Principles of Geometric Optics approximation to wave optics, it is technologically useful for the...
Transcript of Principles of Geometric Optics approximation to wave optics, it is technologically useful for the...
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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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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.
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33Principles of Geometric Optics
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
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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
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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
= −( )
=−
=
In.the.limit.that.s2.→ ∞,.we.have
.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
+ = −( ) −
In.the.limit.that.s1.→ ∞,.we.have
.
1 1 1 1
11
2 1 2
21 2
2 1
sn
R R
sn
R RR R
= −( ) −
=− −
= f .
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37Principles of Geometric Optics
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.)
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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.)
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39Principles of Geometric Optics
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.)
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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.)
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41Principles of Geometric Optics
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.
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