s-108_2110_intoduction_to_matrix_optics.pdf

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Topic 3: Matrix Geometric Optics

3.1 Introduction

In this section we formulate par-axial geometric optics in terms of 2×2 matrices1 and associ-ated rays. These rays can be traced through the optical system by matrix-vector multiplication.This scheme allows complex, multi-element, optical systems to be analysed and simplifiedis a simple algorithmic scheme which is easily calculated, either analytically or numerically.This scheme is then used to identify the main optical reference planes, known as theprinci-pal planes, and the effectivefocal length. The properties of these planes allow us to representcomplex optical system by their equivalent simple lenses. Finally we consider the limitationsof this system and outline the next level of optical analysis, being full geometric ray tracing.

3.2 Ray Vectors and Matrices

In this scheme we represent an optical ray in planeP, a distancez along the optical axis by atwo component vector,

r =

[

rθ

]

wherer is the height from the optical axis, andθ angle made with the perpendicular to the planeP as shown in figure 1.

Optical Axis

Ray

θ

r

0

P

z

Figure 1: Vector notation for an optical ray of heightr, angleθ in plane a distancez along theoptical axis.

We then describe the propagation of this ray by a 2×2 matrix

M =

[

a bc d

]

such that the height and angle of the new ray is given by

r1 = Mr 0 ⇒[

r1

θ1

]

=

[

a bc d

] [

r0

θ0

]

We then define a Matrix OperatorM for each opticalcomponent2 we can trace rays through thesystem by a series of matrix multiplies.

1This formulation is different than that given in Hecht who uses the transpose, the final results are however thesame.

2Each refractive index interface and distance between them.

School of Physics Physics 3 (U03232) (Optics) & Optics (U01358) Revised: 3 January 2006

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If we haveN components with propagation matricesM i for i = 1, . . . ,N, then for an initial rayof r0 the final rayrN is given by a series ofpre-multiplies, giving,

rN = MN MN−1 . . . M2M1 r0

which we can then write as

rN = Msr0 where Ms = MN MN−1 . . . M2M1

so allowing us to form a singlesystem matrixform a combination of optical components by aseries of 2×2 matrix multiplies. This reduces to geometrical optical properties of a potentiallyvery complex optical system to a simple 2×2 matrix. The power of this formulation becomeclearer when we start to define the actual matrices.

3.2.1 Propagation between Two Planes

Consider a rayr0 in planeP0 propagating a distanced to a planeP1 as shown in figure 2. InplaneP1 we have, assuming small angles so that tanθ0 ≈ θ0, giving,

r1 = r0+dθ0 and θ1 = θ0

which, in matrix formulation, can be written as:

r1 = Mr 0 where M =

[

1 d0 1

]

so giving us the matrix for propagation a distanced along the optical axis.

Note: This does not depend on the refractive index of the media it is propagating through.

r

θθ

r

Optical Axis

Ray

P P0

0

0

1

1

1

d

Figure 2: Propagation between two planes separated by a distanced

3.2.2 Dielectric Interface

Consider a rayr0 incident of a flat dielectric interface where the refractiveindex changes fromn0 to n1 as shown in figure 3. After the interface, we have from Snell’sLaw, again assumingsmall angles, then,

r1 = r0 and θ1 =n0

n1θ0

which in matrix formulation can be written as

r1 = Mr 0 where M =

[

1 00 n0

n1

]

so giving us the matrix for a flat dielectric interface.

See tutorial 3.1 as an example of using and combining these initial two matrices.


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Ray

Optical AxisP

n n

r

θ

θ0

0 1

0

1

Figure 3: Dielectric interface between materials of refractive indexn0 andn1

3.2.3 Thin Lens

Consider a rayr0 incident on a thin lens of focal lengthf in planeP1, as shown in figure 4. Ifwe assume the lens isthin, thenr1 = r0. The angle expression is slightly more complex beingdependent on bothr0 andθ0. From figure 4 we have that,

θ0 =r0

uand θ1 = −r0

v

We recognise planesP0 andP2 as theobjectandimageplanes of the lens, tou andv are linkedby theGaussian Lens Formulaof,

1u

+1v

=1f

so but substitution we get that,

θ1 =r0

u− r0

f= θ0−

r0

f


r1 = Mr 0 where M =

[

1 0−1f 1

]

giving the matrix for a thin lens of focal lengthf . This is valid for both positive lenses (wheref > 0) and negative lenses (wheref < 0).

θ1

Ray

P

θ0r0

PPu v0 1 2

Figure 4: Thin lens of focal lengthf .

3.2.4 Spherical Dielectric Interface

The most complex, but also the most useful, surface is the spherical dielectric interface betweentwo materials of refractive indexn0 andn1, with radius of curvatureR as shown in figure 5.


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P R

Optical Axis

Interface

n

r 0

0 n1

θ0 θ1

Figure 5: Curved dielectric interface of radiusR.

r

n1

0

θ0

θ1

ψ

n0c

ϕ0

ϕR

1

Figure 6: Detail of curved dielectric interface of radiusR with added construction angles.

Since the interface is of zero thickness, then clearlyr1 = r0, but the angular relation is muchmore complex. Consider the more detailed drawing of the interface in figure 6.

The anglesφ0 andφ1 are the angles the ray makes with the surface normal which, for a sphericalsurface, is the line fromr0 back to the centre of curvaturec. For a dielectric interfaceφ0 andφ1

are linked by Snell’s Law, so forsmallangles, we have that

n0φ0 = n1 φ1

We also have thatθ0 = φ0−ψ and θ1 = φ1−ψ

whereψ is the angle between the surface normal and the optical axis.From this we can thensolve forθ1 to give

θ1 =n0

n1θ0 +ψ

(

n0−n1

n1

)

Additionally from the geometry, we have for small angles, that ψ = r0/R, so we get a finalexpression forθ1 of

θ1 =

(

n0−n1

n1R

)

r0+

(

n0

n1

)

θ0


r1 = Mr 0 where M =

[

1 0n0−n1n1 R

n0n1

]


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giving the matrix for the spherical dielectric surface.

Note the sign of the radius of curvature of the surface.R> 0 gives a concave surface as shownin the above figures, whileR< 0 gives a convex surface. See tutorial 3.2 to shown that is thisis consistent with thelens makers formulafrom previous.

UsingR the radius of curvature to characterise has numerical problems since for aflat surfacethe radius becomes infinite. It is therefore computationally more convenient to definecurvature,

C =1R

so the propagation matrix becomes,

M =

[

1 0C(n0−n1)

n1

n0n1

]

Now a flat surface simply has curvatureC = 0, which is then identical to the matrix for a flatdielectric surface discussed above.

So given this set of matrices we can now model any par-axial system containing curved dielec-tric surfaces with spaces between them, so in practice any system containing lenses.

Note we actually only needtwo of the matrices, these being (a) for propagation a distancedand (b) refraction at a spherical dielectric interface. Theother matrices being combinations orspecial cases of these two.

3.3 Ray Manipulation

To complete the analysis tools we need to consider three ray geometries. The first is, given aray in planeP0 the location of the planeP1 where it crosses the optical axis as shown in figure 7.

0

P

Optical Axis

rθ

z d

PRay 0 1

Figure 7: Location of the plane where a ray crosses the optical axis.

If the ray in planeP0 has ray vector,

r =

[

rθ

]

then noting thatdownwardangles are negative, the distance fromP0 to P1 is given by

d = − rθ

with d > 0 specifying thatP1 is theright of P0 andd < 0 thatP1 is to theleft of P0.

The second is giventwo rays inP0 the location of the planeP1 where they cross each other asshown in figure 8.


meelis

Highlight

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0

P

Optical Axisz d

P1

θ

θ

r1

20

2

1

rh

Figure 8: Location of the plane where a two rays cross.

If in planeP0 we have two rays,

r1 =

[

r1

θ1

]

and r2 =

[

r2

θ2

]

then from simple geometry. the distance fromP0 to P1 is just

d =r1− r2

θ2−θ1

where againd > 0 specifying thatP1 is theright of P0 andd < 0 thatP1 is to theleft of P0. Thedistanced → ∞ when in the first caseθ = 0 and in the secondθ1 = θ2.

The third is given a ray in planeP0, what is its height in planeP1 separated by a distanced.This is denoted byh in figure 8. Simple geometry shows that if inP0 we have

r =

[

rθ

]

then h = r +θd

Now give this set of tools, we are able to apply them to opticalsystem.

3.4 Image Location and Magnification

Consider the rather complex system as shown in figure 9, wherewe have a three elementlens system defined by six curvatures, (C1 → C6), three refractive indexes (n1 → n3) and fiveseparation distances (d1 → d5). We then place an object of heightr in planeP0 a distanced0

from the front of the lens system.

If we define thelens systemas being between planesP1 → P2, then for each curved dielectricsurface and propagation distance we can form a matrix, in this case we get 11 matrices, 6 fromthe curved surfaces and 5 from the propagations, being

M1 = air/n1 interface with curvatureC1

M2 = propagation distanced1

M3 = n1/air interface with curvatureC2

· · · · · ·M10 = propagation distanced5

M11 = n3/air interface with curvatureC6

we can then form the lenssystem matrixas being,

Ms = M11M10 . . . M2M1


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n n nC C CC CC

dddd

1 2 3

d1 2 3 4 5

1 2 3 4 5 6

d d0 6

r

P PP0 1 P32

MM o s

h

Figure 9: Image location and magnification of a multi-element lens by matrix ray methods.

which will describe propagation between planesP1 andP2, being the front and back surfacesof the lens system, these being know at theinputandoutputplanes of the system.

If we locate an object in planeP0, the propagation fromP0 →P1 is again described by a matrix,Mo, so we can form total matrix

MT = MsMo

which describes propagation from the input planeP0 to P2, being the back surface of the lens.

To find the image location and magnification,

1. Propagate two rays fromP0 to P2 with the same initial height, say unity, butdifferentangles using theMT matrix.

2. The plane where these rays cross will give the location of the image plane,d6 in theabove figure, relative to theoutputplaneP2. d6 > 0 signifies a real image, whiled6 < 0a virtual image.

The condition thatd6 → ∞, the image is formed at infinity, can be identified checking ifthe two rays in planeP2 have the same angleθ.

3. The heighth of the either ray (both the same), in planeP3 will give the system magnifi-cation as

M =hr

with M > 0 signifying an upright image andM < 0 an inverted image.

So once the system matrixMs has been formed, this operation can be repeated for any objectdistance. Such calculations are best performed by computer. There is a accompanying setof JAVA classes to do these calculations allowing you to experiment, these are detailed in theappendixto these notes.

3.5 Properties of the System Matrix

The system matrixMs is formed from a multiplication of the component matrices, in the aboveexample 11 of them to calculate propagation of a ray fromP1 the input planeof the lens, toP2


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theoutput planeof the lens. From the previous section, we note that the determinate of all thecomponent matrices are unityexceptthe two representing a refractive index change at either aflat or curved interface, where for an0 → n1 interface,

|M | = n0

n1

so noting that for matrices, if we have that

C = BA then |C| = |B| |A|Then if we have a system with a series of refractive index changes as show in figure 10, 7 inthis case, the determinant for the whole system,

|Ms| =n0

n1

n1

n2. . .

n6

n7

n7

n8=

n0

n8

so the ratio of the initial to final refractive index.

P P

n n n n n n n1 2 3 4 5 6 7

1 2d1 d2 d3

dd

75d

4 6d

M sn n

0 8

Figure 10: Series of refractive index changes.

In the more common situation, as shown in figure 9, we have the same refractive index in-frontand behind the optical system, usually air. So in these situations,

|Ms| = 1

The one common system where this isnot true in in the human eye, which we will consider inmore detail in Topic 5.

If for a system of lenses we have a optical matrixMs, which describes the propagation fromplaneP1 → P2, as discussed above, then for rayr1 in planeP1, then inP2 we get rayr2, being

r2 = Msr1

so by simple matrix rules, we have that

r1 = M−1s r2

so the inverse ofMs describes propagationbackwardsfrom planeP2 back to planeP1, which isequivalent of going back through the lens by reversing the direction of propagation of the rays.SinceMs is just a 2×2 matrix, the inverse is simply given by,

Ms =

[

a bc d

]

then M−1s =

1|Ms|

[

d −b−c a

]

which for the most common case of a lens system in air where thedeterminant is unity, itbecomes a simple re-arrangement of the matrix elements.


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3.6 Focal Lengths and Principal Planes

When dealing with multi-lens systems it is convenient to think in terms offocal length, howeverthis does require a little additional thought. Consider thecomplex multi-element optical systemshown in figure 11 (a) with input planeP1 and output planeP2.

P P P

P P

1 2 f

P2 f

f

M s

Back Principle Plane

(a)

(b)

θ2r r1 2

PlaneBack Focal

bBack Focal Length

bb

b

s

t

t

Figure 11: Back Principal Plane and focal length of a Multi-element system.

We first calculate the system matrixMs which describes propagation between planeP1 → P2.If we then consider a ray parallel to the optical axis of unit height, then in the output planeP2

we will have a ray

r2 =

[

r2

θ2

]

= Ms

[

10

]

which crosses the optical axis at a distancet from planeP2 in planePfb where

t = − r2

θ2.

Due to the imaging properties of any lens, all input rays parallel to the optical axis will befocused at the same point, also any pair of parallel rays entering planeP1 will also cross inplanePfb. This defines planePfb as theBack Focal Planeof the system, being the plane inwhich an image of an infinite object will be formed.

The whole optical system defined byMs is therefore acting as a single thins lens of focal lengthfb located in a planePb which, in general, is displaced fromP2, the back aperture of the system.This planePb is defined as theBack Principal Planeand thefb as theback focal lengthof thesystem.

The location ofPb is where rayr2 and the inputr1 crosses, so noting that

r2 =

[

r2

θ2

]

and r1 =

[

10

]


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we get from previous,s the distance fromP2 to Pb is given by

s=1− r2

θ2

then noting thats is the distance fromP2 back to Pb, theback focal lengthis then justfb = t−s,so combining there two we get that

fb = − 1θ2

If we write the 2×2 system matrix in its components, being

Ms =

[

a bc d

]

then we immediately have that,

r2 =

[

ac

]

⇒ r2 = a and θ2 = c

so we immediately get from thesystem matrixMs that

t = P2 → Pfb = −ac

position of Back Focal Plane

s= P2 → Pb =1−a

cposition of Back Principal Plane

fb = Pb → Pfb = −1c

Back Focal Length

We can then repeat the operation to find the equivalentfront planes by tracing a ray of unitheight from planeP2 back toP1 as shown in figure 12.

Noting, from above, that the propagating back fromP2 to P1 is equivalent to taking the inverseof the system matrix, then in planeP1 we have ray

r1 =

[

r1

θ1

]

= M−1s

[

10

]

so the ray cross the optical axis a distancet from planeP1 in planePf f , where now,

t = − r1

θ1

wherePf f is theFront Focal Plane. Similarly the distance fromP1 to Pf , theFront PrincipalPlaneis

s=1− r1

θ1

and thefront focal length, is then given by

f f = − 1θ1

Note in this casef f is negative, since it is the distance from theFront Principal Plane back totheFront Focal Plane.


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P P1 2

M s

r r1 2

PPf 1

Pff

s

t

Front Principle Plane

Pff

Front FocalPlane

t

f f

Front Focal Length

(a)

(b)

θ1

Figure 12: Front Principal Plane and focal length of a Multi element system.

The inverse system matrix

M−1s =

1|Ms|

[

d −b−c a

]

=

[

d′ −b′

−c′ a′

]

where for the usual case of the same refractive index on both sides of the system,a = a′ etc.,so we immediately have that

r1 =

[

d′

−c′

]

⇒ r1 = d′ and θ1 = −c′

so giving that get that

t = P1 → Pf f =d′

c′position of Front Focal Plane

s= P1 → Pf =d′−1

c′position of Front Principal Plane

f f = Pf → Pf f =1c′

Front Focal Length

so giving us the locations of the planes and the focal lengthsin terms of the system matrixelements. Not in almost all cases, the determinate is unity,so

f f = − fb

so thefront focal lengthis the same as theback focal length, with the negative signifying thedirection of the front and back focal planes from the front and back principle planes.


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3.7 Properties of Principal Planes

To illustrate the use and properties ofprincipal planes, it is best to start with simple numericalexample. Take athick lenslens with parameters,

Front Surface R1 200 mmThickness t 20 mmBack Surface R2 -100 mmRefractive Index n 1.50

From these we can calculate thesystem matrixfor propagation from input to output planes,numerically it is,

Ms =

[

0.9667 13.333−0.0073 0.9330

]

we can then use this matrix to calculate the planes as shown infigure 13. This gives the focallength to be 136.36mm.

R = 200R = −100

P

n = 1.5d = 131.82

PPt = 20

PP

f = 136.36

9.09 −4.54

b

12

1 2

f b

fb

Figure 13: Locations of plane of a Thick Lens, all length are in mm..

See workshop question 3.5 to do this calculation yourself using the supplied program.

The important point aboutPrincipal Planesis when we come to consider imaging. Take thesame lens as in figure 13, but now use it to image an object as shown in figure 14. Here weplace an object a distancedo from the input plane of the lens,P1. If we trace a ray from thebase of the object at angleθ1, then at the input planeP1 we have rayr1, we know the systemmatrix Ms, we can calculater2 in planeP2 and hence the image distancedi . We can then tracea ray from the top of the object of heightho, and find the image heighthi . For the above lens atypical set of numerical results is shown below.

Object distance d0 300 mmImage distance d1 244.02 mmMagnification hi/ho -0.7894


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n = 1.5

PPt = 20

1 2

bPfP

u v

d d

r r rθ

θ1

12

2f

rb

o i

Object

Image

h

h

o

i

s=9.09 t=−4.54

Figure 14: Location of object and image for a thick lens be considering the Principal Planes.

Now is we define the object and image distances with respect tothePrincipal Planes, as

u = do +s Object toPf distance

v = t +di Pb to Image distance

then we find, that1u

+1v

=1f

where f is the focal length as defined above, and the magnification

M = −vu

which gets us back to theGaussian Lens Formula. The numerical values for the above geometryare:

u = 309.09mm

v = 244.02mm

f = 136.36mm

M = −0.7894

which you can verify yourself! See question 3.6 for the rather difficult analytic calculation ofthis same result.

Also importantly if we take rayr1 in planeP1 and project it to the frontPrincipal Plane, itintersects it at heightr f . If we repeat the process for rayr2 in planeP2 and project it to the backPrincipal Plane, it intersects it at heightrb. It can be shown(see question 3.6), that

r f = rb

so that a ray enteringPf at heightr f will leave Pb at thesameheight, only is direction will bechanged. This characterised thePrincipal PlanesasPlanes of Unit Magnification.


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This means that the propagation between the twoprincipal planesis equivalent to a thin lensof focal length f , the focal length of system. The imaging properties are shown in figure 15.Therefore the imaging between the system input planeP1 and output planeP2 can be consideredare being:

1. Propagation from planeP1 to Pf , the front Principle Plane.

2. Refraction by an ideal thin lens of focal lengthf , which gives output in planePb, theback Principle Plane.

3. Propagation fromPb to P2, the output plane. (Note the expression calculated above forthe location of planePb is the distance fromP2 → Pb).

See question 3.7 for a numerical example of this.

P P P P1 2f b

s t

Object

f

u

v

Imaged

r r

e

Figure 15: Imaging with respect to the Principal Planes.

In all this analysis we havenot discussed the distance between theprincipal planes Pf andPb,e in figure 15,or the distance between the input and output planesP1 andP2, e in figure 15, butonly the position of the principle planesrelativeto the in input and output planes beings andtrespectively. There distances are clearly linked with

d = s+e− t

but neither d or ecan be recovered from the system matrix. So to obtain a full description ofthe optical system is is therefore necessary to know one of theses.

3.8 Applications of Principal Planes

This analysis allows us to characterise complex optical system, as are found in camera lenses,projector lenses, microscope objectives etc., by using simple par-axial geometric optics. Inparticular if the location of theprincipal planesandfocal lengthof a compound lens are knownthen we have fully characterised its imaging properties andthus how it will function in anoptical system. As shown in question 3.7, we can form thesystem matrixfrom the location oftheprincipal planesandfocal length.

In Topic 5, we will discuss how to actually measure the location of principal planesandfocallengthof a compounds lens, so allowing us to apply this analysis to real optical systems.


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3.9 Beyond Matrix Method

This matrix scheme forgeometrical opticsgives an analytically and computationally simpleroute to calculate be basic parameters of complex optical system, however it is limited thepar-axial approximationin particular:

1. rays makesmallangles with the optical axis, so we can take

sinθ ≈ tanθ ≈ θ

This impliessmall object, or more correctly object much smaller than their image orobject distances.

2. spherical surfaces are approximated by parabolic surfaces, which assumes that surfaceshaveshallowcurves, (large radius of curvature), or more correctly, thesmall aperturecompared to there radius of curvature.

These assumptions have the effect of linearising the equations, (removing the sin() and√), soallowing us to applylinear matrix methods to the problem. For optical systems this analysiswill give:

1. correct location on object, image and principal planes,

2. correct focal length and magnification,

3. no information about imagingquality (aberration),

4. no information how the performance of the system changes with object size or lens aper-ture size.

This scheme is howeververy usefulin understanding the operation of pre-assembled systemand also for design of systems from pre-made components. We will be using it in Topic 5 tolook at various practical optical systems. It does not however give a detailed analysis of theimaging quality or the design of actual components.

Full Ray Tracing???

The next step inGeometrical opticsis full ray tracingwhere we use the fullSnell’s Lawex-pression, and the full expression for the shape of the surfaces. This also allow us to deal withnon spherical surfaces, know asaspherical surfaces. The basic geometry for such an analysis isshown in figure 16, where the rays directions are characterised by unit vectorr and the surfacesnormal byu.

In three dimensions the vector formulation ofSnell’s Lawcan be shown to be given by

n1 (r1× u) = n2 (r2× u)

which for reflection,shown in figure 16.a, can be relatively simply expressed as,

r2 = r1−2(r1.u) u

while for refraction, shown in figure 16.b, we get the rather more complex relation that,

n2r2 = n1r1 + u [n2(r2.u)−n1(r1.u)]


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R

u u

n nR

r

r

rr^

^

^

^

^^

1

2

12

1 2

a) b)

Figure 16: Geometry for full ray tracing with ray direction and surface normals characterisedby unit vectors.

where

n2(r2.u) =r1.u|r1.u|

[

n22−n2

1

(

1− (r1.u)2)]12

provided that we are below critical angle. These expressiondo not simplify any further andhave to be calculated digitally at each surface3 .

So with reference to figure 17 can trace a ray from pointp0 with directionr0 by,

1. find intersection point with surfaceS1, beingp1. If S1 is flat, spherical or parabolicsurface the intersection pointp1 has an analytic solution, for other shapes is also requireiteration,

2. find the surface normalu1 at pointp1.

3. calculater2 from the vector form of Snell’s Law,

4. find intersection withS2, beingp2, and repeat the process.

rr

rn n n

u

0

1

2

0 1 2p

p p

0

1 2

S S1 2

u12

Figure 17: Scheme from full ray tracing through surfaces

To analyse an optical system we now tracepencilsof many rays which as shown in figure 18,which shows the typical result for a pencil of parallel rays incident on a positive lens. The par-axial rays are focused to the expected geometrical focus as in the back focal plane, but the outerrays are focusedshortwhich results in a blurred image. This blurring is known as aberrations,in this case the cause isSpherical Aberrationcaused by the spherical surfaces.

3In theearly daysthis was done graphically with angles calculatedby hand.


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f

Outer rays focused short

Figure 18: Schematic of tracing a pencil of parallel rays through a positive lens. The par-axialrays are focused a distancef behind the lens but the outer rays are focused short.

There are a series of aberrations types, and the aim of optical design is to use combinationof surfaces to cancel, or a least reduce the aberrations to anacceptable level, so that the realsystem closely matches the performance of the ideal par-axial system. We will look as somesimple examples in Topic 5, but any significant analysis is beyond this type of optics course.

Optical ray tracing is an ideal computational task, being one of the first computerised numer-ical tasks. Optical ray tracing is usually combined with computer aided optical design in anintegrated package that allows the user to predict the optical performance of a system and alsooptimise the design by iterative modification of the surfaces curvatures and component thick-nesses.

3.10 Summary

In this section we have,

1. described the matrix scheme for the par-axial geometric analysis of optical system

2. shown that by tracing rays, this system can be used to determine the imaging propertiesof complex optical systems,

3. shows that complex optical system can be represented by a simple 2× 2 matrix, thesystem matrix,

4. investigated to properties of thesystem matrixin particular its determinant which givesthe ratio of refractive index on object to image side of the system, and its inverse, whichgives the imaging properties of rays reversed back through the lens,

5. defined theprincipal planesand shown how they can be calculated from thesystemmatrix,

6. shown how a complex optical system can be specified by itprincipal planesand focallength, and how this relates to the simpleGaussian Lens Formula,

7. discussed the limitations ofmatrix rays methods, and outlined the underlying models offull ray tracingas the next level of complexity in geometric analysis of optical system.

The methods covered in this topic will be used in the analysisof practical optical systems inTopic 5.


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Workshop Questions

3.1 Dielectric Block

Calculate the 2×2 matrix operator that corresponds to a block of dielectric material with par-allel sides, lengthl and refractive indexn.

3.2 Matrix Operator for a Thin Lens

Use the matrix operators for a spherical dielectric surfaceto calculate the overall matrix oper-ator for the thins lens as shown below.

1

2

R

nR

Show that your answer is equivalent to the matrix operator for a simple thin lens.

3.3 Adding Two Lenses Together

Use matrix methods to calculate the focal length of two thin lenses of focal lengthf1 and f2placed in contact with each other.

3.4 Beam Expander

A Galilean beam expander consists of a short focal length negative lens and a long focal lengthand positive lens arranges as follows,

r1

r2

f − f2 1

−f1

2f

Usematrix methodsto calculate the magnification ratior2/r1.

3.5 Numerical Calculations for a Thick Lens

Run the supplied JAVA program toexplorethe focal length, location of principal planes andgeometric imaging properties of a thick lens.

To run simply type,


M G O Session: 2005-2006 -19

java /ifp/java/examples/optics/ThickLens

using any of the CPLAB systems.

This program uses the DISPLAY classes from SCIENTIFIC PROGRAMMING to ask you for theparameters and and performs the calculation when you click GO. It involves no program-ming!

3.6 Height in Principal Planesdx

Show that any ray passing through a lens system from input to output planes will be at the sameheight from the optical axis in the front and backPrincipal Planesof the system.

Hence, show analytically, that the this system obeys theGaussian Lens Equation

1u

+1v

=1f

whereu is the object to frontprinciple planedistance,v is the backprincipal planeto imagedistance andf is the focal length of the system.

3.7 System Matrix from Principal Planes

An optical system has focal lengthf , andprincipal planes Pf andPb located a distances andtrespectively from the input and output planes. Derive the system matrix for propagation fromthe input to output planes in terms off , s andt. Check that your answer is numerically correctusing the numbers for the thick lens supplied in the notes.

3.8 Your first Matrix Java Program

Using the JAVA classes forParaxialRay andParaxialMatrix, write a JAVA program calcu-late the imaging properties to a system oftwo thin lenses, as in question 2.13 and 2.14 from theprevious section, where you can specify the focal lengths ofthe both lenses, their separationand location of theobjectplane. Your program should then calculate the location ofimageplane and the linear magnification of the formed image.

Test your program with the parameters from 2.13 and 2.14 and make sure you get the sameanswers!.

3.9 Vector form of Snell’s Lawdx

Show that vector form of Snell’s Law, being

n1 (r1× u) = n2 (r2× u)

is consistent with the normal formulation of Snell’s Law when applied to the system below.


M G O Session: 2005-2006 -20

z

x

u

r

r

^

^

n

n

θ

θ

1

2

1

2

1

2


s-108_2110_intoduction_to_matrix_optics.pdf

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