# THE LASER IDEA Consider two arbitrary energy levels 1 and 2 of a given material, and let N 1 and N 2 be their respective populations. If a plane wave with.

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THE LASER IDEA Consider two arbitrary energy levels 1 and 2 of a given material, and let N 1 and N 2 be their respective populations. If a plane wave with a photon flux F is travelling in the z- direction in the material, the elemental change dF of this flux along the elemental length dz of the material is due to both stimulated absorption and emission processes occurring in the shaded region. Let S be the cross-sectional area of the beam. SdF = (W 2l N 2 – W 12 N 1 )(Sdz) each absorption removes a photon, SdF must equal the difference between stimulated emission and absorption events occurring in the shaded volume per unit time

With the help of Esq.,, nonradiative decay does not add new photons, while photons created by radiative decay are emitted in any direction and thus give negligible contribution to the incoming photon flux F. Equation (1.2.1) shows that: the material behaves as an amplifier (dF/dz>0) if N 2 > g 2 N 1 /g 1 while it behaves as an absorber if N 2 < g 2 N 1 /g 1. At thermal equilibrium populations are described by Boltzmann statistics

Then if N 1 and N 2 are the thermal equilibrium populations of the two levels: where k is Boltzmann's constant and T is the absolute temperature of the material. In thermal equilibrium we thus have the material then acts as an absorber at frequency ν 0 if a nonequilibrium condition is achieved for which, then the material acts as an amplifier. In this case we say that there exists a population inversion in the material.

This means that the population difference is opposite in sign to what exists under thermodynarnic equilibrium the condition for thermodynarnic equilibrium where the material is an obsorber the condition for population inversion where the material is an active A material in which this population inversion is produced is referred to as an active medium.

Example: Calculate the population ratio N 1,N 2 for the two energy levels E 1,E 2 at temperature (300K) if the energy difference is 0.5ev, what is the of the photons prodused by this transition? Solution: This ratio means that at temperature 300K for 10 9 atoms in the ground level E 1 there are only 4 atoms in the exited energy level E 2 The wavelength of the photons:

How to make an oscillator from an amplifier material: it is necessary to introduce suitable positive feedback by placing the active material in a resonant cavity having a resonance at frequency ν 0. In the case of a laser, feedback is often obtained by placing the active material between two highly reflecting mirrors, such as the plane parallel mirrors in Fig. 1.3. In this case a plane em wave travelling in a direction perpen -dicular to the mirrors bounces back and forth between the two mirrors, and is amplified on each passage through the active material.

If one of the two mirrors (e.g. mirror 2) is partially transparent, a useful output beam is obtained from that mirror. It is important to realize that, for masers and lasers, a certain threshold condition must be reached. In the laser case, oscillation begins when the gain of the active material compensates the losses in the laser (e.g., losses due to output coupling). According to Eq. (1.2.1)

the gain per pass in the active material the ratio between output and input photon flux) is exp{σ[N 2 — (g 2 N 1 /g 1 )]Ɩ} where we denote for simplicity σ = σ 21, and where Ɩ is the length of the active material.

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