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
Brief solution
1. Reasoning and answer.
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.
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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.
Brief solution
1. Reasoning and answer.
Metastable state: a. Its long lifetime lets excited particles accumulate, making it suitable as an upper laser level.
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Metastable state: a. Its long lifetime lets excited particles accumulate, making it suitable as an upper laser level.
Brief solution
1. Reasoning and answer.
Growth by stimulated emission: c. Existing photons stimulate more matching photons, which can stimulate still more; unsaturated gain is exponential rather than merely additive.
Show detailed steps
Growth by stimulated emission: c. Existing photons stimulate more matching photons, which can stimulate still more; unsaturated gain is exponential rather than merely additive.
Brief solution
1. Reasoning and answer.
Amplification over 20 cm: e. For small-signal gain coefficient \(g=0.01\ \mathrm{cm^{-1}}\),
2. Key calculation.
\[G=e^{gL}=e^{(0.01)(20)}=e^{0.2}=1.221.\]Show detailed steps
Amplification over 20 cm: e. For small-signal gain coefficient \(g=0.01\ \mathrm{cm^{-1}}\),
\[G=e^{gL}=e^{(0.01)(20)}=e^{0.2}=1.221.\]Brief solution
1. Reasoning and answer.
Steady-state round-trip gain: c. Gain must replace the 2% internal loss and the 1% useful output coupling, or approximately \(3\%\) total.
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Steady-state round-trip gain: c. Gain must replace the 2% internal loss and the 1% useful output coupling, or approximately \(3\%\) total.
Brief solution
1. Reasoning and answer.
Round-trip length in wavelengths: e.
2. Key calculation.
\[N=\frac{2L}{\lambda} =\frac{0.60\ \mathrm m}{442\times10^{-9}\ \mathrm m} =1.36\times10^6.\]Show detailed steps
Round-trip length in wavelengths: e.
\[N=\frac{2L}{\lambda} =\frac{0.60\ \mathrm m}{442\times10^{-9}\ \mathrm m} =1.36\times10^6.\]Brief solution
1. Reasoning and answer.
Adjacent longitudinal wavelengths: a. Near wavelength \(\lambda\), cavity resonances are separated by
2. Key calculation.
\[\Delta\lambda\approx\frac{\lambda^2}{2L} =\frac{(632.8\times10^{-9}\ \mathrm m)^2}{0.30\ \mathrm m} =1.34\times10^{-12}\ \mathrm m=0.00134\ \mathrm{nm}.\]Show detailed steps
Adjacent longitudinal wavelengths: a. Near wavelength \(\lambda\), cavity resonances are separated by
\[\Delta\lambda\approx\frac{\lambda^2}{2L} =\frac{(632.8\times10^{-9}\ \mathrm m)^2}{0.30\ \mathrm m} =1.34\times10^{-12}\ \mathrm m=0.00134\ \mathrm{nm}.\]Brief solution
1. Reasoning and answer.
TEM01 minimum: b. This first-order transverse mode has one internal nodal line separating its two bright lobes.
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TEM01 minimum: b. This first-order transverse mode has one internal nodal line separating its two bright lobes.
Brief solution
1. Reasoning and answer.
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.
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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.
Brief solution
1. Reasoning and answer.
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.
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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.
Brief solution
1. Reasoning and answer.
Cascaded wall-plug efficiency: b. Successive efficiencies multiply:
2. Key calculation.
\[\eta_{\mathrm{wall}}=(0.95)(0.50)(0.60)=0.285=28.5\%.\]Show detailed steps
Cascaded wall-plug efficiency: b. Successive efficiencies multiply:
\[\eta_{\mathrm{wall}}=(0.95)(0.50)(0.60)=0.285=28.5\%.\]Brief solution
1. Reasoning and answer.
Quantum defect: b. With \(E=hc/\lambda\), the useful energy ratio is \(E_l/E_p=\lambda_p/\lambda_l\). Thus
2. Key calculation.
\[q=1-\frac{E_l}{E_p} =1-\frac{975}{1080}=0.0972\approx9.7\%,\]Show detailed steps
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 \(9.75\%\).