optics part i

1

OPTICSPart I

SOLO HERMELIN

Updated: 16.01.10http://www.solohermelin.com

http://www.solohermelin.com/

2

Table of Content

SOLO OPTICS

Maxwell’s Equations

Boundary Conditions

Electromagnatic Wave Equations

Monochromatic Planar Wave Equations

Spherical Waveforms

Cylindrical Waveforms

Energy and Momentum

Electrical Dipole (Hertzian Dipole) Radiation

Reflections and Refractions Laws Development Using the Electromagnetic Approach

IR Radiometric Quantities

Physical Laws of Radiometry

Geometrical Optics

Foundation of Geometrical Optics – Derivation of Eikonal Equation The Light Rays and the Intensity Law of Geometrical Optics The Three Laws of Geometrical Optics

Fermat’s Principle (1657)

3

Table of Content (continue)

SOLO OPTICS

Plane-Parallel Plate

Prisms

Lens Definitions

Derivation of Gaussian Formula for a Single Spherical Surface Lens Using Fermat’s PrincipleDerivation of Gaussian Formula for a Single Spherical Surface Lens Using Snell’s Law

Derivation of Lens Makers’ Formula

First Order, Paraxial or Gaussian Optics

Ray Tracing

Matrix Formulation

4


SOLO OPTICS

Optical Diffraction

Fresnel – Huygens’ Diffraction Theory

Complementary Apertures. Babinet Principle

Rayleigh-Sommerfeld Diffraction Formula

Extensions of Fresnel-Kirchhoff Diffraction Theory

Phase Approximations – Fresnel (Near-Field) Approximation

Phase Approximations – Fraunhofer (Near-Field) Approximation

Fresnel and Fraunhofer Diffraction Approximations

Fraunhofer Diffraction and the Fourier Transform

Fraunhofer Diffraction Approximations Examples

Resolution of Optical Systems

Optical Transfer Function (OTF)Point Spread Function (PSF)

Modulation Transfer Function (MTF)

Phase Transfer Function (PTF)

Relations between Wave Aberration, Point Spread Function and Modulation Transfer Function

Other Metrics that define Image Quality – Srahl Ratio

Other Metrics that define Image Quality - Pickering Scale

Other Metrics that define Image Quality – Atmospheric Turbulence

Fresnel Diffraction Approximations Examples

OPTICSPart II

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5


SOLO OPTICS

References

Optical Aberration

Monochromatic Seidel Aberrations

Chromatic Aberration

Interference

OpticsPart II














6

Optics SOLO

Hierarchy of Optical Theories

• Quantum Light as particle (photon)

Emission, absorption, interaction of light and matter • Electromagnetic Maxwell’s Equations

Reflection/Transmission, polarization

• Scalar Wave Light as wave

Interference and Diffraction

• Geometrical Light as ray

Image-forming optical systems

λ → 0

7

Optics SOLO

Hierarchy of Optical Theories

8

MAXWELL’s EQUATIONSSOLO

SYMMETRIC MAXWELL’s EQUATIONS

Magnetic Field Intensity H

1mA

Electric Displacement D 2 msA

Electric Field Intensity E 1mV

Magnetic InductionB 2 msV

Electric Current Density eJ

2mA

Free Electric Charge Distributione 3 msA

Fictious Magnetic Current Density mJ 2mV

Fictious Free Magnetic Charge Distributionm 3 msV

1. AMPÈRE’S CIRCUIT LW (A) eJ

t

DH

2. FARADAY’S INDUCTION LAW (F)mJ

t

BE

3. GAUSS’ LAW – ELECTRIC (GE) eD

4. GAUSS’ LAW – MAGNETIC (GM) mB

Although magnetic sources are not physical they are often introduced as electricalequivalents to facilitate solutions of physical boundary-value problems.

André-Marie Ampère1775-1836

Michael Faraday1791-1867

Karl Friederich Gauss1777-1855

James Clerk Maxwell(1831-1879)

9

SOLO

The Electromagnetic Spectrum

10

SOLO

Visible Spectrum

11

SOLO

The Infrared (IR) Spectrum of Interest

Return to TOC

12

SOLO ELECTROMAGNETICS BOUNDARY CONDITIONS

Boundary Conditions

2t

1t

h

2H

1H

1

2

C

CS1P2P

3P

4P

b

21ˆ n

ek

ldtHtHhldtHldtHldHh

C

2211

0

2211ˆˆˆˆ

where are unit vectors along C in region (1) and (2), respectively, and 21ˆ,ˆ tt

2121 ˆˆˆˆ nbtt

- a unit vector normal to the boundary between region (1) and (2)21ˆ n- a unit vector on the boundary and normal to the plane of curve Cb

Using we obtainbaccba

ldbkldbHHnldnbHHldtHH e

ˆˆˆˆˆˆ21212121121

Since this must be true for any vector that lies on the boundary between regions (1) and (2) we must have:

b

ekHHn

2121ˆ

S

e

C

Sdt

DJdlH

dlbkbdlht

DJSd

t

DJ e

h

e

S

eˆˆ

0

AMPÈRE’S LAW

1

0lim:

mAht

DJk e

he

13


Boundary Conditions (continue – 1)

2t

1t

h

2E

1E

1

2

C

CS1P2P

3P

4P

b

21ˆ n

mk

ldtEtEhldtEldtEldEh

C

2211

0

2211ˆˆˆˆ

where are unit vectors along C in region (1) and (2), respectively, and 21ˆ,ˆ tt

2121 ˆˆˆˆ nbtt

- a unit vector normal to the boundary between region (1) and (2)21ˆ n- a unit vector on the boundary and normal to the plane of curve Cb

Using we obtainbaccba

ldbkldbEEnldnbEEldtEE m

ˆˆˆˆˆˆ21212121121

Since this must be true for any vector that lies on the boundary between regions (1) and (2) we must have:

b

mkEEn

2121ˆ

S

m

C

Sdt

BJdlE

dlbkbdlht

BJSd

t

BJ m

h

m

S

mˆˆ

0

FARADAY’S LAW

1

0lim:

mVht

BJk m

hm

14



h

2D

1D

1

2

21ˆ n

dS

1n

2n

e

SdnDnDhSdnDSdnDSdDh

S

2211

0

2211 ˆˆˆˆ

where are unit vectors normal to boundary pointing in region (1) and (2), respectively, and

21 ˆ,ˆ nn

2121 ˆˆˆ nnn

- a unit vector normal to the boundary between region (1) and (2)21ˆ n

SdSdnDDSdnDD e 2121121 ˆˆ

Since this must be true for any dS on the boundary between regions (1) and (2) we must have:

eDDn 2121ˆ

dSdShdv e

h

e

V

e 0

GAUSS’ LAW - ELECTRIC

1

0lim:

msAhe

he

V

e

S

dvSdD

15



h

2B

1B

1

2

21ˆ n

dS

1n

2n

m

SdnBnBhSdnBSdnBSdBh

S

2211

0

2211 ˆˆˆˆ

where are unit vectors normal to boundary pointing in region (1) and (2), respectively, and

21 ˆ,ˆ nn

2121 ˆˆˆ nnn

- a unit vector normal to the boundary between region (1) and (2)21ˆ n

SdSdnBBSdnBB m 2121121 ˆˆ

Since this must be true for any dS on the boundary between regions (1) and (2) we must have:

mBBn 2121ˆ

dSdShdv m

h

m

V

m 0

GAUSS’ LAW – MAGNETIC

1

0lim:

msVhm

hm

V

m

S

dvSdB

16


Boundary Conditions (summary)

2t

1t

h

22 , HE

11, HE

1

2

C

CS1P2P

3P

4P

b

21ˆ n

me kk

,21ˆ n

dS

11, BD

22 , BD

me ,

mkEEn

2121ˆ FARADAY’S LAW

ekHHn

2121ˆ AMPÈRE’S LAW 1

0lim:

mAht

DJk e

he

1

0lim:

mVht

BJk m

hm

eDDn 2121ˆ GAUSS’ LAW

ELECTRIC 1

0lim:

msAhe

he

mBBn 2121ˆ GAUSS’ LAW

MAGNETIC 1

0lim:

msVhm

hm

Return to TOC

17

ELECTROMAGNETICSSOLO

ELECTROMGNETIC WAVE EQUATIONS

For Homogeneous, Linear and Isotropic Medium

ED

HB

where are constant scalars, we have ,

Jt

EJ

t

DH

t

t

H

t

BE

ED

HB

Since we have also tt

t

J

t

EE

DED

EEE

t

J

t

EE

2

222

2

2

&

18


ELECTROMGNETIC WAVE EQUATIONS (continue 1)

Define

meme KK

c

KKv

00

11

where

smc /103

1036

1104

11 8

9700

is the velocity of light in free space.

The absolute index of refraction n is

me KKv

cn

0

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Electric Field Intensity is

t

J

t

E

vE

2

2

22 1

19


ELECTROMGNETIC WAVE EQUATIONS (continue 2)

In the same way

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Magnetic Field Intensity is

Jt

EJ

t

DH

t

H

t

BE

t

ED

HB

Since are constant andtt

,

J

t

HH

HHB

HHH

Jt

HH

2

222

2

2

0&

Jt

H

vH

2

2

22 1

Return to TOC

20


Monochromatic Planar Wave Equations

Let assume that can be written as: trHtrE ,,,

tjrHtrHtjrEtrE 00 exp,,exp,

where are phasor (complex) vectors.

rHjrHrHrEjrErE

ImRe,ImRe

We have tjrEjtjt

rEtrEt 00 expexp,

Hence

m

e

m

e

jt

m

e

m

e

B

D

JBjE

JDjH

BGM

DGE

Jt

BEF

Jt

DHA

)(

21


Fourier Transform

The Fourier transform of can be written as: trHtrE ,,,

dttjtrHrHdtjrHtrH

dttjtrErEdtjrEtrE

exp,,&exp,2

1,

exp,,&exp,2

1,

This is possible if:

drHdttrH

drEdttrE

22

22

,2

1,

,2

1,

JEAN FOURIER

1768-1830

22


NoteThe assumption that can be written as: trHtrE ,,,

tjrHtrHtjrEtrE 00 exp,,exp,

is equivalent to saying that has a Fourier transform; i.e.: trHtrE ,,,

dtjrHtrHdttjtrHrH

dtjrEtrEdttjtrErE

exp,2

1,&exp,,

exp,2

1,&exp,,

This is possible if:

drHdttrH

drEdttrE

22

22

,2

1,

,2

1,

00

0

exp

expexpexp,,

rEdttjrE

dttjtjrEdttjtrErE

End Note

23


m

e

m

e

ED

HBm

e

JHjE

JEjH

JHjE

JEjH

JBjE

JDjH

me JJjEkE

2

em JJjHkH

2

22 f

c

c

fk

Using the vector identity AAA

For a Homogeneous, Linear and Isotropic Media:

m

e

ED

HBm

e

H

E

B

D

e

me JJjEkE

22

m

em JJjHkH

22

and

we obtain

Monochromatic Planar Wave Equations (continue)

24


Assume no sources:

we have


0,0,0,0 meme JJ

022 EkE

022 HkH

nkk

n

k

0

00

00

0

rktjtj

rktjtj

eHerHtrH

eEerEtrE

0

0

,,

,,

022

rkj

rkjrkjrkjrkj

ek

ekkeekje

Helmholtz Wave Equations

satisfy the Helmholtz wave equations ,,, rHrE

rkj

rkj

eHrH

eErE

0

0

,

,

Assume a progressive wave of phase rkt ) a regressive wave has the phase ( rkt

For a Homogeneous, Linear and Isotropic Media

k

0E

0H

r t

k

Planes for whichconstrkt

25


To satisfy the Maxwell equations for a source free media we must have:


we haveUsing: 1ˆˆ&ˆˆ sssc

nsk

0

0

H

E

HjE

EjH

0ˆ

0ˆ

ˆ

ˆ

0

0

00

00

Hs

Es

HEs

EHs

sPlanar Wave

0E

0Hr

0

0

0

0

00

00

rkj

rkj

rkjrkj

rkjrkj

ekje

eHkj

eEkj

eHjeEkj

eEjeHkjrkjrkj

0

0

0

0

00

00

Hk

Ek

HEk

EHk

For a Homogeneous, Linear and Isotropic Media:

Return to TOC

26



Spherical Waveforms z

x

y

rcosr

,,rP

sinsinr cossinr

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Electric Field Intensity is

t

J

t

E

vE

2

2

22 1

In spherical coordinates:

cos

sinsin

cossin

rz

ry

rx

2

2

2222

22

sin

1sin

sin

11

rrr

rrr

For a spherical symmetric wave: rErE

,,

Errrr

E

rr

E

r

Er

rrE

2

2

2

22

22 121

27



SourceSourceSource

Spherical Waveforms z

x

y

rcosr

,,rP

sinsinr cossinr

The Inhomogeneous Wave (Helmholtz) Differential Equation for the Electric Field Intensity is assuming no sources

011

2

2

22

2

t

E

vEr

rr

In spherical coordinates:

cos

sinsin

cossin

rz

ry

rx

01

2

2

22

2

Ertv

Err

or:

A general solution is:

waveregressive

waveeprogressiv

tvrFtvrFEr 21

0,0,0,0 meme JJ

r

eEerEtrE

rktjtj

0,,

Assume a progressive monochromatic wave of phase rkt

) a regressive wave has the phase ( rkt

r

eErE

rkj

0, Return to TOC

28




z

x

yr

zrP ,,

sinr

cosr

In cylindrical coordinates:

zz

ry

rx

sin

cos

2

2

2

2

22 11

zrrr

rr

For a cylindrical symmetric wave: rEzrE

,,

r

Er

rrE

12


011

2

2

2

t

E

vr

Er

rr

0,0,0,0 meme JJ

29



Source


z

x

yr

zrP ,,

sinr

cosr

SourceSource

In cylindrical coordinates:

zz

ry

rx

sin

cos


011

2

2

2

t

E

vr

Er

rr

0,0,0,0 meme JJ

Assume a progressive monochromatic wave of phase rkt

) a regressive wave has the phase ( rkt

tjerEtrE ,,

0

12

2

2

Evr

E

rr

E

k

The solutions are Bessel functions which for larger approach asymptotically to: rkje

r

ErE 0,

Return to TOC

30

SOLO

Energy and Momentum

Let start from Ampère and Faraday Laws

t

BEH

Jt

DHE e

EJt

DE

t

BHHEEH e

HEHEEH

But

Therefore we obtain

EJt

DE

t

BHHE e

First way

This theorem was discovered by Poynting in 1884 and later in the same year by Heaviside.

ELECTROMAGNETICS

John Henry Poynting1852-1914

Oliver Heaviside1850-1925

31

SOLO

Energy and Momentum (continue -1)

We identify the following quantities

-Power density of the current density EJe

HEDEt

BHt

EJe

2

1

2

1

BHt

pBHw mm

2

1,

2

1

DEt

pDEw ee

2

1,

2

1

HEpR

eJ

-Magnetic energy and power densities, respectively

-Electric energy and power densities, respectively

-Radiation power density

For linear, isotropic electro-magnetic materials we can write HBED

00 ,

DEtt

DE

ED

2

10

BHtt

BH

HB

2

10

ELECTROMAGNETICS

32

SOLO

Energy and Momentum (continue – 3)

Let start from the Lorentz Force Equation (1892) on the free charge

BvEF e

Free Electric Chargee 3 msA

Velocity of the chargev 1sm

Electric Field Intensity E 1mV

Magnetic InductionB 2 msV

Hendrik Antoon Lorentz1853-1928

e

Force on the free chargeF

Ne

Second way

ELECTROMAGNETICS

33

SOLO


The power density of the Lorentz Force the charge

EJBvEvp e

Bvv

Jve

ee

0

or

HEt

BHE

t

D

Et

DHEEH

Et

DHEJp

t

BE

HEHEEH

Jt

DH

e

e

e

ELECTROMAGNETICS

34

SOLO


HEDEt

BHt

EJe

2

1

2

1

dve

E

B

eJv

,

V

FdF

Fd

Let integrate this equation over a constant volume V

VVVV

e dvSdvDEtd

ddvBH

td

ddvEJ

�

2

1

2

1

If we have sources in V then instead of we must use

E

sourceEE

Use Ohm Law (1826)

sourceee EEJ

VV td

d

t

Georg Simon Ohm1789-1854

sourcee

e

EJE

1

For linear, isotropic electro-magnetic materials HBED

00 ,

ELECTROMAGNETICS

35

SOLO


VVVR

n

V

sourcee dvSdvDE

td

ddvBH

td

ddRIdvEJ

2

1

2

12

V

FieldMagnetic dvBHtd

dP

2

1

V

FieldElectric dvDEtd

dP

2

1 SV

Radiation SdSdvSP

V

sourceeSource dvEJP

V

sourcee

R

n

V

sourcee

L S eee

V

sourcee

L S eee

V

e

dvEJdRI

dvEJdS

dldSJdSJdvEJldSdJJdvEJ

2

11

R

nJoule dRIP 2

RadiationFieldMagneticFieldElectricJouleSource PPPPP

For linear, isotropic electro-magnetic materials HBED

00 ,

R – Electric Resistance

Define the Umov-Poynting vector: 2/ mwattHES

The Umov-Poynting vector was discovered by Umov in 1873, and rediscovered by

Poynting in 1884 and later in the same year by Heaviside.

ELECTROMAGNETICS


36

ElectromagnetismSOLO

EM People


Oliver Heaviside1850-1925

Nikolay Umov1846-1915

1873 “Theory of interaction on final

distances and its exhibit to conclusion of electrostatic and

electrodynamic laws”

1884 1884

Umov-Poynting vector

HES

The Umov-Poynting vector was discovered by Umov in 1873, and rediscovered byPoynting in 1884 and later in the same year by Heaviside.

1873 - 1884

Return to TOC

37

Note:Since there are not magnetic sources the Magnetic Hertz’s Vector Potential is :

0

m

Electrical Dipole (Hertzian Dipole) RadiationSOLO

Given a dipole monochromatic of electric charges defined by the Polarization Vector Intensity

tq

tq

d

r

dqP

dr

tdqdeqaltP tj

e cosRe 00

we want to find the radiation properties.

We start with the Helmholtz Non-homogeneous Differential Equation of the Electric Hertz’s Vector Potential : te

trPtrtc

tr eee ,1

,1

,0

2

2

22

Heinrich Rudolf Hertz1857-1894

- speed of propagation of the EM wave [m/s]00

1

c

- Polarization Vector Intensity eP 2 msA

- Permitivity of space 2122 mNsA

- Electric Hertz’s Vector Potential (1888)e NsA 11

tA e

000 eV

0

Using the Electric Hertz’s Vector Potential we obtain :

The field vectors are given by ee

tcV

t

AE

2

2

200 1

tAH e

000

1

38

SOLO Electric Dipole Radiation

tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

Compute (continue-3) ee

tcE

2

2

2

1

We have

32

0

4

0

2

5

0

2

2

2

2 44

3

4

31

rc

rpr

rc

rprpr

r

rprrp

tcE ee

e

230 44 rc

rp

r

rp

tH e

r

ptre

04,

krtjkrtj epedqp 00

Let use spherical coordinates

zyxr rrr 1111 cossinsincossin

111 sincos00 rz krtjkrtj epepp

krtjeprccr

jr

rc

rprrp

rc

rprrp

r

rprrpE

rr

02

0

2

2

0

3

0

32

0

2

4

0

2

5

0

2

4

sin

4

sincos2

4

sincos2

44

3

4

3

11111

r1

1

1

pckpp

pckjpjp222

39


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

Using we can write

11 0

2

0

2

2sin1

4sin

44

krtjkrtj ep

rk

j

r

kcep

rcr

jH

krtjepr

k

r

kj

r

rccrj

rE

r

rr

0

2

23

0

2

0

2

2

0

3

0

111

11111

sinsincos21

4

1

4

sin

4

sincos2

4

sincos2

We can divide the zones around the source, as function of the relation between dipole size d and wavelength λ, in three zones:

Near, Intermediate and Far Fields

22

: c

f

ck

The Magnetic Field Intensity is transverse to the propagation direction at all ranges, but the Electric Field Intensity has components parallel and perpendicular to .r1

r1

E

However and are perpendicular to each other.H

• Near (static) zone: rd

• Intermediate (induction) zone: ~rd

• Far (radiation) zone: rd

40


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

102sin

4

tj

FieldNear epr

kcjH

tj

FieldNear epr

E r 03

0

11 sincos24

1

Near, Intermediate and Far Fields (continue – 1)

• Near (static) zone: rd

In the near zone the fields have the character of the static fields. The near fields are quasi-stationary, oscillating harmonically as , but otherwise static in character.tje

02

r

rk

41


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

102sin

4

krtj

FieldteIntermedia epr

kcjH

krtj

FieldteIntermedia epr

kj

rE r

023

0

11 sincos21

4

1

Near, Intermediate and Far Fields

• Intermediate (induction) zone: ~rd

• Far (radiation) zone: rd

10

2

sin4

krtj

FieldFar epr

kcH

10

0

2

sin4

krtj

FieldFar epr

kE

r1

FieldFarE

FieldFarH

At Far ranges are orthogonal; i.e. we have a transversal wave.

rHE 1,,

In the Radiation Zone the Field Intensities behave like a spherical wave (amplitude falls off as r-1)

12

r

rk

12010

36

11041

:9

7

0

0

1

0

00

c

FieldFar

FieldFar

cH

EZ

42


http://dept.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_apr_07_2003.shtml#tth_sEc12.1 http://www.falstad.com/mathphysics.html

Electric Field Lines of Force

43


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

The phasors of the Magnetic and Electric Field Intensities are:

10

2

sin4

1

krtjep

cr

j

rH

krtjepcrc

jrc

jrrr

E r

02

2

2

0

11 sin11

cos12

4

1

Poynting Vector of the Electric Dipole Field

The Poynting Vector of the Electric Dipole Field is

The Magnetic and Electric Field Intensities are:

1sincossin4

20

krt

ckrt

rr

pHrealH

11 sinsin

1cos

1cossincos

12

4 2

2

2

0

0 krtrc

krtcr

krtc

krtrrr

pErealE r

1

1

cossincossinsincos1

4

2

sincossinsincos1

42

0

32

2

0

22

2

2

2

0

22

2

0

krtc

krtr

krtc

krtrr

p

krtc

krtr

krtrc

krtcrr

pHES r

The Poynting Vector of the Electric Dipole Field is given by:

44


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1

Let compute the time average < > of the Poynting vector:


Using the fact that:

1

1

cossincossinsincos1

4

2

sincossinsincos1

42

0

32

2

0

22

2

2

2

0

22

2

0

krtc

krtr

krtc

krtrr

p

krtc

krtr

krtrc

krtcrr

pHES r

T

TdttS

TS

0

1lim

2

12cos

1lim

2

11lim

2

1cos

1limcos

0

0

1

00

22

T

T

T

T

T

Tdtrkt

Tdt

Tdtrkt

Trkt

2

12cos

1lim

2

11lim

2

1sin

1limsin

0

0

1

00

22

T

T

T

T

T

Tdtrkt

Tdt

Tdtrkt

Trkt

02sin1

lim2

1cossin

1limcossin

0

00

T

T

T

Tdtrkt

Tdtrktrkt

Trktrkt

rrc

pS 12

23

0

2

42

0 sin42

11 cossin

4sin

1

42

22

0

32

2

02

2

2

2

2

2

2

0

22

2

0

rcrcr

p

rccrcr

pS r

we obtain:

or: Radar Equation

Irradiance

45


tq

tq

d

r

zSS rrdqP 10

dr

sinr

cosr

zyx

r

r

rr

111

1

cossinsincossin

r1

1

1

x1

y1

z1


rrc

pS 12

23

0

2

42

0 sin42

Radar Equation

45 90 135 1800

0

5

10

15

20

25

30

0

45

90

135

180

225

270

315

z

y5.0 0.1

Polar Angle , in degrees

Rel

ativ

e P

ower

, in

db

The Total Average Radiant Power is:

0

22

23

0

2

42

0 sin2sin42

drrc

pdSSP

Arad

22

0

22

120123

0

42

0

3/4

0

3

23

0

42

0 4012

sin16

0

prc

pd

rc

pP

c

c

rad

3

4

3

2

3

2cos

3

1coscoscos1sin

0

30

2

0

3

dd

Return to TOC

46


To satisfy the Maxwell equations for a source free media we must have: Monochromatic Planar Wave Equations

we haveUsing: 1ˆˆ&ˆˆ0 kkknkkk

0

0

H

E

HjE

EjH

0ˆ

0ˆ

ˆ

ˆ

0

0

00

00

Hk

Ek

HEk

EHk

kˆPlanar Wave

0E

0Hr

0

0

0

0

00

00

rkj

rkj

rkjrkj

rkjrkj

ekje

eHkj

eEkj

eHjeEkj

eEjeHkjrkjrkj

22

22&

2

ˆ

2

ˆHwEwwcn

kwwcn

kS meme

Time Average Poynting Vector of the Planar Wave


47

SOLO REFLECTION & REFRACTION

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t

Consider an incident monochromatic planar wave

c

nk

eEkH

eEE

iiii

rktjiii

rktjii

ii

ii

1

00

110011

0

0

The monochromatic planar reflected wave from the boundary is

11

1

1

0

0

&n

cv

vc

nk

eEkH

eEE

rrr

rktjrrr

rktjrr

rr

rr

The monochromatic planar refracted wave from the boundary is

22

2

2

0

0

&n

cv

vc

nk

eEkH

eEE

ttt

rktjttt

rktjtt

tt

tt


48


The Boundary Conditions at z=0 must be satisfied at all pointson the plane at all times, impliesthat the spatial and time variations of

This implies that

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t

Phase-Matching Conditions

yxteEeEeEz

rktjt

z

rktjr

z

rktji

ttrrii ,,,,0

00

00

0

yxtrktrktrktz

ttz

rrz

ii ,,000

ttri

yxrkrkrkz

tz

rz

i ,000

must be the same


49


tri nnn sinsinsin 211

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t


zyxc

nk

zyxc

nk

ttttttt

irirrrr

ˆcossinˆsinsinˆcos


2

1

yyxc

nrk

yxc

nrk

yc

nrk

tttz

t

irrz

r

iz

i

ˆsinsincos

sinsincos

sin

2

0

1

0

1

0

yxrkrkrkz

tz

rz

i ,000

2

tr

ttri

x

y

Coplanar

Snell’s Law

zzyyxxr

zyc

nk iiii

ˆˆˆ

ˆcosˆsin1

Given:

Let find:


50


Second way of writing phase-matching equations

ri 11

22

2

1

1

2

sin

sin

v

v

n

n

t

iRefraction Law

Reflection Law


zzyyxxr

zyc

nk iiii

ˆˆˆ

ˆcosˆsin1

zyxc

nk

zyxc

nk

ttttttt

irirrrr



2

1

ynnync

kkz

ynnync

kkz

ittrti

irrrri

ˆsinsinsinˆcosˆ

ˆsinsinsinˆcosˆ

122

111

ttri

We can see that

tri

tiri kkzkkz 0ˆˆ

tri

tri

tr

nnn sinsinsin

2/

211

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t


51


ri 11

22

2

1

1

2

sin

sin

v

v

n

n

t

iRefraction Law

Reflection Law

Phase-Matching Conditions (Summary)

ttri

tri

tiri kkzkkz 0ˆˆ

tri

tri

tr

nnn sinsinsin

2/

211

iE

iH

rE

rH

ik rk

tH

tE

tk

21n

z

x yi

r

t yxrkrkrk

zt

zr

zi ,

000

yxtrktrktrktz

ttz

rrz

ii ,,000

Vector Notation

ScalarNotation


52


iE

iH

rErH

ik rk

tH

tE

tk

21n

z

x yi

r

t

i r

ttH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

2

2

1

1

2

2

1

1

0

0

ti

i

i

t

nn

n

E

Et

coscos

cos2

2

2

1

1

1

1

0

0

For most of media μ1= μ2 ,

and using refraction law: 1

2

sin

sin

n

n

t

i

ti

ti

i

r

E

Er

sin

sin21

0

0

ti

it

i

t

E

Et

sin

cossin221

0

0

Assume is normal to plan of incidence(normal polarization)

E

xEExEExEE ttrrii ˆ&ˆ&ˆ 000000


Fresnel Equations

See full development in P.P.“Reflection & Refractions”

53


iE

iH

rE

rHik rk

tH

tE

tk

21n

z

x yi

r

t

i r

t

tH

tE

tk

rH

rk

rE

iH

iE

ik

21n

Boundary

Assume is parallel to plan of incidence(parallel polarization)

E

zyEE

zyEE

zyEE

tttt

rrrr

iiii

ˆsinˆcos

ˆsinˆcos

ˆsinˆcos

0||0

0||0

0||0

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

1

1

2

2

1

1

2

2

||0

0||

ti

i

i

t

nn

n

E

Et

coscos

cos2

1

1

2

2

1

1

||0

0||

For most of media μ1= μ2 ,

and using refraction law: 1

2

sin

sin

n

n

t

i

ti

ti

i

r

E

Er

tan

tan21

||0

0|| titi

it

i

t

E

Et

cossin

cossin221

||0

0||


Fresnel Equations


54


ti

ti

i

r

nn

nn

E

Er

coscos

coscos

1

1

2

2

1

1

2

2

||0

0||

ti

i

i

t

nn

n

E

Et

coscos

cos2

1

1

2

2

1

1

||0

0||

ti

ti

i

r

nn

nn

E

Er

coscos

coscos

2

2

1

1

2

2

1

1

0

0

ti

i

i

t

nn

n

E

Et

coscos

cos2

2

2

1

1

1

1

0

0

The equations of reflection and refraction ratio are called Fresnel Equations, that first developed them in a slightly less general form in 1823, using the elastic theory of light.

Augustin Jean Fresnel

1788-1827

The use of electromagnetic approach to prove those relations, as described above, is due to H.A. Lorentz (1875)


Hendrik Antoon Lorentz1853-1928


Return to TOC

a

The Lorentz derivation in 1875 is given in V.Novacu"Electrodinamica", (romanian), p.163. I must find a cross reference.

55

IR Radiometric Quantities SOLO

RTA

DA 2cm 2cm

TARGETSOURCE

DETECTORRECEIVER

Radiation Flux Power W

Spectral Radial Power

m

W

Irradiance

2mc

W

AE

Spectral Radiant Emittance

mmc

WMM

2

Radiant Intensity

str

WI

Spectral Radiant Intensity

mstr

WII

Radiance

strmc

W

A

IL

2cos

Spectral Radiance

mstrmc

WLL

2

Radiant Emittance

2mc

W

AM

Spectral Irradiance

mmc

WEE

2

T

TdttS

TS

0

1lim

Irradiance is the time-average of the Poynting vector

Return to TOC

56

Physical Laws of Radiometry SOLO

Plank’s Law

1/exp

1

2

5

1

Tc

cM BB

Plank 1900

Plank’s Law applies to blackbodies; i.e. perfect radiators.

The spectral radial emittance of a blackbody is given by:

KT

KWk

Wh

kmc

Kmkhcc

mcmWchc

in eTemperaturAbsolute-

constantBoltzmannsec/103806.1

constantPlanksec106260.6

lightofspeedsec/458.299792

10439.1/

107418.32

23

234

4

2

4242

1

MAXPLANCK

(1858 - 1947)Plank’s Law

57

Physical Laws of Radiometry SOLO

Plank’s Law

1/exp

1

2

5

1

Tc

cM BB

Plank 1900

Plank’s Law applies to blackbodies; i.e. perfect radiators.

The spectral radial emittance of a blackbody is given by:

MAXPLANCK

(1858 - 1947)Plank’s Law

58

Physical Laws of Radiometry (Continue -1) SOLO

Wien’s Displacement Law

0

d

Md BB

Wien 1893

from which:

The wavelength for which the spectral emittance of a blackbody reaches the maximumis given by:

m

KmTm

2898 Wien’s Displacement Law

Stefan-Boltzmann Law

Stefan – 1879 Empirical - fourth power law

Boltzmann – 1884 Theoretical - fourth power law

For a blackbody:

42

12

32

45

2

4

0 2

5

1

0

10670.515

2:

1/exp

1

Kcm

W

hc

k

cm

WTd

Tc

cdMM BBBB

LUDWIG BOLTZMANN(1844 - 1906)

WILHELM WIEN

(1864 - 1928)

Stefan-Boltzmann Law

JOSEFSTEFAN

(1835 – 1893)

59

Physical Laws of Radiometry (Continue -1a) SOLO

Black Body Emittance M [W/m2]

M (300ºK) 5.86 121

M (301ºK) - M (300ºK) 0.22 2

M (600ºK) 1,719 1,555

M (601ºK) - M (600ºK) 17 7

3 – 5 µm 8 - 12 µm

60


Emittance of Real Bodies (Gray Bodies)

For real (gray) bodies:

BBMM

- Directional spectral emissivity is a measure of how closely the flux radiated from a given temperature radiator approaches that from a blackbody at the same temperature

,

BBM

M

61


Kirchhoff’s Law

rM

iE aE

tM

Gustav Robert Kirchhoff1824-1887

- Incident IrradianceiE

- Absorbed IrradianceaE

- Reflected Radiant ExcitancerM

- Transmitted Radiant ExcitancetM

Law of Conservation of Energy: trai MMEE

i

t

i

r

i

a

E

M

E

M

E

E11

i

a

E

E: - fraction of absorbed energy (absorptivity)

i

r

E

M: - fraction of reflected energy (reflectivity)

i

t

E

M: - fraction of transmitted energy (transmissivity)

Opaque body (no transmission): 01 Blackbody (no reflection or transmission): 0&01

Sharp boundary (no absorption): 01

62


Kirchhoff’s Law (Continue – 1)

Gustav Robert Kirchhoff1824-1887

Kirchhoff’s Law (1860) states that, for any temperature and any wavelength, the emissivity of an opaque body in an isothermal enclosure is equal to it’s absorptivity.

This is because if the body will radiate to the surrounding less than it absorbs it’stemperature will rise above the surrounding and will be a transfer of energy from acold surrounding to a hot body contradicting the second law of thermodynamics.

TT

222 ,, T2A

111 ,, T1A

63


Lambert’s Law

Johann Heinrich Lambert

1728 - 1777

http://www-groups.dcs.st-andrews.ac.uk/~history/Biographies/Lambert.html

A Lambertian Surface is defined as a surface from which the radiance

L [W/(cm2 str)] is independent of the direction of radiation.

2

sin

r

drdrd

A

cosAAn

z

x

y

0

2

cos, L

AL

coscos, 00 IALI

Lambert’s Law

0

2

0

2/

0

00 sincoscos LddLdLA

M

The Radiant Intensity from a Lambertian Surface is

The Radiant Emittance (Exitance) from a Lambertian Surface is

64


Transfer of Radiant Energy

We have two bodies 1 and 2.

The radiant power (radiance) transmitted from 1 to 2 is:

212

2222

211

122

1

cos&

cos R

Add

strcm

W

AL

1A

1dA

1 12R

Radiating(Source)Surface

2A

2dA

ReceivingSurface

2

2d

2

12

2211112

coscos

R

AdAdLd

The total radiant power (radiance) received at surface A2 from A1 is:

2 1

21212

21112

coscos

A A

AdAdR

L

65


Transfer of Radiant Energy (Continue – 1)

Define the projected areas:

and the solid angles:

222111 cos&cos AdAdAdAd nn 1A

1dA

1 12R


2A

2dA

ReceivingSurface

2

2d

212

2222

12

111

cos&

cos

R

Add

R

Add

1A

1dA

1 12R


2A

2dA

ReceivingSurface

2

1d

then:

212

2112

12

2211112

coscos

R

AdAdL

R

AdAdLd nn

12121112 dAdLdAdLd nn

The Power is the product of the Radiance, the projected Area, and the Solid Angleusing the other area.

66



Optics

R f

ATARGET ADETECTOR

AOPTICS

TO,OD,OT , DO,

For an Optical System define:

ATARGET – Target Area

ADETECTOR – Detector Area

AOPTICS – Optics Area

R – Range from Target to Optics

f – Focal Length (from Optics to Detector)

ΩO,T – solid angle of Optics as seen from the Target2, R

AOPTICSTO

ωT,O – solid angle of Target as seen from the Optics2, R

ATARGETOT

ΩD,O – solid angle of Detector as seen from the Optics2, f

ADETECTOROD

ωO,D – solid angle of Optics as seen from the Detector2, f

ADETECTORDO

67



Optics (continue – 1)

R f

ATARGET ADETECTOR

AOPTICS

TO,OD,OT , DO,

For the Figure we can see that:

ODOT ,,

22 f

A

R

A DETECTORTARGET

Also we found that:

DODETECTOROTOPTICS

ODOPTICSTOTARGET

ALAL

ALAL

OTOD

,,

,,

,,

68




R f

ATARGET ADETECTOR

AOPTICS

TO ,OD ,OT , DO ,

R f

ATARGET ADETECTOR

AOPTICS

TO ,OD ,OT , DO ,

R f

ATARGET ADETECTOR

AOPTICS

TO ,OD ,OT , DO ,

R f

ATARGET ADETECTOR

AOPTICS

TO ,OD ,OT , DO ,

TOTARGETAL ,

ODOPTICSAL ,

OTOPTICSAL ,

DODETECTORAL ,

69




2

,

R

AAL

AL

TARGETDETECTOR

DTDETECTOROpticsNo

2

,

f

AAL

AL

OPTICSDETECTOR

DODETECTOROpticsWith

R f

ATARGET ADETECTOR

AOPTICS

TO,OD,OT , DO,

R

ATARGET ADETECTOR

TD,DT ,

• IR Detector without Optics

• IR Detector with Optics

2

#

/

2

40

44 0#

fAL

f

DAL DETECTOR

Dff

DETECTOR

The Optics increases the energy collected by the Detector

since DTDO ,, 22

#2 4 R

A

ff

A TARGETOPTICS

OpticsNoOpticsWith

70


Targets

The parts of the aircraft that are especially hot are:

• The exhaust nozzle of the jet engine

• The hot exhaust gas area, or the plume

• The areas in which aerodynamic heating is the highest

71


Targets

72


Targets

73


Targets

74

Physical Laws of Radiometry (Continue -16)

a

R.E. Ball

75


Targets (continue – 1)

• The exhaust nozzle of the jet engine

The exhaust nozzle can be regarded as a gray body with ε = 0.9.

Example: Turbojet Engine 4-P&W JT4A-923660 cmANOZZLE

rafterburnewithCT 538 24124 207.22735381067.59.0 cmWTM

We are interested only in the band 3 μm ≤ λ ≤ 5 μm.

By numerically integration or using infrared radiation calculators we obtain: 397.0

811

4

5

3

KT

BB

T

dM

Hence:

2876.0207.2397.053 cmWmmM In a tail-on situation the radiant intensity is:

110203660876.0

53

strWA

MmmI NOZZLE

Lambertian

76



• The plume

The plume is characterized by the radiant emittance of the hot gases that are expanding into the atmosphere after passing through the exhaust nozzle.

The products of combustion are H2O, CO2, some times CO (incomplete combustion),OH, HF, HCl. The infrared emission is produced by changes in the energy contained in the molecularvibrations and rotations, only at certain frequencies..

a

Robert E. Ball "The Fundamentals of Air Combat Survivability Analysis and Design", 2nd Ed.

77



• The plume (continue – 1)

78



• The plume (continue – 2)Breathing engines have exhaust plume temperatures of

K600450 Cruise flight K800600 Maximum Un-augmented Thrust K15001000 Augmented (After burner) Thrust

Rockets have exhaust plume temperatures of

K75002500 Liquid propellant

K35001700 Solid propellant

ExampleAssume:

mm 55.433.45.0

KCCTPLUME

643273370

then: 2222

55.4

33.4

1075.1105.35.0

cmWcmWdMM

For a plume surface of APLUME = 10000 cm2 = 1 m2 the Radiant Intensity is:

142

8.27101075.1

55.433.4

strWA

MmmI PLUME

Lambertian

79



• Aerodynamic Heating The Target body is heated by the compression and friction of the air against it’s surface and by friction. Assuming a negligible friction effect and an adiabatic compression the Target skin temperature is given by:

2

0 2

11, MachrMachHTT

- air temperature at altitude HTARGET and mach number Mach MachHT ,0

- recovery factor r

vp CC / - specific heat ratio = 1.4 for air

ExampleMach = 2.0, HTARGET = 5000 m

27.0,250.2,50000 KMachmHT

then KT 41422

14.182.01250 2

235

3

1066.1414

cmWdKTMM

assume 215mATARGET

143

3.7910151066.1

53

strWA

MmmI TARGET

Lambertian

80


81


82



• Aerodynamic Heating (continue – 1)

Emissivity Reflectance Absorptance Material

.04 .81 .19 Polished Aluminium

.04 .63 .37 Unpolished Aluminium

.18 .43 .57 Titanium

.05 .60 .40 Polished Stainless Steel

.88 .79 .21 White Paint

.92 .05 .95 Black Paint

.27 .71 .29 Aluminum Paint

83


Sun, Background and Atmosphere

84


Sun, Background and Atmosphere (continue – 1)

The spectrum distribution of the sun radiation is like a black body with a temperature of T = 5900 °K

From Wien’s Law the maximum of Mλ is at

mTm 49.0

5900

28982898

This is almost at the middle of the visible spectrum mm 75.040.0

Loss by Scattering

85



Atmosphere

Atmosphere affects electromagnetic radiation by

3.2

11

RkmRR

• Absorption • Scattering • Emission • Turbulence

Atmospheric Windows:

Window # 2: 1.5 μm ≤ λ < 1.8 μm

Window # 4 (MWIR): 3 μm ≤ λ < 5 μm

Window # 5 (LWIR): 8 μm ≤ λ < 14 μm

For fast computations we may use the transmittance equation:

R in kilometers.

Window # 1: 0.2 μm ≤ λ < 1.4 μmincludes VIS: 0.4 μm ≤ λ < 0.7 μm

Window # 3 (SWIR): 2.0 μm ≤ λ < 2.5 μm

86



Atmosphere Absorption over Electromagnetic Spectrum

87



Rain Attenuation over Electromagnetic Spectrum

FREQUENCY GHz

ON

E-W

AY

AT

TE

NU

AT

ION

-Db

/KIL

OM

ET

ER

WAVELENGTH

88



Add scanned Figure from McKenzie

Atmosphere (continue – 1)

89

GEOMETRICAL OPTICSSOLO

http://en.wikipedia.org/wiki/Optics

From “Cyclopaedia” or “An Universal Dictionary of Art and Science”Published by Ephraim ChambersIn London in 1728

Return to TOC

90

SOLO

DERIVATION OF EIKONAL EQUATION

Foundation of Geometrical Optics

Derivation from Maxwell Equations

Consider a general time-harmonic field:

tjrHtjrHtjrHaltrH

tjrEtjrEtjrEaltrE

exp,exp,2

1exp,Re,

exp,exp,2

1exp,Re,

*

*

in a non-conducting, far-away from the sources 0,0 eeJ

No assumption of isotropy of the medium are made; i.e.: rr ,

Far from sources, in the High Frequencies we can write using the phasor notation:

00000 &,&, 00

kerHrHerErE rSjkrSjk

Note

The minus sign was chosen to get a progressive wave:

End Note

SktjSktj erHaltrHerEaltrE 0000 Re,&Re,

James Clerk Maxwell(1831-1879)

See full development in P.P.“Foundation of Geometrical Optics”

91

SOLO

From those equations we have


Sjktj

SjkSjktjSjktjtj

eeESjkE

EeeEeeEeerE0

000

000

000,

Sjk

SjktjSjktjtj

eHjk

eHejeHejerHt

0

00

0

00

0

0

00

000

1

1,

from which

0

00

0000 HjkESjkEF

and

01 0

00

0

00

0

k

Ejk

HES

DERIVATION OF EIKONAL EQUATION (continue – 2)

Derivation from Maxwell Equations (continue – 2)

92

SOLO

From Maxwell equations we also have


from which

and



Sjktj

SjkSjktjSjktjtj

eeHSjkH

HeeHeeHeerH0

000

000

000,

Sjk

SjktjSjktjtj

eEjk

eEejeEejerEt

0

00

0

00

0

0

00

000

1

1,

0

00

0000 EjkHSjkHA

01 0

00

0

00

0

k

Hjk

EHS

93

SOLO




We have Faradey (F), Ampére (A), Gauss Electric (GE), Gauss Magnetic (GM) equations:

0

0

HGM

EGE

EjHA

HjEF

0&0

2

0 00

ee

e

e

J

ck

jt

HB

ED

BGM

DGE

Jt

DHA

t

BEF

André-Marie Ampère1775-1836

Michael Faraday1791-1867


94

SOLO

From Maxwell equations we also have


from which

and



0

,0

000

0000

000

Sjktj

SjkSjktjSjktjtj

eeESjkEE

EeeEeeEeerE

00000 ESjkEEGE

01 0

000

0

k

EEjk

ES

We also have

from which

and

0

,0

000

0000

000

Sjktj

SjkSjktjSjktjtj

eeHSjkHH

HeeHeeHeerH

00000 HSjkHHGM

01 0

000

0

k

HHjk

HS

95

SOLO

To summarize, from k0 → ∞ we have




00

00

0 HESF

00

00

0 EHSA

00 ESGE

00 HSGM

We will use only the first two equations, because the last two may be obtained from the previous two by multiplying them (scalar product) by . S

96

SOLO Foundation of Geometrical Optics



00

00

0 HESF

00

00

0 EHSA

From the second equation we obtain

000

0 HSE

And by substituting this in the first equation

00 000

00

00

000

HHSSHHSS

But

2

00

02

0

0

00

n

HSHSSSHSHSS

97




Finally we obtain

0022 HnS

or

zyxnz

S

y

S

x

SornS ,,0 2

222

22

S is called the eikonal (from Greek έίκων = eikon → image) and the equation is called Eikonal Equation.

Return to TOC

98


THE LIGHT RAYS AND THE INTENSITY LAW OF GEOMETRICAL OPTICS From Max Born & Emil Wolf, “Principles of Optics”, 6th Ed., Ch. 3

00000 &,&, 00

kerHrHerErE rSjkrSjk

We found the following relations

00

00

0 HESF

00

00

0 EHSA

00 ESGE

00 HSGM

We can see that the vectors are perpendicular in the same way as the vectors for the planar waves (where is the Poynting vector).

SHE ,, 00

SHE

,, 00 00 HES

S

0E

0H

99


THE LIGHT RAYS AND THE INTENSITY LAW OF GEOMETRICAL OPTICS (continue – 1)

T

T

T

TT

e

dttjrErErEtjrET

dttjrEtjrEtjrEtjrET

dttjrEalT

dttrEtrET

dttrDtrET

w

0

2**2

0

**

0

2

00

2exp,,,22exp,4

1

exp,exp,exp,exp,4

1

exp,Re1

,,1

,,1

But

0

2

2exp2exp

2

12exp

1

02

2exp2exp

2

12exp

1

00

00

T

TT

T

TT

Tj

Tjtj

Tjdttj

T

Tj

Tjtj

Tjdttj

T

Therefore

rErEerEerEdtT

rErEw rSjkrSjkT

e

*00

*00

0

*

22

1,,

200

Let compute the time averages of the electric and magnetic energy densities

100



In the same way

rErEerEerEdtT

rErEw rSjkrSjkT

e

*00

*00

0

*

22

1,,

200

rHrHdttrHtrHT

dttrBtrHT

wTT

m*

00

00 2,,

1,,

1

Using the relations

000

0 HSEA

000

0 ESHF

since and are real values , where * is the complex conjugate, we obtain

S )**,( SS

e

m

e

wrHSrErHSrErHSrE

rESrHrESrHrHrHw

rHSrErHSrErErEw

*

00

*

0*

00**

0

*

00

*

000

0*

00

*

00

*

000

0*

00

2

1

2

1

2

1

2

1

22

2

1

22

S

0E

0H

101



Therefore *002

1rHSrEww me

Within the accuracy of Geometrical Optics, the time-averaged electric and magnetic energy densities are equal.

*0000*

00 22rHSrErHrHrErEwww me

The total energy will be:

The Poynting vector is defined as: trHtrEtrS ,,:,

T

tjtjtjtj

Ttjtj

T

dterHerHerEerET

dterHerEalT

dttrHtrET

trHtrES

0

**

00

,,2

1,,

2

11

,,Re1

,,1

,,

,,,,4

1

,,,,,,,,4

11

**

0

2****2

rHrErHrE

dterHrErHrErHrEerHrET

Ttjtj

rHrErHrE

erHerEerHerE rSjkrSjkrSjkrSjk

0*

0*

00

)(0

)(*0

)(*0

)(0

4

14

10000

The time average of the Poynting vector is:


102



Using the relations

000

0 HSEA 0

000 ESHF

rHrHSrESrErHrErHrES 0

*

0

*

0000

0*

0*

00 222

1

4

1

we obtain

*

00

0

0*

0

0

0*

0*

0000

22222

1HHSHSHESEEES

*0000*

00 22rHSrErHrHrErEwww me

we obtain

Using

wSn

cwwSS me

200

00 22

1

00

2

00

&1

nc

103



Using 22 nS Eikonal Equation

we obtain nS

Define snSn

S

S

Ss ˆ:ˆ

We have swvwSn

cS

n

cv

ˆ2

1

2 2

s

constS constdSS

s

r0s

0r

A Bundle of Light Rays

104



swvwSn

cS

n

cv

ˆ2

1

2 2

s

constS constdSS

s

r0s

0r

From this equation we can see that average Poynting vector is the direction ofthe normal to the geometrical wave-front , and its magnitude is proportional to the product of light velocity v and the average energy density, therefore we say that defines the direction of the light ray.

S

ss

Suppose that the vector describes the light path, then the unit vector is given by

r

s

sd

rd

rd

rds ray

ray

ray

ˆ

where is the differential of an arc length along the ray pathrayrdsd

105



Let substitute in and differentiate it with respect to s.sd

rd

rd

rds ray

ray

ray

ˆ rayrdsd

Ssd

d

sd

rdn

sd

d

ray

Ssd

rd ray

sd

rdf

sd

zd

zd

fd

sd

yd

yd

fd

sd

xd

xd

fd

sd

zyxfd

,,

SSn

1 S

sd

rdn ray

ABBAABBABA

AB

AAAAAA

2

1

SA

SSSSSSSS 0

2

1 SSn

2

1

2nSS 2

2

1n

n

n

106



Therefore we obtained nSsd

d

and

nsd

rdn

sd

d

ray

We obtained a ordinary differential equation of 2nd order that enables to find the trajectory of an optical ray , giving the relative index and the initial position and direction of the desired ray.

srray

zyxn ,, 00 rrray

0s

s

constS constdSS

s

r0s

0r

We can transform the 2nd order differential equation in two 1st order differential equations by the following procedure. Define

Ssnsd

rdnp ˆ: ray

We obtain 0ˆ0 snpnp

sd

d

0ˆ0 snpnpsd

d

Return to TOC

107

SOLO

The Three Laws of Geometrical Optics

1. Law of Rectilinear Propagation In an uniform homogeneous medium the propagation of an optical disturbance is instraight lines.

2. Law of Reflection

An optical disturbance reflected by a surface has the property that the incident ray, the surface normal, and the reflected ray all lie in a plane,and the angle between the incident ray and thesurface normal is equal to the angle between thereflected ray and the surface normal:

2v

1v

Refracted Ray

21ˆ n

2n

1n

i

t

Reflected Ray

21ˆ n

2n

1n

i r

3. Law of Refraction

An optical disturbance moving from a medium ofrefractive index n1 into a medium of refractive indexn2 will have its incident ray, the surface normal betweenthe media , and the reflected ray in a plane,and the relationship between angle between the incident ray and the surface normal θi and the angle between thereflected ray and the surface normal θt given by Snell’s Law: ti nn sinsin 21

ri

“The branch of optics that addresses the limiting case λ0 → 0, is known as Geometrical Optics, since in this approximation the optical laws may be formulated in the language of geometry.”

Max Born & Emil Wolf, “Principles of Optics”, 6th Ed., Ch. 3


Return to TOC

108


Fermat’s Principle (1657)

1Q

1P

2P

2Q1Q

2Q

1S

SdSS 12

2PS

1PS

2'Q

rd

s

s

The Principle of Fermat (principle of the shortest optical path) asserts that the optical length

of an actual ray between any two points is shorter than the optical ray of any other curve that joints these two points and which is in a certai neighborhood of it. An other formulation of the Fermat’s Principle requires only Stationarity (instead of minimal length).

2

1

P

P

dsn

An other form of the Fermat’s Principle is:

Princple of Least Time The path following by a ray in going from one point in space to another is the path that makes the time of transit of the associated wave stationary (usually a minimum).

The idea that the light travels in the shortest path was first put forward by Hero of Alexandria in his work “Catoptrics”, cc 100B.C.-150 A.C. Hero showed by a geometrical method that the actual path taken by a ray of light reflected from plane mirror is shorter than any other reflected path that might be drawn between the source and point of observation.

a

Hero proof is described in M.V.Klein, T.E.Furtak, "Optics", pp.3-5

109

SOLO

1. The optical path is reflected at the boundary between two regions

0

2121

rd

sd

rdn

sd

rdn rayray

In this case we have and21 nn 0ˆˆ

2121

rdssrd

sd

rd

sd

rd rayray

We can write the previous equation as:

i.e. is normal to , i.e. to the boundary where the reflection occurs.

21 ˆˆ ss rd

0ˆˆˆ 2121 ssn11 sn

21 sn

1121 ˆˆˆ snsn

rd 0ˆˆ 121 rdssn

Reflected Ray

21ˆ n

1n

i r

REFLECTION & REFRACTION

Reflection Laws Development Using Fermat Principle

ri Incident ray and Reflected ray are in the same plane normal to the boundary.

This is equivalent with:

&

110

SOLO

2. The optical path passes between two regions with different refractive indexes n1 to n2. (continue – 1)

0

2121

rd

sd

rdn

sd

rdn rayray

where is on the boundary between the two regions andrd

sd

rds

sd

rds rayray 2

:ˆ,1

:ˆ 21

rd

22 sn

11 sn

1122 ˆˆˆ snsn

0ˆˆˆ 1122 rdsnsn

Refracted Ray

21ˆ n

2n

1n i

t

Therefore is normal to .

2211 ˆˆ snsn rd

Since can be in any direction on the boundary between the two regions is parallel to the unit vector normal to the boundary surface, and we have

rd

2211 ˆˆ snsn 21ˆ n

0ˆˆˆ 221121 snsnn

We recovered the Snell’s Law from Geometrical Optics

REFLECTION & REFRACTION

Refraction Laws Development Using Fermat Principle

ti nn sinsin 21 Incident ray and Refracted ray are in the same plane normal to the boundary.

&Return to TOC

111

SOLO

Plane-Parallel Plate

i

r

ri r

t ld

i

A

C

B

E

2n1n

A single ray traverses a glass plate with parallel surfaces and emerges parallel to itsoriginal direction but with a lateral displacement d.

Optics

irriri lld cossincossinsin

r

tl

cos

r

iritd

cos

cossinsin

ir nn sinsin 0Snell’s Law

n

ntd

r

ii

0

cos

cos1sin

For small anglesi

n

ntd i

01

112

SOLO

Plane-Parallel Plate (continue – 1)

t

r

ii n

ntd

cos

cos1sin

1

2

1n

2n

i

r

r

i

i n

nt

dl

cos

cos1

sin 1

2

l

Two rays traverse a glass plate with parallel surfaces and emerge parallel to theiroriginal direction but with a lateral displacement l.

Optics

irriri lld cossincossinsin

r

tl

cos

r

iritd

cos

cossinsin

ir nn sinsin 0Snell’s Law

n

ntd

r

ii

0

cos

cos1sin

r

i

i n

nt

dl

cos

cos1

sin0 For small anglesi

n

ntl 01

Return to TOC

113

SOLO

Prisms

2i1i1t

11 ti

2t 22 it

Type of prisms:

A prism is an optical device that refract, reflect or disperse light into its spectral components. They are also used to polarize light by prisms from birefringent media.

Optics - Prisms

2. Reflective

1. Dispersive

3. Polarizing

114

Optics SOLO

Dispersive Prisms

2i1i1t

11 ti

2t 22 it

2211 itti

21 it

21 ti

202 sinsin ti nn Snell’s Law

10 n

1

1

2

1

2 sinsinsinsin tit nn

11

21

11

1

2 sincossin1sinsinsincoscossinsin ttttt nn

Snell’s Law 110 sinsin ti nn 11 sin

1sin it n

1

2/1

1

221

2 sincossinsinsin iit n

1

2/1

1

221

1 sincossinsinsin iii n

The ray deviation angle is

10 n

115

Optics SOLO

Prisms

2i1i1t

11 ti

2t 22 it

1

2/1

1

221


116

Optics SOLO

Prisms

2i1i1t

11 ti

2t 22 it

1

2/1

1

221


21 ti

Let find the angle θi1 for which the deviation angle δ is minimal; i.e. δm.

This happens when

01

0

11

2

1

ii

t

i d

d

d

d

d

d

Taking the differentials of Snell’s Law equations

22 sinsin tin

11 sinsin ti n

2222 coscos iitt dnd

1111 coscos ttii dnd

Dividing the equations1

2

1

2

1

1

2

1

2

1

cos

cos

cos

cos

i

t

i

t

t

i

t

i

d

d

d

d

2

22

1

22

2

2

2

2

1

2

2

2

1

2

2

2

1

2

sin

sin

/sin1

/sin1

sin1

sin1

sin1

sin1

t

i

t

i

i

t

t

i

n

n

n

n

11

2 i

t

d

d

21 it

12

1 i

t

d

d

2

2

1

2

2

2

1

2

cos

cos

cos

cos

i

t

t

i

21 ti 1n

117

Optics SOLO

Prisms

2i1i1t

11 ti

2t 22 it

1

2/1

1

221


We found that if the angle θi1 = θt2 the deviation angle δ is minimal; i.e. δm.

Using the Snell’s Law equations

22 sinsin tin

11 sinsin ti n 21 ti

21 it

This means that the ray for which the deviation angle δ is minimum passes through the prism parallel to it’s base.

2i1i

1t

m

11 ti

2t 22 it

21 ti 21 it

Find the angle θi1 for which the deviation angle δ is minimal; i.e. δm (continue – 1).

118

Optics SOLO

Prisms

1

2/1

1

221


Using the Snell’s Law 11 sinsin ti n

21 it

This equation is used for determining the refractive index of transparent substances.

2i1i

1t

m

11 ti

2t 22 it

21 ti 21 it

21 it

21 ti

21 ti

m 2/1 t

12 im 2/1 mi

2/sin

2/sin

mn

Find the angle θi1 for which the deviation angle δ is minimal; i.e. δm (continue – 2).

119

Optics SOLO

Prisms

The refractive index of transparent substances varies with the wavelength λ.

1

2/1

1

221


2i1i1t

11 ti

2t 22 it

120

Optics SOLO

http://physics.nad.ru/Physics/English/index.htm

Prisms

υ [THz] λ0 (nm) Color

384 – 482482 – 503503 – 520520 – 610610 – 659659 - 769

780 - 622622 - 597597 - 577577 - 492492 - 455455 - 390

RedOrangeYellowGreenBlueViolet

1 nm = 10-9m, 1 THz = 1012 Hz

1

2/1

1

221


In 1672 Newton wrote “A New Theory about Light and Colors” in which he said thatthe white light consisted of a mixture of various colors and the diffraction was color dependent.

Isaac Newton1542 - 1727

121

SOLO

Dispersing PrismsPellin-Broca Prism

Abbe Prism

Ernst KarlAbbe

1840-1905

At Pellin-Broca Prism an incident ray of wavelength λ passes the prism at a dispersing angle of 90°. Because the dispersing angleis a function of wavelengththe ray at other wavelengthsexit at different angles.By rotating the prism aroundan axis normal to the pagedifferent rays will exit at

the 90°.

At Abbe Prism the dispersing

angle is 60°.

Optics - Prisms

122

SOLO

Dispersing Prisms (continue – 1)Amici Prism

Optics - Prisms

123

SOLO

Reflecting Prisms

2i

1i

1t

2t

E

B D

G

A

F C

BED 180

360 ABEBEDADE

190 iABE

290 tADE

3609090 12 it BED

12180 itBED

21180 tiBED

The bottom of the prism is a reflecting mirror

Since the ray BC is reflected to CD

DCGBCF Also

CGDBFC CDGFBC

FBCt 901CDGi 90221 it

202 sinsin ti nn Snell’s Law

Snell’s Law 110 sinsin ti nn 21 ti 12 i

CDGFBC ~

Optics - Prisms

124

SOLO

Reflecting Prisms

Porro Prism Porro-Abbe Prism

Schmidt-Pechan Prism

Penta Prism

Optics - Prisms

Roof Penta Prism

125

SOLO

Reflecting Prisms

Abbe-Koenig Prism

Dove Prism

Amici-roof Prism

Optics - Prisms

126

SOLO

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Polarization can be achieved with crystalline materials which have a different index ofrefraction in different planes. Such materials are said to be birefringent or doubly refracting.

Nicol Prism The Nicol Prism is made up from two prisms of calcite cemented with Canada balsam. The ordinary ray can be made to totally reflect off the prism boundary, leving only the extraordinary ray..

Polarizing Prisms

Optics - Prisms

127

SOLO

Polarizing Prisms

A Glan-Foucault prism deflects polarized lighttransmitting the s-polarized component. The optical axis of the prism material isperpendicular to the plane of the diagram.

A Glan-Taylor prism reflects polarized lightat an internal air-gap, transmitting onlythe p-polarized component. The optical axes are vertical in the plane of the diagram.

A Glan-Thompson prism deflects the p-polarized ordinary ray whilst transmitting the s-polarized extraordinary ray. The two halves of the prism are joined with Optical cement, and the crystal axis areperpendicular to the plane of the diagram.

Optics - Prisms

Return to TOC

128

Optics SOLO

Lens Definitions

Optical Axis: the common axis of symmetry of an optical system; a line that connects all centers of curvature of the optical surfaces.

FFL

First FocalPoint

Second FocalPoint

Principal Planes

Second Principal Point

First Principal Point

Light Rays from Left

EFLBFL

Optical System

Optical Axis

Lateral Magnification: the ratio between the size of an image measured perpendicular to the optical axis and the size of the conjugate object.

Longitudinal Magnification: the ratio between the lengthof an image measured along the optical axis and the length of the conjugate object.

First (Front) Focal Point: the point on the optical axis on the left of the optical system (FFP) to which parallel rays on it’s right converge.

Second (Back) Focal Point: the point on the optical axis on the right of the optical system (BFP) to which parallel rays on it’s left converge.

129

Optics SOLO

Definitions (continue – 1)

Aperture Stop (AS): the physical diameter which limits the size of the cone of radiation which the optical system will accept from an axial point on the object.

Field Stop (FS): the physical diameter which limits the angular field of view of an optical system. The Field Stop limit the size of the object that can beseen by the optical system in order to control the quality of the image.

Entrance Pupil: the image of the Aperture Stop as seen from the object through theelements preceding the Aperture Stop.

Exit Pupil: the image of the Aperture Stop as seen from an axial point on theimage plane.

A.S. F.S.

I

Aperture and Field Stops

Entrancepupil

Exitpupil

A.S.

I

xpEnpE

ChiefRay

Entrance and Exit pupils

EntrancepupilExit

pupil

A.S. I

xpE

npE

ChiefRay

130

Optics SOLO


Aperture Stop (AS): the physical diameter which limits the size of the cone of radiation which the optical system will accept from an axial point on the object.

Field Stop (FS): the physical diameter which limits the angular field of view of an optical system. The Field Stop limit the size of the object that can beseen by the optical system in order to control the quality of the image.

Entrance Pupil: the image of the Aperture Stop as seen from the object through theelements preceding the Aperture Stop.

Exit Pupil: the image of the Aperture Stop as seen from an axial point on theimage plane.

Entrancepupil

Exitpupil

A.S.

I

ChiefRay

MarginalRay

Exp Enp

131

Optics SOLO


Principal Planes: the two planes defined by the intersection of the parallel incident raysentering an optical system with the rays converging to the focal pointsafter passing through the optical system.

FFL

First FocalPoint

Second FocalPoint

Principal Planes

Second Principal Point

First Principal Point

Light Rays from Left

EFLBFL

Optical System

Optical Axis

Principal Points: the intersection of the principal planes with the optical axes.

Nodal Points: two axial points of an optical system, so located that an oblique ray directed toward the first appears to emerge from the second, parallel to the original direction. For systems in air, the Nodal Points coincide with the Principal Points.

Cardinal Points: the Focal Points, Principal Points and the Nodal Points.

132

Optics SOLO


Relative Aperture (f# ): the ratio between the effective focal length (EFL) f to Entrance Pupil diameter D.

Numerical Aperture (NA): sine of the half cone angle u of the image forming ray bundlesmultiplied by the final index n of the optical system.

If the object is at infinity and assuming n = 1 (air):

Dff /:#

unNA sin:

#

1

2

1

2

1sin

ff

DuNA

EFL

Du

Last Principal Plane of theOptical System (Spherical)

133

Optics SOLO

Perfect Imaging System

• All rays originating at one object point reconverge to one image point after passing through the optical system.

• All of the objects points lying on one plane normal to the optical axis are imaging onto one plane normal to the axis.

• The image is geometrically similar to the object.

Object ImageSystemOptical



134

Optics SOLO

Lens

Convention of Signs

1. All Figures are drawn with the light traveling from left to right.

2. All object distances are considered positive when they are measured to the left of the vertex and negative when they are measured to the right.

3. All image distances are considered positive when they are measured to the right of the vertex and negative when they are measured to the left.

4. Both focal length are positive for a converging system and negative for a diverging system.

5. Object and Image dimensions are positive when measured upward from the axis and negative when measured downward.

6. All convex surfaces are taken as having a positive radius, and all concave surfaces are taken as having a negative radius.

Return to TOC

135

Optics SOLO

Derivation of Gaussian Formula for a Single Spherical Surface Lens Using Fermat’s Principle


s 's

n 'n

h

l 'l

'

M

T

CAM’

R

The optical path connecting points M, T, M’ is'' lnlnpathOptical

Applying cosine theorem in triangles MTC and M’TC we obtain:

2/122 cos2 RsRRsRl

2/122 cos'2'' RsRRsRl

2/1222/122 cos'2''cos2 RsRRsRnRsRRsRnpathOptical Therefore

According to Fermat’s Principle when the point Tmoves on the spherical surface we must have

0d

pathOpticald

0

'

sin''sin

l

RsRn

l

RsRn

d

pathOpticald

from which we obtain

l

sn

l

sn

Rl

n

l

n

'

''1

'

'

For small α and β we have ''& slsl

and we obtainR

nn

s

n

s

n

'

'

'

Gaussian Formula for a Single Spherical Surface

Return to TOC

a

Hecht & Zajac, " Optics", 4th Ed., McGraw-Hill, 1979, pp.103-104

136

Optics SOLO

Derivation of Gaussian Formula for a Single Spherical Surface Lens Using Snell’s Law

Apply Snell’s Law: 'sin'sin nn

If the incident and refracted raysMT and TM’ are paraxial theangles and are small and we can write Snell’s Law:

'

From the Figure '

'' nn

nnnnnn '''

For paraxial rays α, β, γ are small angles, therefore '/// shrhsh

r

hnn

s

hn

s

hn '

''

or

r

nn

s

n

s

n

'

'

'

Gaussian Formula for a Single Spherical SurfaceKarl Friederich Gauss

1777-1855

Willebrord van Roijen Snell

1580-1626

s 's

n 'n

h

l 'l

'

M

T

CAM’

r

a

Jenkins & White, "Fundamentals of Optics", 4th Ed., McGraw-Hill, 1976, pp.56-57

137

Optics SOLO

Derivation of Gaussian Formula for a Single Spherical Surface Lens Using Snell’s Law

for s → ∞ the incoming rays are parallel to opticalaxis and they will refract passing trough a commonpoint called the focus F’.

r

nn

s

n

s

n

'

'

'

s '' fs n 'n

h

'l

'

T

CA

F’

R

fs 's

n 'n

h

l

F

T

CA

R

'

r

nn

f

nn

'

'

'r

nn

nf

'

''

for s’ → ∞ the refracting rays are parallel to opticalaxis and therefore the incoming rays passes trough a common point called the focus F.

r

nnn

f

n

'' rnn

nf

'

'' n

n

f

f

Return to TOC

a


138

Optics SOLO

Derivation of Lens Makers’ Formula

We have a lens made of twospherical surfaces of radiuses r1

and r2 and a refractive index n’,separating two media havingrefraction indices n a and n”. Ray MT1 is refracted by the firstspherical surface (if no secondsurface exists) to T1M’.

111

'

'

'

r

nn

s

n

s

n

11111 ''& sMAsTA

Ray T1T2 is refracted by the second spherical surface to T2M”. 2222 ""&'' sMAsMA

222

'"

"

"

'

'

r

nn

s

n

s

n

Assuming negligible lens thickness we have , and since M’ is a virtual objectfor the second surface (negative sign) we have

21 '' ss 21 '' ss

221

'"

"

"

'

'

r

nn

s

n

s

n

M’

M

'1f1f

1s

Axis

T1 T2

A1

A2C1

1rC2 F’1F’’2

M’’

F’2F1

''2f'2f'1s

'2s''2s

2r

n 'n ''n

a


139

Optics SOLO

Derivation of Lens Makers’ Formula (continue – 1)

M”

M

f

s

AxisA1

A2C1

1rC2

F”

F

''f

''s

2r

n 'n ''n

111

'

'

'

r

nn

s

n

s

n

Add those equations

221

'"

"

"

'

'

r

nn

s

n

s

n

2121

'"'

"

"

r

nn

r

nn

s

n

s

n

M’

M

'1f1f

1s

Axis

T1 T2

A1

A2C1

1rC2 F’1F’’2

M’’

F’2F1

''2f'2f'1s

'2s''2s

2r

n 'n ''n

The focal lengths are defined by tacking s1 → ∞ to obtain f” ands”2 → ∞ to obtain f

f

n

r

nn

r

nn

f

n

212

'"'

"

"

Let define s1 as s and s”2 as s” to obtain

21

'"'

"

"

r

nn

r

nn

s

n

s

n

f

n

r

nn

r

nn

f

n

21

'"'

"

"

a


140

Optics SOLO

Derivation of Lens Makers’ Formula (continue – 2)

M”

M

f

s

AxisA1

A2C1

1rC2

F”

F

''f

''s

2r

n 'n n

If the media on both sides of the lens is the same n = n”.

21

111

'

"

11

rrn

n

ss

21

111

'1

"

1

rrn

n

ff

Therefore

"

11

"

11

ffss

Lens Makers’ Formula

a


141

Optics SOLO

First Order, Paraxial or Gaussian Optics

In 1841 Gauss gave an exposition in “Dioptrische Untersuchungen”for thin lenses, for the rays arriving at shallow angles with respect toOptical axis (paraxial).


Derivation of Lens Formula

'y

s 's

M’A F’

M

T

F

'ffx 'x

Q

Q’'y

y

S

Axisy From the similarity of the triangles

and using the convention:

''

''~'

f

y

s

yyTAFTSQ

Lens Formula in Gaussian form

f

y

s

yyFASQTS

''~

0' y

Sum of the equations:

'

'

'

''

f

y

f

y

s

yy

s

yy

since f = f’ fss

1

'

11

Return to TOC

142

Optics SOLO

First Order, Paraxial or Gaussian Optics (continue – 1)

Gauss explanation can be extended to the first order approximationto any optical system.


'y

s 's

M’P1 F’

M

T

F'ffx 'x

Q

Q’'y

yAxis

y P2First Focal

Point

First PrincipalPoint

Second FocalPointSecond Principal

Point

Optical SystemObject

Image

Lens Formula in Gaussian form

'y

s 's

M’A F’

M

T

F

'ffx 'x

Q

Q’'y

y

S

Axisy

fss

1

'

11

s – object distance (from the first principal point to the object).

s’ – image distance (from the second principal point to the image).

f – EFL (distance between a focal point to the closest principal plane).

143

Optics SOLO

Derivation of Lens Formula (continue)

'y

s 's

M’A F’

M

T

F

'ffx 'x

Q

Q’'y

y

S

Axisy

From the similarity of the trianglesand using the convention:

f

y

x

yFASQMF

'~

Lens Formula in Newton’s form

f

y

x

yQMFTAF

'

''''~'

0' y

Multiplication of the equations:

2

'

'

'

f

yy

xx

yy

or 2' fxx

Isaac Newton1643-1727


Published by Newton in “Opticks” 1710

a


144

Optics SOLO

Derivation of Lens Formula (continue)

'h

s 's

M’A F’

M

T

F

'ffx 'x

Q

Q’'h

h

S

Axish


Lateral or Transverse Magnification

f

x

x

f

s

s

h

hmT

'''

(-) sign (+) sign Quantityvirtual object real object s

virtual image real image s’

diverging lens converging lens f

inverted object erect object h

inverted image erect image h’

inverted image erect image mT

a


145

Optics SOLO

ConcaveSpherical

ConvexSpherical

Paraboloidal

ConicEllipsoidal

GeneralAspherical

Plane

Converging : General use

Diverging : General use

Accurately focuses a parallel beamor produces a parallel beam froma point source

Refocuses a diverging bundle atanother point (P) displaced fromthe point of origin (O)

Change the direction of beam

Used mostly in combination systems of twoor more components

BASIC MIRRORS FORMS

146

Optics SOLO

Convex

PlanoConvex

Meniscus

Concave

PlanoConcave

Meniscus

Doublet

Multi-Element

Aspheric

Converging: General Use, Magnification

Converging: Used often in opposed doubles to reduce spherical aberration

Converging: reduced spherical aberration

Diverging: General Use, Demagnification

Diverging: Used in multi-element combinations

Diverging: reduced spherical aberration

Corrected for chromatic aberration

High order of aberration correction used incomplex systems

Corrected for spherical aberrationused in condenser systems

BASIC LENS FORMS

Return to TOC

147

Optics SOLO

Ray Tracing

F CO

I

Object VirtualImage

ConvexMirror

R/2 R/2R

F

F’CO

I

Object

RealImage

ConvergingLens

FCO

I

Object

RealImage

ConcaveMirror

F

F’CO I

Object

VirtualImage

DivergingLens

Ray Tracing is a graphically implementation of paralax ray analysis. The constructiondoesn’t take into consideration the nonideal behavior, or aberration of real lens.

The image of an off-axis point can be located by the intersection of any two of thefollowing three rays:

1. A ray parallel to the axis that isreflected through F’.

2. A ray through F that is reflectedparallel to the axis.

3. A ray through the center C of thelens that remains undeviated andundisplaced (for thin lens).

148

Optics SOLO

InfinityPrincipalfocus

SUMMARY OF SIMPLE IMAGING LENSES

f f2f2 f 0

's

'ss

fs 2 fsf 2'

fs 2 fs 2'

fsf 2 fs 2'

's

's

s

s

fs 's

s

s's

fs fs '

s's

fsf 2 fs '

Real, invertedsmall

Telescope

Real, invertedsmaller

Camera

Real, invertedsame size

Photocopier

Real, invertedlarger

Projector

No image Searchlight

Virtual, erectlarger

Microscope

Virtual, erectsmaller

Various

Figure ObjectLocation

ImageLocation

ImageProperties Example

L.J. Pinson, “Electro-Optics”, John Wiley & Sons, 1985, pg.54

Return to TOC

149

Optics SOLO

Matrix Formulation

The Matrix Formulation of the Ray Tracing method for the paraxial assumption was proposed at the beginning of nineteen-thirties by T.Smith.

Assuming a paraxial ray entering at some input plane of an optical system at the distancer1 from the symmetry axis and with a slope r1’ and exiting at some output plane at the distancer2 from the symmetry axis and with a slope r2’, than the following linear (matrix) relationapplies:

PrincipalPlanesInput

planeOutputplane

Ray path

1h2h

1r2r

'1r

'2r

Symmetryaxis

''' 1

1

1

1

2

2

r

rM

r

r

DC

BA

r

r

DC

BAMwhere ray transfer matrix

When the media to the left of the input planeand to the right of the output plane have thesame refractive index, we have:

1det CBDAM

150

Optics SOLO

Matrix Formulation (continue -1)

Uniform Optical Medium

In an Uniform Optical Medium of length d no change in ray angles occurs:

Ray path

d

1r2r

'1r

'2r

Symmetryaxis

1 2

''

'

12

112

rr

rdrr

10

1 dM

MediumOpticalUniform

Planar Interface Between Two Different Media

Ray path

1r 2r

'1r '2r

Symmetryaxis

1 2

1n 2n

12 rr

'' 1

2

12

12

rn

nr

rr

Apply Snell’s Law: 2211 sinsin nn

paraxial assumption: tan'sin r

From Snell’s Law: '' 1

2

12 r

n

nr

21 /0

01

nnM

InterfacePlanar

1det2

1 n

nM

InterfacePlanar

1det MediumOpticalUniformM

The focal length of this system is infinite and it hasnot specific principal planes.

151

Optics SOLO


A Parallel-Sided Slab of refractive index n bounded on both sides with media of refractive index n1 = 1

Ray path

d

21 rr 43 rr

'1r '4r

Symmetryaxis

'2r

'3r

nn 211 n 11 n

We have three regions:• on the right of the slab (exit of ray):

'/0

01

' 3

3

124

4

r

r

nnr

r

• in the slab:

'10

1

' 2

2

3

3

r

rd

r

r

• on the left of the slab (entrance of ray):

'/0

01

' 1

1

212

2

r

r

nnr

r

Therefore:

'/0

01

10

1

/0

01

' 1

1

21124

4

r

r

nn

d

nnr

r

21

21

122112 /0

/1

/0

01

/0

01

10

1

/0

01

nn

nnd

nnnn

d

nnM

mediaentranceslabmediaexit

S labSidedParallel

10

/1 21 nndM

SlabSidedParallel

1det SlabSidedParallelM

152

Optics SOLO


Spherical Interface Between Two Different Media

Ray path

21 rr '1r

'2r

Symmetry axis

1n2n

i r

1

12 rr

Apply Snell’s Law: rnin sinsin 21

paraxial assumption: rrii sin&sin

From Snell’s Law: rnin 21

2

1

2

1

2

1

12

21

0101

n

n

n

D

n

n

Rn

nnMInterfaceSpherical 1det

2

1 n

nM

InterfaceSpherical

12

11

'

'

rr

ri From the Figure:

122111 '' rnrn

111 / Rr

12

121

2

112

''

Rn

rnn

n

rnr

1

12

11

1122

12

''

n

rn

Rn

rnnr

rr

1

121 :

R

nnD

where: Power of the surface If R1 is given in meters D1 gives diopters

153

Optics SOLO


Thick Lens

21 rr

43 rr

'1r

i

2 1

'2r '3r

r

2R

1R

f

'4r1C2F IO 1F

2C

Principal planes

2n

1n

s 's

d

We have three regions:• on the right of the slab (exit of ray):

'

01

' 3

3

1

2

1

2

4

4

r

r

n

n

n

Dr

r

• in the slab:

'10

1

' 2

2

3

3

r

rd

r

r

• on the left of the slab (entrance of ray):

'

01

' 1

1

2

1

2

1

2

2

r

r

n

n

n

Dr

r

Therefore:

'

101

'

01

10

101

' 1

1

2

1

2

1

2

1

2

1

1

2

1

2

1

1

2

1

2

1

1

2

1

2

4

4

r

r

n

n

n

D

n

nd

n

Dd

n

n

n

Dr

r

n

n

n

Dd

n

n

n

Dr

r

2

2

21

21

1

21

2

1

2

1

1

1

n

Dd

nn

DDd

n

DD

n

nd

n

Dd

MLensThick

2

212 R

nnD

1

121 :

R

nnD

2

1

21

21

1

21

2

1

2

2

1

1

1

n

Dd

nn

DDd

n

DD

n

nd

n

Dd

MLensThick

1det LensThickM

or21 DD

154

Optics SOLO

Matrix Formulation (continue -5)Thick Lens (continue -1)

21 rr

43 rr

'1r

i

2 1

'2r '3r

r

2R

1R

f

'4r1C2F IO 1F

2C

Principal planes

2n

1n

2R

1R

2f

1C 2F I

O

1F

2C

Principal planes

2n

1n

1h2h

s

s 's

's

d

Ray 2

Ray 1

1f

Let use the second Figure where Ray 2 is parallelto Symmetry Axis of the Optical System that is refractedtrough the Second Focal Point.

'1

1

' 1

1

2

2

21

21

1

21

2

1

2

1

4

4

r

r

n

Dd

nn

DDd

n

DD

n

nd

n

Dd

r

r We found:

2141 /'&0' frrr Ray 2:

By substituting Ray2 parameters we obtain:

1

2

1

21

21

1

214

1' r

fr

nn

DDd

n

DDr

1

21

21

1

212

nn

DDd

n

DDf

frrr /'&0' 414 Ray 1:

We found:

'1

1

' 4

4

2

1

21

21

1

21

2

1

2

2

1

1

r

r

n

Dd

nn

DDd

n

DD

n

nd

n

Dd

r

r

4

1

4

21

21

1

211

1' r

fr

nn

DDd

n

DDr

2

1

21

21

1

211 f

nn

DDd

n

DDf

155

Optics SOLO


Thin Lens

21 rr

43 rr

'1r

i

2 1

'2r '3r

r

2R

1R

f

'4r1C2F IO 1F

2C

Principal planes

2n

1n

s 's

d

For thick lens we found

2

2

21

21

1

21

2

1

2

1

1

1

n

Dd

nn

DDd

n

DD

n

nd

n

Dd

MLensThick

21

21

1

211

nn

DDd

n

DD

f

For thin lens we can assume d = 0 and obtain

11

01

f

MLensThin

1

211

n

DD

f

2

212 R

nnD

1

121 :

R

nnD

211

2

1

21 111

1

RRn

n

n

DD

f

21 rr

43 rr

2R

1R

f

'4r1C2F IO 1F

2C

Principal planes

2n

1n

s 's

'1r

156

Optics SOLO


Thin Lens (continue – 1)

For a biconvex lens we have R2 negative

211

2 111

1

RRn

n

f

For a biconcave lens we have R1 negative

211

2 111

1

RRn

n

f

11

01

f

MLensThin

157

Optics SOLO


A Length of Uniform Medium Plus a Thin Lens

f

d

f

dd

f

MMMMediumUniform

LensThin

LensThinMediumUniform 1

1

1

10

1

11

01

21 rr

43 rr

2R

1R

f

'4r1C2F IO 1F

2C

Principal planes

2n

1n

s 's

'1r

d

Combination of Two Thin Lenses

2n

1d

1f

2d

2f

2n

21

21

2

2

2

1

1

1

21

2

21

1

2121

2

2

1

1

1

1

22

2

111

1

11

1

11

1

1122

ff

dd

f

d

f

d

f

d

ff

d

ff

f

dddd

f

d

f

d

f

d

f

d

f

d

MMMMMdMedium

UniformfLens

ThindMedium

UniformfLens

Thin

LensesThinTwo

The Focal Length of the Combination of Two Thin Lenses is:

21

2

21

111

ff

d

fff

158

Optics SOLO


Mirrors r

Spherical Mirror

i

i

ii i

i

iy

RSpherical MirrorCenter of Curvature

r

Ryiii /tan

Consider a Spherical Mirror of radius R.From the geometry:

For small angles:

Ryiii /

also: iri 2 2/rii Ryiri /2

Define by n the index of reflexion of the medium:

Rynnn

yy

iir

ir

/2

i

i

r

r

n

y

Rnn

y

1/2

01

Therefore:

1/

01

1/2

01

fnRnM

MirrorSpherical

159

Optics SOLO


Cavity of two Mirrors

d

12M

21M2MirrorM

1MirrorM

Spherical Mirror M1

Radius R1

Spherical Mirror M2

Radius R2

O

10

1

1/2

01

10

1

1/2

01

121221

12

d

Rn

d

RnMMMMM

MMirrorSpherical

MMirrorSphericalCavity

Figure shows two spherical mirrorsfacing each other forming an opticalcavity.

Light leaves point O, traverse the gapin the positive direction, is reflected byMirror M1, retraces the gap in thenegative direction, and is reflected byMirror M2. The System Matrix is:

21221

221

12

1

1122 /21/21/2/4/2/2

/22/21

/21/2

1

/21/2

1

RdnRdnRdnRRdnRnRn

RdndRdn

RdnRn

d

RdnRn

d

2122

21212

21

12

1

/4/4/21/4/2/2

/22/21

RRdnRdnRdnRRdnRnRn

RdndRdnM Cavity

160

Go to OPTICS Part II

Optics SOLO


April 13, 2023 161

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

http://upload.wikimedia.org/wikipedia/en/a/ae/RAFAEL_logo.png

optics part i

Science

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