PY3101 Optics - Physics · PY3101 Optics Revision ColáistenahOllscoileCorcaigh, Éire University College Cork, Ireland ROINN NA FISICE Department of Physics PY3101 Optics Course

1

M.P. Vaughan

PY3101 Optics

Revision

Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland

ROINN NA FISICE

Department of PhysicsPY3101 Optics

Course overview

Wave Optics

Electromagnetic Waves

Geometrical Optics

Crystal Optics

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Wave Optics

• General physics of waves with application to

optics

• Huygens-Fresnel Principle

• Derivation of Laws of Optical Propagation

• Rectilinear motion

• Reflection

• Refraction

• Diffraction

• Diffraction gratings (use in spectroscopy)


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Electromagnetic Waves

• The wave equation

• The electric susceptibility tensor• Light propagation in isotropic media

• Refractive index and dispersion

• Optical loss

• Polarisation• Polarising optical elements (linear, retardation

plates)

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Geometrical Optics

• Fermat’s Principle (least time)

• Derivation of Laws of Optical Propagation

• Imaging with lenses and mirrors

• Perfect imaging

• Spherical lenses and mirrors

• Paraxial approximation

• Aberrations

• Systems of lenses and mirrors


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Crystal Optics

• The index ellipsoid

• Birefringence

Crystal Optics – light propagation in anisotropic

media

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Huygens-Fresnel Principle


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• You should be able to

• Write down the form of a spherical wave

• Define a wavefront

• State Huygens’ Principle

• Using Huygens’ Principle

• Derive the Law of Rectilinear Propagation

• Derive the Law of Reflection

• Derive the Law of Refraction

Huygens’ Principle

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Spherical waves


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Spherical waves

Since the intensity of an EM wave is proportional to the squared modulus of the amplitude, by the conservation

of energy, the amplitude must vary as 1/r.

Moreover, the requirement that the amplitude be finite at r= 0 means that the spherical wave must be of the form

( ) .sin, krer

EtrE tir ω−=

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Geometric wavefront

A geometric wavefront is the surface in

space containing all points in an optical

field that have the same phase.

A ray is a path through space that is

everywhere perpendicular to the

wavefront.


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Geometric wavefront - spherical

Wavefronts –contours of

constant phase

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Geometric wavefront - spherical


constant phaseRays –

everywhere perpendicular to wavefronts


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Geometric wavefront - plane


constant phase

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Geometric wavefront - plane


constant phaseRays –

everywhere perpendicular to wavefronts


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Huygens’ Principle

Each point on a wavefront acts as a source

of secondary, spherical wavelets.

At a later time, t, a new wavefront is

constructed from the sum of these

wavelets.

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Huygens’ Principle – rectilinear propagation

All points on the wavefront act as sources of spherical wavelets

z0

constant phase over surface of sphere


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Huygens’ Principle – rectilinear propagation

Since all points on the spheres must have the same phase, the tangent to the leading edge of all the spheres must also be at a constant phase.

z0

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Huygens’ Principle - reflection


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Huygens’ Principle - refraction

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The Huygens-Fresnel Principle


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• Explain what is meant by coherence and

interference

• State the Huygens-Fresnel Principle

• Explain how this is different to Huygens’

Principle


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Coherence

If two beams of light are coherent with

each other, then there is a fixed relation

between their phases


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Interference

Two coherent beams may add together via

the Principle of Linear Superposition to

obtain interference

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The Huygens-Fresnel Principle

For light of a given frequency, every point

on a wavefront acts as a secondary source

of spherical wavelets with the same

frequency and the same initial phase.

The wavefront at a later time and position

is then the linear superposition of all of

these wavelets.

Diffraction

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• Describe the diffraction regimes (near-field

and far field)

• Derive the single-slit diffraction pattern

• Generalise to the multiple-slit case

• Explain the Rayleigh criterion

• Analyse diffraction gratings

• Discuss the applications of diffraction

• Sketch and explain basic monochromator

designs



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What is diffraction?

• Diffraction is the ‘bending’ of

waves around objects or through

apertures

• It is an interference effect

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Light passing through a narrow aperture



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Light passing through a narrow aperture

Maximum possible path difference

.max DABBPAP ==−=∆

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Limiting cases: λλλλ >> D

∆max always less than λ – wavelets add constructively in all directions.

Emergent field looks like point

source.


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Limiting cases: λλλλ << D

Both constructive and destructive interference outside shaded region

Wavelets add constructively in this region

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Fresnel and Fraunhofer diffraction

Near field (Fresnel diffraction)

Far field (Fraunhofer diffraction)


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Single slit diffraction

EL is the field strength per unit

length

EP is the total field a the point P

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Field at x

.dxEdE L=

Contribution to field EP due to dE

( )( )[ ] .sin dxxkrt

xr

EdE L

P −= ω

Total field EP

( )( )[ ] .sin

2/

2/∫− −=D

D

LP dxxkrt

xr

EE ω


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r(x) is given by the cosine rule

( ) ( )θπ −−+=2

222 cos2RxxRxr

x

or

( ) .sin2

1

2/1

2

2

−+= θ

R

x

R

xRxr

To find a closed form solution, we must approximate this expression.

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Taylor series expansion of r(x)

The Taylor series expansion for a function (1 + ξ)1/2 is

( ) K+−+=+82

112

2/1 ξξξ

Hence,

( )

++−= Kθθ 2

2

2

cos2

sin1R

x

R

xRxr

and

( ) .cos2

sin 22

K++−= θθR

kxkxkRxkr


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The Fraunhofer condition

The third term in the expression for kr(x) takes its maximum

when x ± D/2 and θ = 0. That is

.48

cos2 2

2

2

22

2

R

D

R

kD

R

kx

λπ

θ =→

The condition that this term makes a negligible contribution to the phase is

.4 2

2

πλπ

<<R

D

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The Fraunhofer condition

Neglecting the factor of 4 in the denominator of the condition just found, it may be re-written as

.DR

D λ<<

This is the Fraunhofer condition for far field diffraction.


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Far field approximations

Assuming that the Fraunhofer condition is valid, the third term in the expression for kr(x) may be neglected and we have

( ) .sinθkxkRxkr −≈

The 1/r(x) factor appearing in the integral for EP is less

sensitive to changes in r(x) than the phase and we may simply put

( ).

11

Rxr≈

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Integrating over x

Using these approximations, the expression for the total field EP becomes

To perform this integral, we note that

[ ] ( ){ }.Imsinsin sinθωθω kxkRtiekxkRt +−=+−

[ ] .sinsin2/

2/∫− +−=D

D

LP dxkxkRt

R

EE θω


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The total field EP

Integrating over the x-dependent part

where

.sin2

θβkD

=

,sin

sin

2/

2/

sin2/

2/

sin

ββ

θ

θθ D

ik

edxe

D

D

ikxD

D

ikx =

=

−−∫

Hence, the total field EP is

( ).sinsin

kRtR

DEE L

P −= ωββ

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Intensity profile for a single slit

Averaging EPover time gives

The squared modulus of this will be proportional to the intensity, i.e.

.sin

2 ββ

R

DEE L

P =

( ) ( ) .sin

0

2

ββ

θ II =


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The zeros of the peaks occur at values of

where m is an integer. Hence, the first zeros around the central peak are given by

,sin2

πθβ mkD

==

.sinD

λθ =

Note that this result is only valid for λ < D. In other cases,

there are no zeros from –π to π.


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Intensity profile for a circular aperture

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The Airy disc

Airy pattern by Sakurambo. A computer-generated image of an Airy disk.URL: http://en.wikipedia.org/wiki/File:Airy-pattern.svg

Airy disc


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The Airy disc

Laser Interference by Petrov Victor. Diffraction of red laser beam by a circular aperture.

URL: http://en.wikipedia.org/wiki/File:Laser_Interference.JPG

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The Rayleigh criterion

The first zero of the intensity profile for diffraction from a circular aperture occurs at

.22.1sinD

λθ ≈

This represents the minimum angular separation that two

points can be so that they may be separately resolved.

Using the small angle approximation, this becomes

.22.1D

λθ ≈

This is known as the Rayleigh criterion.


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Diffraction limited imaging

Intensity profiles for two resolvable distant point sources.

Merged intensity profiles for unresolvable distant point sources.

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Intensity profile for multiple slits

( ) ( ) .sin

sin

sin0

22

=ββ

αα

θN

NII


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Intensity profile for multiple slits

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Diffraction condition

Note that the condition for constructive interference is

.sin λθ ma =


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We can re-write this as

.sin2

mka

πθ =

But this is just,mπα =

( ) ( ) .sin

sin

sin0

22

=ββ

αα

θN

NII

which gives the condition for the local maxima of the intensity

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Diffraction – wavelength dependence

.sin λθ ma =

Red (longer wavelength) light is diffracted to a greater extent than

blue (shorter wavelength).

(Yellow arrow is incident light and specular reflection)


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Incident and diffracted wavevectors:

,zk k=

( ).cosˆsinˆ' θθ zxk += k

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Diffraction condition: off-axis incidence

.sin,sin mi aOBaAO θθ ==


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The grating equation

( ) .sinsin λθθ ma im =+

For off-axis transmission, the diffraction condition is now

This reduces to the case of normal incidence when .0=iθ

This result may be further generalised by taking the incident angle around to the front of the grating – i.e. making the grating into a reflection grating.

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Reflection gratings


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Reflection gratings

Light strikes the reflection grating at an angle θi. For certain angles θm, the diffraction condition will

be met:

The path lengths of rays from the incident

wavefront via the successive rulings of the

grating and leaving at the same angle must

differ only be integral multiples of the

wavelength λ.

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Derivation of the reflection grating equation

Incident wavefront AC

Path from A to wavefront BD

.sin maAB θ=

Path from C to wavefront BD

.sin iaCD θ=

( ).sinsin imaCDAB θθ −=−Path difference


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The reflection grating equation

The diffraction condition for a reflection grating may then be expressed mathematically as

( ).sinsin imam θθλ −=

This is known as the reflection grating equation.

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Dispersion

The dispersion of a grating is defined as

.λθ

θd

dD m=

Differentiating the grating equations, we have

.cos m

m

ad

dm θ

θλ

=


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Number of orders

Recall that

So the highest order m is the largest integral value of

( ) .sinsin λθθ ma im =±

( ).sinsin im

aθθ

λ±

.2

max λa

m <

Hence

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Resolving power

The resolving power of a grating is defined as

,λλ∆

=R

where, via Rayleigh’s criterion, ∆λ is the minimum resolvable wavelength between the peaks of two wavelengths with midpoint λ.


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Common example of a grating

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Diffraction around a razor blade


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X-ray diffraction (non-optical)

(See PY3105)

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Maxwell’s equations


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• Define the electric susceptibility tensor

• Derive the wave equation for an isotropic

medium

• Write down plane-wave solutions of the wave

equation

• Explain the phenomena of dispersion

• Describe the mechanism of optical loss

• Starting with a complex wave vector, derive

the light intensity with distance and the

absorption coefficient


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.1

and

,

,

,0

,

0

0 MHPED

DjH

BE

B

D

−=+=

∂∂

+=×∇

∂∂

−=×∇

=⋅∇

=⋅∇

µε

ρ

t

t

f

f


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The electric susceptibility tensor

The electrical polarisation P of a medium is given by

,0 EP Eχε=

where χE is the electric susceptibility tensor. χE

characterises the frequency response of the medium to an applied electric field E.

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Linear, isotropic and homogeneous media

In a linear, isotropic and homogeneous (LIH) medium

.

00

00

00

0

0

0

=

χχ

χχE


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Wave equation in a LIH medium

In a dielectric the free charge density and free current are zero, so, from Maxwell’s equations

0=⋅∇ D

and

.t∂

∂=×∇D

H

( ) ,00 EEID εεχε =+= E

where ε is the relative permittivity.

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Similarly

,1

0

BHµµ

=

where µ is the relative permeability. Hence

00 =⋅∇=⋅∇ ED εεand

.

,1

00

0

0

t

tt

∂∂

=×∇→

∂∂

=∂∂

=×∇=×∇

EB

EDBH

µεµε

εεµµ


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From Maxwell’s equations

.2

2

00tt ∂

∂−=

∂×∂∇

−=×∇×∇EB

E µεµε

Using the vector identity

,0=⋅∇ E

and

( ) EEE2∇−⋅∇∇=×∇×∇

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we have

.2

2

00

2

t∂∂

=∇E

E µεµε

The wave speed is

( ) 2/1εµ=n

where

( ) ,2/1

00n

cv == −µεµε

is the refractive index.


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Plane-wave solutions

We look for solutions of the form

where E0 is a Jones vector containing information about the polarisation.

.

,

2

2

2

222

ω−→∂∂

=−→∇

t

kk

Now

( ) ( ),, 0

tiet ω−⋅= rkErE

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Plane-wave solutions

So

which gives

.nk

vω

=

,00

22

2

2

00

2EE

EE µεµεωµεµε =→

∂∂

=∇ kt

.k

vω

=

We may put where k0 is the free space wave-vector, so

,0nkk →


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Frequency dependence

A simple analysis yields

showing that χE is frequency dependent. This gives rise to the phenomenon of dispersion (different frequencies of light travelling at different speeds in an optical medium).

( )∑ +−=

i ii

iiE

i

mq,

22

2

0 τωωωχε

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Frequency dependence

The relative permittivity is

This means

.1 21 εεχε iE −=+=

21 innn −=and

,21 ikkk −=

where .0nkk =


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Relative permittivity

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Optical loss

Using

Substituting for v using the complex refractive index,

( ) ,exp, 0

−=v

ztitz ωEE

( ) .expexp, 210

−

−=c

zn

c

zntitz ωEE


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The absorption coefficient

Now the intensity of the radiation I(z) is proportional to the squared modulus of the field

So

( ) .2

exp 22

0

2

−=∝c

znzI

ωEE

( ) ( ) ,0 zeIzI α−=

where

c

n22ωα =

is the absorption coefficient.

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Polarisation


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• Define the polarisation of light

• Describe and analyse

• plane-polarisation

• circular polarisation

• elliptical polarisation

• Describe

• linear polarisers

• retardation (wave) plates

• Apply the Jones calculus to states of

polarisation and optical elements

Polarisation

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Linear polarisation

( ),0

rkEE ⋅−= tie ω

A plane-wave may be written

where E0 is a Jones vector, containing information about the polarisation.


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Linear polarisation

.sin

cos00

=

θθ

EE

For linear polarisation at an angle θ to the x-axis E0 is given by

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Linear polarisation


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Linear polarisation – special cases

,0

100

= EE x

,1

000

= EE y

,0=θ

,2

πθ =

x-linearly polarised

y-linearly polarised

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y-linearly polarised light

(x-linearly polarised aligned with x-axis)



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The dichroic sheet

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The linear polariser


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• The most general form of polarisation

is elliptical polarisation

• the electric field spirals around the

propagation axis tracing out an ellipse.

• This may be understood by resolving

the electric field into orthogonal

components.

• So long as these components remain in

phase, the polarisation will be linear.

Elliptical polarisation

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• If, however, a phase shift is introduced

on to one of the components, the

polarisation will become elliptical.

• This is illustrated in the next slide...



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Elliptically polarised light

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Retardation

( ),sin

cos0

rkEE

⋅−

= ti

i

i

ee

ey

x

ωφ

φ

θθ

.xy φφ −=Γ

This phase shift is known as the between orthogonal components is known as the retardation Γ. For the polarisation

This is


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Retardation

( ).sin

cos0

rkEE

⋅−Γ

−

= ti

i

ie

ee x ωφ

θθ

.sin

cos00

= Γ θ

θie

EE

We may re-write the polarisation as

Since the x-phase factor is arbitrary,

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Circular polarisation

iee ii ±== ±Γ 2/π

.1

2

0

0

±

=i

EE

,4

πθ =

In the case

and

we have


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Circular polarisation – x component

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Circular polarisation – ΓΓΓΓ = ππππ/2

,2/ iee ii ==Γ π

( ).1

2

0 rkE

E⋅−

= tie

i

ω

[ ] ( )( )

.sin

cos

2Re

0

⋅−−

⋅−=

rk

rkEE

t

t

ωω

In the case

we have

The real part is then


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Circular polarisation – y component

2/πixy eEE =

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Circular polarisation – left polarised

2/πixy eEE =


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Circular polarisation – ΓΓΓΓ = –ππππ/2

,2/ iee ii −== −Γ π

( ).1

2

0 rkE

E⋅−

−

= tiei

ω

[ ] ( )( )

.sin

cos

2Re

0

⋅−

⋅−=

rk

rkEE

t

t

ωω

In the case

we have

The real part is then

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Circular polarisation – y component

2/πixy eEE −=


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Circular polarisation – right polarised

2/πixy eEE −=

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Elliptical polarisation: general case

.sincos2 2

00

2

0

2

0

Γ=Γ−

+

x

x

y

y

x

x

y

y

E

E

E

E

E

E

E

E


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Elliptical polarisation: general case

xE

yE

E

α

. arbitrary, is 00 yx EE ≠Γ

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• A retardation or wave plate is an optical

element that produces some

retardation between the orthogonal

components of the wave.

• The physical origin of this retardation

is due to the phenomenon of

birefringence

Wave plates


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• Birefringence is a phenomenon in

which the components of the wave see

a different refractive index depending

on the orientation of the polarisation

within some anisotropic material

• This leads to light with different polarisation

directions having different phase velocities

Wave plates

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Fast and slow axes


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• The speed of the waves is determined

by the refractive index that it sees.

• This is determined by a construction

known as the index ellipsoid

The index ellipsoid

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• Quarter wave plate

• Introduces a phase shift of

• Produces circularly polarised light from

linearly polarised light

• Half wave plate

• Introduces a phase shift of

• Reverses sign of y-component

• Hence, for linearly polarised light at an angle θto the wave plate axis, the light is rotated by

2θ.

Types of wave plate

2/π±

π±


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• Polarised light may be analysed by

passing it through a linear polariser

• In the case of initially linearly polarised

light, the emergent intensity follows

Malus’ Law

• Here, we consider the general case

The analysis of polarised light

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The analysis of polarised light

( ) ( ),cossincos2sincos 222

0 Γ++= θθθθθ rrII

.0

0

x

y

E

Er =

The power intensity through an analyser is given by

where


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The analysis of linearly polarised light

( ) .cos20 θθ II =

,00 =yE

For x-linearly polarised light, we have

which gives

This is known as Malus’ Law.

63


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The analysis of circularly polarised light

( ) ( ) .sincos 0

22 II =+= θθθ

.2

and1π

±=Γ=r

For circularly polarised light, we have

which gives

In other words, the time-averaged intensity is

constant.


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Jones vectors

.sin

cos00

=

θθ

EE

Linear polarisation

General case

x-linearly polarised

,0

100

= EE x


,1

000

= EE y

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Jones vectors


General case

left-circularly polarised

.1

2

0

=+

i

EE

right-circularly polarised

.1

2

0

−

=−i

EE

.sin

cos00

= Γ θ

θie

EE


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Jones matrices

Polarisers

General case

x-linear polariser

.00

01

=xP

y-linear polariser

.10

00

=yP

.sinsincos

sincoscos2

2

=

θθθθθθ

θP

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Jones matrices

Retardation plate

General case

Quarter wave-plate

.0

012/

±

=±i

πM

Half wave-plate

.10

01

−

=πM

.0

01

= ΓΓ ie

M


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Jones matrices

The effect of a series of optical elements may be modelled by multiplying the

corresponding Jones matrices together

to form a combined element.

66

Fermat’s Principle


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• You should be able to• Define the optical path length

• State Fermat’s Principle

• Use Fermat’s Principle to

• derive the Law of Reflection

• derive the Law of Refraction

• Demonstrate perfect imaging


67


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Geometrical wavefront

These points are all in phase with one another and constitute a geometric wavefront.


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The optical path length

Putting T = t - t0, we may then multiply T by c to express the propagation time in dimensions of space

( ) ( ).0ttcr −=Λ

The quantity Λ(r) is known as the optical path length and is a function of distance.

68


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Now, if

,n

cv

dt

dS==

then

( ) ,ndScdtrd ==Λ

so

( ) ( ) .,,0∫=Λr

dSzyxnr


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Homogeneous medium

In a homogeneous medium

( ) ,0,, =∇ zyxn

that is:

the refractive index is the same everywhere.

69


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Isotropic medium

In an isotropic medium

( ) ,,, 0nzyxn =

that is:

the refractive index is the same in all directions


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Optical path length – LIH medium

We also make the usual assumption that the response of

the medium is linearly proportional to the applied field.

For a linear, isotropic homogenous (LIH) medium

becomes

( ) .0∫=Λr

dSnr

( ) ( )∫=Λr

dSzyxnr0

,,

70


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The path taken between two points by

a ray of light is the path that can

traversed in the least time

or, equivalently in terms of optical path length,

Light traverses the route between two points for which the optical path

length is a minimum.


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Rectilinear propagation

Using the calculus of variations, we can use Fermat’s Principle to derive Law of Rectilinear Propagation

In a LIH medium, light propagates in straight lines.

71


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Reflection

We may also apply Fermat’s Principle under constraint.

For instance, we may impose the constraint that light

travelling between A and B in a medium of refractive index

n1 must touch some point on the interface between this

medium and another of refractive index n2.

This is the required constraint for reflection.


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Reflection

72


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Reflection

.11 BA SnSn +=Λ

( )[ ]( ) .

,

2/122

2/122

BBB

AA

yxxS

yxS

+−=

+=

Applying Fermat’s Principle, we have

.01 =

∂∂

+∂∂

=∂Λ∂

x

S

x

Sn

x

BA


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Reflection

( )

[ ]( ).

,

2/122

2/122

B

B

BB

BB

AA

A

S

xx

yxx

xx

x

S

S

x

yx

x

x

S

−=

+−

−=

∂∂

=+

=∂∂

But

.sinandsin r

B

Bi

A S

xx

S

xθθ =

−=

73


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Reflection

So Fermat’s Principle implies

( ) .0sinsin1 =−=∂Λ∂

rinx

θθ

This is satisfied when

ri θθ sinsin =

or

ri θθ =


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Reflection

Thus, Fermat’s Principle reproduces the Law of Reflection

In a LIH medium, the angle of reflection equals the

angle of incidence.

74


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Refraction


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Refraction

.21 BA SnSn +=Λ

( )[ ]( ) .

,

2/122

2/122

BBB

AA

yxxS

yxS

+−=

+=

Applying Fermat’s Principle, we have

.021 =∂∂

+∂∂

=∂Λ∂

x

Sn

x

Sn

x

BA

75


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Refraction

( )

[ ]( ).

,

2/122

2/122

B

B

BB

BB

AA

A

S

xx

yxx

xx

x

S

S

x

yx

x

x

S

−=

+−

−=

∂∂

=+

=∂∂

But

.sinandsin t

B

Bi

A S

xx

S

xθθ =

−=


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Refraction

So Fermat’s Principle implies

.0sinsin 21 =−=∂Λ∂

ti nnx

θθ

This is satisfied when

,sinsin 21 ti nn θθ =

i.e. by Snell’s Law.

76


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Refraction

Thus, Fermat’s Principle reproduces the Law of Refraction

In a LIH medium, the Law of Refraction is given by Snell’s Law

,sinsin 21 ti nn θθ =


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Perfect imaging – hyperbolic lens

77


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Perfect imaging – elliptical lens


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Perfect imaging – elliptical mirror

78


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Perfect imaging – elliptical mirror


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Perfect imaging – parabolic mirror

79

Spherical lenses and mirrors


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• You should be able to• State the paraxial approximation

• Define the focal length

• Recall

• the thin lens equation

• the lens-maker’s equation

• the Gaussian lens formula

• the expression for a series of thin lenses in close

combination

• Recall and apply the rules for image construction

• Calculate transverse magnification

• Define the optical power of a lens

Spherical lenses and mirrors

80


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• Cannot obtain perfect imaging

• Reasonable approximation possible

Imaging by a spherical lens


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Lens sign conventions

81


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Lens sign conventions

• Light is always taken to

propagate from left to right.

• If A is to the left of B, then so is

taken to be positive (and vice

versa).

• If C is to the right of B, then si is

taken to be positive (and vice

versa).

• If the centre of the sphere is to the right of B, R is taken

to be positive. This is a convex lens.

• If the centre of the sphere is to the left of B, R is taken

to be negative. This is a concave lens.


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• Requirement for perfect imaging is that

all the rays have equal optical path-

length.

• That is, we require Λ to be a constant


82


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.1221

o

o

i

i

io l

sn

l

snR

l

n

l

n−=

+

However

( )( ),,

,,

iii

ooo

sll

sll

φ

φ

=

=

so no closed form solutions.

83


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The paraxial approximation

Small angle approximation

.1cos,sin ≈≈ φφφ

22

oo sl →

So

and

.22

ii sl →


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The paraxial approximation

With these approximations, we obtain

( ).112

21 nnRs

n

s

n

io

−=+

84


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The focal length

In the case

,0 ∞→s

From this, we may define the focal length within the lens fi

( ).112

2 nnRs

n

i

−=

.12

2ii fR

nn

ns ≡

−=


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The focal length

Similarly, when

,∞→is

we may define the focal length outside the lens fi

.12

1oo fR

nn

ns ≡

−=

85


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A thick lens


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A thick lens

86


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A thick lens

For the surface with radius of curvature R1,

( ).112

11

2

1

1 nnRs

n

s

n

io

−=+

( ).112

22

2

2

1 nnRs

n

s

n

oi

−−=+

For the surface with radius of curvature R2,


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A thick lens

Combining these results

( )

+−−

−=

+

2

2

1

212

2112

1

11111

oioi s

n

snnn

RRssn

Inserting

,12 io sds −=

87


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The thin lens equation

( )( )

.1111

11

212

2112

1

iioi sds

dnnn

RRssn

−−−

−=

+

.1111

211

12

−

−=+

RRn

nn

ss oi

Taking the limit and putting,0→d ,, 22 ooii ssss ==

This is the thin lens equation.


ROINN NA FISICE


The lens maker’s formula

.111

211

12

−

−=

RRn

nn

f

If either or ∞→is ∞→os

.111

or111

211

12

211

12

−

−=

−

−=

RRn

nn

sRRn

nn

s io

But the term on the RHS is a constant, which we define to be the focal length f

This is the lens maker’s formula.

88


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The Gaussian lens formula

.111

fss oi

=+

We must also have

This is the Gaussian lens formula.


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Thin lenses in close combination

In general

.11

∑=i iff

89


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Mirror sign conventions


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• Light is always taken to propagate from

left to right.

• The object distance so is positive when it

is to the left of the mirror surface.

• The image distance si is positive when it

is to the left of the mirror surface (real

image).

Mirror sign conventions

• The image distance si is negative when it is to the

• right of the mirror surface (virtual image).

• The radius R is positive if the mirror surface is to the right of the

centre of the sphere (convex mirror)

• The radius R is negative if the mirror surface is to the left of the

centre of the sphere (concave mirror)

90


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Spherical mirrors

From the Law of Reflection

.φαθ +=

( ) .φββπφπθ −=−−−=

From inspection of the figure

Hence

.2 αβφ −=


ROINN NA FISICE


Spherical mirrors

Hence

.211

Rss io

−=+

Taking limits as before and defining the focal length f, we then have

,111

fss io

=+

which is the same expression as the Gaussian lens

formula.

91


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• Sketch a ray from the tip of the object parallel to the

horizontal (the principle axis) to the centre line of the

lens. From there, sketch another ray passing

through the focus associated with the left-hand lens

surface.

• Sketch a ray from the tip of the object directly

through the centre of the lens without deviation.

• Sketch a ray from the tip of the object passing

through the focus associated with the right-hand

lens surface to the centre line of the lens. From there

sketch a line parallel to the principle axis towards

the image.

Image construction for convex lenses


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Convex lens

92


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Convex lens

The image at i is real and inverted.

The magnification of the image M is given by

.o

i

y

yM =

Since yi is negative, so is M (inverted image).


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• Sketch a ray parallel to the optical axis of the

lens. Since f < 0, this must pass through f on

the left of the lens (as a virtual ray).

• Sketch a ray passing through the centre of

the lens without deviation

• Sketch a ray following the line through f on

the right of the lens (this extension is virtual

on the right) and emerging parallel to the

optical axis. The parallel line is then

extended to the left as a virtual ray

Image construction for concave lenses

93


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Concave lens


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Concave lens

In this case, xo is the distance

between the object and f on the right of the lens.

xi is the distance between the image and f on the left of the lens.

94


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Other types of lenses

plano-convex plano-convex doublet


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Optical power

.1

f=P

The optical power P of a lens is a measure of the degree to which it converges or diverges light.

P is defined as the reciprocal of the focal length.

The SI unit of P is called the dioptre (m-1).

95

Optical instruments


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• You should be able to• Discuss the anatomy and function of the

human eye

• Describe common visual impairments

• Derive the angular magnification for a

magnifying glass

• Describe different types of refracting

telescope and their designs

• Describe the design of reflecting

telescopes

Optical instruments

96


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The anatomy of the eye


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The lens

The curvature of the lens may be changed by muscle contractions in the eye.

97


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Near and far points

• The near point is the closest distance for

which the lens can focus light on the retina

• Typically at age 10, this is about 18 cm

• It increases with age, ~ 25 cm for an adult

• The far point of the eye represents the

largest distance for which the lens of the

relaxed eye can focus light on the retina

• Normal vision has a far point of infinity


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Farsightedness – hyperopia

Distant objects may be focussed but not nearby objects.

98


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Correcting farsightedness


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Nearsightedness – myopia

• axial myopia

• Lens too far from retina

• refractive myopia

• Lens-cornea system too powerful to focus properly onto the

retina

99


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Correcting nearsightedness


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Angular size of unaided image

The object at o subtends an angle of αu at the viewing point.

100


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Magnifying glass - aided image

The image i of an object placed at

o within the focal length of a convex mirror subtends an angle of αa at the viewing point.


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Angular magnification

The angular magnification Mα is defined as the ratio of the aided and unaided viewing angles

.u

aMαα

α =

We shall employ the paraxial approximation to obtain an expression for this.

101


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.,i

i

o

oa

ou

d

h

d

h

D

h=== αα

From the diagrams

So

.ou

a

d

DM ==

αα

α

where D is the distance to the object in the unaided case.


ROINN NA FISICE



.fd

h

f

h

i

io

+=

From the diagram

,i

o

i

o

d

d

h

h=

.111

oi ddf=+

From the expression for αa,

Leading to

102


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From

,od

DM =α

.11

+=

idfDMα

We then have

Thus the shorter the focal length, the greater the angular

magnification.


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• Terrestrial

• Produces upright image

• Employs a concave lens for the eyepiece

• Astronomical

• Inverts the image

• Employs a convex lens for the eyepiece

Types of refracting telescope

103


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Galilean (terrestrial) telescope

Gives upright image. Note the focal points coincide.


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Keplerian (astronomical) telescope

Gives inverted image. Note the focal points coincide.

104


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Galilean telescope Keplarian telescope

From the system matrix of each telescope, both are found to have an angular magnification of

.e

o

f

fM −=α


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Newtonian telescope

105


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Newtonian telescope

.e

o

f

fM −=α

Essentially, we have the same magnification system as for the astronomical telescope.

Hence, the angular magnification is

Aberrations

106


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• You should be able to• Discuss what is meant by third order

aberration

• List and describe common forms of

monochromatic aberration

• Explain the origin of chromatic aberration

and strategies to correct it

Aberrations


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Third order correction

θθ ≈sin

L++−=!5!3

sin53 θθ

θθ

Previously, we employed the paraxial approximation

In reality

The second term is referred to as the third order correction. The second term in the expansion of cos is also used in corrections to this order.

and.1cos ≈θ

107


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.1111

2

21

22

1221

−+

++

+−=

+

iioo

io

sRs

n

sRs

nR

nnRs

n

s

n

φ

108


ROINN NA FISICE



Note that the Rφ2 term gives a measure of the displacement of the intersection of the ray with the lens from the optical axis.

Thus, in the third-order treatment, the new term increases in proportion with the square of the angular displacement.


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• Spherical Aberration

• Coma

• Astigmatism

• Field Curvature

• Distortion

Third order corrections

109


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Spherical Aberration

• Due spherical curvature of lens


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Longitudinal and transverse spherical aberration

The longitudinal spherical aberration LSA is defined as the distance between the intersection of a ray with the optical axis and the paraxial focus.

'.ooSA ssL −=

The transverse spherical aberration LSA is defined as the perpendicular distance above (or below) the paraxial focus that a ray actually passes.

110


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Spherical Aberration

• Due spherical curvature of lens


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Soft image focusing

111


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Coma

• Due to off-axis object points

• Transverse magnification is a function of ray height

• Pattern looks like a comet


ROINN NA FISICE


Coma in a lens

112


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Coma in a parabolic mirror


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• Corrective lenses for Newtonian

telescopes with f numbers less than f/6

have been designed

• These employ a dual lens system of a

plano-convex and a plano-concave lens

fitted into an eyepiece adaptor

• An example of a correction strategy for

coma is Baader Rowe Coma Correction

Coma

113


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Baader Rowe Coma Correction

Comparison of the coma in an uncorrected f/3.9 Newtonian telescope vs the affects of coma with the Baader Rowe Coma Corrector.


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Astigmatism

• Vertical plane is the ‘tangential’ plane

• Horizontal plane is the ‘sagittal’ plane

• Astigmatism results in different focal length in each plane

114


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Field Curvature

• A thin lens images a spherical surface onto a spherical surface

• Image is distorted in the image plane

• Important in lens design for close objects


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Distortion

• All points in the object plane are imaged to pointsin image plane

• Distortion arises when the magnification of off-axis image is a function of the distance to the lens center

115


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Barrel distortion

• magnification decreases with distance from the optical axis


ROINN NA FISICE


Pincushion distortion

• magnification increases with distance from the optical axis

116


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Moustache distortion

• Moustache distortion, in which initially the magnification decreases with distance from the optical axis, whilst at further distances, the magnification increases with distance.


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Chromatic Aberration

• Blue refracts more than red (greater refractive index for normal dispersion

117


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Achromatic doublet

• An achromatic doublet (achromat) is often used to compensate for the chromatic aberration.


ROINN NA FISICE


Achromatic doublet

011

)(11

)(32

12

21

11 =

−−+

−−

RRnn

RRnn RBRB

BR

BRff

ff11

=→=

We require

Thus, we need to choose parameters such that

118


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Example of the use of an achromatic doublet, using a doublet as the objective.

Achromatic doublet

The index ellipsoid

119


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• You should be able to• Describe the modes of vibration of the light

• Explain the modes of vibration in terms of the index ellipsoid

and the k-vector direction

• Explain how the optic axes of the crystal are determined.

Hence, describe the different optical classes

• Derive the expression for the index ellipsoid from the energy

density

• Explain birefringence and apply to problems in uniaxial

crystals

• Describe the use of birefringence in wave plates

• Explain double refraction in anisotropic crystals

The index ellipsoid


ROINN NA FISICE


The wave equation

We look for solutions of the form

( ) .2

2

00

2

t

EE

x

iii

i ∂∂

−=∇−⋅∇∂∂

µεεµE

( ).ti

ieEE ω−⋅= rk

120


ROINN NA FISICE


The wave equation

We obtain an eigenvalue problem with the characteristic equation

.ω

κ ii

ck=

where we have defined

,0222

222

222

=

−−−−

−−−−

−−−−

zzzyzx

zyyyyx

zxyxxx

nn

nn

nn

κκκκκκκκκκκκκκκ


ROINN NA FISICE


The wave equation

In the general case

and

( ),222222

zzyyxx ananana ++−=

( ) ( ) ( )222222222 111 xzyyzxzyx annannannb −+−+−=

.222

zyx nnnc −=

121


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• The solutions for n2 correspond to two

modes of vibration

• Different components of the

polarisation see different refractive

indices

• These modes of vibration may be

visualised by means of the index

ellipsoid

The index ellipsoid


ROINN NA FISICE


The index ellipsoid

122


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• Consider some arbitrary wavevector k

• Taking the intersection of the plane

perpendicular to k with the index

ellipsoid defines an ellipse

• The semi-axes of this ellipse give

refractive indices n’ and n’’, which

correspond to the two modes of

vibration D’ and D’’

The index ellipsoid


ROINN NA FISICE


• In general, an ellipsoid has two circular

cross-sections

• In the case of just two distinct semi-

axes, we have a spheroid and there is

just one circular cross-section

• The normals to these cross-sections

are known as the optic axes of the

crystal

Optic axes

123


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Optic axes


ROINN NA FISICE


• Biaxial crystals

• two optic axes (these are shown as N1 and N2

in the previous Fig.)

• Uniaxial crystals

• only one optic axis (taken, convention, to be

along the z-axis)

• Isotropic crystals

• No optic axis – refractive index the same in all

directions

Optic axes – optical classes

124


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Optic axes

For certain special k directions, the quadratic will have repeated roots.

.042 =− acb

In these cases, the optical field will only see one refractive index (the cross-section with the index ellipsoid is circular)

These directions are therefore the optic axes of the crystal and determined by the condition


ROINN NA FISICE


Uniaxial crystals

125


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Uniaxial crystals

In a uniaxial crystal, we have (by convention) nx = ny = no

and nz = ne. These are known as the ordinary and extraordinary refractive indices respectively. The coefficients of the quadratic equation for n2 are then

and

( )[ ],2222

zoeo annna −+−=

( ) ( )[ ]2222

0

2

0 11 zez anannb −+−=

.24

eo nnc −=


ROINN NA FISICE


Optic axes

The discriminant is

( ) ( ) .142222242

eozo nnanacb −−=−

For

the discriminant is zero when az = 1, i.e. when k is parallel

with the z-axis. Thus, this is the optic axis of the crystal

,22

eo nn ≠

126


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Uniaxial crystals

The solutions for n2 are then

and

1

2

2

222 11

−

−+= z

o

ee a

n

nnn

.202 nn =


ROINN NA FISICE


Uniaxial crystals

k

127


ROINN NA FISICE


Uniaxial crystals

we may obtain the explicit angular dependence

( ) .cos11

1

2

2

222

−

−+= θθ

o

ee

n

nnn

Putting

,cosθ=za


ROINN NA FISICE


Index ellipsoid (easier derivation)

The energy density due to the electric field is given by

This may be re-written

( ).21

21

zzyyxxE EDEDEDu ++=⋅= ED

.2

1

0

2

0

2

0

2

++=

z

z

y

y

x

xE

DDDu

εεεεεε

128


ROINN NA FISICE


Index ellipsoid (easier derivation)

Making the change of variable (associated with a scaling)

,2 0

22

εE

ii

u

Dx =

we then have

.12

2

2

2

2

2

=++zyx n

z

n

y

n

x

This is the equation of the index ellipsoid.


ROINN NA FISICE


Uniaxial crystal

For a uniaxial crystal, we have

Taking x = 0 with no loss of generality, from the figure,

.12

2

2

2

2

2

=++eoo n

z

n

y

n

x

( )( ) .sin

,cos

,0

θθθθ

nz

ny

x

=

−=

=

129


ROINN NA FISICE


Uniaxial crystal

Substituting these expressions into the index ellipsoid

Rearranging this, we obtain

( ) ( ).1

sincos2

22

2

22

=+eo n

n

n

n θθθθ

( ) ,cos11

1

2

2

222

−

−+= θθ

o

ee

n

nnn

as found earlier.


ROINN NA FISICE


Birefringence in a uniaxial crystal

The birefringence is defined as

For a wave with extraordinary and ordinary components, Ee

and Eo, propagating in a direction r, we may write

( ) .0nnn −=∆ θ

( )

−=c

rntiEEe

θωexp0

.exp0

−=c

rntiEE o

o ω

and

130


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Birefringence in a uniaxial crystal

The second of these equations may be re-written

( ) ( )[ ] .expexp0

−

−= rnnc

ic

rntiEE oo θ

ωθω

Hence, after a distance r, the ordinary wave acquires a retardation

( ) ( ).r

cr

θω∆=Γ

This provides the physical basis for retardation plates (see

Polarisation).


ROINN NA FISICE


• Further definition according to the

relative sizes of ne and no.

• A negative uniaxial crystal has ne < no

• E.g. calcite CaCO3 and ruby Al2O3.

• For ne > no, we have a positive uniaxial

crystal

• E.g, quartz SiO2

Uniaxial crystal

131


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• In negative uniaxial crystal, the

extraordinary axis is aligned with the

fast axis of the plate, since c/ne > c/no.

• For a positive uniaxial crystal, we have

the opposite case and the extraordinary

axis is aligned with the slow axis.

Uniaxial crystal – wave plates


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Birefringence - double refraction

132


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Double refraction – example: calcite

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PY3101 Optics - Physics · PY3101 Optics Revision ColáistenahOllscoileCorcaigh, Éire University College Cork, Ireland ROINN NA FISICE Department of Physics PY3101 Optics Course