PHY 108: Optical Physics. Solution to Midterm Test

Transcription

1 PHY 108: Optial Physis Solution to Midterm Test TA: Xun Jia 1 May 14, jiaxun@physis.ula.edu

2 Spring 2008 Physis 108 Xun Jia (May 14, 2008) Problem #1 For a two mirror resonant avity, the resonane requirement is nλ 2L. (a). (5pt) Derive an expression for frequeny spaing, ν axial, for the suessive resonanes of the optial avity. (b). (5pt) Calulate ν axial for L 1, 10 and 100 m with vauum in the avity. (). (5pt) Calulate ν axial for L 100 m with the avity filled with a medium with refrative index of 2.3 and 3.5. Hint: Veloity of light in a medium depends on the refrative index, n, of the medium and is given by /n. From the resonane ondition mλ 2L. (a). Sine λν, the resonane ondition an be equivalently written ν m/2l, for m being interger. When m inreases by one, the resonane frequeny ν inreases by ν axial, i.e.: ν axial 2L. (1) (b). For L 1, 10 and 100 m with vauum in the avity, ν axial is alulated to be , , and Hz. (). In a medium with refrative index of n the speed is redued to /n. Therefore, substitute /n, we obtain: ν axial 2nL. (2) for L 100 m with the avity filled with a medium with refrative index of 2.3 and 3.5, ν axial is alulated to be , and Hz. Problem #2 Consider the resonator geometry shown in Fig. with eah leg being L m long: L (separation between eah of the three lossless mirrors) 100 m, R 1 R 2 R , Intput 1 W, L α 10 m. Assume steady state has been reahed. (a). For α 0 (5pt) Calulate avity resonane separation in MHZ. (5pt) Calulate round trip avity gain at resonane. 1

3 Spring 2008 Physis 108 Xun Jia (May 14, 2008) Out 2 Loss Medium Input Refletion Cir Out 1 1 R 1, T 1 R 2, T 2 Cir 3 Cir 2 R 3, T 3 Figure 1: The illustration of the laser resonator. (b). Now if α 0.01 m 1 (5pt) Calulate avity resonane separation in MHZ. (5pt) Calulate round trip avity gain at resonane. (). At some time after steady state having been reahed, we turn off the input. (5pt) Calulate Out 1, and Out 2 and their deay time. (5pt) Calulate Refletion and its deay time. (a). For α 0, there is no energy loss while propagating inside the avity. The resonane ondition in this ase is mλ 3L for integer m. Sine λν, the resonane ondition an be equivalently written ν m/3l. Therefore, the avity resonane separation is: ν 100 MHZ. (3) 3L For a round trip, the light is refleted by three times, eah by one mirror; at the same time, it aquires a phase fator of 2π 3L λ in a round trip. Thus: g rt r 1 r 2 r 3 e 6πL/λ R 1 R 2 R (4) 2

4 Spring 2008 Physis 108 Xun Jia (May 14, 2008) where we used the fat that on the resonane, the phase fator is a multiple of 2π. (b). Now if α 0.01 m 1 : The presene of absorption will not affet the resonane ondition, thus the avity resonane separation is still 100 MHZ. Now, in addition to the loss on the mirror refletion, the intensity is also redued by a fator of e αlα due to the absorption. Note the eletri field only gains a fator of e αlα/2. So the round trip gain: g rt r 1 r 2 r 3 e αlα/ (5) (). At some time after steady state having been reahed, we turn off the input. The intensity of Out 2 is always zero due to the lokwise propagation diretion of the laser in the avity. To alulate Out 1 intensity, onsider at the steady state, whih gives: so the intensity of Out 1 is: E Cir1 jt 1 E Input + g rt E Cir1, (6) I Cir1 T 1 (1 g rt ) 2I Input. (7) I Out1 I Cir1 e αlα T 2 T 1T 2 e αlα (1 g rt ) 2 I Input. (8) This is also the expression of Out 1 intensity at the moment when input is shut off. With the parameters given in the problem, I Out W and I Out W for α 0 and α 0.01 m 1. For the deay, in a round trip time 3L/, the intensity of the light dereased by a fator of 1 R 1 R 2 R 3 due to the leak at mirrors and a fator of 1 e αlα αl α due to the absorption. Therefore: di Cir1 dt whih gives a deay time of: τ (1 R 1R 2 R 3 + αl α ) I Cir1, (9) 3L 3L (1 R 1 R 2 R 3 + αl α ). (10) This is also the deay time of Out 1, sine the light intensity of Out 1 is always proportional to that of the Cir 1, and therefore they have the same deay time. In the ases of α 0 and α 0.01 m 1, we obtain the deay time of τ se and τ se, respetively. 3

5 Spring 2008 Physis 108 Xun Jia (May 14, 2008) For the refletion light, one the input light is turned off, its intensity is: I Refl I Cir1 e αlα R 2 R 3 T 1 R 2R 3 T 2 1 e αlα (1 g rt ) 2 I Input. (11) The values are I Refl W and I Refl W for α 0 and α 0.01 m 1. The deay time for the refletion light is the same as in the previous part, sine the refletion light is also always proportional to I Cir1. Problem #3 Consider a light pulse with following parameters: Pulse width 10 femtosetond, (1 fs se). Center wavelength 600 nm. Pulse energy 1 millijoule. Shape is approximately retangular. (a). The pulse is propagating through vauum (5pt) Calulate the peak power in the pulse at the start. (5pt) Calulate the pulse width (in time domain) after it has traveled through 100 meters of vauum. (b). The pulse is propagating through normal air having its refrative index, n, given by: ) (1 + Bλ2 (n 1) A with A and B and λ is in mirometers. (12) (15pt) Calulate the approximate pulse width (in time domain) after it has propagated following distane through normal air: (a) 10 m, (b) 100 m, () 100 m. (10pt) Peak power in the pulse for eah of the three ases above. Hint: Veloity of light through air at any frequeny, ν, an be alulated by using Eqn. (12). Approximate the width of the pulse in the frequeny domain as 1/pulse width. From propagation times for the enter frequeny and extremes, realulate the pulse width at the desired loation. 4

6 Spring 2008 Physis 108 Xun Jia (May 14, 2008) (a). When the pulse is propagating through vauum, The peak power in the pulse at the start is simply: P E t 1 millijoule 10 femtosetond 1011 W. (13) The pulse width (in time domain) after it has traveled through 100 meters of vauum is still 10 femtosetond, sine all omponents of wave travel at the same speed, and the pulse shape is thus preserved. (b). When the index depends on the wavelength, waves with different wavelength will propagate at different speed. For the wave with enter frequeny ν 0 / λ 0, where λ 0 is the enter wavelength, its veloity is: v n ( 1 + A 1 + b ). (14) λ 2 0 So the time it takes to travel a distane of L is: [ ( )] t L v L 1 + A 1 + b λ 2 0. (15) Now for the omponent with extreme frequeny, its wavelength is: λ e ν e ν 0 + ν 2 + 1, (16) λ 0 4π t where t is the initial pulse width, and ν 1/2π t. With the expression for λ e, we ould alulate its veloity, and then the time t e it takes to travel a distane of L: [ ( )] t e L v e The final pulse width t f therefore approximately: L 1 + A 1 + b λ 2 e. (17) in time domain after traveling L distane is t f 2 t e t + t 2L ( 1 AB 1 λ 2 e λ 2 ) + t. (18) Aording to the data given, λ µm, λ e µm. Therefore, after traveling a distane of L, the final pulse width: 5

7 Spring 2008 Physis 108 Xun Jia (May 14, 2008) L 10 m: t f fs. L 100 m: t f fs. L 100 m: t f fs. (10pt) For eah of the above ases, the peak power P E/ t f is alulated as: L 10 m: P W. L 100 m: P W. L 100 m: P W. Problem #4 (20pt)An optial beam propagating through a medium of length L has the following parameters: L 10 m. I out 0.9I in. (a). What is the absorption oeffiient of the medium. (b). Determine I out for L 20 m. (). Determine I out for L 200 m. (d). Determine I out for L 5000 m. (a). The absorption oeffiient of the medium is: α 1 L ln I out I in m 1. (19) (b). For L 20 m: (). For L 200 m: (d). For L 5000 m: I out I in e αl 0.81I in. (20) I out I in e αl 0.121I in. (21) I out I in e αl I in. (22) 6