principles of geometric optics approximation to wave optics, it is technologically useful for the...

29

2.1 Introduction

In this chapter, we consider geometric optics, which is an approximation to wave optics that can be used when considering an optical system composed of elements that are much larger than the wavelength of light going through the system. Then, we can ignore the wave nature of light, aside from its color, and assume that it will travel in a straight line, which is often called a ray. Although geometric optics is only an approximation to wave optics, it is technologically useful for the design and modeling of the adaptive optical systems that will be considered here. It greatly simplifies the calculations to a point that allows intuition to guide the design. This is not usually the case when considering the Huygens integral!

To determine the direction in which a light ray will pass through an optical system, we can apply Fermat’s principle of least time or shortest optical path length that was discussed in Chapter 1. Fermat’s principle is that the optical path distance OPD between points A and B given by

OPD A,B n x x

A

B

( )= ( )∫ d

(2.1)

is shorter than the optical path length of any other curve that joins these points and lies in its certain regular neighborhood (Born and Wolf, 2006).

2.2 Reflection

We consider the application of Fermat’s principle to two simple optical surfaces: a mirror that reflects light and an interface between two media that refracts light. With these two simple optical surfaces, we can understand the most important aspects of geometrical optics for the design of optical systems. In Figure 2.1, we show a mirror surface where a light ray starting from point A is reflected off the mir-ror surface to reach point B. The question is what path will the light ray take? If we assume that the

2Principles of

Geometric Optics

Joel A. KubbyUniversity of California at Santa Cruz

2.1 Introduction ........................................................................................292.2 Reflection .............................................................................................292.3 Refraction .............................................................................................312.4 Paraxial Lens Equation ......................................................................322.5 Thin Lens Equation ............................................................................342.6 Magnification ......................................................................................392.7 Aberrations ..........................................................................................40

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© 2013 by Taylor & Francis Group, LLC

30 Principles

horizontal.distance.between.points.A.and.B.is.L,.then.at.what.point.x.on.the.mirror.will.the.light.beam.be.reflected?.We.see.that.the.light.will.take.a.path.that.is.a.distance.d1.from.point.A.to.the.mirror.and.a.path.that.is.a.distance.d2.from.the.mirror.to.point.B..We.assume.that.the.light.is.traveling.in.a.vacuum.at.a.speed.c.

The.time.t.required.for.the.light.to.travel.from.point.A.to.B.along.this.path.is.given.by.the.following.equation:

.

t x d xc

d L xc c

d x d L x

ch

( ) ( ) ( ) ( ( ) ( ))= + − = + −

=

1 21 2

2

1

1 ++ + + −( )( )x h L x2 2 2

To.find.the.minimum.time,.we.take.the.derivative.of.t.with.respect.to.x.and.set.it.equal.to.zero,.which.gives

.

dt xx c

h x x h L x L( ) (d

= +( ) ( ) + + −( )( ) −− −1 1

22 1

222 2 1

2 2 212 −−( )

=

=+

− −

+ −( )

= − −

x

xh x

L x

h L xxd

L

) 0

2 2 2 2

1

xxd2

1 2

1 2 1 2

= −= → =sin( ) sin( )

sin( ) sin( ) .θ θ

θ θ θ θ .

(2.2)

We.see.that.for.the.light.to.take.the.path.of.least.time.to.get.from.point.A.to.B.by.reflecting.off.the.mirror.surface.requires.that.the.angle.of.incidence,.θ1,.is.equal.to.the.angle.of.reflection,.θ2..This.simple.formula.allows.us.to.calculate.the.path.light.will.take.when.reflected.from.a.mirror..As.we.shall.see.in.Chapter.9,.Section.3,.it.is.interesting.to.consider.mirrors.that.are.not.flat..In.this.case,.the.angle.of.inci-dence.is.equal.to.the.angle.of.reflection,.where.the.angles.of.incidence.and.reflection.are.defined.by.the.local.normal.to.the.curved.surface.

A

h

B

h

d1

x L – x

L

d2

θ1 θ2

FIGuRE 2.1 Reflection.of. the. light. from.a.mirror.surface..The.light. travels. from.point.A. to.point.B.by.being.reflected.off.the.mirror.surface.at.a.horizontal.distance.x. from.point.A..Point.B. is.a.horizontal.distance.L. from.point.A..The.distance.that.the.light.travels.from.point.A.to.the.mirror.is.d1.and.the.distance.that.it.travels.from.the.mirror.to.point.B.is.d2..To.find.the.angle.of.reflection,.θ2,.given.the.angle.of.incidence,.θ1,.we.minimized.the.travel.time.along.this.path.according.to.the.Fermat’s.principle.of.least.time.

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31Principles of Geometric Optics

2.3 refraction

We.can.also.calculate.the.path.a.ray.of.light.will.take.when.it.passes.through.an.interface.that.separates.media.with. two.different. indices.of. refraction..The.speed.of. light. inside. the.media.with.an. index.of.refraction.n.is.c/n,.where.c.is.the.speed.of.light.in.vacuum..Since.n.>.1,.the.speed.of.light.is.always.slower.in.media.other.than.vacuum..Given.this.speed.of.light.within.the.media,.we.can.then.calculate.the.time.that.light.takes.to.travel.from.a.point.A.in.the.first.medium,.with.index.n1,.to.a.point.B.in.the.second.medium,.with.index.n2,.as.shown.in.Figure.2.2.

The.time.t.required.for.the.light.to.travel.from.point.A.to.B.along.this.path.is.given.by.the.following.equation:

.

t x ncd x n

cd L x

cn d x n d L( ) ( ) ( ) ( ( ) (= + − = + −1

12

2 1 1 2 21 xx

cn h x n h L x

))

= + + + −( )( )11

2 22

2 2

To.find.the.minimum.time,.we.take.the.derivative.of.t.with.respect.to.x.and.set.it.equal.to.zero,.which.gives:

.

dt xx c

n h x x n h L x( )d

= +( ) ( )+ + −( )(−1 12

2 121

2 2 12

22 2 )) − −( )

=

=+

− −

−12

1 2 2 2

2 0( )L x

n xh x

n L xhh L x

n xd

n L xd

n n

2 2

11

22

1 1 2 2

+ −( )

= − −

= −sin( ) sin(θ θ ))sin( ) sin( )n n1 1 2 2θ = θ . .

(2.3)

We.recognize.this.as.Snell’s.law.from.Chapter.1.

A

h1n1

n2h2

d2

d1

θ1

θ2x

L − x B

FIGuRE 2.2 Refraction.of.the.light.through.an.interface..The.light.from.point.A.is.incident.at.an.angle.θ1.on.an.interface.between.two.media..The.upper.media.has.an.index.of.refraction.n1.and.the.lower.media.has.an.index.of.refraction.n2..The.change.in.the.index.bends.the.path.of.the.light.at.an.angle.θ2,.causing.it.to.pass.through.point.B.

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32 Principles

2.4 Paraxial Lens equation

It.is.also.useful.to.consider.refraction.from.a.curved.surface,.such.as.the.circular.arc.shown.in.Figure 2.3..When.the.arc.separates.two.regions.of.different.indices.of.refraction,.n1.and.n2,.a.ray.of.light.will.be.refracted.at.the.interface.according.to.Snell’s.law..If.we.choose.the.correct.shape.for.the.interface,.we.can.cause.light.that.originates.from.a.point.O.along.the.axis.of.the.arc.to.pass.through.a.point.I.that.is.also.along.the.axis..This.is.the.geometry.of.a.lens.and.it.enables.an.object.at.one.point.in.an.optical.system.to.be.imaged.at.a.different.point.in.the.system.

To.find.the.equation.that.describes.the.action.of.the.lens,.we.consider.the.fact.that.a.light.ray.traveling.from.position.O.to.position.I.must.take.the.same.amount.of.time.regardless.of.the.path..This.means.that.the.time.taken.for.light.to.travel.from.point.O.to.P.and.then.from.point.P.to.I.must.equal.the.amount.of.time.taken.for.light.to.travel.directly.from.point.O.to.I..Of.course,.the.spatial.distance.OP + PI. is.greater than.the.distance.OI,.but.if.n1.is.smaller.than.n2,.then.light.will.travel.slower.inside.the.media.with.index.n2.and.thus.will.take.the.same.amount.of.time.along.the.direct.path.OI.as.it.would.along.the.path.OP.+.PI..This.is.because.more.of.the.path.of.light.is.inside.the.media.with.higher.index.

To.estimate.these.travel.times,.it.is.useful.to.make.an.approximation.for.light.rays.that.are.close.to.the.optical.axis.(Feynman.1966)..These.are.called.paraxial.rays..Consider.the.right.triangle.with.sides.of.length.s > d > h.shown.in.Figure.2.4.

From.the.Pythagorean.theorem,.we.know.that

. s h d h s d s d s d2 2 2 2 2 2= + → = − = − +( )( )

To.simplify.this,.we.can.approximate.that.s ≈ d,.so.that.s + d ≈.2s.and.substitute.Δ.=.(s.–.d).to.obtain

.h s d s d s h

s2

22

2= − + ≈ → ≈( )( ) ∆ ∆

.(2.4)

O I

S1

n1 n2

S2

P

R

V Q C

FIGuRE 2.3 Refraction.at.a.curved.interface..The.light.traveling.from.point.O.(object).to.point.P.is.refracted.at.the.interface.between.two.media.with.indices.of.refraction.n1.and.n2..The.refracted.light.intersects.the.optical.axes.at.point.I.(image)..The.curved.surface.has.a.radius.of.curvature.R.

h

Δ

s

d

FIGuRE 2.4 The.paraxial.approximation.to.find.the.difference.in.optical.path.difference.for.two.rays.that.are.close.to.the.optical.axis.




We.can.use.this.paraxial.approximation.to.find.the.travel.times.for. light.along.the.different.paths.shown.in.Figure.2.3..The.travel.times.along.the.paths.from.point.O.to.P.and.from.point.P.to.I.are.given.by

.t n

cOP t n

cPIOP PI= =1 2

The.travel.times.along.the.path.from.point.O.to.I.are.given.by

.t n

cOV t n

cVQ t n

cQC t n

cCIOV VQ QC CI= = = =1 2 2 2

Then,.the.total.travel.times.along.the.paths.OPI.and.OI.are.as.follows:

.

t nc

OP nc

PIs

t nc

OV nc

VQ QC CI

OPI

OI

= +

= + + +( )

1 2

1 2

We.can.then.use.the.paraxial.equation.to.simplify.this.equation:

.

OP OQ OV VQ

OV VQ hs

PI QI QC CI

= + = + +

= + +

= + = +

∆ ∆

∆

1 12

1

2

2++

= + +

∆22

22QC CI h

s

The.difference.in.the.travel.times.along.the.two.routes.would.be

.

t t nc

OP OV nc

PI VQ QC CI

nc

VQ

OPI OI− = −( )+ − − −( )

= +

1 2

1 hhs

nc

hs

VQ2

1

22

22 2

+ −

For.the.travel.times.along.the.two.routes.to.be.equal,.we.need

.

nc

VQ hs

nc

VQ hs

nchs

12

1

22

2

12

1

2 2

2

+

= −

++ = −

+ =

nc

hs

VQ nc

nc

ns

ns

VQh

n

22

2

2 1

1

1

2

22 2

22 −−( )n1



34 Principles

We.can.apply.the.paraxial.approximation.(Equation.2.4).to.find.the.length.VQ:

.

R QC

QC hR

R QC VQ hR

= +

= +

− = =

∆32

22

2

Substituting.for.VQ.then.gives

.

ns

ns

n nR

1

1

2

2

2 1+ =−( )

Consider.what.happens.when.the.distance.s1.becomes.very.large,.that.is,.for.an.object.at.a.very.long.distance..In.the.limit.that.s1→∞,.we.have

.

ns

n nR

s n Rn n

f2

2

2 12

2

2 1=

−( ) → =−( ) = '

Here,.the.light.comes.to.a.focus.at.a.set.distance.f′.into.the.media.with.index.n2..This.is.defined.as.the.focal.point.within.medium.2..Since.the.light.is.coming.from.infinity,.the.light.rays.would.be.parallel.and.the.wavefronts.would.be.plane.waves..If.s2→∞,.we.have

.

ns

n nR

s n Rn n

f1

1

2 11

1

2 1=

−( ) → =−( ) =

Here,.the.light.comes.to.a.focus.at.a.set.distance.f.into.the.media.with.index.n1..This.is.the.focal.point.within.medium.1.

2.5 thin Lens equation

In.most.optical.systems,.we.would.like.to.have.the.light.pass.from.one.point.s1.to.another.point.s2,.both.of.which.are.in.air.rather.than.inside.some.material.such.as.glass..This.is.accomplished.using.a.thin.lens.that.has.two.curved.surfaces,.as.shown.in.Figure.2.5..Since.a.lens.is.usually.used.in.air,.we.have.set.n1.=.1.

When.describing.the.object.and.image.distances,.and.the.radii.of.curvature.of.the.two.surfaces.of.the.lens,.the.following.conventions.for.the.signs.of.the.distances.and.radii.of.curvature.are.used.(Feynman.1966):

O

PS1 S2

IV Q

n2

V’ C

R

FIGuRE 2.5 Refraction.at.two.curved.interfaces.to.form.a.biconvex.lens.

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. 1.. The.object.distance.s1.is.positive.if.the.point.O.is.to.the.left.of.the.lens.and.is.negative.if.the.point.is.to.the.right.of.the.lens.

. 2.. The.image.distance.s2.is.positive.if.the.point.I.is.to.the.right.of.the.lens.and.is.negative.if.the.point.is.to.the.left.of.the.lens.

. 3.. The.radius.of.curvature.of.the.lens.is.positive.if.the.center.of.the.radius.of.curvature.is.to.the.right.of.the.lens.and.is.negative.if.the.center.is.to.the.left.of.the.lens.

For.the.example.shown.in.Figure.2.5,.both.s1.and.s2.are.positive..The.radius.of.curvature.on.the.left-hand.side.of.the.lens.is.positive.and.that.on.the.right-hand.side.of.the.lens.is.negative..Since.this.lens.has.two.convex.surfaces,.it.is.called.a.biconvex.lens.

To.solve.the.lens.equation,.we.again.calculate.the.travel.times.for.two.different.paths—direct.from.point.O.to.I.and.from.point.O.to.P.and.from.P.to.I—and.equate.them..The.travel.time.along.the.straight.path.between.O.and.I.is.given.by.the.sum.of.the.travel.times.along.segments.OV,.VQ,.QV ,́.and.V´I:

.t

cOV t n

cVQ t n

cQV t

cV IOV VQ QV V I= = = =1 1

' '' '

Then.the.total.travel.times.along.paths.OPI.and.OI.are

.

tc

OPc

PI

tc

OV V I nc

VQ QV

OPI

OI

= +

= +( )+ +( )

1 1

1 ' '

We.can.then.use.the.paraxial.approximation.(Equation.2.4).to.simplify.this.equation:

.

OP OQ OV VQ

OV VQ hs

PI QI QV V

= + = + +

= + +

= + = +

∆ ∆

∆

1 12

1

2

2' ''

' '

I

QV V I hs

+

= + +

∆22

22

The.difference.in.the.travel.times.along.the.two.different.routes.would.be

.

t tc

OP PI OV V I nc

VQ QV

chs

OPI OI− = + − −( )− +( )

=

1

12

2

' '

11

2

221+

− − +( )hs

nc

VQ QV '

If.the.magnitude.of.the.radius.of.curvature.on.both.sides.of.the.lens.is.the.same,.then.we.have

.VQ QV h

R= ='

2

2



36 Principles

The.difference.in.the.travel.time.can.then.be.written.as

.t t

chs

hs

nc

hROPI OI− = +

− −

12 2

12

1

2

2

2

For.the.travel.times.along.the.two.routes.to.be.equal,.we.need

.

12 2

1

12

12

2

1

2

2

2

1 2

chs

hs

nc

hR

s s

+

= −

+

= −( )

+ = −( )

nR

s sn

R

1 1

1 1 1 21 2 .

(2.5)

In.the.limit.that.s1.→ ∞,.we.have

.

1 1 2

11 2

2

2

sn

R

sn

R f

= −( )

=−

=


.s

nR f1

11 2

=−

=

Therefore,.the.focal.lengths.would.be.the.same.on.either.side.of.the.lens,.and.the.light.that.comes.in.from.infinity.is.brought.to.a.focus.at.a.distance.f.from.the.lens..If.the.radii.of.curvature.R1.and.R2.on.either.side.of.the.lens.are.not.equal,.and.using.the.convention.that.the.radius.of.curvature.is.positive.if.the.center.of.the.radius.of.curvature.is.to.the.right.of.the.lens.(R1).and.is.negative.if.the.center.is.to.the.left.of.the.lens.(R2),.then.we.would.have

.VQ h

RQV h

R= = −

2

1

2

22 2'

and

.

1 1 1 1 11 2 1 2s s

nR R

+ = −( ) −


.

1 1 1 1

11

2 1 2

21 2

2 1

sn

R R

sn

R RR R

= −( ) −

=− −

= f .

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Substituting.for.the.focal.length.f.in.Equation.2.5,.we.have

.

1 1 12s s f1

+ =.

(2.6)

This.is.called.the.lensmaker’s.equation,.which.provides.a.relationship.between.the.focal.length.f.and.the.distances.to.the.object.and.the.image..If.both.the.object.and.the.image.distances.s1.and.s2,.respec-tively,.are.equal,.then.for.s1.= s2.=.s:

.

1 1 2 1 2s s s f

s f+ = = → =

In.general,.the.object.and.the.image.may.not.be.points,.as.shown.in.Figure.2.5,.but.rather.extended.objects,.as.shown.in.Figure.2.6..Here.the.object.is.shown.as.an.arrow.that.extends.above.the.optical.axis..We.can.find.out.where.the.image.is.positioned.by.considering.two.principal.light.rays,.one.from.the.tip.of.the.object.that.is.parallel.to.the.optical.axis.and.one.from.the.tip.of.the.object.that.passes.through.the.center.of.the.lens..This.is.called.ray.tracing..The.light.ray.from.the.tip.of.the.object.that.is.parallel.to.the.optical.axis.is.equivalent.to.a.light.ray.from.infinity,.and.therefore,.after.passing.through.the.lens,.it.will.intersect.the.optical.axis.on.the.right-hand.side.of.the.lens.at.the.focal.point.f..The.light.ray.that.passes.through.the.center.of.the.lens.has.a.symmetric.optical.path.through.the.thin.lens,.and.therefore.it.can be.drawn.as.a.straight.line..This.ray.will.be.refracted.by.a.certain.amount.in.traveling.from.air,.with.n.= n1 =.1,.into.the.glass.with.n.=.n2,.with.the.deflection.being.given.by.Snell’s.law,.as.described.in.Equation.2.3..This.light.ray.will.be.refracted.a.second.time.when.it.exits.the.lens,.and.therefore.it.will.continue.along.the.same.direction.after.passing.through.the.lens..For.a.thin.lens,.we.can.draw.that.ray.as.a.straight.line..Where.these.two.rays.intersect.on.the.right-hand.side.of.the.lens,.an.image.will.be.formed..Since.this.image.is.on.the.opposite.side.of.the.lens.from.the.object,.it.is.called.a.real.image..If a.screen.were.to.be.placed.at.this.location,.an.image.of.the.object.would.be.seen,.although.the.image.is.inverted.(i.e.,.upside.down),.and.the.arrow.in.the.image.may.be.different.in.length.from.the.arrow.shown.as.the.object..We.will.discuss.this.in.Section.2.6,.when.we.consider.the.magnification.of.the.lens..

Object

S1 S2

ff

Real image

FIGuRE 2.6 Ray. tracing. to. find. the. object. and. the. image. for. a. thin. biconvex. lens.. The. object. is. located. at. a.distance.S1.to.the.left.of.the.lens,.and.the.image.is.located.at.a.distance.S2.to.the.right.of.the.lens..The.position.of.the.image.can.be.found.by.finding.the.intersection.of.two.light.rays.emanating.from.the.tip.of.the.object..Three..principal.light.rays.are.shown..One.passes.through.the.focal.point.on.the.left-hand.side.of.the.lens,.while.another.passes.through.the.focal.point.on.the.right-hand.side.of.the.lens..These.light.rays.appear.to.come.in.from.infinity..A.third.light.ray.passes.through.the.center.of.the.lens..Since.the.lens.is.considered.to.be.a.thin.lens,.this.light.ray.is.drawn.as.a.straight.line.from.the.tip.of.the.object.to.the.tip.of.the.image..(Credit:.Wiki.)



38 Principles

We.note.there.is.a.third.ray.that.can.be.traced,.as.shown.in.Figure.2.6..This.ray.passes.through.the.focal.point.on.the.left-hand.side.of.the.lens..From.Equation.2.6,.we.know.that.a.light.ray.coming.from.infinity.toward.the.lens.from.the.right-hand.side.would.pass.through.the.focal.point.on.the.left-hand.side.of.the.lens,.and.therefore.we.can.draw.this.light.ray.parallel.to.the.optical.axis.on.the.image.side.of.the.lens..We.see.that.this.light.ray.also.intersects.the.real.image.where.the.first.two.light.rays.intersect..Since.we.only.need.to.find.the.position.of.the.real.image.at.the.intersection.of.any.of.these.lines,.any.two.light.rays.are.sufficient.for.finding.the.object.location.and.size.

Consider.what.happens.when.the.object.distance.S1.is.closer.to.the.lens.than.the.focal.length.f..This.situation.is.shown.in.Figure.2.7..We.can.draw.the.light.ray.from.the.tip.of.the.object.that.is.parallel.to.the.optical.axis..It.intersects.the.optical.axis.at.a.distance.f.to.the.right.of.the.lens..We.can.also.draw.the.light.ray.that.goes.through.the.center.of.the.lens,.but.we.see.that.these.two.light.rays.do.not.intersect.at.any.point.on.the.right-hand.side.of.the.lens..If.we.extend.these.two.lines.to.the.left.of.the.object,.as.shown.by.the.dashed.lines.in.Figure.2.7,.we.see.that.they.intersect.at.a.distance.S2.to.the.left.of.the.lens..From.convention.2.above,.S2.has.a.negative.value.since.it.is.to.the.left.of.the.lens..The.light.rays.from.the.object.appear.as.though.they.were.coming.from.a.virtual.image.at.a.distance.S2.to.the.left.of.the.lens..This.image.is.called.a.virtual.image.since.it.would.not.be.visible.if.a.screen.were.to.be.located.at.that.point.

In.addition.to.having.convex.surfaces,.a.lens.can.also.be.ground.to.have.concave.surfaces,.as.shown.in.Figure.2.8..A.lens.with.two.convex.surfaces.is.called.a.“biconcave”.lens..Using.ray.tracing.from.an.object.

Object

S1S2

ff

Virtualimage

FIGuRE 2.7 A.virtual.image.formed.from.an.object.at.S1.that.is.closer.than.one.focal.length.f.to.the.surface.of.the.lens..In.this.case,.a.real.image.is.not.formed.to.the.right.of.the.lens,.but.rather.a.virtual.image.is.formed.to.the.left.of.the.lens..The.light.rays.on.the.right-hand.side.of.the.lens.appear.to.be.emanating.from.this.virtual.image..(Credit:.Wiki.)

Object

S1

S2

ff

Virtual image

FIGuRE 2.8 Biconcave.lens.composed.of.two.concave.lens.surfaces..(Credit:.Wiki.)




located.at.a.distance.S1.to.the.left.of.the.lens,.which.is.further.than.one.focal.length.f.to.the.left.of.the.lens,.will.form.a.virtual.image.on.the.same.side.of.the.lens.at.S2,.which.is.closer.than.the.focal.length..It.is.also.possible.to.have.a.lens.with.one.convex.surface.and.one.concave.surface..This.form.of.a.lens.is.called.a.meniscus.lens..It.is.also.possible.to.have.a.lens.with.one.side.planar.and.the.other.side.either.convex.or.concave..These.are.called.plano-convex.and.plano-concave.lenses,.respectively.

2.6 Magnification

Again. from. Figure.2.6,.we.can.determine. the. ratio.of.heights.of. the.object. and. the. image.. We.have.redrawn.this.figure.with.some.similar.triangles,.where.the.height.of.the.right.triangle.that.includes.the.object.is.y,.and.the.height.of.the.right.triangle.that.includes.the.image.is.y´.(Figure.2.9).

From.the.similar.right.triangles.on.the.left-hand.side.of.the.lens,.we.can.find.the.magnification.M:

.

yS f

yf

M yy

fS f1 1−

= ′ → ≡ ′ =−

We.can.also.use.similar.triangles.on.the.right-hand.side.of.the.lens.to.find.that

.

yf

yS f

M yy

S ff

= ′−

→ ≡ ′ = −2

2

Since.y′.is.below.the.optical.axis,.it.is.a.negative.quantity.by.convention,.we.say.that.the.magnification.is.negative,.and.it.results.in.an.inverted.image..The.virtual.image.in.Figure.2.8.is.positive,.and.therefore,.in.this.case,.the.magnification.is.positive.and.the.virtual.image.is.not.inverted.

Combining.these.two.equations,.we.find

.

M yy

fS f

S ff

S f S f f

S S S f S

≡ ′ =−

= − → −( ) −( ) =

− −1

21 2

2

1 2 1 222 2

1 2 1 2 1 2

1 2

1

1f f f S S S f S f f S S

S SS S

f

+ = → = + = +

+= →

( )

SS S f1 2

1 1+ =

We.recover.Equation.2.6,.the.lensmaker’s.formula.

Object

S1 S2

ff

y

y′

Real image

FIGuRE 2.9 Magnification.of.a.lens..The.height.of.the.object.is.y.and.that.of.the.image.is.y′..(Credit:.Wiki.)



40 Principles

2.7 Aberrations

The.restriction.of.the.thin.lens.analysis.to.monochromatic.paraxial.rays.leads.to.optical.aberrations.for.real.lenses..For.an.extended.lens.that.admits.light.rays.away.from.the.axis,.the.paraxial.approximation.used.in.Equation.2.4.is.no.longer.valid..In.general,.the.optical.surface.obtained.from.the.principle.of.least.time.is.no.longer.a.spherical.surface,.but.rather.a.higher-order.surface.than.a.sphere..Nonetheless,.a.spherical.surface.is.much.easier.to.fabricate.by.grinding.and.polishing.than.an.aspherical.surface,.and.therefore,.lenses.with.spherical.surfaces.are.often.used..As.the.light.rays.become.further.removed.from.the.optical.axis,.they.no.longer.come.to.a.focus.at.one.point,.but.rather.focus.at.different.points.depending.on.their.distance.from.the.axis..The.resulting.aberration.is.known.as.spherical.aberration,.since.it.results.from.the.spherical.surface.of.the.lens..An.example.is.shown.in.Figure.2.10..The.light.rays.at.the.edge.of.the.lens.come.to.a.focus.closer.to.the.lens.than.those.from.the.paraxial.region.near.the.optical.axis.

In.addition.to.the.spherical.shape.of.the.lens,.spherical.aberration.can.arise.from.index.mismatches.between.the.lens.and.the.sample..When.the.light.travels.between.regions.of.different.indices.of.refrac-tion,.the.light.rays.can.be.bent.at.the.interface.according.to.Snell’s.law,.as.described.in.Equation.2.3..This.aberration.can.be.the.dominant.aberration.when.imaging.deeply.into.a.specimen.since.it.increases.with.depth.into.the.sample.

Another.aberration.is.caused.by.the.wavelength.dependence.of.the.index.of.refraction.n..Chromatic.aberration.arises.when.multiwavelength.light.is.refracted.at.an.interface.between.two.regions.with.dif-ferent.indices.of.refraction..The.different.colors.of.light.can.be.refracted.by.differing.amounts,.as.shown.in. Figure. 2.11.. A. familiar. example. is. the. decomposition. of. white. light. by. a. prism. or. water. droplet,.which.forms.the.familiar.colored.rainbow.when.sunlight.is.decomposed.into.its.spectral.components..The.cause.for.the.spectral.decomposition.is.the.wavelength.dependence.of.the.index.of.refraction..The.

Spherical aberration

FIGuRE 2.10 Spherical.aberration..The.light.rays.that.are.farther.from.the.optical.axis.come.to.a.focus.at.different.distances.from.the.lens..(Credit:.Wiki.)

Chromatic aberration

FIGuRE 2.11 Chromatic.aberration..Different.colors.of. light.are.refracted.by.different.amounts.when.passing.through.regions.with.different.indices.of.refraction..This.is.because.the.index.of.refraction.depends.on.the.wave-length.of.light..(Credit:.Wiki.)




wavelength.dependence.of. the. index.of.refraction.is.called.dispersion,.since. it.causes. light. to.be.dis-persed.into.its.spectral.components.

A.solution.to.overcome.chromatic.aberration.is.to.use.two.different.lenses.with.two.different.shapes.(e.g.,.convex.and.concave.radii.of.curvature).made.with.optical.materials.with.different.dispersion.char-acteristics.(e.g.,.crown.and.flint.glasses)..An.example.is.shown.in.Figure.2.12.

In.addition,.lenses.can.have.other.optical.aberrations.such.as.coma,.astigmatism,.field.curvature,.and.distortions.(barrel.and.pincushion)..Misalignment.between.optical.elements.can.cause.further.aberra-tion.as.discussed.in.Chapters.4.and.9..While.the.primary.goal.of.adaptive.optics.in.biological.imaging.is.to.overcome.sample-induced.aberrations,.as.discussed.in.Chapter.3,.the.adaptive.optical.system.can.also.overcome.optical.system–induced.aberrations.due.to.aberrations.in.the.optical.components.and.misalignments.between.components..In.some.situations,.system.aberrations.can.dominate.specimen-induced.aberrations..Here,.adaptive.optics.can.be.used.to.relax.system.tolerances. to.bring.down.the.cost.of.high-performance.optical.systems.or.to.improve.the.performance.of.lower-performance.optical.systems.

references

Feynman,.R..P.,.R..B..Leighton,.and.M..Sands,.The Feynman Lectures on Physics,.Addison-Wesley,.Reading,.MA,.1966.

Born.M..and.E..Wolf,.Principles of Optics,.7th.ed.,.Cambidge:.Cambridge.University.Press,.2006.

Achromatic doublet

FlintCrown

FIGuRE 2.12 Achromatic.doublet. lens. to.compensate. for.chromatic.aberration.due.to. the.wavelength.depen-dence.of.the.indices.of.refraction..Here.one.lens.is.biconvex.and.is.made.of.crown.glass.with.one.index.of.refraction,.and.the.other.lens.is.plano.convex.and.is.made.of.flint.glass.with.a.different.index.of.refraction..The.lens.pair.tends.to.compensate.for.the.dispersion.of.multiwavelength.light.



principles of geometric optics approximation to wave optics, it is technologically useful for the...

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