IZhO 2019, Теория — Задача 3. Laser

XV International Zhautykov Olympiad/Theoretical Competition · 11 януари 2019 г. · 10 т.

Please read this first:

THEORETICAL COMPETITION

January 11, 2019

  1. The time available for the theoretical competition is 4 hours. There are three questions.

  2. Use only the pen provided.

  3. You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.

  4. You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures and plots.

  5. Use only the front side of Writing sheets**.** Write only inside the boxed area.

  6. Begin each question on a separate sheet of paper.

  7. Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.

  8. At the end of the exam, arrange all sheets for each problem in the following order:

    • Used Writing sheets in order.
    • The sheets you do not wish to be evaluated.
    • Unused sheets.
    • The printed problems.

    Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper out of the room.

SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

Attention. Points in grading are not divided!

Условие

According to quantum theory, a molecule can only be found in certain states characterized by discrete energy values E0,E1,E2,…E_0, E_1, E_2, \ldots. These states are represented by horizontal segments on the vertical energy scale and are numbered in order of increasing energy, starting from 0. In the absence of external influence, the molecule is found in the state with the number 0 and the lowest possible energy E0E_0. This state is then called a ground state, and the rest are then called excited states. The molecule can be driven from one energy state to another by absorbing or emitting light quanta, i.e. photons. The intensity of the light flux is characterized everywhere below by the density of the photon flux II, i.e. the number of photons passing perpendicularly through the unit area per unit of time, the dimension of this quantity is obviously equal to [I]=m−2⋅s−1[I] = \mathrm{m^{-2}\cdot s^{-1}}.

Energy scale: horizontal level segments labelled E0 (lowest), E1 and E2, with dots above indicating further levels.

For further consideration, it is necessary to take into account the following processes that take place when the light flux passes through a medium.

Absorption. If a molecule is found in the ground state 0, then it can absorb a photon and be raised to the excited state 1. Such a transition is possible if the photon energy is equal to the difference between the energies of the excited and ground states, hν01=E1−E0h\nu_{01} = E_1 - E_0, where ν01\nu_{01} is the photon frequency and h=6.63⋅10−34 J⋅sh = 6.63\cdot 10^{-34}\ \mathrm{J\cdot s} stands for the Planck constant. If N0N_0 molecules are found in the ground state 0, then the number of molecules dNdN that have absorbed photons and have been elevated to the excited state in a very short time interval dtdt is equal to

dN=Iσ01N0dt,(1)dN = I\sigma_{01}N_0dt,\qquad (1)

where the quantity σ01\sigma_{01} is called the absorption cross section and is determined by the properties of the molecules.

Absorption: a wavy photon arrow hits a molecule (cloud), a block arrow leads to the excited molecule (dotted cloud); level diagram with an upward arrow from level 0 to level 1.

Stimulated emission. If a molecule is found in the excited state 1, then under the influence of a photon with the frequency ν01\nu_{01}, it can be forced to move to the ground state 0 with the emission of another photon that is completely identical to the first one and propagates in the same direction, has the same energy and polarization. The number of such transitions dNdN for a very small time interval dtdt is described by the formula similar to formula (1) as

dN=Iσ10N1dt,(2)dN = I\sigma_{10}N_1dt,\qquad (2)

where N1N_1 denotes the number of molecules in the excited state 1 with energy E1E_1, and σ10\sigma_{10} signifies the cross section of the stimulated emission.

Stimulated emission: a wavy photon arrow hits an excited molecule (dotted cloud), a block arrow leads to the molecule and two identical wavy photon arrows; level diagram with a downward arrow from level 1 to level 0.

Spontaneous emission. A molecule in the excited state can spontaneously drop to its ground state with the emission of a photon. In contrast to the stimulated emission, the direction of emission and the polarization of the photon are both random, and the energy can slightly vary, i.e. the spontaneous emission cannot contribute to an increase in the light flux. The number of spontaneous transitions dNdN from the excited state 1 to the ground state 0 for a very small time interval dtdt is written as

dN=AN1dt=1τN1dt,(3)dN = AN_1dt = \frac{1}{\tau}N_1dt,\qquad (3)

where N1N_1 still designates the number of molecules in the excited state 1 with energy E1E_1, AA is the transition probability or the Einstein coefficient, and its inverse value τ=A−1\tau = A^{-1} is referred to as the lifetime of the excited state.

Spontaneous emission: an excited molecule (dotted cloud), a block arrow, then the molecule emitting a wavy photon in a random direction; level diagram with a wavy downward arrow from level 1 to level 0.

To describe each state kk of the molecule, it is convenient to use not the total number NkN_k of molecules in the state, but its reduced value divided by the total number of molecules NN in the medium

nk=NkN,(4)n_k = \frac{N_k}{N},\qquad (4)

This value is called the population of the state. For populations, the normalizing condition is satisfied: the sum of the populations of all the states of the molecules is equal to unity, i.e.

n0+n1+n2+…=1.(5)n_0 + n_1 + n_2 + \ldots = 1.\qquad (5)

If the number of molecules in a certain excited state exceeds the number of molecules in the ground state, then this state of the medium is called the population inversion, and the emitted radiation may prevail over absorption, which results in an increase in the intensity of the light flux propagating in such a medium. This phenomenon is used in optical quantum light generators, called lasers. The population inversion is conventionally created using an external energy source, called pumping. In this problem, we consider the operation of an optically pumped laser, when the population inversion is generated by an external light flux. Unlike this pumping light flux, the laser light flux is monochromatic, coherent, polarized and narrowly directed.

Population inversion: two-level system

Consider a medium that is subject to monochromatic pumping light flux I0I_0. The incident light flux leads to transitions of molecules only between two states, i.e. the ground state 0 and the excited state 1. The absorption σ01\sigma_{01} and stimulated emission σ10\sigma_{10} cross sections are equal, i.e. σ10=σ01=σ\sigma_{10} = \sigma_{01} = \sigma and the lifetime in the excited state is found to be τ\tau.

Two-level diagram: levels 0 and 1 with an upward arrow, a downward arrow and a wavy downward arrow between them.

3.1 Write down an equation describing the change in the population of the excited state 1 over time, i.e. express the derivative dn1/dtdn_1/dt in terms of n1n_1, I0I_0, σ\sigma and τ\tau.

3.2 Find the population of the excited state and the difference of the populations between the excited and ground states under steady conditions as functions of the pumping light flux I0I_0. Express your answer in terms of the parameter I0στI_0\sigma\tau.

3.3 Is it possible to achieve an amplification of the laser light flux in this case?

Population inversion: three-level system

Let three states of the molecule be involved in possible transitions: the ground state 0 and the two excited states 1, 2. Under the action of the pumping light flux I0I_0, the molecule can be raised from the ground state 0 to the first excited state 1. The absorption cross section of this transition is equal to σ\sigma. As a result of intramolecular relaxation, the molecule that has resided in state 1 almost instantly drops to the lower energy state 2, whose lifetime is τ\tau. In this case, assume that the stimulated emission is completely absent.

Three-level diagram: an upward arrow from level 0 to level 1, a short wavy arrow from level 1 down to level 2, and a long wavy arrow from level 2 down to level 0.

3.4 Write down an equation describing the change in the population n2n_2 of the excited state 2 over time.

3.5 Find the population nˉ2\bar{n}_2 of the excited state 2 and the population difference (nˉ2−nˉ0)\left(\bar{n}_2 - \bar{n}_0\right) of the excited and the ground states under steady conditions as functions of the pumping light flux I0I_0. Express your answer in terms of the parameter I0στI_0\sigma\tau.

3.6 At what minimum value of the parameter I0στI_0\sigma\tau is it possible to amplify the laser light flux with the frequency equal to the frequency of the transition 2→02 \to 0?

Population inversion: four-level system

Let four states of the molecule be involved in possible transitions. Under the action of the pumping light flux I0I_0, the molecule can be raised from the ground state 0 to the first excited state 1. The absorption cross section of this transition is equal to σ\sigma. As a result of intramolecular relaxation, the molecule that has resided in state 1 almost instantly drops to the lower energy state 2, whose lifetime is τ\tau. From this state, the molecule undergoes a transition to the intermediate state 3, which results in the emission of a photon. In this case, assume that the stimulated emission is completely absent. As a result of intramolecular relaxation, the molecule that has fallen into state 3 almost instantly drops to the ground energy state 0.

Four-level diagram: an upward arrow from level 0 to level 1, wavy arrows from level 1 down to 2, from 2 down to 3, and from 3 down to 0.

3.7 At what minimum value of the parameter I0στI_0\sigma\tau is it possible to amplify the laser light flux with the frequency equal to the frequency of the transition 2→32 \to 3?

Resonator

The four-level system is usually implemented in dye solutions. Dyes consist of complex molecules with multiple energy states. Therefore, the possible energy states are grouped into bands: the ground state band S0S_0 contains an almost continuous spectrum of sub-levels, and the same is true for the first excited state band S1S_1. Thus, there are two bands of possible states. The absorption of the pumping light flux I0I_0 leads to transitions from the sublevels of the ground state band S0S_0 to different sublevels of the excited state band S1S_1. Transitions between the sub-levels of state 1 occur almost instantaneously, so the stimulated emission of the laser light flux I1I_1 occurs at lower frequencies, and the stimulated emission due to the pumping light flux I0I_0 can be neglected. In the approximation described, it is sufficient to know the population of the ground and excited state bands. Rhodamine 6G is used as a dye, for which: the absorption cross section in the transition S0→S1S_0 \to S_1 is σA=3,90⋅10−16 cm2\sigma_A = 3{,}90\cdot 10^{-16}\ \mathrm{cm^2}; the stimulated emission cross section S1→S0S_1 \to S_0 is σE=2,20⋅10−16 cm2\sigma_E = 2{,}20\cdot 10^{-16}\ \mathrm{cm^2}; the lifetime of the molecule in the state S1S_1 is τ=4,20⋅10−9 s\tau = 4{,}20\cdot 10^{-9}\ \mathrm{s}.

To generate light, cuvette 1 with the solution of rhodamine 6G is placed between two parallel mirrors 2 and 3, thus forming a resonator. The solution is excited by a uniform pumping light flux 4 of intensity I0I_0, whose frequency strictly corresponds to the maximum absorption of the solution. The pumping light flux 4 is directed perpendicular to the axis of the resonator and fully illuminates the entire cell. The intensity of this flux, of course, decreases as it passes through the solution; however, for carrying out our estimations, consider I0I_0 constant in the entire bulk of the resonator, assuming it to be averaged over the solution volume. The laser light flux 5 generated in the resonator propagates along the resonator axis, and its amplification occurs due to multiple reflections from the resonator mirrors. Consider mirror 2 fully reflective, and the second mirror 3 translucent with the reflectance ρ\rho. The absorption of light in the mirrors, the solvent-body, as well as the scattering of light and other losses can be completely ignored. The resonator has the following parameters: the cuvette length is l=3,00 cml = 3{,}00\ \mathrm{cm}; the rhodamine 6G concentration is γ=1.30⋅1016 cm−3\gamma = 1.30\cdot 10^{16}\ \mathrm{cm^{-3}}; the reflection coefficient of the translucent mirror is ρ=0,90\rho = 0{,}90; the refractive index of the solution of rhodamine 6G is r=1,50r = 1{,}50. The speed of light is denoted as c=3,00⋅1010 cm/sc = 3{,}00\cdot 10^{10}\ \mathrm{cm/s}.

For a simplified description of the laser light flux propagating along the resonator axis, one can consider the intensity of the light fluxes averaged over the length of the resonator. We denote the average intensity of the laser light flux propagating to the translucent mirror as IGI_G, and the intensity of the laser light flux propagating in the opposite direction as IG′I'_G. Since the transmittance of mirror 3 is small, then we can assume that the average intensities of these fluxes be approximately equal IG≈IG′I_G \approx I'_G.

Band diagram of the dye: ground band S0 and excited band S1, each drawn as a stack of closely spaced sub-levels; an upward arrow I0 from S0 to S1, a downward arrow I1 from the bottom of S1 to S0, and short wavy relaxation arrows inside each band.
Resonator sketch: cuvette 1 between two parallel plate mirrors 2 (left) and 3 (right); pumping light flux 4 drawn as vertical arrows falling on the cuvette from above; laser light flux 5 as a horizontal block arrow along the axis inside the cuvette, with a dashed arrow leaving through mirror 3.
Schematic resonator between two mirrors: arrow to the right labelled I_G and arrow to the left labelled I′_G inside; a dashed arrow leaving through the right mirror labelled (1 − ρ)I_G.

3.8 Let a laser light flux IGI_G be created in the resonator. Show that in the absence of absorption and stimulated emission, the change in the intensity of the laser flux over time is described by the equation

dIGdt=−1TIG,(6)\frac{dI_G}{dt} = -\frac{1}{T}I_G,\qquad (6)

where TT stands for the so-called photon lifetime in the resonator. Express the parameter TT in terms of the parameters of the resonator. Calculate its numerical value.

3.9 Show that in the absence of the photon losses through mirror 3, the laser light flux variation over time obeys the following equation

dIGdt=KnIG,(7)\frac{dI_G}{dt} = KnI_G,\qquad (7)

where nn designates the population of the excited state of rhodamine 6G, and KK is the resonator gain. Express the resonator gain KK in terms of the parameters of the resonator and the stimulated emission cross section σE\sigma_E of rhodamine 6G. Calculate its numerical value.

Stationary generation mode

In this part, we assume that the pumping light flux is constant and does not depend on time. In the stationary mode, all quantities remain constant: the population of the excited state and the laser light flux. Assume that the population of the excited state is low, i.e. n≪1n \ll 1.

3.10 Write down a set of equations describing the change in the population dndt\frac{dn}{dt} of the excited state and the laser light flux dIGdt\frac{dI_G}{dt} in the resonator.

3.11 Obtain the formula and calculate a minimum (threshold) value nthn_{th} of the population of the excited state at which the amplification (generation) of laser light in the resonator occurs. Express it in terms of the parameters of the resonator K,TK, T.

3.12 Derive the formula and calculate a minimum (threshold) value of the pumping light flux I0,thI_{0,th} at which the laser light amplification in the resonator starts. Let the wavelength of the pumping light flux be λ=520nm\lambda = 520\mathrm{nm}. Calculate the pumping flux in energetic units of W/cm2\mathrm{W/cm^2}.

3.13 Find the laser flux at the output of the resonator as a function of the pumping light flux I0I_0, and express it in terms of the ratio η=I0/I0,th\eta = I_0 / I_{0,th}, which is called the threshold overrun, and the characteristics of molecules. Draw a graph of the laser flux at the output of the resonator as a function η\eta.

3.14 Find the quantum output f=NE/NAf = N_E / N_A, i.e. the ratio of the number of photons NEN_E leaving the resonator per unit of time to the number of photons NAN_A absorbed in the resonator per the same unit of time as a function of the parameter η\eta.

Отговори

Решение

Mathematical hint for the theoretical competition

You may need to know the following integrals:

∫dxax+b=1aln⁡∣ax+b∣.\int\frac{dx}{ax+b} = \frac{1}{a}\ln|ax + b|.
∫xndx=xn+1n+1, where n is an integer number\int x^n dx = \frac{x^{n+1}}{n+1},\ \textit{where } n \textit{ is an integer number}

Оригинал в Архива: IZhO-2019-Theory_eng.pdf · официални решения: IZhO-2019-Theory_eng_sol.pdf