II.3 Reflection and refraction of EM Waves. Aims. To describe the reflection and refraction of waves from pure dielectric surfaces.

Transcription

1 II.3 Reflecton and efacton of M Waves Ams To descbe the eflecton and efacton of waves fom pue delectc sufaces Objectves At the end of ths secton you should maste the bounday condtons fo and H felds, be able to calculate the eflecton and efacton angles, the eflected and tansmtted powe and be awae of the Fesnel fomulae. 4 II.3.0 Bounday condtons n s a unt vecto nomal to the bounday, pontng fom medum to medum ρ s and J s ae suface chage and suface cuent denstes on the bounday suface 43

2 The tangental components of the feld ae contnuous acoss the nteface The dffeence of the tangental components of the H feld s equal to the suface cuent densty The dffeence of the nomal components of D s equal to the suface chage densty The nomal components of B ae contnuous acoss the nteface In the absence of suface chages o cuents, the electc and magnetc felds ae contnuous. 44 How to deve them?- xamples Fom Gauss Law B d S = S 0 B d S = B d S + B d S + B d S = 0 S Uppesuface Lowesuface Sdesuface h 0 B nds B nds + 0 = 0 Uppesuface Lowesuface 45

3 dl h Fom Maxwell-Faaday c dl = B ɺ d S s c h 0 S=dl h 0 dl = dl dl = 0 46 II.3. Reflecton of ncdent wave nomal to the plane of eflecton t H H t H Ths s a specal case, whch could be deved by analogy wth eflecton of V and I. We now deve the same esults by matchng the bounday condtons: 47

4 Bounday Condtons In the absence of suface chages o cuents, the electc and magnetc felds ae contnuous the total feld n medum s equal to the total feld n medum x H y = x /η xt x H yt = xt /η H y =- x /η 48 x + x = xt H y + H y = H yt By defnton of η n medum and we have: η = and H x y = H x y η = H xt yt 49

5 lmnatng H y, H y, H yt usng η, η, gves: x x xt = η η η snce x + x = xt lmnatng xt gves: ρ x = = x η η η + η Reflecton Coeffcent (6.a) lmnatng x gves: ρ t xt = = x η η + η Tansmsson Coeffcent (6.b) 50 Incdent powe densty Tansmtted powe densty x P = η xt P t = η Reflected powe densty x P = η η η P = P ρ P η η = + 4η η P P ρ P = t ( ) ( η + η) = NOT ρ t P 5

6 II.3. Reflecton of a wave at oblque ncdence on a delectc bounday The behavou of the eflected and tansmtted waves wll depend on the oentaton of wth espect to the bounday We theefoe splt the ncdent waves nto two polased pats ) Pependculaly polased - electc feld at ght angles to ncdent plane ) Paallel polased - electc feld n ncdent plane Fesnel Fomulae (89) 5 II.3.. Snell s law of efacton (Snellus 6/ Ibn Sahl 984) Befoe calculatng the eflecton and efacton coeffcents we need to know the angles at whch the eflected and efacted waves wll be tavellng θ Medum C θ θ A θ t D θ t B Medum 53

7 The wave tavels n medum fom C to B n the same tme as t does fom A to D n medum CB AD = v v Hence But CB = AD = ABsn θ ABsnθ t snθ snθ = t v v Snce v = µ 54 snθ snθ t µ = = µ n n Snell s Law of Refacton (6.) Whee n s called efactve ndex Fo delectcs we assume µ =µ =µ 0 Ths assumpton wll be used n the est of the lectues Then snθ η = snθ t = η Snce η = µ Do not confuse n wth η 55

8 Bent pencl Boken pencl xamples of efacton 56 II.3.. Snell s law of eflecton θ Medum C θ θ θ A θ t D θ t B Medum 57

9 By a smla agument t can be shown that the angle of eflecton s equal to the angle of ncdence A=CB but CB = ABsn θ A = ABsnθ snθ = snθ θ = θ Angle of eflecton s equal to angle of ncdence 58 II.3..3 Incdent and eflected powe The next step s to consde the powe stkng the suface AB and equate t to the powe leavng that suface θ Medum C θ θ θ A θ t D θ t B Medum 59

10 P = cosθ η P = cosθ η t P t = cosθt η We get: θ θ θ η η η t cos = cos + cos t Fo delectcs we assume µ =µ =µ 0 Remembeng that η = η cosθ = cosθ t cosθt = CAMBRIDG UNIVRSITY cos LCTRONIC NANOMATRIALS θ DVICS AND (6.3) 60 II.3.3 Pependculaly polased waves In these waves the electc feld s pependcula to the plane of ncdence.e. paallel to the bounday between the two meda. Lght wth pependcula polazaton s called s-polazed (s=senkecht, pependcula n Geman). Plane of ncdence H (n plane of ncdence) (pep to plane of ncdence) H H t t Reflectng Plane 6

11 Fom equaton 6.3: t cos t = cos Whee k But, n ths case = θ = k θ cosθt cosθ + = t t We then get the followng: ( + ) = k 6 ( + k) + ( k) + ( k ) = 0 Ths has the same fom as ax bx c The postve soluton s Thus o + + = 0 = k + k cosθ cosθ θ θ / / t = / / cos + cos t = n cosθ n cosθ DPARTMNT nof NGINRING cosθ + n cos AND SPCTROSCOPY MATRIALS θ GROUP 63 t t

12 Ths expesson contans the angle of the tansmtted wave We can use Snell s law to obtan a moe useful expesson whch contans only the angle of ncdence Fom Snell s Law: snθ snθ t = But cosθ = ( snθ ) = ( snθ ) t t Then cosθ sn = cosθ + θ / n n cosθ n snθ n / = CAMBRIDG UNIVRSITY sn θ n DPARTMNT OF NGINRING n cosθ + nand SPCTROSCOPY MATRIALS sn GROUP θ n (6.4) 64 II.3.4 Paallel polased waves Lght wth paallel polazaton s called p-polazed (p=paallel). Plane of ncdence (n plane of ncdence) H (pep to plane of ncdence) H H t H t Reflectng Plane 65

13 In ths case s no longe paallel to the eflectng plane. The bounday condton apples to the component of paallel to the eflectng plane: cosθ cosθ = cosθ t Followng though the algeba ou expesson fo the ato of eflected to ncdent waves becomes: cosθ cosθ n cosθ n cosθ θ θ n θ n θ / / t t = = / / cos + cos t cos + cos t (6.5) / n cosθ sn θ n cosθ n snθ = = n / cosθ + sn θ n n cosθ + n snθ CAMBRIDG UNIVRSITY n 66 Note: nomal ncdence coesponds to θ =0 θ = 0 n θb n n n = n + n (6.5) s smla to equaton (6.4), but n (6.5) the numeato can become zeo,.e. no eflected wave The angle at whch ths occus s known as the Bewste Angle Settng the numeato of (6.5) equal to zeo we get Remembeng that ( ) tan( θb) = = / tan θ cosθ = sn θ sn θ = + tan θ s called Bewste Angle Note (6.4) can go to zeo only fo n =n :.e. tavel n the same medum 67

14 Zeo eflecton at the Bewste angle explans why polased sunglasses cut down eflectons Let us consde the a (n =) wate (n =.33) nteface θb 53 o 53 0 NANOMATRIALS AND A ule of thumb: CAMBRIDG polazed UNIVRSITY fltes lmt the LCTRONIC glae fom DVICS calm wates fo a sun alttude between 30 and 60 degees 68 69

15 II.3.5 Compason between eflecton of paallel and pependculaly polased waves Gaphs of as a functon of angle of ncdence θ B ~55 0 θ B ~7 0 0 = = 70 The gaphs show the values of As the ato )The Bewste angle nceases ) The value of fo the pependculaly polazed wave tends fo two pemttvty atos nceases thee effects can be seen to -,.e. pefect antphase at all angles 3) The value of pefect phase at all angles 4) When fo the paallel polazed wave tends to,.e. total eflecton occus at all angles of ncdence. 7

16 II.3.6 Total ntenal eflecton The pevous secton showed gaphs fo >. I.e. ou wave s movng fom a lowe efactve ndex medum to a hghe one If nstead < then the phenomenon known as total ntenal eflecton can occu The tem sn θ n equatons (6.4) o (6.5) can be negatve when / n snθ = n The ctcal angle s defned as snθ c / = = n n 7 θ = θ = Fo c θ > θ Fo c the tem sn θ ts squae oot s magnay and A B = A + B * A + B = A B * n (6.4) o (6.5) s negatve = = s complex.e. the magntudes of the ncdent and eflected powe ae equal. 73

17 Ths gves se to total ntenal eflecton: the ncdent wave s eflected at the bounday and no wave emeges fom the hghe efactve ndex medum Ths s shown n the gaph below at all angles of ncdence whee: Sn θ (wth 0.5 θ 45 = o ).5 Pependcula ( θ, 0.5) θ c =45 0 Paallel( θ, 0.5) Gaph of as a functon of angle of ncdence 0 θ B ~ θ. 80 π 90 b) Pemttvty Rato of / 74 Total ntenal eflecton undepns optcal fbe technology 84 Colladon 854 Tyndall 965 Chales K. Kao Nobel Pze

18 II.3.7 xample: eflected powe Damond has = 5.84 & µ =. What powe facton of lght s eflected off an a/damond suface fo nomal ncdence? Recallng that fo tansmsson lnes: ρ L V B Z L Z0 V F Z L + Z0 = = Smlaly fo -M waves, fo the case whee and H ae paallel to the eflectng plane: ηdamond x ηdamond η A η A ρ = = = µ η wth η = x ηdamond + η A Damond + η Then µ µ 0 η Damond µ ηa µ A 0 = = = CAMBRIDG 0 UNIVRSITY ρ 0.4 = 7.5o o R LCTRONIC NANOMATRIALS 0.4 DVICS + AND 76 III Antennae and Rado Tansmsson Ams To gve a qualtatve descpton of antennae and ado tansmsson Objectves At the end of ths secton you should be able to ecognse the man classes of antennae, and do smple calculatons of adaton esstance, effectve aea, gan. 77

19 III. Antennae The am of an antenna s to get sgnal powe fom the tansmtte to the eceve ccut as effcently as possble Tansmtte Receve 78 III.. Slot and apetue antennae Almost any gude cayng an electomagnetc wave wll adate pat of the wave f ts end s open The electomagnetc waves n a gude wll adate f you chop ts end off (vey neffcent) The deal antenna sends as much adaton as possble n the desed decton and wth the mnmum of ntenal eflecton. 79

20 III... Hon antennae Moe wave s adated f the end of the gude s flaed Z 0 ~50Ω η 0 ~377Ω Bgge apetue so -Less dffacton and moe gan (dected powe, see III.) -The launched wave s moe smla to a plane wave -Thee s a gadual change between the electcal wave, wth chaactestc mpedance Z 0, and the adated wave, wth ntnsc mpedance η 0, hence less eflecton. 80 III... Lase Hon antennae such as these wok vey well but they ae bulky, theefoe unweldy 965 Penzas & Wlson used ths hon antenna to detect the backgound mcowave adaton of the unvese. 978 Nobel Pze Physcs A lase can be thought of as an apetue antenna and the output oughly appoxmates to a plane wave. 8

21 III.. Dpole antennae III... Half-wave dpole If the end of a tansmsson lne s left open ccut, the cuent at the end s 0 and the eflecton coeffcent s ρ f L = Z Z L L L + Z Z 0 0 Z then ρ L In ths case, the telegaphe s equatons show that ¼ of a wavelength back fom the load the voltage must be 0 Z λ b= 4 = Z Z 0 L =0 V = Z I = 0 λ λ b= b= a cuent souce s needed to dve the lne x But H dl = ( J ) d S = I + ( I) = 0 c s No adated feld 83

22 Openng out the two bas of the tansmsson lne has lttle effect on the cuent dstbuton and the exposed oscllatng cuent adates an electomagnetc wave I zx λ/4 λ/ Hetz GHz half-wave dpole Antenna 85

23 Powe dstbuton. lectc feld 86 III... Shot dpole At long wavelengths a half-wave dpole becomes mpactcally bg. A shote dpole stll adates, but moe of the wave s eflected back down the wavegude, snce thee s an nput mpedance L <<λ/ 87

24 Powe dstbuton lectc feld 88 lectc and magnetc felds 89

25 III...3 Half dpole Conductos eflect ado waves because =0 at the conducto (that s why mos ae shny) A sngle dpole placed above a conducto adates lke one half of a dpole pa Long wave ado masts ae lke ths, and can be fomed by coveng the gound wth we mesh 90 III..3 Loop antennae The loop antenna s lke the half-wave dpole except that t behaves lke a tansmsson lne whose end s a shot ccut and s theefoe dven by a voltage souce λ/4 Shot Ccut a λ/4 tansmsson lne.. If the end of a tansmsson lne s a shot ccut, Z L =0 ρ = L The telegaphe s equatons show that ¼ of a wavelength back fom the load the cuent must be 0. 9

26 Z λ Z b= 0 4 λ λ b= Z b= 4 L Z 4 λ b= 4 = I = 0 so a voltage souce s needed to dve the wave V λ/4 Shot Ccut a λ/4 tansmsson lne.. V then open t out.. To get a loop antenna x 9 Insde a potable ado you wll fnd a fete od wound wth coppe we. Ths s the long wave antenna. It has seveal loops and a fete coe. The coe concentates the electomagnetc waves nto the antenna and the whole aangement s essentally half of a tansfome F e te R o d Receve 93

27 III..4 Reflecto antennae Anothe way of concentatng the electomagnetc waves s usng a paabolc mo Placng a souce, S, at the focus of a paabalod can esult n a vey dectve beam S 94 III..5 Aay antennae Antennae can also be joned nto an aay The aay must be coectly desgned so that the sgnals combne n phase.e. that they add up athe than cancel out a b e jα e jβ Supeposton effects esult n a hghly dectonal LCTRONIC NANOMATRIALS wave DVICS AND as long as the spacngs DPARTMNT (a,b) and OF NGINRING the phases (eand jα SPCTROSCOPY,e jβ MATRIALS ) ae coect GROUP 95

28 96 III. Rado tansmsson III.. Radaton esstance Antennae emt powe, so they can be modelled as esstos: The adaton esstance, R a, of an antenna s that esstance whch n place of the antenna would dsspate as much powe as the antenna adates R = a P ant ( I ) ms (8.) whee P ki d S k I d S ant = = s ant ant s (8.) { } and ki = Re H * DPARTMNT ant OF NGINRING 97

29 The powe s calculated by ntegatng ove a suface fa fom the adatng antenna whch encloses the cuent cayng poton Fo example, a sphee at a dstance away fom the axs of a we cayng a cuent I ant As the dstance nceases vaes as 0 / and H as o /η These ae temed the fa feld values of the electomagnetc adaton fom the antenna The ntensty of powe fom a eal antenna s not unfom. It has a 3 dmensonal patten, whch s temed the adaton patten xample: consde a half dpole antenna wth: I P = I I = ant 36.5 ant; ms ant R a 36.5Iant = = 73 Ω. I ant LCTRONIC NANOMATRIALS DVICS AND AND SPCTROSCOPY MATRIALS GROUP III.. Gan The gan, G, of an antenna s the facto by whch ts maxmum adated ntensty exceeds that of an sotopc antenna f they emt equal powe fom an equal dstance { } * ant θ ϕ H ant θ φ { } * so θ ϕ H so θ φ Re (,, ) (,, ) Max ki G = = Re (,, ) (,, ) RaI / 4 ant so π Povded that antenna sotopc s Int ds = Int ds s (8.3) 98 An sotopc antenna s a hypothetcal devce whch adates equally n all dectons. 99

30 III..3 ffectve aea The effectve aea, A eff, of an antenna s that aea of wavefont whose powe equals that eceved fom the wavefont by the antenna A eff = Powe collected by antenna Wave ntensty (.e. powe / aea ) nto antenna Note: even though the aea of ado wave ntecepted by a paabolc dsh antenna s obvous, t s not n othe cases Fo example: n pncple a half wave dpole could have no aea at all and yet stll eceve powe fom a ado wave. Hence the need to defne an effectve aea. 00 III..4 xample: Powe tansmsson If two half-wave dpoles ae km apat and one s dven wth 0.5 amps (RMS) at 300 MHz, what powe s eceved by the othe? G =.64, R a =73 Ω, A eff =0.3 m Intensty metes fom an sotopc antenna = Tansmtted powe/(4π ) Intensty metes fom ths antenna = G Iso R 4π a Powe eceved by ecevng antenna = Intensty A eff = G Iso R 4π = =0.3 µw. a A eff 4π ( 000) 0

31 Optonal xamples O-Negatve efactve ndex O-Invsblty cloak O3-Had dsk dves O4-Tanspaent conductos 0 O-Negatve efactve ndex In 967 Vkto Veselago theoetcally nvestgated the popetes of meda wth a negatve pemttvty togethe wth negatve pemeablty µ n the same fequency ange He pedcted that the wave vecto of a wave popagatng though such a medum s antpaallel to ts Poyntng vecto 03

32 Ths has fa-eachng consequences: A wave mpngng fom vacuum onto the suface of such a medum unde an angle wth espect to the suface nomal wll be efacted towads the "wong" sde of the nomal,.e., we obtan negatve efacton 04 Postve Refactve Index Negatve Refactve Index Dollng et al. 05

33 06 Negatve pemttvty s not unusual, and occus n metals fom zeo fequency to the plasma fequency ω p 4π ne ( ω) = ω p = Plasma Fequency ω m Wth n electon chage densty and m electon mass Howeve a lage magnetc esponse, n geneal, and a negatve pemeablty µ at CAMBRIDG optcal UNIVRSITY fequences, n patcula, LCTRONIC NANOMATRIALS do DVICS not AND occu n natual mateals 07

34 A negatve ndex of efacton can be mplemented by an aay of metallc splt-ng esonatos (SRR) SRR act as LC-oscllatos. Snce the capactance and nductance ae detemned by the dmensons of the SRR, scalng of the stuctue allows to tune the esonance fequency fom mcowave to teahetz to the nfaed Peequste fo the magnetc esponse s the exctaton of a cculatng cuent n the ndvdual SRR by the ncdent feld. Ths nduces a magnetc feld whch can lead to an effectve negatve pemeablty µ 08 O-Invsblty cloak Susumu Tach 003 Howeve: Optcal Camouflage! Poject n font DPARTMNT what seen OF NGINRING n back. Not genune AND SPCTROSCOPY MATRIALS see GROUP though 09

35 Invsblty cloak (genune!) Ths s anothe example of a new optcal devce whch could esult fom ou ablty to talo an optcal mateal popetes It eles on a contolled spatal vaaton of pemttvty and pemeablty gudng lght aound the cental pat of the cloak The plane wave mpngng fom left s "flowng" LCTRONIC NANOMATRIALS aound DVICS AND the cloak wthout beng DPARTMNT dstubed OF by NGINRING the metallc cylnde AND SPCTROSCOPY MATRIALS n the GROUP mddle 0 Pendy-Smth 000 onwads

36 Ths cloakng devce s pactcally nvsble If you see the wold n mcowaves wth a wavelength of 3.5cm Schug et al. Scence 006 Smulatons xpement Schug et al. Scence 006 3

37 4 5

38 Paallel Recodng 6 Pependcula Recodng 7

39 8 9

40 Mn-Had Dve 0 O4-Tanspaent conductos Flexble, Foldable AMOLD Dsplay Touch Sceen Anode (ITO) TFT OLD Cathode Substate Font Plane : Touch Sceen, OLD Back plane : TFTs

41 Touch sceen dsplays lectonc pape Photovoltac cells Sensos Rado fequency tags Smat textle ITO (Indum-Tn-Oxde) dawbacks Inceasng cost due to Indum scacty Pocessng equements, dffcultes n pattenng Senstvty to acdc and basc envonments Bttleness Wea esstance 3

42 d Gaphene Cabon Allotopes 0d d 3d Buckyball Cabon Nanotube Gaphte 4 Bendablty of lectonc Mateals Mateal Factue Stan Mateal Factue Stan Slcon ~0.7% Poly- ZnO 0.03% ITO 0.58~.5% Polymde 4% Au 0.46% Gaphene >5-0% 5

43 ITO eplacements 00 Tansmttance (%) ITO ZnO/Ag/ZnO TO /Ag/TO Ac dschage SWNTs Wavelength (nm) Metal gds, metallc nanowes, metal oxdes and nanotubes (SWNTs) have been exploed as ITO altenatve 6 00 Tansmttance (%) Gaphene ITO ZnO/Ag/ZnO TO /Ag/TO Ac dschage SWNTs Wavelength (nm) Gaphene flms have hghe T ove a wde wavelength ange wth espect to SWNT flms, thn metallc flms, and ITO 7

44 NANOMATRIALS AND LCTRONIC DVICS AND MATRIALSGROUP GROUP SPCTROSCOPY NANOMATRIALS AND LCTRONIC DVICS AND MATRIALSGROUP GROUP SPCTROSCOPY 8 9

45 30 3