Optics and Non-Linar Optics I - 007 Non-linar Optics Tutorial Sht Novmbr 007 1. An altrnativ xponntial notion somtims usd in NLO is to writ Acos (") # 1 ( Ai" + A * $i" ). By using this notation and substituting into th inducd non-linar polarisation quation (us only th χ (1) and χ () trms), show that th trms gnratd ar th sam as thos w obtaind in lcturs..a. Explain what is mant by cohrnc lngth. b. Calculat th cohrnc lngth for a fundamntal wav of 1064nm gnrating a scond harmonic wav of 53nm in a KTP crystal. (n ω = 1.74, n ω =1.78) c. Sktch how th nd harmonic convrsion fficincy varis with distanc along th crystal with th cohrnc lngths markd on th x-axis. d. Calculat th maximum convrsion fficincy for SHG for a mm long non-phas matchd crystal (d ff =3.18pm/V) for th following pump powrs (P ω = 50mW, 5W, 100W, 1000W) focussd to a spot siz of 0µm.. What input powr is rquird to obtain an SHG fficincy of 10% aftr 1 cohrnc lngth in th scnario outlind abov? 3. a. Sktch th rfractiv indx as a function of angl for th ordinary and xtraordinary rays in a ngativ uniaxial crystal. Mark on your diagram th valus of n o and n. b. Starting from th llipsoidal quation: # 1 & % ( $ n (")' = 1 # & # % ( cos " + % $ n o ' $ 1 & ( sin " n ' Show that th phas matching angl for scond harmonic gnration from frquncy ω to ω, θ m, is givn by: o n " m = sin #1 $ ( ) # o # ( n $ ) # ( n $ ) # o # ( n $ ) # c. In ordr to work out th rfractiv indx for a particular wavlngth, w mak us of formula known as th Sllmir Equations. For th common non-linar matrial Bta-Barium Borat (BBO) w hav: 0.01878 n o =.7359 + (" - 0.018) - 0.014 0.01354" and n =.3753 + (" - 0.01667) - 0.01516" Whr th wavlngth, λ, is writtn in µm. Calculat n o and n for wavlngths of 1064nm and 53nm and hnc work out th phas matching angl for scond harmonic gnration.
d. It is now dcidd to frquncy doubl th 53nm radiation in a scond crystal. What is th nw phas matching angl? Will th fficincy of gnration b th sam?. What limits th rang of wavlngths that a particular non-linar optical can b usd for SHG? 4. In our labs w hav grown a nw ngativ uniaxial crystal. Th crystal mlts at 140 o C is it suitabl to us this crystal in a non-critical phas matching gomtry for scond harmonic gnration from a fundamntal wavlngth of 800nm? Data: n o 800nm = 1.501 and n 400nm = 1.415 (Both masurd at 0 o C) dn o /dt = 0.001K -1 dn /dt = 0.00K -1. 5. a. For a rang from 0 to 10xl c, sktch th scond harmonic output powr vrsus distanc along th crystal for unphasmatchd LiNbO 3. Mark th x axis in units of l c. b. In ordr to obtain quasi-phasmatching, w now priodically pol th crystal by invrting a distanc of with = l c with a grating priod = l c, sktch on you diagram th gnratd scond harmonic vrsus distanc. c. For som wavlngths poling vry cohrnc lngth is impossibl. Instad w hav to mov to highr ordr priods. Sktch on your graph th gnratd scond harmonic vrsus distanc for third ordr (i width of pold rgion = 3l c, priod = 6l c ) priodic poling. Commnt on th fficincy of this mthod of SHG. d. Commnt on th proprtis rquird from a crystal to allow lctric fild poling for fficint SHG.. Using th data providd in lcturs, calculat th grating priod rquird for first-ordr QPM to obtain SHG from λ incidnt =946nm. 6.a. Show that whn considring χ (3) intractions and starting from E 3 0 χ (3) cos 3 (ωt) th gnratd trms can b writtn in th form: " (3) ( ( ) + cos(3#t) ). 4 E 3 0 3cos #t b. Starting from n = ((1 + χ (1) + χ (3) I(ω)), show that th rfractiv indx can b approximatd such that: n " n 0 + n I(#) whr n = " (3). n 0 7. a. Dscrib th origin of th stoks and antistoks photons. b. An 800nm lasr is focussd onto a sampl of intrst and stoks lins ar obsrvd at 847nm, 876nm, 947nm and 1030nm. Calculat υ R for ach transition in units of wavnumbrs. c. Using th simpl diagram shown in lcturs for th origin of antistoks radiation, calculat th xpctd antistoks wavlngths from this sampl. d. Assuming a 1064nm lasr was usd to xcit th sampl, calculat th stoks wavlngths you would xpct to obsrv.
1. An altrnativ xponntial notion som tim usd in NLO is to writ Acos (") # 1 ( Ai" + A * $i" ). By using this notation and substituting into th inducd non-linar polarisation quation (us only th χ (1) and χ () trms), show that th sam gnratd frquncy trms ar obtaind as obtaind in lcturs. So w now hav: E = E " cos("t) = 1 (E " i"t + E " * #i"t ) Substituting into: P = ε 0 χ (1) E +ε 0 χ (1) E givs: P = " 0# (1) ( E $ i$t + E * $ %i$t ) + " 0# () ( E $ i$t + E * $ %i$t )( E $ i$t + E * $ %i$t ) 4 = " 0# (1) ( E $ i$t + E * $ %i$t ) + " 0# () E $ i$t + E $ E * $ + E * * $ E $ + E $ %i$t 4 Now : EE * = EE * = E ( ) ( ) So : P = " 0# () E $ + " 0# (1) ( E $ i$t + E * $ %i$t ) + " 0# () 1 E $ i$t * + E $ %i$t & " 0# () E $ + " 0 # (1) cos($t) + " 0# () cos($t) Giving th sam trms as gnratd in lcturs (without nding to know th trig id s!).a. Explain what is mant by cohrnc lngth. Th cohrnc lngth is th ffctiv intraction lngth in non-linar optics whn k ω k ω. Th gnratd output rachs a maxmimum aftr on cohrnc lngth bfor dstruciv intrfrnc rducs it to 0 aftr anothr cohrnc lngth. b. Calculat th cohrnc lngth for a fundamntal wav of 1064nm gnrating a scond harmonic wav of 53nm in a KTP crystal. (n ω = 1.74, n ω =1.78) l 0 = " #k = $ fundamntal 4[n % & n % ] = 1.064x10-6 / (4 (1.78-1.74)) = 6.65µm c. Sktch how th nd harmonic convrsion fficincy varis with distanc along th crystal with th cohrnc lngths markd on th x-axis. d. Calculat th maximum convrsion fficincy for SHG (1.064µm 53nm) for a mm long non-phas matchd crystal (d ff =3.18pm/V) for th following pump powrs focussd to a spot siz of 0µm. P ω = 50mW, 5W, 100W, 1kW Maximum convrsion fficincy occurs aftr ach odd intgr cohrnc lngth. Tak xprssion from lcturs and substitut Δk = π/l 0 : " SHG = C L I # sinc ($kl /) = " 0 sinc (%L /l 0 )
So aftr on cohrnc lngth: sinc (π/) = (/π) = 0.405 So η SHG = 0.405 x η 0 Powr/W Intnsity = η 0 η SHG powr / πω 0 [W/m ] 0.05 3.978E+07 8.016E-06 0.0003% 5 3.978E+09 8.016E-04 0.03% 100 7.957E+10 1.603E-0 0.65% 1000 7.957E+11 1.603E-01 6.49%. What input powr is rquird to obtain an SHG fficincy of 10% aftr 1 cohrnc lngth in th scnario outlind abov? In this cas w want η SHG = 0.1 η 0 = 0.1/0.405 = 0.47 0.47 = C L I ω I ω =0.47/C L = 0.47 / (5.036x10-8 x 4x10-6 ) = 1.x10 1 Wm - P ω = I ω A = I ω x πω 0 = 1533W 3. i. Sktch th rfractiv indx as a function of angl for th ordinary and xtraordinary rays in a ngativ uniaxial crystal. Mark on your diagram th valus of n o and n. From lctur nots: Optical Axis n o (θ) n o n n (θ) ii. Starting from th llipsoidal quation: # 1 & % ( = 1 # & % ( cos " + 1 # & % ( sin " $ n (")' $ n o ' $ n ' Show that th phas matching angl for scond harmonic gnration from frquncy ω to ω, θ m, is givn by:
o n " m = sin #1 $ ( ) # o # ( n $ ) # ( n $ ) # o # ( n $ ) # For scond harmonic gnration w hav to st n ω(θ) = n o ω. Substituting into th abov qution: # 1 & # 1 & # % o ( = % o ( cos 1 & # ) + % ( sin 1 & ) = % o ( 1* sin ) $ n " ' $ n " ' $ n " ' $ n " ' Thrfor : ( ) sin ) o n * o " * n * " = n * o " * n * " So : sin ) = n o * o " * n * " n * o " * n * " o * o n Thrfor :) = sin *1 " * n * " n * o " * n * " ( ) + # 1 & % ( sin ) $ ' n " iii. In ordr to work out th rfractiv indx for a particular wavlngth, w mak us of formula known as th Sllmir Equations. For th common non-linar matrial Bta-Barium Borat (BBO) w hav: n o =.7359 + 0.01878 (" - 0.018) - 0.01354" and n =.3753 + 0.014 (" - 0.01667) - 0.01516" Whr th wavlngth, λ, is writtn in µm. Calculat n o and n for wavlngths of 1064nm and 53nm and hnc work out th phas matching angl for scond harmonic gnration. n o 1064 = 1.655, n o 53 = 1.674, n 1064 = 1.539, n o 53 = 1.555 Substitut back into th answr drivd in part ii to obtain θ m =.8 o iv. It is now dcidd to frquncy doubl th 53nm radiation in a scond crystal. What is th nw phas matching angl? Will th fficincy of gnration b th sam? Scond harmonic of 53nm is 66nm. n o 53 =1.674, n o 66 = 1.759, n 53=1.555, n 66=1.613 Givs phas matching angl of 47.6 o. Efficincy will not b th sam as χ cofficint dpnds on dirction. v. What limits th rang of wavlngths that a particular non-linar optical can b usd for SHG? Limitd by crystal transparncy rang. For BBO 190nm to 3500nm. 4. In our labs w hav grown a nw ngativ uniaxial crystal. Th crystal mlts at 140 o C is it suitabl to us this crystal in a non-critical phas matching gomtry for scond harmonic gnration from a fundamntal wavlngth of 800nm? Data: n o 800nm = 1.501 and n 400nm = 1.415 (Both masurd at 0 o C) dn o /dt = 0.001K -1 dn /dt = 0.00K -1. For NCPM, us 90 o phas matching and us tmpratur to tun n ω = n o ω. So: n o 800+ dn o /dt x ΔT = n 400 + dn /dt x ΔT
Givs ΔT = 86K so crystal should b hatd to 106 o C and things should b ok! 5. a. For a rang from 0 to 10xl c, sktch th scond harmonic output powr vrsus distanc along th crystal for unphasmatchd LiNbO 3. Mark th x axis in units of l c. b. In ordr to obtain quasi-phasmatching, w now priodically pol th crystal by invrting a distanc of with = l c with a grating priod = l c, sktch on you diagram th gnratd scond harmonic vrsus distanc. c. For som wavlngths poling vry cohrnc lngth is impossibl. Instad w hav to mov to highr ordr priods. Sktch on your graph th gnratd scond harmonic vrsus distanc for third ordr (i width of pold rgion = 3l c, priod = 6l c ) priodic poling. Commnt on th fficincy of this mthod of SHG. Sktch is shown on nxt pag. Th procss of priodic poling is obviously lss fficint that BPM. In gnral th fficincy of QPM can b shown to b (s Kochnr p.65) /(πw ) x E bpm whr w is th numbr of cohrnc lngths pold and E bpm is th fficincy of BPM. So whn w pol 3l c w hav a factor 9 rduction in fficincy. d. Commnt on th proprtis rquird from a crystal to allow lctric fild poling for fficint SHG. In ordr for lctric fild poling to b succssful, th crystal must b frrolctric (i, a prmannt chang in polarisation must b causd by th incidnt lctric fild.) Th crystal must also hav a good non-linar cofficint in th dsird propagation dirction and also b transparnt for th fundamntal and SHG wavlngths.. Using th data providd in lcturs, calculat th grating priod rquird for first-ordr QPM to obtain SHG from λ incidnt =946nm. So λ incidnt =946nm λ SHG = 473nm From lcturs: ( n o ) = 4.9048 + 0.11768-0.07169 " " # 0.04750 So: n 946 =.404 n 473 =.3593 (rmmbr to us λ in µm in th quation!)
So following th quations givn: " = 4" [ # g $ n % & n % ] Thrfor : $ % # g = (n % & n % ) W can obtain Λ g = 0.946 / (.404-.3593) = (-) 3.98µm this is vry hard to do! 6.a. Show that whn considring χ (3) intractions and starting from E 3 0 χ (3) cos 3 (ωt) th gnratd trms can b writtn in th form: " (3) 3 ( 3cos (#t) + cos(3#t) ). 4 E 0 W hav trms arising from: E 0 3 χ (3) cos 3 (ωt) E 0 3 χ (3) cos 3 (ωt) = E 0 3 χ (3) cos(ωt)cos (ωt) Using th idntity: cos x = ½ + ½ cos(x) E 3 0 χ (3) cos 3 (ωt) = E 3 0 χ (3) cos(ωt)( ½ + ½ cos(ωt)) Using th idntity: = E 0 3 χ (3) ( ½cos(ωt) + ½cos(ωt)cos(ωt)) cosacosb = ½ (cos(a-b) + cos(a+b)) E 0 3 χ (3) cos 3 (ωt) = E 0 3 χ (3) ( ½cos(ωt) + ¼ (cos(-ωt)+cos(3ωt))) Now cos(-x) = cos(x) so: E 0 3 χ (3) cos 3 (ωt) = E 0 3 χ (3) ( ¾ cos(ωt) + ¼cos(3ωt)) = " (3) 4 E 3 0 3cos #t ( ( ) + cos(3#t) ) QED b. Starting from n = ((1 + χ (1) + χ (3) I(ω)), show that th rfractiv indx can b approximatd such that: n " n 0 + n I(#) whr n = " (3) n 0. n = = 1+ " (1) 1+ " (1) + " (3) I(#) $ ( ) 1 1+ " (3) I(#) & % $ = n 0 1+ " (3) I(#) ' & ) % n 0 ( $ * n 0 1+ " (3) I(#) ' & ) % n 0 ( 1+ " (1) 1 * n 0 + " (3) I(#) n 0 = n 0 + n I(#) Q.E.D ' ) ( 1 Making us of th binomial xpansion of (1+x) 1/ 1+x/+. 7. a. Dscrib th origin of th stoks and antistoks photons.
Stoks and antistoks radiation arrivs as a consqunc of th Raman ffct. In th cas of stoks light, som of th incidnt radiation is coupld into a vibrational mod rducing th nrgy of th photon and gnrating a longr wavlngth photon. In th cas of antistoks, an xcitd vibrational mod givs up som nrgy to th incoming photon gnrating radiation of a shortr wavlngth than th pump. b. An 800nm lasr is focussd onto a sampl of intrst and stoks lins ar obsrvd at 847nm, 876nm, 947nm and 1030nm. Calculat υ R for ach transition in units of wavnumbrs. In this cas, w nd to calculat th frquncy of th radiations in wavnumbr units. So: 800nm 800x10-7 cm 1500cm -1 Similarly 847nm 11806cm -1, 876nm 11415cm -1, 947nm 10560cm -1, 1030nm 9708cm -1 In ach cas w can obtain ν R from ν P - ν S : 847nm ν R = 1500 11806 = 694cm -1, 876nm ν R = 1085cm -1, 947nm ν R = 1940cm -1, 1030nm ν R = 79cm -1. c. Using th simpl diagram shown in lcturs for th origin of antistoks radiation, calculat th xpctd antistoks wavlngths from this sampl. From th simpl diagram ν AS = ν R +ν P, so w hav antistoks wavs at frquncis of: 1500 + 694 = 13194cm -1 λ AS =(1/1319)cm = 757nm 1500 + 1085 = 13585cm -1 λ = 736nm 1500 + 1940 = 14440cm -1 λ = 69nm 1500 + 793 λ = 654nm d. Assuming a 1064nm lasr was usd to xcit th sampl, calculat th Stoks wavlngths you would xpct to obsrv. λ p =1064nm 9398cm -1, thrfor w ll hav stoks wavs at: 9398 694 cm -1 1149nm, 9398 1085 10nm, 9398 1940 1341nm and 9398 793 1514nm.