Last Latexe: Feruary 1 11 at 15:11 1 Physics 54 Lecture 1 Fe. 4 11 1 Geometrical Fier Optics The wave guies consiere so far containe their fiels within conucting walls ut we know from stuying total internal reflection in elementary optics that it is also possile to contain fiels y changes in the inex of refraction. n extremely important application is fier optics. fier optic cale is a silica fier with an inex of refraction which varies as a function of raius. The simplest form is a core of raius a an inex of refraction n 1 surroune y a shell of outer raius anhavinganinexof refraction n < n 1. If the angle of incience α>α c =sin 1 n /n 1 there will e total internal reflection an the light will e confine. It is more convenient in iscussing optical fiers to use the angle θ θ α α a which the light makes with the axis of the fier so the conition for total internal reflection is θ<θ max =cos 1 n /n 1. Of course this iscussion was in terms of geometrical optics suitale if the wavelength of the light is negligile compare to the geometrical istances a. Optical fiers come in two categories multimoe an single moe. For multimoe fiers a 5 µm 75 µm an the light use is generally near infrare.85 µm so geometrical optics is a reasonale approach though we shall see that interference effects are still relevant. Single moe fier is smaller a µm an we nee to treat these as waveguies. Consier the simple multimoe fier an efine = n 1 n n 1 1 n n 1 which is often aout.1. s it is small cos θ max 1 1 θ max =1 so θ max. There is an uncertainty principle etween the localization of a wave an its wave numer which limits the numer of moes that can e transmitte. We may think of this quantum mechanically where the ensity ofmoesisgiveny pq/π h D for each moe in D imensions. The 54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 coorinate integral is q = πa anas k k z tan θ max = k z p = h k =π h kz. There are polarizations so the numer of moes that can propagate is roughly N = kza = 1 V wherev := ka is calle the fier parameter. For a multimoe fier N is aout 1 ut for a single moe it is one for each polarization. There is a prolem with this simple arrangement as the istance light travels to get a istance z own the fier is z sec θ so light with ifferent θ values will travel ifferent optical istances to get to the same θ point an will interfere. This can e ameliorate y having more than one transition or even a continuum. In fact for next week you will fin how it can e fixe with a continuous istriution of nr. To analyze such a situation where ɛ x varies smoothly an assuming µ = µ as the fier is not magnetic we may write Maxwell s equations having fourier transforme time to iscuss a particular frequency as ɛe = = ɛ E + ɛ E E H = µ t = iµ ωh H = ɛ E = iωɛe t H =. So E = E + E = iµ ω H = µ ω ɛe = E 1 ɛ E ɛ H = H + H = iω ɛe or H = iω ɛ E iωɛ E = iω ɛ E + µ ω ɛ H. E + µ ω ɛe + 1 ɛ E = ɛ H + µ ω ɛh iω ɛ E =
54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 3 54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 4 We can simplify these equations if we assume that ɛ varies slowly compare to a wavelength so ɛ ɛ = ɛω c. The other terms in the equations are of orer ω /c times E or H respectively ɛe H/c sothe ɛterms may e roppe as small. Then the components of oth E an H satisfy 1 + ω c n r ψ r =. 1 This equation which escries the rapily oscillating function ψ can e replace y a more graually varying function y using the Eikonal writing ψ r =e iωs r/c so ψ = iω Se c [ iω ω S ] iωs r/c = c S i e iωs r/c. c This is ω n /c e iωs/c from 1 so n r S S = i c ω S. c/ω soas S varies on the same scale as n r which is slowly compare to 1/ the right han sie can e roppe an we have the eikonal approximation S S = n r. This equation oesn t tell us the irection in which S changes ut it oes tell us that the rate is just n r so in following a particular ray s path we can efine a unit vector ˆk r such that S = n rˆk r. Near a point r we may expan ψ r =e iω S r + r r S /c = e iωs r /c e iωˆk r r n r/c 1 Jackson uses x instea of r in most of these equations to emphasize that n epens only on the transverse location an not on the axial coorinate z. But clearly the wave function ψ an the eikonal S o epen on z along a given ray. so it is locally a plane wave with k = ωn r/c suitale for a wave with spee c/n r as one might expect. If s is the istance measure along the path of the ray r/s = ˆk n r r/s = S so s n r r s = S s = S = n r. s 3 Γ In the penultimate expression Γ represents the ray s path. In general a ray will e irecte in a irection with a large axial component a raial component an an aximuthal component. If the latter is zero the ray will pass through the axis an the ray is calle meriional. Rays with a nonzero azimuthal component will never pass irectly through the axis an are calle skew rays. In terms of wave functions such rays correspon to azimuthal quantum numers m an have vanishing intensity at ρ = on the axis. Following Jackson we will avoi complication an iscuss only meriional rays which is also equivalent for geometric optics at least to iscussing a plane sla. So the ρ irection will e rename x an we consier a ray confine to the xz plane where z is the overall irection of motion. We assume n oesn t epen on z. The x an z components of Eq 3 are nxsinθ =nx s x nxcosθ =n r =. s z Thus nxcosθ is a constant along the path an if θ<θ max the path will turn aroun at x max atwhichcosθ = 1 so the constant value of nxcosθ is just n := nx max ut it is also of course n cos θ. Note that z/s =cosθ = n/n so the path is etermine y rewriting the x component of Eq. 3 as n n x nx = n nx z nx z nx z n x = n z x so n x z = nxnx x = 1 x n x. Whether y inspiration from mechanics viewing 1 n x as a potential or otherwise if we multiply this equation y x := x/z we get 1 n z x = 1 z z n x = n x = n x n
54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 5 where the constant of integration n is etermine y x =atx max nx max = n. So the axial istance travele from the point the ray crosse the axis to reaching raius x is zx = x z x x x = n x n x n. The path is important ecause if ifferent rays have ifferent path lengths to get to the same large isplacement z own the fier they will interfere. From one crossing of the axis to the next the ray moves in z aistance xmax x Z = n n x n. The istance travelle is not important ut the optical istance nxs is ecause that etermines the change in phase an the optical istance corresponing to the axial isplacement Z is L opt = xmax nx s z xmax z x x = xmax n x = n x n x. nx nx n n n x n x The phase ifference in a single half-perio is not likely to e important ut when the rays travel a large istance z the total optical path will e L opt z/z. Thus it is ieal if L opt /Z is inepenent of n that is it is the same for all rays. In prolem 8.14 you will show how to accomplish that. In aition to geometrical ispersion coming from not having L opt /Z inepenent of n there can also e frequency ispersion ue to the variation of n x with frequency. This will not effect a very narrow anwith signal one with a very narrow range of ω ut the rate at which information can e carrie is proportional to the anwith so we woul like to have little ispersion in frequency as well. We also want minimal asorption. These two issues for silica encourage using 1.4µm. We will skip Jackson 8.11. 54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 6 Sources of Waves in Wave Guies We now turn our attention to the sources of waves in waveguies. With given sources i.e. ignoring ack reactions the equations are still linear inhomogeneous in the fiels an time-inepenent so we can assume everything has a e iωt time epenence. The normal moes of free waves in a waveguie form a complete set of states for the source-free solutions to Maxwell s equations in the guie though we nee to inclue all the moes incluing those whose cutoff frequency is aove ω. Far from the sources however only the real values of k formoeswithω <ω will have appreciale amplitue. We will expan our fiels in the normal moes which will e inexe y a composite inex which inclues information aout whether the moe is TE or TM or of other nature e.g. TEM an also the inices which specify which moe roughly the angular an raial quantum numers For a given moe we have two values of k either positive right moving an negative left moving or i k for cutoff moes. We will lump the +i k moes in with the positive k ones an write the + part of the moe as E + x y z = [ E x y+ẑe z x y ] e ik z H + x y z = [ H x y+ẑh z x y ] e ik z where E an H are purely transverse an were given in terms of E z or H z for TM or TE moes respectively earlier. There are also moes travelling in the negative z irection. The equations are symmetric uner z z uner which the transverse E x y is unchange ut E z x y changes sign. The magnetic fiel is a pseuo-vector so for it the transverse H x y changes sign ut H z x y is unchange. Thus E x y z = [ E x y ẑe z x y ] e ik z H x y z = [ H x y+ẑh z x y ] e ik z The transverse fiels E form a asis which we can choose to e real an normalize E E µ = δ µ.
54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 7 54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 8 That this can e one an what this normalization requires for the asis of normal moes for E z an H z is elaorate in the slies. With this normalization an from H = Z 1 ẑ E we also have H H µ = 1 Z δ µ an in the time-average power expression P = 1 E H µ ẑ = 1 Z δ µ E H ẑ we have The aove normalization for the E comes from requiring that the z components satisfy orthogonality conitions an normalization conitions ajuste appropriately. For TM waves E = ik γ E z so δ µ = E E µ = k k µ E γ z E zµ = k k µ E γ µ γ z E zµ γ µ = k k µ E γ z E zµ where in the integration y parts thir = the surface term vanishes as E S = an for the fourth E zµ = γµe zµ so E z E zµ = γ δ k µ whilefortewavesthesameargumentforh gives H z H zµ = γ δ Z µ. k For a rectangular waveguie x a y the moes are laelle y integers m an n with TM waves: ψ S = TE waves: E zmn = ψ = iγ mn sin k a E xmn = πm γ mn a a cos E ymn = ψ n S = πn γ mn a sin mπx a mπx a mπx H zmn = ψ = iγ mn cos k Z a E xmn = πn γ mn a cos a E ymn = πm γ mn a a sin a a sin cos sin nπy nπy nπy mπx nπy cos a mπx nπy sin mπx nπy cos where m γmn = π a + n. The functional form of ψ is immeiately apparent from the ounary conitions an for TM moes E = ik ψ/γ anfortemoese = izkẑ ψ/γ. The overall constants are etermine from the normalization E x + Ey = 1 except that for TE moes we nee an extra factor of 1/ foreach n or m which is zero as cos mπx/a =a1 + δ m /..1 Expansion of free waves In our waveguie an aritrary fiel configuration in the asence of sources can e escrie y expaning in normal moes with positive an negative or +i an i wave numers as escrie aove. Thus a total fiel E = E + + E H = H + + H
54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 9 54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 1 with E = E H = H can escrie an aritrary fiel configuration in a region that has no sources with the s constant inepenent of z in such a section of the wave guie. The coefficients are etermine y the transverse components E an H along any cross section for E has expansion coefficients + eikz + e ikz while H has expansion coefficients + eikz e ikz. This gives the expansion coefficients taking z = as = 1 E E Z H H.. Localize Sources We have escrie the waves that can propagate in the waveguie ut what actually prouces such waves? We will consier a localize source with current ensity J xe iωt confine to some region z [z z + ] at the ens of which we imagine cross sections enote y S S +. We will assume there are no sources or ostacles to the right of S + or to the left of S soats + there is no amplitue for any moe with negative k or with i k which woul represent left-moving waves or exponential low up as z +. The reverse is true at S so E = + E + H = + H + at S + E = E H = H at S In etween we have the full Maxwell equations with sources E = B t = iωµ H H = J + ɛ E t = J iωɛ E while the normal moes oey Maxwell equations without sources: If we apply the ientity H = iωɛ E E = iωµ H. B = B B we fin E H E H = E H E H E H + E H = iωµ H H + iωɛ E E iωµ H H + E J iωɛ E = J E If we integrate this over the volume etween S an S + using Gauss theorem an the ounary conition that E = at the conuctor s surface E H E H ˆn = J E S where S consists of S + with ˆn =ẑ ans with ˆn = ẑ. Let s take the upper sign. The contriution from S + is can e reuce to an integral over at z =: E + H + E + H + + S + = + E H E H S + z eik +k z = + E Z 1 ẑ E E Z 1 ẑ E = 1 + E Z E 1 E Z E e ik +k z = 1 + δ 1 e ik +k z =. Z Z On the other han the contriution from S is E H S + E + H ẑ = S E H + E H ẑe ik k z so = Z δ = Z = Z V z J E +. V z eik +k z
54: Lecture 1 Last Latexe: Feruary 1 11 at 15:11 11 The same argument for the lower sign as spelle out in the ook gives the equation with the superscript signs reverse so oth are = Z V J E. In aition to sources ue to currents we may have contriutions ue to ostacles or holes in the conucting ounaries. These can e treate as aitional surface terms in Gauss law y excluing ostacles from the region of integration V ut this requires knowing the full fiels at the surface of the ostacles or the missing parts of the waveguie conuctor. This is treate in 9.5B ut we won t iscuss it here. So finally we are at the en of Chapter 8.