Tentative Schedule: Date, Place & Time Topics Sep.4 (Mo) No classes Labor Day Holiday Exam 1 Exam 2 Over Chapters 4-6

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1 Tentative Schedule: Date, Place & Time Topics 1 Aug. 8 (Mo) 394; 5:00-6:15 Introduction, Spontaneous and Stimulated Transitions (Ch. 1) Lecture Notes Aug. 30 (We) 394; 5:00-6:15 Spontaneous and Stimulated Transitions (Ch. 1) Lecture Notes Homework 1: PH481 Ch.1 problems 1.4 &1.6 PH581 Ch.1 problems 1.4, 1.6 & 1.8 due Sep.6 before class Sep.4 (Mo) No classes Labor Day Holiday 3 Sep. 6 (We) 394; 5:00-6:15 Optical Frequency Amplifiers (Ch..1-.4) Lecture Notes Problem solving for Ch.1 4 Sep. 11 (Mo) 394; 5:00-6:15 Optical Frequency Amplifiers (Ch ) Lecture Notes Homework : PH481 Ch. problems. (a,b),.4 &.5 (a,b) PH581 Ch. problems. (a,b),.4 &.5 (a,b,c,d) due Sep.0 before class 5 Sep. 13 (We) 394; 5:00-6:15 Problem solving for Ch. Introduction to two Practical Laser Systems (The Ruby Laser, The Helium Neon Laser) (Ch. 3) Lecture Notes 6 Sep. 18 (Mo) 394; 5:00-6:15 Review Chapters 1 & Lecture Notes 7 Sep. 0 (We) 394; 5:00-6:15 Exam 1 Over Chapters 1-3; Grades for exam 1 8 Sep. 5 (Mo) 394; 5:00-6:15 Exam 1 problem solving. Passive Optical Resonators (Lecture notes) 9 Sep. 7 (We) 394; 5:00-6:15 Passive Optical Resonators (Lecture notes). 10 Oct. (Mo) 394; 5:00-6:15 Passive Optical Resonators (Lecture notes). Physical significance of and (Ch..8-.9). Homework 3: read Ch. & notes. Work out problems. Due Oct Oct. 4 (We) 394; 5:00-6:15 Optical Resonators Containing Amplifying Media (4.1-). 1 Oct. 9 (Mo) 394; 5:00-6:15 Optical Resonators Containing Amplifying Media (Ch ) Homework 4: Ch. 4 problems 4.7 and 4.9. Due Oct Oct. 11 (We) 394; 5:00-6:15 Laser Radiation (Ch ) 14 Oct. 16 (Mo) 394; 5:00-6:15 Control of Laser Oscillators ( ) Homework 5: Ch. 5 problems 5.1 and 5.5. Due Oct Oct. 18 (We) 394; 5:00-6:15 Control of Laser Oscillators ( ) and exam review 16 Oct. 3 (Mo) 394; 5:00-6:15 Optically Pumped Solid State Lasers ( ) 17 Oct. 5 (We) 394; 5:00-6:15 Optically Pumped Solid State Lasers ( ) 18 Oct. 30 (Mo) 394; 5:00-6:15 Exam Over Chapters 4-6 Grades for exam Exam correct solution; Homework 6 Due Nov.6; Article on Cr:CdSe 19 Nov. 1 (We) 394; 5:00-6:15 Optically Pumped Solid State Lasers ( ) 0 Nov. 6 (Mo) 394; 5:00-6:15 Optically Pumped Solid State Lasers ( ) Homework 7 Due Nov Nov. 8 (We) 394; 5:00-6:15 Optically Pumped Solid State Lasers ( ) Supplemental material for - Homework 6 diode pumped LiF:F laser Nov. 13 (Mo) 394; 5:00-6:15 Spectroscopy of Common Lasers and Gas Lasers (Ch and class material) Voluntary supplemental homework 6A design of RT CW Fe:ZnSe laser Due Nov.7 3 Nov. 15 (We) 394; 5:00-6:15 Gas lasers (Ch ); Molecular Gas lasers I (Ch ) Nov.0 (Mo) No classes Thanksgiving - no classes held Nov. (We) No classes Thanksgiving - no classes held 4 Nov. 7 (Mo) 394; 5:00-6:15 Molecular Gas lasers I (Ch ) Homework 8 Due Dec. 4 5 Nov. 9 (We) 394; 5:00-6:15 Molecular Gas Lasers II (Ch ) and review for exam 3 (Ch ) Homework 9 Due Dec 6 6 Dec. 4 (Mo) 394; 5:00-6:15 Exam 3 Over Chapters 7-10 Grades; Exam 3 Correct solution 7 Dec. 6 (We) 394; 5:00-6:15 Review for Final 8 Dec. 11 (Mon) in CH 394 FINAL EXAM Over Chapters 1-10 (4:15-6:45pm) in CH 394 Final Grades 1

2 Laser Physics II (PH58 Spring 018) Tentative Schedule: # Date Text Topics 1 Jan 8 (Mo) LEO Ch , Class Lecture Tunable Lasers, Organic Dye Lasers Jan 10 (We) LEO Ch , Class Lecture Tunable Solid State Lasers, Alexandrite & Ti-Sapph. Lasers, TM:II-VI Lasers(Class Lecture) Homework 1, Due January 4, Jan 15 (Mo) MLK Holiday No classes 4 Jan 17 (We) LEO Ch , Class Lecture Tunable Solid State Lasers, Alexandrite & Ti-Sapph. Lasers, TM:II-VI Lasers(Class Lecture) Jan (Mo) Class Lecture Color Center Lasers 5 Jan 4 (We) Exam 1 - Grades Exam 1 over chapter 11 Correct Solution 6 Jan 9 (Mo) Ch and class lecture Semiconductor Lasers, Semiconductor Physics Background 7 Jan 31 (We) Ch and class lecture Semiconductor Lasers, Semiconductor Physics Background, Homework, Due February 7, Feb 5 (Mo) (Ch ) Semiconductor Lasers 9 Feb 7 (We) (Ch ) (Class Lecture, Ch.13) Semiconductor Lasers Ray Tracing in an Optical System Homework 3, Due March 5, Feb 1 (Mo) (Class Lecture, Ch.13) Semiconductor problem Solving Ray Tracing in an Optical System 11 Feb 14 (We) (Class Lecture or Ch ) Gaussian Beams 1 Feb 19 (Mo) (Class Lecture or Ch ) Gaussian Beams 13 Feb 1 (We) (Class Lecture or Ch.15 and Optical Cavities Ch ) 14 Feb 6 (Mo) (Class Lecture or Ch.15 and Optical Cavities; Three and four Mirror Focused Cavities Ch ) 15 Feb 8 (We) (Ch ) Optical Cavities; Cavities for Producing Spectral Narrowing of Laser Output Sample Problems for Test 16 Mar 5 (Mo) Exam Grades Exam over chapters Correct Solutions 17 Mar 7 (We) (LEO Ch. 17) Optics of Anisotropic Media Homework #4 problems 17.; 17.3; 17.4; 17.5 ch.17 LEO Due March 16. Mar 1 (Mo) Spring Break No classes Mar 14 (We) Spring Break No classes 18 Mar 19 (Mo) (LEO Ch. 17) Optics of Anisotropic Media 19 Mar 1 (We) (LEO Ch. 19, 0) Wave Propagation in Nonlinear Media Homework # 5 du March 8 0 Mar 6(Mo) (LEO Ch. 19,0) nd Harm. Generation. Up and Down-Conversion, Optic Parametr. Amplification; 1 Mar 8 (We) (LEO Ch. 19,0) nd Harm. Generation. Up and Down-Conversion, Optic Parametr. Amplification Apr (Mo) (LEO Ch. 18) The Electro-Optics and Acousto-Optic Effects and Modulaton of Light Beams Homework #6 due April 9 Homework #5 review 3 April 4 (We) (LEO Ch ) Detection of Optical Radiation 4 April 9 (Mo) (LEO Ch ) Noise in Photodetectors Homework #7 due April 16 5 April 11 (We) (LEO Ch ) Photodiode Arrays and CCDs; Sample problems for test 6 April 16 (Mo) Exam 3 Grades Exam 3 over chapters 18- Correct Solutions 7 April 18 (We) Review for Final Review for Final 8 April 3 (Mo) FINAL GRADES FINAL EXAM Over Chapters 11- and class notes 4:15-6:45pm CH 394

3 Review for Final 3

4 4

5 What we want to find is as follows: 1) Saturation intensity at pump and lasing wavelength ) Effective length of the crystal 3) Optimal output coupler 4) Survival (photon) factor S for the passive cavity 5) Threshold power for a mode diameter of 00 m 6) Pump power necessary for providing 30W of 4400nm output power Solution: 1) Saturation intensity at 4400 nm m/ s Js 9 hv m e Is W / cm cm s I sp e 1 hvp ; p 1 (branching factor) ( v ) p a p m/ s Js m I.0710 W / cm sp cm s 5 5 5

6 ) Effective length of the crystal Assume absorption utilization efficiency a p ano cm Assume pump intensity is 3 times above pump saturation intensity I a p 0.9 deff 33mm o 1 I 9cm p sp 1) Optimal output coupler Assuming high gain medium approximation T 1 ln 0deff lossdeff 1T 1T o en cm 110 cm 11cm T ln 11cm 0.3cm0.03cm 0.3cm3.91 1T 1T Assume T 0.50; Actual 1.69 Assume T 0.70; Actual 3.53 Assume T 0.65; Actual.91 Assume T 0.68; Actual 3.65 Optimal transmission of the output coupler is 68%. 6

7 4) Survival (photon) factor S for the passive cavity R 1 R lossdeff S e (1 T ) e ) Threshold power for a mode diameter of 100 m Is 1 I pthr ( ) ln p a q s p 800 p 1; a 0.9; q 0.636; qe e I pthr ( ) 5 ( W / cm ) 1 ln.710 W / cm (0.01) Ppthr ( ) Ipthr ( ) A.710 W / cm 1W 4 7

8 6) Pump power necessary for providing 30W of 4400nm output power P 30W out P P P out p a c p p( thr) T 0.0 1S c 0.53 ( ) ( ) P P P 0.30 P P 30 Pp Pp( thr) 100W 0.30 q out p p thr p p thr Pp W - realistic from the first glance 7) Experimental verification of spinning disk Fe:ZnSe laser of radius 11 mm rotating with rev/min and pumped by 40 W at 800 nm radiation from spinning ring Cr:ZnSe laser demonstrated no lasing at all. The reason of the failure is that the experimental arrangement was not able to sustain room temperature of the active zone of the crystal. Our estimation of the steady state temperature of the active zone in spite of the using compressed air for cooling of the spinning Fe:ZnSe element the temperature of the pumped zone exceeded 100C which stimulated temperature quenching of Fe:ZnSe up to 0 ns. 8

9 At this temperature the lifetime of Fe:ZnSe decreased to 0 ns due to temperature quenching. I W cm sp 6 increased by factor 19 and reached /. At pump intensity 40W Ip W cm /, whic 4 the gain medium, and hence, don't have lasing. I sp h is ~8 times smaller than we don't saturate 9

10 8) What is the reason of temperature increase in the active zone of a spinning Fe:ZnSe gain element. Let us calculate time t necessary for active zone of the spinning element with radius R=1.1 cm spinjing at 10000rev/min to pass through the pump causitcs of diameter 100 m. rt cm cm t rev 1min cm min60sec1rad During this time the pumped volume of Fe:ZnSe of diameter 100um and length 3 mm m V (310 m).3610 m will absorb energy Q Pt 40W s Since QcVT s 0.35mJ 4 Q J T 8.3 C cv 3 J kg m kg C m 10

11 This temperature increase of the pump volume happens in 8.7 us of pumping. 1 The period of one complete revolution is 6ms rev 1min min 60sec During this time the heat cannot dissipate completely from the pumped zone and after one revolution the pumped zone will be further heated to 16.6 C and so on until the pumped zone will reach 100C steady state temperature. Possible soution - use spining cavity approach with effective cooling of the Fe:ZnSe element through a headspreader. 11

12 LASER PHYSICS I PH 581-VT (MIROV) Exam 3 (1/04/17) STUDENT NAME: key STUDENT id #: Opened textbook WORK ONLY 3 questions ALL QUESTIONS ARE WORTH 50 POINTS NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.) 1. A Nd YAG rod 5 mm in diameter, 6 cm long, with 100% and 90% reflectivity mirrors deposited on its facets is pumped by a pulsed flashlamp with a pulse duration much shorter than the lifetime of Nd ions. The average wavelength of excitation is 810 nm and all the pump radiation is completely absorbed by a rod. Nd YAG cross section of emission of is 9x10-3 m and branching ratio is equal to 1. Assume the distributed loss in the cavity is 0.01 cm -1. a) Calculate threshold population inversion. b) Calculate the absorbed pumped energy corresponding to threshold. c) Calculate the electrical pump energy of a lamp at threshold, assuming that the rod is uniformly pumped with an overall pump efficiency=efg=1% ) Threshold gain th ln R1R 0.01 ln cm 1.9m l 6 th 3 ) Population inversion Nth.110 m Nth hvp 3 3 3) The absorbed pump energy density Uth 5.10 J / m () v 4) The absorbed pump energy U V 6.1mJ 5) The electrical pump energy 6.1 mj / J th 1 1 1

13 . Consider the rigid rotation of biatomic molecule, made of two atoms with masses M 1 and M at intermolecular distance R o. The moment of inertia I about an axis passing through the center of mass and M 1M perpendicular to the internuclear axis can be obtained as I Ro M rro. Recalling the M 1 M quantization rule of angular momentum, L J ( J 1) and the facts that rotational kinetic energy of a rigid body rotating around a given axis can be written as E=L /I and rotational energies of the biatomic molecule can be expressed as E J rot=j(j+1)bhc, express the rotational constant B of the molecule as a function of the reduced mass and intermolecular distance. JJ 1 E JJ 1Bhc I h B I hc 8 Ic 4Ic 4cMrRo 13

14 3. Consider a laser system made of a cascade of three lasers: a laser, emitting at 500 nm that pumps Ti:Al O 3 laser, that pumps a Nd:YAG laser. Suppose that the green laser has a threshold power P th1 =0.75 W and a slope efficiency s1 =13%, the Ti:Al O 3 laser has a threshold power P th =1.7 W and a slope efficiency s =15% and the Nd:YAG laser has a threshold power P th3 =1 W and a slope efficiency s3 =1%. Calculate the pump power that must be provided to the green laser to get an output power P out3 =0.75 W from the Nd:YAG laser. The expression for the output power from the Nd:YAG laser is P P P out3 s3 p th3 Eq 1 can be rewritten by expressing the output power from the Ti-sapphire laser P P P P P out3 s3 s p1 th th3 out3 s3 s s1 p th1 th th3 Eq can be rewritten by expressing the output power from the green laser P P P P P where P is the pump power provided to the green laser. p P One can readily solve eq 3 to obtain P a a function of other variables p (1) () (3) p1 p P p P P P P out3 s3 th3 s3 s th s3 s s1 th1 s1 s s3 (4) Pp 385.6W W 14

15 4. Consider a Nd:YLF laser (n=1.448) to be Q-switched as shown in Figure depicted below. The Nd:YLF rod has a cross-sectional area of 0. cm, is 8 cm in length, and is placed in a cavity 0 cm long. The Q-switch is 1 cm long and its index of refraction is The transmission of the cavity output coupler is 30% and the internal loss per pass is 5%. Assuming a stimulated emission cross section 1.9x10-19 cm of Nd:YLF at the laser wavelength and that the energy of the pump pulse is twice the threshold. (a) Find the maximum peak output power of this laser. (b) Find the output energy in the Q-switched pulse. (c) Estimate the pulse width. Shutter Pump Laser medium l 1) Compute the threshold gain coefficient. Round trip gain must be greater than 1 for an oscillator and equal to 1 for threshold: 1 l l s s RR TTTT e e th a b c d 1; 1 1 l s 1 1 ln ln l l l th s R1 TaTbTcTd R cm 1 d ln 0 ln th 0.087cm ) The threshold inversion ( N N 1) th cm cm 3) The total # of inverted atoms in the cavity at threshold is n th N N A l cm 0.cm 8cm.410 atoms 1 th 17 no nth atoms ) Because of the specification of being pumped to two times threshold we also know the initial inversion

16 Problem 4 continuation. 5) Photon lifetime of the passive cavity. First let us calculate the cavity round trip time lair nsls ngl ns c RT ns 1 RR TTT l sl s s T e e RT o cpl 0 1 a b c 6) Maximum photon # in the cavity d no nth n th n o Φ max ln nth ln ) Calculate coupling efficiency max max 34 cpl h o photons coupling loss per round trip l total loss per round trip e s ) The output power at the maximum of the pulse P MW 9) Assume that the fraction of the initial inversion converted to photons x n o n n 0 f nhv 6 o 10) The output energy W out cpl x 4mJ W out ) The pulse width t 18ns 6 P max

17 5.

18 Exam 1 (09/0/17)

19

20 x10-10 J

21

22

23

24 LASER PHYSICS I PH 581-VTA (MIROV) Exam II (10/30/17) STUDENT NAME: Key STUDENT id #: Opened textbook, opened notes ALL QUESTIONS ARE WORTH 37.5 POINTS (WORK OUT ANY 4 PROBLEMS) NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.) 1. A Fabry-Perot interferometer consisting of two identical mirrors, air-spaced by a distance L, is illuminated by a monochromatic em wave of tunable frequency. From a measurement of the transmitted intensity versus the frequency of the input wave we find that the free spectral range of the interferometer is 3x10 9 Hz and its resolution is 30 MHz. Calculate the spacing L of the interferometer, its finesse, and the mirror reflectivity. 1) For a Fabry-Perot interferometer made of air-spaced mirrors, the free spectral range is: c vfsr. Hence, the mirror spacing in our case is given by: L 11 1 c 310 mms L 50mm 9 1 vfsr 310 s ) The finesse of the interferometer, i.e. the ratio of free spectral range to width of the 9 vfsr 310 Hz transmission peak, is : F v 3010 Hz 3) The finesse is a function of mirror reflectivity, in the case of equal mirrors R we have F, which gives the equation 1 R R R10, F the solution of which is R 0.968

25

26 3. A laser (=10.6 m, =3x10-5 m, =4 s) measured to have an intensity of 0.3 W/cm emerging from one end of the laser, which has two identical mirrors each with a transmission of 10%. The gain of the laser is also measured to be 0.5. What is the optimum output mirror transmission? hv Is 1.56 W / cm Iout Is (1 AR) 1.56 (1-A-0.9). ( 0Lln R) 0.3= (0.5 ln 0.9) 1 R A.510 Topt ol 3. 1; Topt A ol A % A A 6

27 4. A helium-neon laser transition (0.638 m) is Doppler broadened with a FWHM of 1.5 GHz. Assume the pumping and the saturated signal gain coefficient are four times the threshold value, and the cavity is 100 cm long. (a) Find the number of longitudinal modes that can oscillate simultaneously. (Hint: ( v v o ) ln v D th ( v ) ( v o ) e ). (b) Suppose all the modes are locked together: (1) What is the pulse spacing? () Estimate the pulse width. 6.7 ns

28 5. (a) What would be the minimum pulse duration of a mode-locked chromium doped ZnS laser (gain bandwidth is 600 nm, central wavelength o is 350 nm)? (b) What would be the coherence time and coherence length of the output beam? (c) If the separation between mirrors is 1.8 m, the ZnS gain element is 1 cm long, and the index of refraction of ZnS crystal is approximately.3. What would be the separation between mode-locked pulses? a) the mode-locked pulse width (bandwidth limited) 1, where is the width of the gain profile 8 9 c 310 (60010 ) Hz 9 (35010 ) ~30fs 1 b) c ~ 30 fs; Lc cc 9m c) Pulse spacing t 1 ( ) (.30.01) ns

29 6. Consider the active medium Cr:ZnSe (refractive index n=.49, gain bandwidth (FWHM) =860 nm, central wavelength o =400 nm. (a) Consider first a resonator with length L=0 cm, employing a rod of length l=1 cm. Find the number of longitudinal modes falling within the FWHM gain bandwidth. (b) Consider then a resonator made upon coating the end mirrors directly onto the active material surfaces (microchip laser). What is the maximum thickness l that allows oscillation of only one longitudinal mode? 8 9 c 310 (86010 ) (40010 ) 13 Hz 4.48x10^ m.49x4.48x10^13

30 Three important methodological problems related to diode pumped solid state lasers 1. Internal loss estimation R=1 R=1 Assume ring cavity & clockwise traveling wave R=1 Exper. 1: R 1 <1; Exper. : R <1;

31 Laser slope efficiency

32 P out, W P out tan P thr P pump P pump, W Laser slope efficiency= Pout s tan p a qe c P P pump thr 3

33 3. A continuous wave Ho:YAG (090 nm) laser is longitudinally pumped at 1908 nm. The laser mode has a spot size of 1 mm and the length of the cavity is 10 cm; the stimulated emission cross-section is e = 1.9x10-1 cm and the upper level lifetime is =8.5 ms. Assume that an output coupler with a transmission T=30% is used, passive cavity losses are % per pass, branching ratio p =1, and a pump utilization efficiency a =80%. Calculate the threshold pump power, slope efficiency, as well as the pump power required to obtain an output power P out =30W from this laser ) Threshold gain th ln R1R ln cm l th 18 3 ) Population inversion Nth cm N v th hvp 3 3) The absorbed threshold pump power per unit volume P th 5.9 W / cm () v v 4) The absorbed pump power P P V / W out p a qe c p th 0.0 p p slope [1 ( e 0.7)] hv s 864 W/ cm ; P / s IsA W (1 0.7) 30 5) P P P P.0 ; P.0 47 W; =66.9%; Alternatively: 6) I th th qe a Ps 1 Ps ) Pth ln (1 S) [1 ( e 0.7)] 3.0W 1908 p qea S p qea (1 0.7) 30 8) P P P P 3.0 ; Pp 3.046W out p a qe c p th 0.0 p [1 ( e 0.7)] 33

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