hhlu."il ,lr'

{--l,rrr ri ff"i':--

ffiFrlnpcig*lms ffi$FHher ffig$frrw

its oi,S : 0.8: 0.5 ,um isir difli'acted

:----__f{

If, ln FigurehS-\, =- 0.6ir..cion (first

sses between

CHAT}T'F]R OBJECTIVES .

This chapir-r tliscusses liht pi'opagation in a step-index fiber. Total internalreflection, lvhich is ihe basic underlyi:rg principle, and some details of lighbpropagation in fiber arr prescnted, in,:luding the rnodes of'p",:opagar-ion, theangle of propagation. end the acceptance angle. An important term, numericalaperture, which is connected to the rel..tiv+: r'efractive index dift'erencebetween cor,: and cladding, is discussetl. You will be able to c.rlculate thescp.arameters numerically and be introtiuced^ to the relarions among t,henurnerical erperture, t're line width, ancl ttre data rate of the fiber.

3.-1 INTRODUCTTON

Iiigure 3-1 iltustrat,rs r step-index liber. In this fiber the refractive indexchanges in step fashion, from ihe cente,- of thr: 6ber, the core, to the outer shell,the claddirrg. It is high in the core and lo';er in the clndding. The ligtrt in theliber prtrpagates by bouncing back anci fc:th from the core-cladding interface.

?o sinrplify the Ciscussion of :rt::;lgation, you will use ray-tracingtechniques. That is, yott will follow a sample ray through the fiber. Mostlv, yourvill assume that the sample ray passe$ ,, rrough the center of the fiber. Suchrays are called meridional rays. Section 3-2-3 briefly covers nonmeridional,rtr skeu'rays.

The ray pr:opagating in the fiber must be launched into the fiber at oneend. The condit.ions necessary to inject ;iuch rar*s efficiently rlepend on the fiber

"l

/\' n,

42

FIGURE LT

Chapter 3

structure, as well as on the characteristics of the light source. Note that incommunication applications the power introduced into the fiber is typically10-100 ^cW with a light-emitting diode (LED) source and approximately 1 mWwith a laser s_ource.

3_2 LIGHT PROPAGATION

3-?-l Total Internal ReflectionA typical step-index frber is shown in Figure 3-1. Two rays are shown inFigure 3-1. One (the solid line) is injected at a lower angle than is the other(the dashed line). Follow the dashed ray first (the dashed line).

At interface A, between air and the core, refraction takes place, and theray continues at a smaller angle, closer to the center line; that is, 0n) 0r. Theray then gets to the core-cladding interface at point B. Again, refraction takesplace and the ray bends and continues in the cladding. Finally, the ray bendsagain, as it exits the fiber at the cladding-air interface, at point C. However,this time the ray leaves the fiber. This ray is not confi.ned and does notpropagate through the fiber"

Nw, follow the second ray (the solid line). Again, refraction takes placeat point A. At point B', the core-cladding interface, total internal reflectionoccurs. This ray is confined to the fiber core. For convenience, assume that theangle of incidence at the core-cladding interface is the critical angle and call ita" (a specific case of 0" in Equation z-lL for a fi.ber, where the index for thecladding is n2 and for the core is n1). From Equation 2--Lla, d, : Sin-L (nrln).An incident ray with an angle larger than a. will propagate in the fiber.

The critical ray (the solid line) in Figure 3-1 makes an angle 0" with thefiber center. Rays with propagation angles larger than f,l" will not propagate.Note that 0r) 0", and that 01ray exits the fiber and is not confined to the fiber.

C .l'

The angle fJas 0" in Ch

It is importrtravels fi'on

EXAMPLI

Find f/,. in e

Solution

The rincident, nfgirren in Jilx(single-rnodshallow tn 0t . The:fi'ar:tion takesthe ray bends

rt C. However,and does not

on l,al',es placenral reflectionjs:rrJ.r[ that, theagL nfl call itr inriex for: the= sin--L qnrlnl').r ttre fiber.gie fl.'r'ith tenot propagabe.,ed to the fiber'.

\\tI

il,- CoreII

If-

_!'.lndexprotile

I

tI

-T

44

FIGURE 3-3 Production of high- and(b) Low-order modes. (c) Off axis launch:

Chapter 3

that the total number of modes increases as the relative refractive indexdifference (n, - nr)ln, increases.

It is common to distinguish between high-order modes, those withpropagation angles close to the critical angle d", and low-order modes, thosewith propagation angles much lower than the critical angle. The high-or.dermodes tend to send light energy into the cladding. This energy is ultimatelylost, particularly at fiber bends.

Mode Conversion (Mode Coupling). Whether the light energy propagatesmostly in high-order modes,'io*-or.r modes, o, " puiticular mix of modesdepends on launch conditions (the angle of incidence of the rays entering thefiber end) and on the extent-that mode coupling (the transfer of light energ-yfrom one mode to another) takes place. If the light source to fiber connectincauses a large part of the light energy to be coupled to the fiber at relativelylarge angles, high-order modes will be set up (Figure 3-3(a)). This tends tocause losses, particularly at fiber bends. Figure 3-3(b) shows a light sourcethat couples the light at shallow angles, and thus low-order modes are set upand energy loss is reduced. In Figure 3-3(c), the light source is misaligned antends to set up higher and leaky modes. It is most efficient to avoid thesituations shown in Figure 3-3(a) and 3--3(c). The mode distribution (therelative amount of energy carried by each mode) initially set up in the fiber issubstantially altered by mode coupling (or mode conversion).

The mode distribution after about 1 or 2 km of fiber reaches what is calleda steady-state mode distribution. This means that the distribution of lightenergy among the modes is relatively constant from there on. Each mode is

Narrowbeam

Lightsource (c)

low-order modes. (a) High-order modes.high-order modes.

Lightsource

FIGUITII }macrobend.

carrying ihthe fiber, n

Modebends, largshows wharWhen thelarger tharcritical anEa leaky moangle 02 is,Note that ntakes placeC. The incirconversionsions to hi

It is sta short fibefibel is prrmixing blc

FIGURE 3-microbends.

(b)

ractive index

:s, those withr modes, thoselhe high-order is ultimately

:gy propagatessrix of modess entering the:f light energyber connection,r at relativelyThis tends toa light sourcerdes are set upaisaligned and; to avoid thetribution (ther irr the fiber is

rwhat is calledbution of lightEach mode is

::--.--------:-

b)

3-2 Light Propagation 45

FIGURE 3-4 Effects ofmacrobend.

carrying its fair share of light. (Although mode coupling continues throughoutthe fiber, mode distribution remains relatively unchanged.)

Mode coupling (the conversion of one mode to another) is caused by fiberbends, large and small, macrobends and microbends, respectively. Figure 3-4shows what happens to two rays as they pass through a macrobend in the fiber.When the ray with angle )1 reaches point A, its propagation angle becomeslarger than 0. In Figure 3-4, this angle is assumed to be larger than 0" (thecritical angle), and the ray exits the fiber. The 91 mode has been converted toa leaky mode (very high order mode) and thus lost. The mode propagating atarrgle 0, is converted to a higher-order mode 02' due to the bend. Here, 0z' ) 02.itlote that mode conversion to both higher-order and lower-order modes usuallytahes place. (To prove this point, see what happens to a ray incident at pointC. The incident and reflected angles mrLst be the same.) Figure 3-b shows modeconversion caused by a srnall indentation in the fiber, a microbend. Conver-sions to high-order and low-order modes aie shown.

It is sonrebines desirable to set up the steady-state mode distribution overr short fiber length by deliberately introducing minute bends in the fiber. Thefiber is pressed between two blocks covered with fine sandpaper (a modeirnixing block). The sandpaper introduces indentations in the fiber and causes

FIGURE l'i-5 Effects ofmicrobends. /r"':'"*

Low-ortlermode

High-ordermode

Low-orrjermode

n =t Q,

I

I

t-\- __

Rny in cladtling

46 Chapter 3

increased mode mixing. This method also causes increased loss because someof the modes become leaky.

A common question regarding light.propagation in the fiber is r,vhether aray ever travels directly along the fiber, parallel to the fiber axis. The ansrveris that sch a mode (0 : 0") would very quickly be converted to higher-orclermodes because of fiber bends.

3-2-3 Skew WavesThus far, all rays have been'arru*".I to be meridional, passing through thefiber center. In reality, a large number of rays travel throgh the fiber withoutgoing through the fiber's center line. Skew waves (also called skew rays)represent a significant part of the total light transmission. Fortunately, theanalysis of meridional rays glves a close approximation of what actually takesplace so that it is not necessary to include the complex analysis of skew waves.Skew waves are a result of the way the light is rnjected into the fiber, and it isnearly impossible and also unnecessary to avoid them.

3-2,4 Acceptance Angle and Numerical ApertureThe propagation angle must be equal to or iess than the critical angle. Thismeans that the light entering the fiber must be shallow enough to maintainthis condition. Figure 3-6 traces two entering rays that become the criticalrays in the fiber. If you follow the solid line ray, there is refraction at pointAso that f/o does not equal 0". The refractive indices involved are thos of air,ft : L, and the core nl. Only rays that enter the fiber edge within the angle 20,,will be accepted by the fiber. The angl e 20nis the

"""uftance angte. In thredimensions, it is an acceptance'cone, limite,l by theLngl e 20n.It is useful to relate the angle 0oto the refractive indices oitho fiber. By

Snell's law, at point A (Figure 8-6),

and

sin 0o/sin 0" : nrlnir : rz1

sin 0o : ftLx sin g"The term sin 0o is called the numerical aperture (N.A.) and

(3-2)

(3-3)

)lo I,t I-+-I

\\

To obtain N.A. in terms of the refractive indices n, and n2, where n, is the coreindex (n"o,.) and n2 is the cladding index (n"o), ,rr Eqrrutio.r, B-1 and 3-3 andthe trigonometric identity

cos2o-1-sinz,you get

N.A. : sin 0n: ftL sin d"

N.A.:(nt2_n22)Lt2

FIGURII 3_6

The "half acc

You cardifference A

From Eqtratic

EXAMPLE

A fiber has thas a decimal

Solution

EXAMPLE :

A step-index f(3) the accept;Solution

l. From(3-4)

;\

3-2 Light Propagation

FIGURE 3-6 Accepance angle.

The 'ihalf acceptance" angle 0o is given by'

% , :ll-l[),]l n,,),,, (B-5)You can express the N.A. in ternrs of tle relave refractive index

difference A, which ls defined as

A: (n12 - nrz)l{zx nrz)- (N.A.)z1(2 x nrz)

From Equations 3-4 and 3-6,

(N..A.)' = nr' - nr' : 2 x nrz x A,N.A. : nL x (2 x 6trz

EXAMPLE 3_2

z\ liber has the follorving characteristics: n1 : 1.35 (core index) and A : 2Vo (or,rs a decimal ratio, 0.02). Find the N.A. and the acceptance angle.

Solrrtion

N.A. : ftr x (2 x Atrz : 1.35 x (2 x O.OZ)Ltz: 0.2'l

0., : sin-1 N.A. : sin-t 0.27 : 15.66'Acceptance angle : 2 x 0o : 31.33o

EXAMPLE 3_3

A step-index fiber has n".,." = L.44 and n"- : 1.40. Find (1) the N.A., (2) A, and(3) the acceptance angle.

Solution ,1. From Equation 3-4,

N.A. - l(1.44)' - (1.40)\lL/2: 0.38?

47

i because some

er is whether a.is. The allswer.o higher'-order'

r1g through therc iiber: withoutlr:d skew t'ays)ortunately, the. aci;uaily takesof'sl

Chapter 3

2. A:(L.442 - 1.402)/(2 x 1.44\ (approximatemethod):0.027: 2.7o/o

3. on: sin-t 0.33? (Equation 3-5): 19.7o

Acceptance angle -- 2 x 0,,: 39.n"

It is often convenient to simplify the expression for A. An approximationfor A is obtained by rewriting

L:(nt2-nr\l(2xrr')- [(nr + n"r) x (nr - nr)U(Z x nr').

lVhen nr is approximately equal to n2,

A : [2 X n, X (n, - n)ll(2 x nr'): (n-, - nr)ln,Remember that the N.A. represents the acceptance angle. A large N.A.

represents a large acceptance angle and vice versa. A large N.A. also implies'large A, a large difference in refractive index. As you will see in Section 5-1,a large N.A. produces a large number of modes and presents some seriousperformance problems. Typically, A is of the order of 0.01-0.03 (1-37o).

3-3 LINE WIDTHThe actual sources of light used in fiber optics produce a light that has a bandof frequencies. Typically, they are not monoehromatic. That is, they are not

Lr*\,, r. !n.a"1 -b itidi'ff'llHl':i"""Ji,'lnl#i;:TT",1l :"1T5"1"x3';"Tf,,1:wavelengths between the two points where the light energy drops off to one

' half its maximum power. In Figure 3-7, the pcwer is maximum at \ : 820 nmand drops to half its maximum at ). : 810 nm and tr : 830 nm. The line widthis, therefore, 830 - 810 : 20 nm. We will use the notation A), to denote linewidth. Al, corresponds to a bandwidth Af. A/can be expressed for narrow linewidths as

Lf : (Ar/ro) x fr, (3-g)where Af : f" - fr,ft and f"are the half-power frequencies of the light source,and Af is the bandwidth of the light source.

tro : center wavelengthf6 : center frequency

That is, the bandwidth is the product, of the relative line width A,L/).' and thecenter frequencY. (This can be derived directly, by realizing that Lflf : AI/I.The relative bandwidth and relative line width are equal.)

48

(3-8)

FIGURE 3

The liof the fiLenumber oflsystem.

EXAMP{,]

A typical Irelative lin

Solution

l. .,

In percent,,

,. t, ..,

/o is the cer

Note that i

3-4 PRThe refractThis makesu : c/n, inwavelengtl

i

it

l___

rocl)

i rppfo.\('imaLi0n

(3--8)

lc.. A large N.A.i.A. also irnplies, in Section 5-1,Ls some serioust:_i ( L-:1q0.

tlitt has a brnd:s, i,irey are not

ol' rval'elengths.; Lhe width inrh'ops ofI'to orrcil ;-it, tr : 820 nm.'flie line width,,\ to denote linet fbr narrorv line

(3-e)

the light source,

th AL/\,, and the''hat L'flf : A\itr'

34 Propagation Velocities .r9

FIGURE 3-7 Line width.

/l--l- I

- I l---r>l.(nl)

800 rt l0 ll20 830 840

The iine.width of the source has serious effecs on the overail performanceof the fiber optic transmission system. A large line width yields a largernumber of modes fclr the same N.A. and lowers the maximum data rate of thesrstern.

EXAMPLE 3--l

A typicat LllD emits light at tro : 0.82 rm with A). : 40 nm. Fintl (1) therela.tive iine width in percent and Q)

^f.Solution

1. av^o I [13nrto-'y(8zo x 10-e)

In percent,

0.0488x100:4.88a/o

2. fo: c/Lu: 300 x 106/(0.82 x 10-o): 0.3658 x 1015

fu is the center frequency, corresponding to trs : 0.82 r,m.Ar : l/31'," r3'?'-?l

x 1015- L t.o(t ,\ IU nz

Note that in his example, an enormous bandwidth Ai is involved.

3_4 PROPAGATION VELOCITIES

The refractive index n of most materials varies with the wavelength involved.This makes the speed of light u depend on wavelength. (Remernber the relationu : cln, in a vacuurn, where r : l, u : c.) For silicon glass, in the range ofwavelengths used in fiber optics, ther variations in and u wiLh wavelength

^=

1.0

L1)

^

.3 o.-r{)

{.) t-I

I

u------l

Chapter 3

FIGURE3-8 nanduversus L (for silicon glass).(Arrows indicate vertical scaleto be used.)

(t'

I]J

.t)

E

ro3 iX

102

A(;cttP'r"\Nr;Ervill enter tAccRpr'.rNcr*formed by r'BANDWTD'ru.

width" anclCnltrcrrl pncritical angf.,lNU WID'l'lI.

emission of

MIRrntonL

Mone coNV[

Monu c()t;rt-

Moor,: DISl'tufiber, usual.Monu MrxrN(

Moue l,ttotAplarce at disrmodes.

Moxocrrrronthas a very lNunrirucnl Pn.6 .8 L0 l.l I .4 L6 l.tl 2.0

1p.rn)

change direction. For wavelengths from about 0.6 to 1.3 rm, n decreases withincreasing wavelengbh, indicating a negative slope (AnlA\ is negative). For thesame wavelengths, u increases with wavelength, indicating a positive slope(Au/AI is positive). For 1.3-1.7 Fn, ft increases with increasing wavelengthwhile u decreases, a reversal in the slopes of and u.

Figure 3-8 shows tiris relationship amot1 tu, u, and increasing wave-length I. At about 1.3 pm, the slope of both n and u is approximately 0. Thismeans that there are no variations in or u as the wavelength changes (forsmall Atr). Because changes in u with wavelength greatly reduce the data ratethe fiber can carry, it is advantageous to operate at about 1.3 pm. That is, it isdesirable to use sources that operate at close to 1.3 ^tm. If the line width of thesgurce is very narrow, Say, 1-Or 2 nm, and you operate near 1.3',tm, you Canexpect AulAL to be near 0 and the bandwidth (or data rate that h fiber cancarry) to be extremely high. These systems can operate at frequencies in excessof 2 GHz. (See the discussion of dispersion in Section 4-3.)

SUMMARY AND GLOSSARY

The terms defined here reflect what you have learned in this chapter. Theseterms are used throughout this text and in industry. You should recognizethem and understeud what they mean. Use this glossary to review thematerial you have just read.

,aJ

[t"u "'

\

Summary and Glossary

Accnt'r.tNcE ANGLE. The range of angles rvithin which an injected light beamwill enter the frber. (This angle is related to the numerical aperture.)AccuprexcE coNE. The accepance angle in three dimensions. (The cone isformed by rotating the acceptance angle, with the fiber center line as the axis.)B,rNnwrptH. Frequency range corresponding to the line width Ar. (See "linewidth" and Equation 3-9.)Cntttcal PRoI'AcArIoN ANGLtt. Rays with propagation angles larger than thecritical angle are not confined to the fiber. (They leave the fiber.)LrNu wIDTH. The range of wavelengths between the two points of half-poweremission of a light source. (See "relative line width.")M;RrororAL RAy. A ray.that passes through the fiber center line.Moon coNVERsIoN. The transfcr of lighi energy from one mode io another.Mooe coupLING. See "mode conversion."Moon DISTIuBUTIoN. The amount of energy carried by each mode in an opticalfiber, usually given in relative terms.Moon MIXING BLocK. A device designed to cause mode conversion in the fiber.Monp PRoPAGATIoN. The propagation of light energy in an optical fiber takesplace at distinct angles of propagation called modes of propagation, or simrlymodes.

MoxoctrRoMArtc. Of single color or single frequency. (A light source whichhas a vely nlrro\r line width is monochromatic.)NunpnrcL ApER'ruRE. The sine of one-half the acceptance angle.Pnop.qcarloN ANcLE. The angle a beam inside a fiber makes with the fiberaxis. (See "critical propagation angle.")Rulertvu REF'RAcrIVE INDlx DrrFErrENcE. Approximately the ratio of therefractive index difference over the core index. (See Equations 8-6 and g-g.)Rel,artvn LINE wIDrH. The ratio of the line width to the center wavelength ofthe source, AL/tr.Sxnw nnY. A ray that propagates in the fiber without crossing the fiber centerline. (Skew ray is the same as skew wave.)Srnenv-srA'rn MoDE DtsrRIBUrroN. After a certain length of fiber, the powercarried by each mode does not change any more. The mode distribution is saidto be in the steady-stabe distribution. :Srnp-lxnEx I'IBER. A fiber made of a core.and cladding with two refractiveindices, D"o.o and n"o.Tot1, INTERNAL REFLEcTIoN. Rays traveling at shallow angles (below thecritical propagation angle) from a high-index material to a low-index materialundergo total internal reflection and do not cross into the low-index material.This behavior is the same as that of a reflected ray.

5l

106

--------t

E

r0-r E

102

-+1.0

riecreases withgative). For ther positive slopeng wavelength

creasing wave-imately 0. Thisth changes (for:r: the data rate.n;'Ihat is, it isine width of the.3 r*, you cana[ the fiber canencies in excess

chapter. Theserould recognizeto review the

52 Chapter'3

FORMULAS

In the following formulas, r, core index and no, cladding index.

sin a" : cos 0" : n2ln,The relation between the refroctiue indices and the criti,cal angles.

sin 0,, : tlr X sin 0.,I)efinition of half acceptancL angle 0n.

N.A. : sin 0o: n1 sin 0,.Definition of N.A

N.A.:(nr2_nr")lt,zRelation between N.A. and the refractiue ndices.

u" l:l-lIf ) -= sin-'(nr,' -- nr'\'''Half acceptance angle.

^ : [ltl^; ,i,1,'Y"n,f;,n")Definition of the relatiue refractiue difference b'.

N.A. : nL x (2 x LtrzRelation between L, and N.A.

6:(nr-n2)ln,Approximation for A,.

(3-l)

(3-2)

(3-.3)

(3-4)

(3-5)

Relation lte

QUESTII

1. What is2. Light bt

angle. \3. Would I

phenonr4. What ar5. What is6. The sim

easier tacceptar

7. How arra. Prop;b. Accec. Line

8. What issteady-s

9. Are lowmodes r

10. FIow ismodes?

11. WiU ligExplain.

PROBLE

f. A step-ipropaga

2. For thea. The Ib. The

3. In a steD"la.l : Ia. The b. The cc. The Id. The

(M)

(3-7)

(3-8)

Iles.

(3-1)

(ii-z)

(3- 3)

(i-.1)

(3-5)

(&-6)

(3-7)

($-8)

Problenrs 58

af : (ar/r,,) x ft (3_g)Relatio betuteen the frequency bandwidth

^f and the line wiclt aI.

QUESTIONS

I-. What is the diff'erence betlveen meridional and the skerv waves in a fiber?2. Light beams are injected into the fiber at angles larger than the acceptance

augle. Wiil they propagate in the fiber? nxplain.3. lVould light be confined in an optical fiber if the total internal reflection

phenomenon ciid not exis? Explain4. What are propagation modes?5. what is nrode coupling? Give causes for mode mixi^g.6. The sinrple flashlight sends beams at a relatively wide angle. Would it be

ea-sier to couple it to a fiber with a low acceptance angle or a highaccr.pbrnce angle? Explain.

7- H.w are the followi'g related to the liber data rate?a. Propagation modesb. Accepta.nce anglec. Line width

8. lYhat is the steady-state mode distribution? Does mocle cor.rpling stop aftersready-state mcde distribtrtion is reached?

f. it.r:e low-order modes or high-order modes more likel-v to become leakyniodes and to lea.ve the fiber.?

10- l{ow is hc relative refractive index difference related to propagationmodes?

11. Will light propagate in a glass iubing (air inside with a glass shell)?Explain.

PROBI,EMS

1. A step-index fiber has ru"o.o : 1.,11 and n"- : 1.82. Find the range ofpropagation angles (all angles below the critical angle).

?. raor the fiber in Problem 1, finda. 'Ihe N.A.b. The acceptance angle

3. In a step-index fiber, the relative refractive index difference is 296 andzclad : 1..40. Finda. The ft"u,ub. The critical propagation anglec. The N.A.d. The acceptance angleIi

t

]-

54 Chapter 3

4. Find A for the step-index fibers listed here, using the exact andapproximate formulas. Compare the resulting values.

n"o;"(n1) n"1o(n)a. L.42 1.415b. - 1.38 1.36c. 1.68 1.34An optical fiber is being designed. It must have an acceptance angle of 75'.Its cladding index is 1.38. Find its core index.A fiber is made of a core with an index of 1.40 and no cladding (aircladding). Finda. The N.A.b. The acceptance angle.For the fiber in Problem 6, what is the highest-order mode (criticalpropagation angle)?

8. A light source has the response in Figure 3-9. Find its line width.

b.

7.

FIGURE 3-9 Figure forProblem 8.

mW)

t.0

0.5

9. For'the source of Problem 8, find Af.10. A source has a 2Vo line width at a center wavelength of 1.3 prn. F'ind Af.11. Repeat Problem 10 for a source with a Lo/o line width.12. A source with a center wavelength of 0.8 .r.m and a line width of 80 nm

(*40 nm) is used with a fiber made of silicon glass (Figure 3-S). Finda. The An involvedb. The Au involved

4Fhe

CHAPT'E]

After studyirenerg'y lossdispersion hwavelength idispersion, arpreferred ove

Ll INTIThe charactecomposition adiameter of tfiber directlyaffect losses i:when discussand the data

l_