Understanding Lasers: Chapter 3 Quiz

Source: Jeff Hecht, Understanding Lasers: An Entry-Level Guide, fourth edition (2019), Chapter 3 quiz, printed pages 91–93. The questions are paraphrased.

Quick answers

Question

Answer

1

b

2

a

3

c

4

e, \(1.221\)

5

c, \(3\%\)

6

e, \(1.36\times10^6\) wavelengths

7

a, about \(0.0013\ \mathrm{nm}\)

8

b, one nodal minimum

9

c

10

a

11

b, \(28.5\%\)

12

b, about \(9.7\%\)

Worked reasoning

  1. Four-level advantage: b. Its lower laser level is above the ground state and empties rapidly. A population inversion therefore needs far fewer excited particles than in a three-level system.

  2. Metastable state: a. Its long lifetime lets excited particles accumulate, making it suitable as an upper laser level.

  3. Growth by stimulated emission: c. Existing photons stimulate more matching photons, which can stimulate still more; unsaturated gain is exponential rather than merely additive.

  4. Amplification over 20 cm: e. For small-signal gain coefficient \(g=0.01\ \mathrm{cm^{-1}}\),

    \[G=e^{gL}=e^{(0.01)(20)}=e^{0.2}=1.221.\]

  5. Steady-state round-trip gain: c. Gain must replace the 2% internal loss and the 1% useful output coupling, or approximately \(3\%\) total.

  6. Round-trip length in wavelengths: e.

    \[N=\frac{2L}{\lambda} =\frac{0.60\ \mathrm m}{442\times10^{-9}\ \mathrm m} =1.36\times10^6.\]

  7. Adjacent longitudinal wavelengths: a. Near wavelength \(\lambda\), cavity resonances are separated by

    \[\Delta\lambda\approx\frac{\lambda^2}{2L} =\frac{(632.8\times10^{-9}\ \mathrm m)^2}{0.30\ \mathrm m} =1.34\times10^{-12}\ \mathrm m=0.00134\ \mathrm{nm}.\]

  8. TEM01 minimum: b. This first-order transverse mode has one internal nodal line separating its two bright lobes.

  9. Heating cannot create the inversion: c. Thermal equilibrium follows a Boltzmann distribution with fewer particles at higher energy. Selective optical or electrical pumping can drive a nonequilibrium inversion.

  10. Atmospheric absorption: a. It reduces power after the beam leaves the laser, not the laser’s electrical-to-optical conversion efficiency. The other choices waste excitation inside the conversion chain.

  11. Cascaded wall-plug efficiency: b. Successive efficiencies multiply:

    \[\eta_{\mathrm{wall}}=(0.95)(0.50)(0.60)=0.285=28.5\%.\]

  12. Quantum defect: b. With \(E=hc/\lambda\), the useful energy ratio is \(E_l/E_p=\lambda_p/\lambda_l\). Thus

    \[q=1-\frac{E_l}{E_p} =1-\frac{975}{1080}=0.0972\approx9.7\%,\]

    which rounds to the listed \(9.75\%\).