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LASER PHYSICS I PH 481/581-VT (MIROV) Exam II (10/26/15) STUDENT NAME: _____Key______STUDENT id #: ______------

ALL QUESTIONS ARE WORTH ------37.5 POINTS FOR------GRADUATES (WORK------OUT ANY 4 PROBLEMS)------AND 1.NOTE: A Clearly Fabry-Perot write out interferometer solutions50 POINTS and answers consistingFOR (circleUNDERGRADUATES of th twoe answers) identical by sectionmirrors, (WORK for air-spaced each partOUT (a., byANY b., a c.,distance 3 etc.) PROBLEMS)

L, is illuminated by a monochromatic em of tunable frequency. From a 9 Hz and its resolution is 30 MHz. measurement of the transmitted intensity versus the frequency of the input wave we find Calculate the spacing L of the interferometer, its finesse, and the reflectivity. that the free spectral1) range For of the a interferometer Fabry-Perot is 3x10 interferometer made of air-spaced , the free spectral range is: c v . Hence, the mirror spacing in our case is given by: FSR 2L cmms31011 1 Lmm91 50 22310vsFSR 2) The finesse of the interferometer, i.e. the ratio of free spectral range to width of the v 310 9 Hz transmission peak, is : F FSR  100 vHz30 106 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 2 2  RR210,  F the solution of which is R  0.968

3. A (  =2.09  m,  =1.15x10 - 2 0 cm 2 ,  =8 ms) measured to have an intensity of 100 2 emergingW/cm from one end of the laser, which has two identical mirrors each with transmission of 15%. The gain of the laser is also measured to be 0.5. a) What is the loss parameter “A” in the cavity? b) What is the optimum output mirror transmission? IIout s (1 AR ) aLR)  o ln 22(1) R hkW6.62 10 34 1.435 10 14 I  1.03 s 1.15 1020 8 10 3cm 2 c 310 8   1.435  1014 Hz  2.09 10 6 2 RIWcmL0.85; out100 / ; o  0.5 Everything is given. Let us find A 100 1030 10.8A 5  0.5 ln 0.85 22(1 0 .8 5 ) (0.15 A )  50 515 0.337; A 0.107 0.15 T  L bA) U se opt o 1 to find AA L 0.5 TAo A0.107 0.107 0.124 12.4% opt A 0.107 4. A helium-neon laser transition (0.6328  m) is Doppler broadened with a FWHM o f 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.  2  ( v  v o )   ln 2  2  (a) Find the number of longitudinal modes v that can oscillate simultaneously. (Hint:  D  th ( v )   ( v o ) e ). (b) Suppose all the modes are locked together: (1) What is the pulse spacing? (2) Estimate the pulse width.

6.7 ns 5. (a) What would be the minimum pulse duration  of a mode-locked chromium doped ZnS laser (gain bandwidth is 600 nm, central  o is 2350 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 2.3. 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   89 c3 10 (600 10 )   3.26  1013 Hz  292 (2350 10 ) ~30fs   1 b) ccc~ 30fs ; L c 9 m  c) Pulse spacing  2 1 (1.8 0.01)  (2.3 0.01) t 12.1ns 310 8 6. Consider the active medium Cr:ZnSe ( n=2.49, gain bandwidth (FWHM)  =860 nm, central wavelength  o =2400 nm. (a) Consider first a with length L=20 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  89 materialc surfaces3 10 (microchip (860 laser). 10 )What is the maximum13 thickness l that allows   3.26  10 Hz oscillation292 of only(2350 one longitudinal 10 ) mode?

3.26x10^13 4.7

2.49x3.26x10^13 3.7 m