Laser Physics: HW3

Due 5pm Wednesday 23 Sept. 2015
Most of these problems are short except for the last one.

Problem 1: relative importance of stimulated emission

a. A two-level atom is placed within a cavity and is allowed to come into equilibrium with blackbody radiation of temperature T.  Show that the condition for the rate of stimulated emission from the upper level being equal to the rate of spontaneous emission is 
(Subscript[k, B])T=((h (Subscript[ν, 21]))/(ln 2)), 
where h (Subscript[ν, 21])is the energy spacing of the levels.

b. Find the temperature in electron volts (eV) and Kelvin when this condition is met for the transition in the following regions of the EM spectrum:

- radio frequencies at 50 Mhz
- microwaves at 1 GHz
- visible light at 500nm
- X-rays of energy 1 keV

c. Under this condition, what is the ratio of the populations per state in the upper and lower level? Explain your answer in terms of the spontaneous and stimulated transition rates between the two levels.

Problem 2.

a. What fraction of the total emission of a normalized Lorentzian lineshape of FWHM Δν occurs between (Subscript[ν, 0])-Δν and (Subscript[ν, 0])+Δν ? Numerical calculation is fine.

b. Repeat this calculation for a Gaussian lineshape.

Problem 3.

a. What is the stimulated emission rate if: (Subscript[A, 21])=(10^8)(s^(-1)), the line is homogeneously broadened with Δν=1 GHz, the input radiation is “white” with ρ(ν)=1 J (m^(-3))(Hz^(-1))?  Take the line center frequency to be (10^14)Hz.

b. What would be the stimulated emission rate if the input radiation were monochromatic, one FWHM from line center, and had intensity 1 W (m^(-2))?

Problem 4. 
Svelto problem 2.5.

Problem 5. 
Calculate the Doppler broadened linewidth for the following two cases:

a. the 488nm transition of an argon ion laser, given that the temperature of the discharge is around 6000K.

b. the 633nm transition of a helium-neon laser, given that the temperature of the discharge is around 400K.

Problem 6. 
Svelto problem 2.12. The saturation intensity is for saturation of absorption.

Problem 7.
A laser-pumped amplifier, pumped longitudinally, has an inversion density distribution 
(SuperStar[N])(r,z)=(SuperStar[(Subscript[N, 0])])exp[−2(r^2)/(Subscript[w, p]^2)−κ z]

a. The total stored (i.e. extractable) energy is (Subscript[E, stor]), the crystal length is L, and the photon energy is (Subscript[ℏω, L]). Calculate an expression for the peak inversion density (SuperStar[(Subscript[N, 0])]) by integrating over the volume of the crystal.

b. Now consider the pumping process to establish this inversion density. Suppose the available pump energy is (Subscript[E, 0])=50mJ, the crystal length is L=1cm, the pump radius is (Subscript[w, p])=2mm, and the absorption coefficient is κ = 3/cm. The pump wavelength is 532nm and the laser wavelength is 800nm. Calculate the peak inversion density (SuperStar[(Subscript[N, 0])]) in (cm^(-3)).

c. The small-signal single-pass gain in the amplifier is ultimately independent o the axial (z) variation of the inversion density. The small signal gain can be calculated as G(r)=exp[(Subscript[σ, 21])(∫_0^L)(SuperStar[N])(r,z)dz]=exp[(Subscript[Γ, stor])(r)/(Subscript[Γ, s])], where (Subscript[Γ, stor])(r) is called the stored energy fluence, and (Subscript[Γ, s])=((Subscript[ℏω, L])/(Subscript[σ, 21]))=0.85J/(cm^2) is the saturation fluence. Using the inversion density calculated i part b, calculate (Subscript[Γ, stor])(r).

d. Our input seed pulse (to be amplified) has an energy of (Subscript[E, in]), and the input energy fluence distribution is Γ(r)=(Subscript[Γ, 0])exp[-2(r^2)/((Subscript[w, L])^2)], with (Subscript[Γ, 0]) calculated so that integrating over the beam area gives (Subscript[E, in]). On one graph, plot t input an output energy fluences for (Subscript[w, L])=5mm  and (Subscript[w, L])=1mm . Make your plots two ways: first without rescaling (the amplified beam will be larger in height), and also by normalizing to make each peak = 1. This second plot illustrates how the beam profile is reshaped by the gain. The phenomena you will see is called spatial gain narrowing.