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Page 1: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

Review for Final

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Page 2: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

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LASER PHYSICS I PH 481/581-VT (MIROV) Exam 3 (11/02/13)

STUDENT NAME: ________________________ STUDENT id #: ___________________________ ------------------------------------------------ -------------------------------

Graduate -ALL QUESTIONS ARE WORTH 50 POINTS. Undergraduate- 75 ------------------------------------------------------------------------------------------------------------------------------------------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 depositedon its facets is pumped by a pulsed flashlamp with a pulse duration much shorter than thelifetime of Nd ions. The average wavelength of excitation is 810 nm and all the pump radiationis completely absorbed by a rod. Nd YAG cross section of emission of is 9x10-23 m2 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%.

Opened textbook

Graduate fellows WORK ONLY 3 questions Undergraduate – 2 questions

1 11 2

22 3

3 3

1 11) ln 0.01 ln 1 0.9 0.019 1.96

2) 2.1 10

3) 5.2 10 /( )

4)

th

thth

th pth

Threshold gain R R cm ml

Population inversion N m

N hvThe absorbed pump energy density U J m

vThe absorbed pump energy

6.15) 6.1 / 0.01 0.61

thU V mJThe electrical pump energy mJ J

Page 3: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

3

2. Consider the rigid rotation of biatomic molecule, made of two atoms with masses M1 and M2 at intermolecular distance Ro. The moment of inertia I about an axis passing through the center of mass and

perpendicular to the internuclear axis can be obtained as 2 21 2

1 2o r o

M MI R M RM M

. Recalling the

quantization rule of angular momentum, 2 2 ( 1)L J J and the facts that rotational kinetic energy of a rigid body rotating around a given axis can be written as E=L2/2I and rotational energies of the biatomic molecule can be expressed as EJ

rot=J(J+1)Bhc, express the rotational constant B of the molecule as a function of the reduced mass and intermolecular distance.

2

2

2 2

11

2

2 8 4 4 r o

J JE J J Bhc

IhB

I hc Ic Ic cM R

Page 4: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

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3. A continuous wave Nd:YAG (1064 nm) laser is longitudinally pumped at 808 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= 2.8x10-19 cm2 and the upper level lifetime is =230 s. Assume that an output coupler with a transmission T=10% is used, passive cavity losses are 3% per pass, branching ratio p=1, and a pump utilization efficiency a=75 %. Calculate the threshold pump power, slope efficiency, as well as the pump power required to obtain an output power Pout=20W from this laser.

11 2

16 3

v 1 3

1 0.03 11) ln ln 1 0.90 0.008310 10

2) 2.96 10

3) P 4.23 10 /( )

4)

th

thth

th pth

qe a

Threshold gain R R cml

Population inversion N cm

N hvThe absorbed threshold pump power per unit volume W cm

v

2

2 0.03

834

42.3 0.1 / 4 10 3.3

(1 0.9) 205) 1 0.75 0.759 3.3 ; 3.3 57 ; =37.3%; Alternatively:[1 ( 0.9)] 0.373

3 106.6 10 16)

vth th

out p a qe c p th p p slope

s

The absorbed pump power P P V W

P P P P P We

hvI

93 2 3 2

19 6

2 0.03

2 0.

064 10 2.9 10 / ; 2.9 10 0.1 / 4 22.82.8 10 230 10

1 22.87) ln (1 ) [1 ( 0.9)] 6.18081 0.751064

(1 0.9)8) 1 0.75 0.759[1 (

s s

s sth

p qe a p qe a

out p a qe c p th

W cm P I A W

P PP S e WS

P P Pe

03

206.1 ; 6.1 600.9)] 0.373p pP P W

Page 5: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

5

4.

0

0

Page 6: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

6

1.946 um

Page 7: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 8: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 9: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 10: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 11: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 12: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

Laser Physics I PH 481/581-3A (Mirov)Exam II (10/24/11)

Page 13: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 14: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 15: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft
Page 16: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

=12 m

Page 17: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft

Two important methodological problems related to diode pumped solid state lasers

Assume ring cavity & clockwise traveling wave

R=1

R=1R=1

Exper. 1: R1<1; Exper. 2: R2<1;

1. Internal lossestimation

Page 18: Review for Final - University of Alabama at Birminghammirov/Exam 3 - fall... · 10.75 1064 (1 0.9) 8) 1 0.75 0.759 [1 (ss ss th pqe a pqe a out p a qe c p th Wcm P IA W PP ... Microsoft