d σ0 (μn 0 = + μp q n p 0 p n p no p n o w l v i +δq n n...

Light

w

d

lV Iphoto

A semiconductor slab of length l, width w and depth d isilluminated with light of wavelength λ.

n = no + Δnp = po + Δp

© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

)( 000 pnq pn μμσ +=

δσσδμδμσ +=+++= 000 )]()([ ppnnq pn

))(( pnpq μμδδσ +=

pp

nn

pGnGτδ

τδ

== &

EJJJ L )()( 00 δσσ +=+=

Enn

p

n

pn

tt

Gain μμμτ

l=+= )1(

Photoconductors

p +SiO 2

Electrode

ρ net

– eN a

eN d x

x

E (x )

R

E max

e –h +

I ph

h υ > E g

W

En

Depletion region

( a )

( b )

( c )

Antireflectioncoating

V r

(a) A schematic diagram of a reverse biased pn junctionphotodiode. (b) Net space charge across the diode in thedepletion region. N d and N a are the donor and acceptorconcentrations in the p and n sides. (c). The field in thedepletion region.

Electrode

V out


Photodiodes

e–h+

iph(t)

Semiconductor

(a)

V

x

(b)

(a) An EHP is photogenerated at x = l. The electron and the hole drift in oppositedirections with drift velocities vh and ve. (b) The electron arrives at time te = (L − l)/ve andthe hole arrives at time th = l/vh. (c) As the electron and hole drift, each generates anexternal photocurrent shown as ie(t) and ih(t). (d) The total photocurrent is the sum of holeand electron photocurrents each lasting a duration th and te respectively.

E

l L − l

t

vevhvh

0 Ll

t

e–h+

th

te

t

0

th

te

iph(t)

i(t)

t

0

th

te

evh/L + eve/Levh/L

ie(t)

ih(t)

(c)

(d)Charge = e

evh/L eve/L


Ev ee μ=

Ev pp μ=

ee v

lLt −=

pp v

lt =

dxtViqEdxwork e )(==

dtdxvLVE e /&/ ==

Lqvti e

e =)(

pp

pee

e ttL

qvtitt

Lqvti <=<= )(&)(

qL

tqvL

tqvdttidttiQ ppt t

eepecollected

e p

=+=+= ∫ ∫0 0

)()( So, the collected charge is one electron charge

0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8

Wavelength (μm)

In0.53Ga0.47As

Ge

Si

In0.7Ga0.3As0.64P0.36

InPGaAs

a-Si:H

12345 0.9 0.8 0.7

1×103

1×104

1×105

1×106

1×107

1×108

Photon energy (eV)

Absorption coefficient (α) vs. wavelength (λ) for various semiconductors(Data selectively collected and combined from various sources.)

α (m-1)

1.0

Figure 5.3

)1()( )(0

xepXP λα−−=

xepXP )(0)( λα−=

Power generation

Power absorption

xepXP )(0)( λα−=

)1()( )(0

xepXP λα−−=

xs

sexG αα −Φ= 0)(

Power generation

Power absorption

∫ ∫ −−− −−

=−Φ=Φ==w

wfww

xsp

sss eAhRP

qAeqAdxeqAdxxGqAI0

00

00 ]1[

)1(]1[)( ααα

να

νAhRP f )1(0

0

−=Φ

νη

hPqI

photonsofincidentatedofEHPgener p

ext //

##

0

==

hcq

PI

WticalpowerincidentopAntphotocurreR ext

p λη===0)(

)(

External Q. Efficiency

EHP generation rate Incident photon flux/unit area

Responsivity of the photodiode

E

CB

VB

k–k

Direct Bandgap Eg Photon

Ec

Ev

(a) GaAs (Direct bandgap)

E

k–k

(b) Si (Indirect bandgap)

VB

CB

Ec

Ev

Indirect Bandgap, Eg

Photon

Phonon

(a) Photon absorption in a direct bandgap semiconductor. (b) Photon absorptionin an indirect bandgap semiconductor (VB, valence band; CB, conduction band)


p +SiO 2

Electrode

ρ net

– eN a

eN d x

x

E (x )

R

E max

e –h +

I ph

h υ > E g

W

En

Depletion region

( a )

( b )

( c )

Antireflectioncoating

V r

(a) A schematic diagram of a reverse biased pn junctionphotodiode. (b) Net space charge across the diode in thedepletion region. N d and N a are the donor and acceptorconcentrations in the p and n sides. (c). The field in thedepletion region.

Electrode

V out


Photodiodes

νη

hPqI

photonsofincidentatedofEHPgener p

ext //

##

0

==

External Q. Efficiency

∫ ∫ −−− −−

=−Φ=Φ==w

wfww

xsp

sss eAhRP

qAeqAdxeqAdxxGqAI0

00

00 ]1[

)1(]1[)( ααα

να

xs

sexG αα −Φ= 0)( EHP generation rate νAhRP f )1(0

0

−=Φ Incident photon flux/unit area

hcq

PI

WticalpowerincidentopAntphotocurreR ext

p λη===0)(

)(Responsivity of the photodiode

Responsivity (R) vs. wavelength (λ) for an idealphotodiode with QE = 100% (η = 1) and for a typicacommercial Si photodiode.

0 200 400 600 800 1000 12000

0.10.20.30.40.50.60.70.80.9

1

Wavelength (nm)

Si Photodiode

λg

Responsivity (A/W)

Ideal PhotodiodeQE = 100% ( η = 1)


p+

i-Si n+

SiO2Electrode

ρnet

-eNa

eNd

x

x

E(x)

R

Eo

E

e-h+

Iph

hυ > Eg

W

(a)

(b)

(c)

(d)

Vr

The schematic structure of an idealized pin photodiode (b) The netspace charge density across the photodiode. (c) The built-in fieldacross the diode. (d) The pin photodiode in photodetection isreverse biased.

Vout

Electrode


pin photodiode

hυ > Eg

p+ i-Si

e– E

h+

Wl

Drift

Diffusion

A reverse biased pin photodiode is illuminated with a shortwavelength photon that is absorbed very near the surface.The photogenerated electron has to diffuse to the depletionregion where it is swept into the i-layer and drifted across.

Vr


š p+

SiO2Electrode

ρnet

x

x

E(x)

R

Ehυ > Eg

p

Iph

e– h+

Absorptionregion

Avalancheregion

(a)

(b)

(c)

(a) A schematic illustration of the structure of an avalanche photodiode (APD) biasedfor avalanche gain. (b) The net space charge density across the photodiode. (c) Thefield across the diode and the identification of absorption and multiplication regions.

Electrode


n+Avalanche photodiode (APD)

h+

E

šn+ p

e–

Avalanche region

e–

h+

Ec

Ev

(a) (b)

E

(a) A pictorial view of impact ionization processes releasing EHPs andthe resulting avalanche multiplication. (b) Impact of an energeticconduction electron with crystal vibrations transfers the electron'skinetic energy to a valence electron and thereby excites it to theconduction band.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

SiO2

Guard ring

ElectrodeAntireflection coating

nn n+

p+

š

p

SubstrateElectrode

n+

p+

š

p

SubstrateElectrode

Avalanche breakdown

(a) (b)

(a) A Si APD structure without a guard ring. (b) A schematic illustration of thestructure of a more practical Si APD© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

E

N n

Electrode

x

E(x)

R

hυ

Iph

Absorptionregion

Avalancheregion

InP InGaAs

h+

e–E

InP

P+ n+

Simplified schematic diagram of a separate absorption and multiplication(SAM) APD using a heterostructure based on InGaAs-InP. P and N refer top and n -type wider-bandgap semiconductor.

Vr

Vout


P+–InP Substrate

P+–InP (2-3 μm) Buffer epitaxial layerN–InP (2-3 μm) Multiplication layer.

Photon

n–In0.53Ga0.47As (5-10μm) Absorption lay

Graded n–InGaAsP (<1 μm)

Electrode

Electrode

Simplified schematic diagram of a more practical mesa-etched SAGM layeredAPD.


hυ

h+

e–

n+

Ec

Ev

10–20 nm

p+

E

Eg1

Eg 2ΔEc

Energy band diagram of a staircase superlattice APD (a) No bias. (b) Withan applied bias.

(a) (b)


n

hυ

Base Collector

h+

e–

Emitter

pn+

E

e–

SCLSCL Iph

VBE VBC

VCC

The principle of operation of thephotodiode. SCL is the space chargelayer or the depletion region. Theprimary photocurrent acts as a basecurrent and gives rise to a largephotocurrent in the emitter-collectorcircuit.


Iph

Photoconductore–

h+

Iph Iph Iph Iph

A photoconductor with ohmic contacts (contacts not limiting carrier entry) can exhibit gain. Asthe slow hole drifts through the photoconductors, many fast electrons enter and drift through thephotoconductor because, at any instant, the photoconductor must be neutral. Electrons drift fasterwhich means as one leaves, another must enter.

(a) (b) (c) (d) (e)


Vou

Current

Time

Id

Vr

In pn junction and pin devices the main source of noise is shotnoise due to the dark current and photocurrent.

pn

Po

Dark

IlluminatedId + IphId + Ip h + in

R A


00.10.20.30.40.50.60.70.8

0.5 1 1.5 2Wavelength(µm)

The responsivity of a commercial Ge pnjunction photodiode

Responsivity(A/W)


0

0.1

0.2

0.3

0.4

0.5

0.6

200 400 600 800 1000 1200Wavelength(nm)

A B

The responsivity of two commercial Si pinphotodiodes

Responsivity(A/W)


0

0.2

0.4

0.6

0.8

1

800 1000 1200 1400 1600 1800Wavelength(nm)

The responsivity of an InGaAs pinphotodiode

Responsivity(A/W)


x

R

E

e–h+

iph

hυ > Eg

W

Vr

An infinitesimally short light pulse is absorbed throughout thedepletion layer and creates an EHP concentration that decaysexponentially

Photogenerated electron concentrationexp(−αx) at time t = 0

BA

vde


Drift velocity vs. electric field for holes and electrons in Si.

102

103

104

105

107106105104

Electric field (V m-1)

Electron

Hole

Drift velocity (m s -1)


x

y

z

Ey

Ex

−yEy^

xEx^

(a) (b) (c)

EPlane of polarization

x

y

E

(a) A linearly polarized wave has its electric field oscillations defined along a lineperpendicular to the direction of propagation, z. The field vector E and z define a plane ofpolarization. (b) The E-field oscillations are contained in the plane of polarization. (c) Alinearly polarized light at any instant can be represented by the superposition of two fields Exand Ey with the right magnitude and phase.

E


Light Polarization and Applications

z

Ey

Ex

EEθ = kΔz

Δz

z

A right circularly polarized light. The field vector E is always at rightangles to z , rotates clockwise around z with time, and traces out a fullcircle over one wavelength of distance propagated.


E

y

x

Exo = 0Eyo = 1φ = 0

y

x

Exo = 1Eyo = 1φ = 0

y

x

Exo = 1Eyo = 1φ = π/2

E

y

x

Exo = 1Eyo = 1φ = −π/2

(a) (b) (c) (d)

Examples of linearly, (a) and (b), and circularly polarized light (c) and (d); (c) isright circularly and (d) is left circularly polarized light (as seen when the wavedirectly approaches a viewer)


E

y

x

Exo = 1Eyo = 2φ = 0

Exo = 1Eyo = 2φ = π/4

Exo = 1Eyo = 2φ = π/2

y

x

(a) (b)E

y

x

(c)

(a) Linearly polarized light with Eyo = 2Exo and φ = 0. (b) When φ = π/4 (45°), the light isright elliptically polarized with a tilted major axis. (c) When φ = π/2 (90°), the light isright elliptically polarized. If Exo and Eyo were equal, this would be right circularlypolarized light.


Field distributions in plane E&M waves

Adding two linearly polarized waves

Elliptically polarized light

Adding linearly polarized waves

Polarizer 1

TA1

Polarizer 2 = Analyzer

TA2

θ Light detectorE

Ecosθ

Unpolarized light

Linearlypolarized light

Randomly polarized light is incident on a Polarizer 1 with a transmission axis TA1. Lightemerging from Polarizer 1 is linearly polarized with E along TA1, and becomes incidenton Polarizer 2 (called "analyzer") with a transmission axis TA2 at an angle θ to TA1. Adetector measures the intensity of the incident light. TA1 and TA2 are normal to the lightdirection.


ΔI ∝−φ +

23

φ3 − ...

7.6 Modulation by Malus’s law Suppose that a linearly polarized light is passed through a polarizer placed with its transmission axis at angle π/4 (45°) to the incoming optical field. Suppose now that we “rotationally modulate” the transmission axis of the polarizer by small amounts φ about π /4 (45°). Show that the change in the transmission intensity is

where φ is in radians. What is the extent of change (in degrees) in φ so that the second term is only 1% of the first term? What is your conclusion.

A line viewed through a cubic sodium chloride (halite) crystal(optically isotropic) and a calcite crystal (optically anisotropic).

Optically isotropic materials like glasses, and cubic crystals

Optically anisotropic materials (birefringent) like noncubic crystals, calcite (CaCO3, Quartz, ice, tournaline,…

Two polaroid analyzers are placed with their transmission axes, alongthe long edges, at right angles to each other. The ordinary ray,undeflected, goes through the left polarizer whereas the extraordinarywave, deflected, goes through the right polarizer. The two wavestherefore have orthogonal polarizations.

Biaxial crystals like mica have 3 distinct principal indices.

Uniaxial crystals like calcite, and quartz have two of their indices equal to each other.

The velocity of light and state of its polarization depends on the direction of propagation of light.

Light (EM wave) entering an isotropic crystal splits into two ┴ linearly polarized waves traveling with different phase velocities. In uniaxial crystals are called ordinary (o) and extraordinary (e) waves.

The o-wave behaves like an ordinary light. Has the same phase velocity in all direction and the field is perpendicular to propagation direction.

The e-wave has a phase velocity that depends on its direction of propagation and the electric field is not necessarily perpendicular to the propagation direction.

The two waves propagate with same velocity only along a special direction called the optic axis.

The o-wave is always perpendicularly polarized to the optic axis and obeys the Snell’s law.

n2Opticaxis

n1

x

y

zn3 O

B

A

P

B′

A′

z

k

O

(b) An EM wave propagating along OP at anangle θ to optic axis.

(a) Fresnel's ellipsoid

θ


no = n1

x

yz = Optic axis

ne (90°) = n3

z = Optic axi

(b)Eo

Ee

x

z

Eo

Ee

z = Optic axis

ne (0°) = n2 = n1

no = n1

y

(a)

Eo = Eo-wave and Ee = Ee-wave (a) Wave propagation along the optic axis. (b)Wave propagation normal to optic axis© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

(b)

Optic axiskz

o-wavee-wave

Ee

O kx

(a)

E oscillations ⊥ to paperWavefronts (constant phase fronts)

S e = Power flow

Q

P ko

k e

k e

k e

Ee

Eo

(a) Wavevector surface cuts in the xz plane for o- and e-waves. (b) An extraordinarywave in an anisotropic crystal with a ke at an angle to the optic axis. The electric fieldis not normal to ke. The energy flow (group velocity) is along Se which is differentthan ke.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

E oscillations // to paper

e-wave

o-wave

Optic axis(in plane of paper)

Optic axisPrincipal section

e-ray

o-ray

Principal section

A calcite rhomb

E⊥

E//Incident ray

Incident wave

An EM wave that is off the optic axis of a calcite crystal splits into two waves calledordinary and extraordinary waves. These waves have orthogonal polarizations andtravel with different velocities. The o-wave has a polarization that is alwaysperpendicular to the optical axis.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Optic axis

x , no

y

Ee-wave

Eo-wave

z , ne

Optic axisy , no

E

z

Ee-wave

Eo-wave

x , no

(a) (b)

(a) A birefringent crystal plate with the optic axis parallel to the plate surfaces. (b) Abirefringent crystal plate with the optic axis perpendicular to the plate surfaces.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

• 7.7 Birefringence Consider a negative uniaxial crystal such as calcite (ne < no) plate that has the optic axis (taken along z) parallel to the plate faces. Suppose that a linearly polarized wave is incident at normal incidence on a plate face. If the optical field is at an angle 45° to the optic axis, sketch the wavefronts and the rays through the calcite plate.

x = Fast axis

z = Slow axis

E//

E⊥

E//

E⊥

E

L

yno

ne = n3α

Optic axis

L

yno

ne = n3

φ

A retarder plate. The optic axis is parallel to the plate face. The o- and e-waves travelin the same direction but at different speeds.


x

α = arbitrary

(b)

Input

z

x E

z

(a)

Output

Optic axis

α

α

Half wavelength plate: φ = š Quarter wavelength plate: φ = š/2

x

α

0 < α < 45°

E

z

x

E

E

x

z z

α = 45°

45°

Input and output polarizations of light through (a) a half-wavelengthplate and (b) through a quarter-wavelength plate.


Optic axis

Optic axisd

D

Wedges can slid

Plate

E1

E2

Soleil-Babinet Compensator© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

))((2 dDnn oe −−=λπφ

)(21 Dndn oe +=

λπφ

)(22 Dndn eo +=

λπφ

• 7.8 Wave plates Calculate and compare the thickness of quarter -wave plates made from calcite, quartz and LiNbO3 crystals all operating at a wavelength of λ ≈ 590 nm. What is your conclusion? Assuming little relative change in the indices, what are the thicknesses at double the wavelength?

• 7.9 Soleil Compensator: Consider a Soleil compensator as in Figure 7Q9 that uses a quartz crystal. Given a light wave with a wavelength λ≈ 600 nm, a lower plate thickness of 5 mm, calculate the range of dvalues in Figure 7Q9 that provide a retardation from 0 to π (half-

wavelength).

Optic axis

Optic axisd

D

Wedges can slide

Plate

E1

E2

Soleil-Babinet Compensator

Optic axis

e-ray

o-rayA

B

Optic axis

e-ray

o-ray

Optic axis A

B Optic axisθ

E1

E2

E1

E1

E2

E2

The Wollaston prism is a beam polarization splitter. E1 is orthogonal to the plane ofthe paper and also to the optic axis of the first prism. E2 is in the plane of the paperand orthogonal to E1.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

• 7.10Quartz Wollaston prism Draw a quartz Wollaston prism and clearly show and identify the directions of orthogonally polarized waves traveling through the prisms. How would you test the polarization states of the emerging rays? Consider two identical Wollaston prisms, one from calcite and the other from quartz. Which will have a greater beam splitting ability? (Explain).

E

Optic axis

θE ′

z

LQuartz

z

Dextro

z

Levo

E′ E′

An optically active material such as quartz rotates the plane of polarizationof the incident wave: The optical field E rotated to E′. If we reflect thewave back into the material, E′ rotates back to E.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Optical Activity

xInput

y

x

y

EL

y

x

ER

xOutput

y

x

y

E′L

y

x

E′R

E

E′

α βα = β

β′α′ α ≠ β

Slow Fast

θ

Vertically polarized wave at the input can be thought of as two right and lefthanded circularly polarized waves that are symmetrical, i.e. at any instant α = β.If these travel at different velocities through a medium then at the output they areno longer symmetric with respect to y, α ≠ β., and the result is a vector E′ at anangle θ to y.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Lnn LR )( −=λπθ

xz Ean1 = no

y′

(a)

x′n2 = no n′1

n′2

z

(b)

x

45°

(c)

xz

KDP, LiNbO 3 KDP LiNbO 3

n′1

n′2

y Eay

(a) Cross section of the optical indicatrix with no applied field, n1 = n2 = no (b) Thapplied external field modifies the optical indicatrix. In a KDP crystal, it rotates theprincipal axes by 45° to x′ and y′ and n1 and n2 change to n′1 and n′2 . (c) Appliedfield along y in LiNbO2 modifies the indicatrix and changes n1 and n2 change to nand n′2 .© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

aErnnn 22311

'1 )2/1(+≈

aErnnn 22322

'2 )2/1(−≈

VdLrnLnLn

o 223

'2

'1 222

λπ

λπ

λπφ =−=Δ

Pockels Effect

Outputlight

Δφ

z

xEx

d

EyV

z

Ex

Eyy

45°Inputlight Ea

Tranverse Pockels cell phase modulator. A linearly polarized input light into anelectro-optic crystal emerges as a circularly polarized light. Ea is the applied fieldparallel to y.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Transmission intensity

V

Io

Q

0 Vλ/2

V

45°

Inputlight

P ADetector

Crystal

zx

y

QWP

Left: A tranverse Pockels cell intensity modulator. The polarizer P and analyzer A havetheir transmission axis at right angles and P polarizes at an angle 45° to y-axis. Right:Transmission intensity vs. applied voltage characteristics. If a quarter-wave plate (QWP)is inserted after P, the characteristic is shifted to the dashed curve.


y

z

x no

no

ne

Ea z

x

yEa

Outputlight

Ez

Inputlight

Δφ

Ex

E

(a) (b)

(a) An applied electric field, via the Kerr effect, induces birefringences in anotherwise optically istropic material. (b) A Kerr cell phase modulator.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

2

22 2

dLKVKEn a

πφλ =Δ=ΔKerr Effect

V(t)

Ea

Cross-section

LiNbO3

d

Thin buffer layerCoplanar strip electrodes

EO Substratez

y

x

Polarizedinputlight

WaveguideLiNbO 3

L

Integrated tranverse Pockels cell phase modulator in which a waveguide is diffusedinto an electro-optic (EO) substrate. Coplanar strip electrodes apply a transversefield Ea through the waveguide. The substrate is an x-cut LiNbO3 and typically thereis a thin dielectric buffer layer (e.g. ~200 nm thick SiO2) between the surfaceelectrodes and the substrate to separate the electrodes away from the waveguide.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Phase & Polarization Modulation

V(t)

LiNbO3 EO Substrate

AB

InOutC

DA

B

Waveguide

Electrode

An integrated Mach-Zender optical intensity modulator. The input light issplit into two coherent waves A and B, which are phase shifted by theapplied voltage, and then the two are combined again at the output.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

Cross-sectionCoupled waveguides

B

E

x

EA EB

d

nBnA

Ans

A

B

Top view

Lo

InputPA(Lo)PA(0)

z

PB(Lo)

z

PA(z)

PB(z)(a) (b)

x

(a) Cross section of two closely spaced waveguides A and B (separated by d)embedded in a substrate. The evanescent field from A extends into B and vice versaNote: nA and nB > ns (= substrate index).(b) Top view of the two guides A and B that are coupled along the z-direction. Lightis fed into A at z = 0, and it is gradually transferred to B along z. At z = Lo, all thelight been transferred to B . Beyond this point, light begins to be transferred back toA in the same way.


Δβ ∝ V

PB(Lo)/PA(0)

100%

(π√3)/Lo0

Transmission power ratio from guide A toguide B over the transmission length Lo as afunction of mismatch Δβ.


V(t)

LiNbO3

In

Electrode

Waveguides

Fibers

Ea

Cross-section

LiNbO3

V(t)

Coupled waveguides

A B

dA B

Lo

An integrated directional coupler. Applied field Ea alters the refractive indices of thtwo guides and changes the strength of coupling.


Interdigitally electrodedtransducerModulating RF voltage

Piezoelectriccrystal

Acousticwavefronts

Induced diffractiongrating

Incidentlight

Diffracted light

θ

Through light

Acoustic absorber

2θ

Traveling acoustic waves create a harmonic variation in the refractive indexand thereby create a diffraction grating that diffracts the incident beam throughan angle 2θ.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

θ

A

B

Incident optical beam Diffracted optical beam

O

O'

P Q

B'

A'

Λsinθ ΛsinθΛ Acoustic

wave fronts

θ

nmax

nmax

nmin

nmin

nmin nmax

x

nmin nmax

x

nn

Simplified Actual

Acousticwave

vacoustic

Consider two coherent optical waves A and B being "reflected" (strictly,scattered) from two adjacent acoustic wavefronts to become A' and B'. Thesereflected waves can only constitute the diffracted beam if they are in phase. Theangle θ is exaggerated (typically this is a few degrees).© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

E θ E ′

θ E ′2θ E ′′

Light

Reflected light

Reflector

Faraday mediumy

Polarizer

Source

The sense of rotation of the optical field E depends only on the direction of themagnetic field for a given medium (given Verdet constant). If light is reflectedback into the Faraday medium, the field rotates a further θ in the same sense tocome out as E′′ with a 2θ rotation with respect to E.© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)

B

vBL=θ

Magneto-Optic Effects (Faraday Effect)

d σ0 (μn 0 = + μp q n p 0 p n p no p n o w l v i +δq n n...

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