光电子器件Ch-4.doc

Chapter 4Problem 4-1.4.2The He-Ne Laser A particular He-Ne laser operating at 632.8 nm has a tube that is 50 cm long. The operating temperature is 130 CaEstimate the Doppler broadened linewidth (Dl) in the output spectrum.bWhat are the mode number m values that satisfy the resonant cavity condition? How many modes are therefore allowed? cWhat is the separation Dum in the frequencies of the modes? What is the mode separation Dlm in wavelength.dShow that if during operation, the temperature changes the length of the cavity by dL, the wavelength of a given mode changes by dlm,Given that typically a glass has a linear expansion coefficient a 10-6 K-1, calculate the change dlm in the output wavelength (due to one particular mode) as the tube warms up from 20 C to 130 C, and also per degree change in the operating temperature. Note that dL/L = adT, and L = L1 + a(T - T). Change in mode wavelength dlm with the change dL in the cavity length L is called mode sweeping.eHow do the mode separations Dum and Dlm change as the tube warms up from 20 C to 130 C during operation? fHow can you increase the output intensity from the He-Ne laser?4.2 SolutionaThe central emission frequency isuo = c/lo = (3108 m s-1) / (632.810-9 m) = 4.741014 s-1. The FWHM width of the frequencies Du1/2 observed will be given by Eq. (3) = 1.515 GHzTo get FWHM wavelength width Dl1/2, differentiate l = c/u so that Dl1/2 Du1/2|-l/u| = (1.515109 Hz)(632.810-9 m) / (4.741014 s-1) orDl1/2 2.0210-12 m or 0.00202 nm.This width is between the half-points of the spectrum. bFor l = lo = 632.8 nm, the corresponding mode number mo is,mo = 2L / lo = (20.5 m) / (632.810-9 m) = 1580278.1and actual mo has to be the closest integer value to 1580278.1, that is 1580278Consider the minimum and maximum wavelengths corresponding to the extremes of the spectrum at the half-power points:= 632.798987and= 632.801012One can now increase and decrease mo step by step and calculate the corresponding wavelength lm for each choice of m, and examine whether it is within lmin lm lmaxlm lminlm lminlm lminlm lminlm lminlm lmaxlm lminlm lmaxlm lminlmin lm Dl1/2 = 2.02 pm.As the tube warms up, the modes sweep across the output spectrum. This is called mode sweeping. eConsider the mode separation in frequency,since a = L-1dL/dT.Thus,= -300 Hz K-1.For dT = 110 K, dDum = (-300 Hz K-1)(110 K) = -33 kHz.Change in the mode separation is 0.014%. Consider the mode separation in wavelength,since a = L-1dL/dT.Thus,= 4.00410-19 m K-1.For dT = 110 K, dDlm = (4.00410-19 m K-1)(110 K) = 4.4110-17 mChange in the mode wavelength separation is 0.011%.fIncrease the power input (to excite more atoms), decrease tube diameter to encourage collisions with walls to increase the rate of return to the ground state, and increase the active volume by increasing length (not by increasing the tube diameter).Problem 4-2.4.3The Ar ion laser The argon-ion laser can provide powerful CW visible coherent radiation of several watts. The laser operation is achieved as follows: The Ar atoms are ionized by electron collisions in a high current electrical discharge. Further multiple collisions with electrons excite the argon ion, Ar+, to a group of 4p energy levels 35 eV above the atomic ground state as shown in Figure 4Q3. Thus a population inversion forms between the 4p levels and the 4s level which is about 33.5 eV above the Ar atom ground level. Consequently, the stimulated radiation from the 4p levels down to the 4s level contains a series of wavelengths ranging from 351.1 nm to 528.7 nm. Most of the power however is concentrated, approximately equally, in the 488 and 514.5 nm emissions. The Ar+ ion at the lower laser level (4s) returns to its neutral atomic ground state via a radiative decay to the Ar+ ion ground state, followed by recombination with an electron to form the neutral atom. The Ar atom is then ready for pumping again.aCalculate the energy drop involved in the excited Ar+ ion when it is stimulated to emit the radiation at 514.5 nm. bThe Doppler broadened linewidth of the 514.5 nm radiation is about 3500 MHz (Du) and is between the half-intensity points.(i)Calculate the Doppler broadened width in the wavelength; Dl(ii)Estimate the operation temperature of the argon ion gas; give the temperature in C cIn a particular argon-ion laser the discharge tube, made of Beryllia (Beryllium Oxide), is 30 cm long and has a bore of 3 mm in diameter. When the laser is operated with a current of 40 A at 200 V dc, the total output power in the emitted radiation is 3 W. What is the efficiency of the laser?Figure 4Q34.3 SolutionaGiven wavelength lo = 514.5 nm or 514.5 10-9 m, the change in the energy isDE = hc/lo = (6.626 10-34 J s)(3.0 108 m s-1)/(514.5 10-9 m)i.e.DE = 3.86 10-19 J or 2.41 eVbThe center frequency of the lasing emission is = 5.831 1014 s-1The relationship between wavelength and frequency linewidths isDl = 3.09 10-12 m or 0.00309 nmThe frequency linewidth Du = 3500 106 s-1 is due to Doppler broadening. From Du we can find the operating temperature by using,in which M is the mass of the Ar ion, that is, M = 39.95 g mol-1 or 6.63610-26 kg.= 2810 K = 2537 Cwhich is the gas temperature. The tube gets very hot. Indeed, beryllia (BeO) is used because the operating temperature of the argon ion laser is high and beryllia has a high melting temperature and good thermal conductivity.cThe output power from the laser is 3 W. The input power is IV or (40 A)(200 V). = 0.0375%Problem 4-3.4.4 Einstein coefficients and critical photon concentration r(hu) is the energy of the electromagnetic radiation per unit volume per unit frequency due to photons with energy hu = E2 - E1. Suppose that there are nph photons per unit volume. Each has an energy hu. The frequency range of emission is Du. Then,Consider the Ar ion laser system. Given that the emission wavelength is at 488 nm and the linewidth in the output spectrum is 5109 Hz (between half intensity points, that is Du = 25109 Hz), estimate the photon concentration necessary to achieve more stimulated emission than spontaneous emission.4.4 SolutionSuppose that there are nph photons per unit volume. Each has an energy hu. The frequency range of emission is Du as is the linewidth between half intensity points. Then,Now, = 1.4310-13 J s m-3nph =3.51015 Photons m-3.Note that this represents the critical photon concentration for stimulated emission to just exceed spontaneous emission in the absence of any photon losses. It does not represent the photon concentration for laser operation. In practice, the photon concentration is much greater during laser operation.Problem 4-4.4.5 Photon concentration in a gas laser The Ar ion laser has a strong lasing emission at 488 nm. The laser tube is 1 m in length, and the bore diameter is 3 mm. The output power is 1 W. Assume that most of the output power is in the 488 nm emission. Assume that the tube end has a transmittance T of 0.1. Calculate the photon output flow (number of lasing photons emitted from the tube per unit time), photon flux (number of lasing photons emitted per unit area per unit time), and estimate the order of magnitude of the steady state photon concentration (at 488 nm) in the tube (assume that the gas refractive index is approximately 1).4.5SolutionThe 488 nm photon energy isEph = hu = hc/l = 4.0710-19 J or 2.54 eV.Tube cross section areaA = p(Diameter/2)2 = p(0.003/2)2 = 7.0710-6 m2.If Po is the output power then photon output flow is= 2.46 1018 Photons s-1This lasing photons emitted per unit time from the tube.If Gph is the photon flux then, by definition of photon flux,Thus,Gph =3.481023 Photons s-1 m-2We can estimate the photon concentration as follows. It takes Dt seconds for a photon to cross the laser tube, then Dt = L/c. We assumed that the refractive index is roughly 1. Suppose that nph is the 488 nm photon concentration in the tube. Then at every Dt seconds a fraction T will escape - transmitted through to outside. Number of photons transmitted every Dt seconds isEquating this to output photon flowwe find 1 1016 (488 nm lasing photons) m-3.This is the steady state 488 nm photon concentration. Problem 4-5.4.11InGaAsP-InP Laser Consider a InGaAsP-InP laser diode which has an optical cavity of length 250 microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is 4. The optical gain bandwidth (as measured between half intensity points) will normally depend on the pumping current (diode current) but for this problem assume that it is 2 nm.aWhat is the mode integer m of the peak radiation?bWhat is the separation between the modes of the cavity? cHow many modes are there in the cavity?dWhat is the reflection coefficient and reflectance at the ends of the optical cavity (faces of