

the initial plane of the polarizer is set at some gen-
eral angle e which is not equal to 450 •

The Z-matrix is then

N N

L:A~cos2e i L:A~sinecose
pp

ZN

-i L:A~sinecose

p

301

cos26 iSin6cos~

NIf2 [ - i s i n 6 c o s 6 Jsin26

For this Zeit-handed elliptical state of polarization,

::::::O:::,t:: [no::~::z]ed.:,tok[esc::~]umnand Maxwell

-Si~26 -isin6

As with type (3) light, if the quarter-wave plate is

rotated through 900 then it is right-handed elliptic-

ally polarized light that will be generated, and the

::::::p::::n:en[o~::]zed'::Oke~~c::~umn]and Maxwell

. 26 ~s~n6

s~n

302

Appendix D

Derivation of
Mueller Matrices

D.l THE POLARIZER
Let the pass-plane of a linear polarizer (for example

a sheet of polaroid or a Nicol prism) make an angle e
with the x-axis of an arbitrary rectangular coordinate

system. Let C1 = cose, C2 = cos2e, 31 = sine and

32 = sin2e. Let the matrix be

xBT D

E F GH

Z=

J KLM

NPR3

(This 4 x 4 matrix Z is not to be confused with the
2 x 2 complex I coherency matrix I Z used in appendix C.)

(a) The device polarizes unpolarized light at an angle
e with the x-axis, that is it turns type (1) light
into type (2c) light (see the list of Stokes para-
meters in appendix C). (We shall use the symbol
W = I/2, where I represents the intensity of the
unpolarized input beam.) Therefore,

W(Column matrix of Stokes Z x 2W(Column matrix of

parameters of type 2c) Stokes parameters

of type 1)

W X B T D 2W

E F GH o

J KL M o

o NPR3 o

303

that is W = 2WX + B.O + T.O + D.O, WC2 = 2WE +
W82 = 2WJ + •.• , 0 = 2WN + ••• (A string of dots

means some zero terms in the sum.) On cancelling

W, we obtain 2X = 1, X = 1/2. Similarly, E = (;2/2,

J = 82/2, N = o.

(b) The device leaves light already polarized at angle
6 with the x-axis unaffected, that is it turns type

(2c) light into type (2c) light.

W 1/2 B T D W

WC2 C2/2 F G H WC2

W8 2 82/2 K L M W82

0 0 PR8 0 (0.1)
that is

W= W/2 + BWC2 + TW82 + D.O

WC2 WC2/2 + FWC2 + GW82 + H.O (D.2)
(0.3)
W82 W82/2 + KWC2 + LW82 + M.o

o 0.0 + PWC2 + RW82 + 8.0 (0.4)

From equation (0.4) PC2 = - R82 for aZZ values of
C2 and of 82. Put C2 = 0; then R = O. Put 82 = 0;

then P = o. Equation (0.1) gives BC2 + T82 = 1/2,

which is used later.

(c) The device turns light of amplitude A polarized

along Ox into light of amplitude Acos6 = AC1, pol-

arized along an~le 6 (by resolution). For the

original beam H = A2 , K = 0, so that I = Q = A2 ,
U = V = O. The component of emergent beam along

Ox is AC12; along ~ it is AC181' Therefore,

H2 = A2Cl'+' K2 = A2Cl sl, b. = O. These lead to

I = A2C{, Q = A2Clc2' U = A2Cl82 , V = 0 (as for

type 2c light). Thus, cancelling A2 throughout,

~ 1/2 B T D 1

ClC2 C2/2 F G H 1

cl82 82 /2 K L M 0
0 00S0
o

304

that is B Cf - 1 /2 = C2 /2

C12 1/2 + B, F C2(Ct - 1/2) = C2 xC2 /2
Ct C2 C2/2 + F,
cf /2
Ct S2 S2/2 + K,
K 2CfSl - S2/2 = 2C13S1 - C1Sl
C1Sl (2Ct - 1) = S2C2/2

Using this B, equation (D.l) now gives

cl/2+ TS 2 1/2
Therefore, T = S2 /2.

(d) Consider circular light of amplitude A (= radius).

The component amplitudes parallel to Ox and to py
are both A, so (see the Stokes column for light of
type 3) I = 2A2, Q = U = 0 and V = 2A2• The device

turns this into a beam which is linearly polarized

at angle 8 with Ox and of amplitude A, so that

H = AC1, K = AS1, {:;. = 0, whence I = A2,
Q = A2(C{ - st) = A2C2' U = 2ACl • ASl • 1 = A2S2'

V = 0 (light of type 2c).

On cancelling A2 throughout, we obtain

1 1/2 C2/2 S2/2 D 2

C2 C2/2 Cl /2 G H 0

S2 S2/2 C2S2/2 L M 0

00 0 0 S2

that is D0
1 1 + 2D,

C2 C2 + 2H, H 0

S2 S2 + 2M, M= 0

0 2S, S =0

305

Using known F, equation (D.2) becomes
C2 C2 /2 + Cl/2 + GS2

C2/2 - Cl/2 = (C2/2) (l - Cl)

Using known K and M, equation (D.3) becomes

S2 S2 /2 + S2 C} r: + LS2

siLS2 S2/2 - S2Cl/2 = (1-Cl)S2/2 /2

L sf /2

Thus, the Mueller matrix of the polarizer is

1 C2 S2 0

C2 Cl C2S2 0
1/2 C2S2 0
Sf
S2

00 00

D.2 THE HALF-WAVE PLATE (optic axis at angle 8 with
the x-direction)

This device introduces a phase shift of TI or 1800
between the 0- and E-vibrations passing through it.
We shall assume that it advances the extraordinary
vibration by TI with respect to the ordinary vibration.
The geometrical effect of the plate is to turn the
plane of vibration of plane-polarized light through
an angle 28, where 8 is the angle between the pass-
plane and the optic axis. No energy is absorbed, and
the vibration-plane is shifted as if by reflection,
to its own mirror image in the optic axis.

Let the matrix of the device be

WB Y D

E F GH

Z=

K L MN

PRXT

306

We shall use the abbreviated code: Cl cos8,
C2 cos28, C4 = cos48, 31 = sin8, 32 sin28,
34 = sin48.

(a) It converts light of unit intensity whose vibration-
plane is in the x-direction into a vibration-plane

at angle 28 with x-direction without loss of inten-

sity, that is it turns type (2a) light into light
of type (2c), except that 28 replaces 8 in the

latter Stokes column to give I = 1, Q = (Cl _ ~2) ,

U = 232C2, and V O.

1 WB Y D 1

cl-sl E F G H 1

2S2C2 KL MN 0

0 PR XT 0

that is (0.5)

W+ B 1

E+ F ci - 3i (0.6)

K+ L 2S2C2 (0.7)

P+ R 0 (0.8)

(b) The device does not affect unpolarized light, that

is type (1) light is turned into type (1) light.

Therefore,

:l1 W B Y 1 that is W = 1
0
0 EF G E =0
=
:J0 K L M 0 K0
0 P0
0 PRX

Combined with equations (0.5) to (0.8) above, these
give

307

B0

F C1 - S1

L

R0

(c) The device leaves light vibrating at angle e with

x-direction unchanged, that is it turns type (2c)
light into type (2c) light. So,

1 1 0 YD1

C2 0 C1-S1 G 8 C2

S2 0 2C2S 2 M N S2

0 0 0 XT 0

that is Y =0
1 1+ YS2,

C2 (Ci - S1) C2 + GS2 , G = [C2 - (1 - 2si) Ci]lS2

= 2S 2C 2 = S ..

S2 2cls2 + MS2, M = 1 - 2cl

sl+ ci - 2cl = S1 - ci = - C..

0 XS2, X = 0

(d) Consider right-handed circularly polarized light

(type 3). The intensity is unaffected, but the
phase difference ts , originally 'IT /2, becomes 3'IT /2,

or equivalently - 'IT/2. This corresponds merely
to a circle described in the opposite sense, so 8 2

and K2 are unaffected. Cos/:' = 0, still, but sin/:'

is now - L The Stokes parameters are thus I = 2A2 ,

Q = 0, U = 01' V = - 2A2 , in accordance with the

left-handed Stokes column for type (3) light. On
cancelling 2A2 , therefo~e, we obtain

308

1 10 0 V1

0 0 cf-sf 2C282 H 0

0 0 2C282 sl-cl N 0

-1 0 0 0 T1

that is V0
1 1+ V,

0 H, H0

0 N, N0

-1 T, T -1

The matrix of the half-wave plate is thus

10 0 0

o C.. S.. 0

o 8.. -C.. 0

o0 0 -1

D.3 QUARTER-WAVE PLATE: GENERAL ORIENTATION (optic

axis at angle e with x-direction)

For a negative uniaxial crystal, the E-vibration in
the plane of the optic axis travels faster than the
O-vibration. We shall assume that the plate has a
thickness such that the O-wave is retarded by 'IT /2

relative to the E-wave. Let C1 = cose, 81 = sine,
C2 = cos2e, 82 = sin2e. Let the Mueller matrix be

WB y V

EFGH

Z=

K L MN

P RXT

309

(a) The device leaves unpolarized light unchanged, that
is it turns type (1) light into type (1) light.
Therefore,

1 1 WB Y D 1 W

0 0 EF GH0 E

=z = K

0 0 KL MN 0

0 0 P RXT 0 P

that is

W= 1

E=O

K =0
P =0

(b) The device leaves alone vibrations along its own

optic axis, at an angle e with the X-direction,

that is it turns type (2c) light into type ( 2c)

light. (For convenience we assume unit intensity.)
Therefore,

1 1 BYD1 Y52.

C2. 0 F G H C2.

52. 0 L M N 52. LC2. + M52.

0 0 RXT 0 RC2. + X52.

that is (0.9)
(0.10)
1 = 1 + BC2. + Y52. (0.11)
(0.12)
C2 = FC2 + G52
52 = LC2 + M52
0 = RC2 + X52

310

(c) To investigate the action of the device on a vibra-
tion parallel to the x-direction, it is very useful
to use the rotation matrices, which convert the
coordinates of a point, referred to one set of rect-
angular axes, into the coordinates referred to a
.pair of axes inclined at an angle 6 with the orig-
inal pair (see Figure D.l). We need to be able to
go both ways. In the diagram:

y

x

Figure D.l

1) Rsin (0 - 6) x Rcos(6 + w)

n Rsinocos6 - Rcososin6 x Rcos6cosw - Rsin6sinw

n ycos6 - xsin6 x ~cos6 - nsin6

~ Rcos (0 - 6) Y Rsin(6 + w)

E; Rcosocos6 + Rsinosin6 y Rsin6cosw + Rcos6sinw

~ xcos6 + ysin6 y ~sin6 + ncos6

In matrix form, we write

[l[z][,] [COS' Sin,] [,]and
[~S'n = -sin6 cos6 y y -Sin~]

sl.n6 cos6 n

The optic axis is now along the ~-direction, so the
E-ray vibrates along the ~-direction, the a-ray along
the n-direction. The original vibration is

x = Asinpt, y = a

311

Therefore,

[:~:::::::::] [-:: Sl] [ASinpt] = [AC1Sinpt]"
Cl 0 -ASlsinpt

In the phase plate, the O-vibration is retarded by
'IT /2:

sin (pt - 'IT (2) = sinptcos'IT /2 - cosptsin'IT /2 = - cospt

Therefore, after phase plate,

E- vi br ati on]

[ O-vibration =

To find the Stokes parameters, we need Hand K, the

components of the final vibration parallel to the

x- and y-directions. Using the rotation matrices

again,

-Sll [Sinpt]

SlCIj cospt

Now suppose ix Hsin (pt + cP), Y = Ksin (pt + S), so

that Hand K are the components of the resultant

vibration in the x- and y-directions and ~ = (S - cP)

is the phase difference between the resultant vib-

rations in the x- and y-direc~ions.

Then x = HsinptcoscP + HcosptsincP

y = KsinptcosS + KcosptsinS

That is HSincP] [Sinpt]
KsinS cospt
X] = fHcos¢
[Y LKcosS

312

In the last two equations, we have [:}rodUCed

frem [SiOFt] in two different ways. The matrices

[sinpt] [x]mUltiP:;::: must therefore
to produce

cospt y

be the same in the two equations, that is (setting
A = unity, for convenience>

HCOS</>

[ Kcos8

We now postmultiply each of these equal matrices

by its transpose (the matrix obtained by inter-

changing the rows and the columns of the original
:1that is the transpose of [:
is

This gives

[HCOS. HSin.] [HCOS. KcosaJ
Kcos8 Ksin8 Hsin</> Ksin8

J[ cf -sf [ cl S,c,]

= SICI SICI -Sf SICl

that is

r2
H

LHKcoS (</>-8>

Therefore, using the fact that corresponding ele-
ments in the two matrices are equal, we find

I = H2 + ,K2 = Cl~ + si + 2slcf = (Cf + Sf> 2 = 1

313

U = 2HKcosb.

giving Sl

V2 = I 2 _ Q2 _ U2 = 1 - ~4 - Sl~2 = 1 - C1 (~2 + sf)

Here, however, V could be either +S2 or - S2. To

find which, we return to the equation

cr -Sf]HCOS¢ . ; [

[ KcosS KsinS = SlCl SlCl

If we equate the determinants of these two matrices,
we find that

HK(cos¢sinS - sin¢cosS) = (ClSl + 8J,3Cl)

Hence immediately,

HKsin(S - ¢) = SlCl

Therefore,

V = 2HKsinb. = 2HKsin(S - ¢) = 2S1Cl = S2

The four Stokes parameters that we have just calculated
are the result of operating on type (2a) light (normal-
ised to unit amplitude), so that

1 1 BYD1

f';BCf 0 F G H 1

=

C2S2,

l:S2
0 L MN 0
0 R XT 0

that is
B 0, from equation (0.9) Y = 0

F cl, from (0.10) GS2 C2, - ci = C2 (1 - cll

= C2,S~ G = C2,S2

314

L (1 Sl
M
x = - C2
R = S2, from (D.12) XS2

(d) Action of the device on right-handed circularly

polarized light, for which, originally, for a beam

of intensity 2 units, x = coswt, Y = - sinwt. As
in (c) H = K = 1, ~ = TI/2, I = V = 2, Q = U = a

In the plate, the a-vibration along the n-direction
is retarded by TI /2:

sin (wt - TI /2) = - coswt, cos (wt - TI /2) = sintllt

So, after the plate,

CCll] ••
coswt

Therefore, final o: and y are given by

Cl] [Sinwt]

C1 coswt

Thus, after the plate,

HCOS¢ HSin¢] fSinwt] =

[Kcos/3 Ksin/3 Lcoswt

as in (c). Again, as in (c), mUltiplying the two-

by-two matrix on each side by its transpose, we get

H2 HKCOS~]

[ HKcos~ K2

315

Therefore, V0

I = H2 + K2 2cf + 2sf 2

Q = H2 - K2 - 4C1Sl = 2S2

U = 2HKcosl:!. 2(Cf - sf> = 2C2

V2 = I 2 _ Q2 _ U2 = 4 - ~2 - 4cf = 0,

Thus 1 B Y D2 2+2D

2

-2S2 0 F GH 0 2H
2C2
=

0 L MN0 2N

0 0 R X T 2 2T

That is D 0, H S2, N = C2, T = 0

Therefore, the Mueller matrix of the quarter-wave
plate is

10 0 0

0 cl C2S2 -S2

0 C2S2 sl C2

0 S2 -C2 0

D.4 GENERAL PHASE PLATE, PRODUCING PHASE RETARDATION 0
IN ORDINARY VIBRATION AND HAVING OPTIC AXIS AT

ANGLE e WITH x-DIRECTION

Let the Mueller matrix be
WB y D

E F GH

Z=

K L MN

p R XT

316

and let Cl = case, C2 = cos2e, 8 1 = sine, 82 = sin2e,
= =S cos~, V sin~.

(a) The device leaves unpolarized light unaffected,
that is it turns type (1) light into type (1)
light. Therefore,

1 W B Y D 1 W that is W = 1

a E F GHa E Ea
K
= K =a

a K L MN a

a P RXT a P Pa

(b) The device leaves unchanged any vibrations along
its own optic axis, that is it turns type (2c)
light into type ( 2c) light:

1 1 B Y D 1 1+BC2+YS2

C2 a F G H C2 FC2 + G82

82 0 L M N 82 LC2 + M82

a a R X T a RC2 + X82

that is 82,

BC2 + Y8 2 = a, FC2 + G82 = C2, LC2 + M82

RC2 + X82 a.

(c) Action of the device on vibration of unit amplit-
ude along the x-direction that is for which

x = sinwt, y = a, I = Q = 1, U = V = a. u;ing the

argument on rotation of axes, decribed in connec-
tion with the quarter-wave plate, if E-vibration
is represented by ~ and the a-vibration by n, then

~ = Clsinwt and n = - 81sinwt.

The phase plate retards the a-vibration by 0 with
respect to the E-vibration. Therefore, after the
phase plate,

n - 81sin(wt - 0) = - 81sinwtcoso + 81coswtsino

317

Thus the final components along the x- and y-axes

are

JCfsinwt + Sf6sinwt - St~coswt

[ Sl'Clsinwt - SlCl6sinwt + SlCl~coswt

lCf + Sf6 - Sf~J [sinwt

[SlCl (1- 6) SlCl~ coswtJ

where 8 = coso and ~ = sino.

By reference to the work on the quarter-wave plate,

if, finally, ta = Hsin (wt + <1» and y = Ksin (wt + a),
=so that 6 (a - <1», then, also,

x ] = fHcoS<l> HS~n<l>] [sinwt]

[ y ~cosa Ks~na coswt

so that

HCOS<l> HSin<pl [Cl + sfs

[ Kcosa KSinaJ = SlCl (1 - 8)

By multiplying each matrix by its transpose, we get,
as before,

H2 JS2C2(1 - 8) 12
[ HKcos6
2sfcf (1 - 8)

This gives I=H2 + K2=1,Q=H2-K2=Cl+ S18,

U = 2HKcos6 = S2C2(l - 6), so that (from I 2 =

Q2 + ,U2 + V2) V2 = Sl~2. It follows that V is

either + S2~ or - S2\.!. To find which, we return

to the equation

HCOS<P HSin<l>]
[ Kcosa Ksina

318

If we equate the determinants of these two matrices,
we find that

HK(sina.cos</J - cososLne) = lJSICl (Cl + SlS) + lJSlCl (1 - I

lJS I Cl

Hence, immediately,

Therefore, 2HKsin (a. - </J) 2S1Cl lJ S2lJ

v = 2HKsint:.

Thus

1 1 BYD1 l+B
0 F GH 1 F
cI + sIs 0 L MN0 L
S2C2 u-s:

S2lJ 0 R XT 0 R

that is B = 0, F cl+sls,L

R = S2lJ.

Combining these with the four equations in section

(b) of this derivation gives Y = 0, G = C2S2(1 - S),

M = sI + cls, X = - C2lJ •

(d) Action of the plate on circularly polarized light,

for which x = coswt, Y = - sinwt, and Stokes para-

meters are (2, 0, 0, 2). As in section (c), we

transform to ~~-axes, along and perpendicular to

the optic axis of the plate, retard the O-vibration
by 0 relative to the E-vibration, then return to

the axes along the x- and y-directions. The result

is

IX]~ [Cl -SlJ [1 OJ [Cl sll [0 IJ [sinwtJ
= Sl 0 coswt
Cl lJ e -Sl CIJ -1

Cll [sinwtJ

-sxj coswt

319

s i nwt ] say
G,
[ coswt

::~:xs::::::t~::'I~e]:::r:~~w~]O ::l:::l:r:::pose.
Ly Lcoswt
It is here convenient to use a standard theorem of
matrix algebra, on the transpose of a product (see
section 1.8). If the sYmbol T denotes transposi-

tion, it is easy to verify that (AB)T = (B)T(A)T,

that is that the transpose of the produce of two
matrices is equal to the product of the transposes
of the matrices separately, but with the order
reversed. Thus, (AB) (AB) T = (A) (B) (B) T (A) T
= (A) (BBT) (A)T (since matrix multiplication is
associative.
Applying this result

c~ [-51
-51J C1

I H2 HKCOStl] • Thus I = H2 + K2 '" 2,
K2
lttKCOStl

Q = H2 - K2 + 2~S2, U = 2VH2Kc=ost4l8=2 , 2~C2' V Then
I 2 = Q2 U2 may be
+ V2 leads to but

+ 28 or - 28. To find out which, we again equate

the two forms that we have found for the two-by-

320

X

rlYtwo matrix which links the column matrices ]

and [sinwt] The equation reads
coswt

HCOS<P
[ Kcosa

Equating the determinants of this matrix on the
left to the product of the determinants on the
right, we obtain

HK(sinacos<p - cosasin<p) (Cf 8-11 S8 1C1 +8f S+11 S8 1C1 ) (St+Ct>

S
Hence, immediately,

HKsin(a - <P) = 8, so that =V 2HKsin~ 2HKsin (a - <P)

2S. Thus:

2 1BYD2 2+2D

o F GH0 0+2H

o L MN 0 0+2N

28 oRXT 2 0+2T

that is D = 0, H 1182, N = I1C2, T 8.

Therefore, the matrix of the general phase plate is

10 0 0

0 cl + 8~8 C28 2 (1 - S) - 8 211
0 8 2C2 (1- 8) st+ 8cl C211
0 8211 -C211 8

321

Notes

(1) This matrix reduces to the matrices representing

plain glass, half-wave plate and quarter-wave plate

on putting 0 = 0, TI and TI/2 respectively (see

Table 3 in section IV. 3) •
[0(2) The matrix
1] was inserted to interchange

-1 0

coswt and sinwt in the right-hand column.

(3) The matrix [: : ] ha' the effect of retarding the

::::::nb:c: ,W::hw::::ec[~]t:[t~; E-w~~]e·[s~~w~]n

n Cl -SI coswt

before and [SI Cl ] [sinwt] after the

VS1+SCl VC1-SSI coswt

phase plate. This matrix is obtained from the
previous one on substituting (wt - 0) for wt

in the expression for n. Also,

322

Appendix E

Derivation of
Jones Matrices

E.1 THE POLARIZER, SUCH AS A SHEET OF POLAROID
(a) First we will consider an ideal linear polarizer,

whose pass-plane is horizontal, that is parallel
to the x-direction. This allows only vibrations
parallel to the x-direction to pass through the
device, so that in the general equations describing
the behaviour of an ideal device H2 is equal to HI
and K2 is zero, no matter what the values of HI
and Kl may be. Thus, the equations become

for aZZ values of HI, Kl and ~1'

Set Kl = O. The equations become HI = JIIHI' that
is Jll = 1, and 0 = J21Hl for all HI' that is

J21 = O.

Set HI = O. =The equations become 0 JI2Klexp(i~1)
and 0 = J22Klexp(i~l) for all Kl and 61' that is

J12 and J22 are both zero. The JO~S matrix for

this ideal polarizer is thus [1 0
o0

In practice, of course, a sheet of polaroid intro-
duces some attenuation, even for the preferred
plane of vibration, and its optical thickness will
be sufficient to introduce at least several hundred
wavelengths of retardation. Fbr some purposes, for
example in an interferometer calculation, the above
matrix needs to be multiplied by a complex scalar,

323

which represents the complex amplitude transmittance

of the polarizer. If only one Maxwell column is in-

volved, information about the absolute phase of the

vibration is unlikely to be needed. As a matter of

convenience, however, a given Jones matrix can often

be converted into a simpler or more symmetrical form

[~:-Pl:]fby multiplying it by a suitable phase factor. The
::o::P:::::::d :: ::"[: e-:j":fW:::sbe

seemed desirable.
It is just as easy to show that, for a polarizer
with its pass-plane vertical, that is parallel to
the y-axis, the Jones matrix can be represented by

[: :] (or [: e:.] for exemple) •

(b) We now consider the more general case of a polarizer

with its pass-plane making an angle e with the x-

direction (see Figure E.l). Suppose we have incid-
ent on the polarizer a plane-polarized vibration
with its vibration direction making an angle a with
the x-axis. Suppose the amplitude of this incident
vibration is Ai then by our definition Xl = Acosa

and Y1 = Asina. Only the component of this vibra-

tion along the direction of the pass-plane of the
polaroid will pass through the device. The ampli-
tude of this component is

r
Vibration"

Pass-plane

Figure E.l

324

U = Acos(a - S) = AcosacosS + AsinasinS

= X1COSS + YlsinS

The component of this vibration parallel to the
x-axis is
X2 = UcosS = X1COS 2S + YlsinScosS

The component of the emerging vibration parallel
to the y-axis is

Y2 = UsinS = XlcosSsinS + Ylsin2S

The last two equations can be written in matrix
form

so that the matrix of the device is

cos2S sinscoss]
[ sinScosS s i n 2S

E.2 JONES MATRIX OF THE GENERAL LINEAR RETARDER

We consider a crystal plate with its optic axis at
an angle a to the x-axis, and we suppose that this
plate ~eta~ds the phase of the ordinary vibration (per-
pendicular to the optic axis) by 0 relative to the
vibration along the optic axis, that is to the E-
vibration. The device makes no difference to the
Maxwell column of a plane-polarized vibration parallel
to the optic axis. Suppose the amplitude of this vib-

ration is A, then, as in the section on the polaroid,

we know that its components are Xl = Acosa and
Y1 = Asina. Since this vibration is unaffected by
the device, the emerging components are the same as

the entering ones, that is X2 = Acosa and Y2 = Asina.

Substituting in the general equation for the behav-
iour of the device, we find the equations

Acosa

Asina

325

which lead immediately to

tana. = -scions-aa.. 1 - Jll

= J12

(given that cosa. does not vanish). We now consider

the action of the device on a beam of plane-polarized

light with its vibration plane parallel to the x-axis

and of amplitude A. For this beam the Maxwell column

is

E, [:]

We now need to consider the components of this vibra-

tion along and parallel to the optic axis. To find

these components we use the rotation matrices which we

introduced while developing the Mueller matrix of a

quarter-wave plate. The result is J [VI] (say)
Comp one nt along optic axis VI

[Component perpendicular to optic axis

[AJ [ Acosa cosa.J
sina]

[ -sino. cosa 0 -Asino.

The component perpendicular to the optic axis is now

retarded by a phase angle 0, that is to say VI is

mUltiplied by exp(-io). The matrix of components along

l [ Jand perpendicular to the optic axis thus becomes
V2 Acoso.

[ V2J = -Asinaexp(-io)

We now need to return to the former axes. To do this,

we again use the rotation matrix, but this time in

reverse to get from our VV-axes to our former XY-axes.

The final form of the Maxwell column is thus

cos a -sina.] [ Acosa J

E2 = [ sina cosa -Asina.exp(-io)

326

JAcos2a + Asin2aexp(-io)

[Acosasina Asinacosaexp(-io)

( cos 2 a + sin2aeXP(-iO»A]
[cosasina(l - exp(-io»A

Now, here,

J12] [A] = [J11A]

J22 a J21A

Therefore, comparing expressions for E2,

Jll = cos 2a + sin 2aexp(-io) and J21 = cosasina(l- exp(-io»

Use of the two equations developed above for tana now
leads, by a very easy manipulation, to

J12 = sinacosa(l - exp(-io» and J22 = sin2a + cos 2aexp(-iol

so that the Jones matrix of the general phase plate in
general orientation is

cos 2a + sin2aexp(-io) cosasina(l - (eXP-iO»]

[ cosasina(l - exp(-io» sin2a + cos2aexp(-io)

It is left as an exercise to the student to prove that
the Jones matrix for a simple rotator, that is a device
which twists the plane of vibration of a beam of plane-
polarized light through an angle e counterclockwise,
like certain organic liquids, is

cos e -Sine]
cose
=R(-e)

[ sine

In general, if J represents the Jon~s matrix already
calculated for a particular device, the new matrix for
the same device rotated through an angle e will be
given by the triple matrix product

R(-e) x J x R(e)

327

As an example of this rule, the student should verify

that the matrix which has already been given above for

a general linear retarder at angle e is equal to the
product

-.cos e -sine] [1 o ] [ ccss
[ sine cose 0 e-i6 -sine coseJ

See Table 4 in section IV.5 for several particular
cases of this type of matrix. It should be noted that

it is the azimuth of the fast axis which is specified
in each case, so that, for a wave plate cut from a
~egative uniaxial crystal, the 6-values refer to the

.etardation of the ordinary vibration.