Understanding Lasers: Chapter 10 Quiz

Source: Jeff Hecht, Understanding Lasers: An Entry-Level Guide, fourth edition (2019), Chapter 10 quiz, printed pages 395–398. The questions are paraphrased.

Quick answers

Question

Answer

1

b, direct-bandgap semiconductor

2

a, InGaAsP

3

c, 40% Ga, 10% Al, 50% As

4

d

5

b

6

e, all listed structures

7

d, VCSEL

8

a, InGaN diode

9

c, \(827\ \mathrm{nm}\)

10

d, distributed feedback

11

e, AlGaN

12

b, AlGaInP

13

e, VCSEL

14

d, stacked arrays

15

e, \(3.1\ \mathrm{eV}\)

Worked reasoning

  1. Brief solution

    1. Reasoning and answer.

    Efficient diode-laser material: b. A direct bandgap lets an electron and hole recombine while conserving crystal momentum and emitting a photon. Indirect-gap materials usually lose energy nonradiatively through phonons.

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    Efficient diode-laser material: b. A direct bandgap lets an electron and hole recombine while conserving crystal momentum and emitting a photon. Indirect-gap materials usually lose energy nonradiatively through phonons.

  2. Brief solution

    1. Reasoning and answer.

    Quaternary III–V compound: a. InGaAsP contains four elements, all drawn from periodic-table groups III and V. GaAlAs is ternary and GaAs is binary.

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    Quaternary III–V compound: a. InGaAsP contains four elements, all drawn from periodic-table groups III and V. GaAlAs is ternary and GaAs is binary.

  3. Brief solution

    1. Reasoning and answer.

    Atomic fractions in Ga0.8Al0.2As: c. One formula unit contains \(0.8+0.2+1=2\) atoms in normalized proportions. Therefore

    2. Key calculation.

    \[x_{\mathrm{Ga}}=\frac{0.8}{2}=40\%,\quad x_{\mathrm{Al}}=\frac{0.2}{2}=10\%,\quad x_{\mathrm{As}}=\frac{1}{2}=50\%.\]
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    Atomic fractions in Ga0.8Al0.2As: c. One formula unit contains \(0.8+0.2+1=2\) atoms in normalized proportions. Therefore

    \[x_{\mathrm{Ga}}=\frac{0.8}{2}=40\%,\quad x_{\mathrm{Al}}=\frac{0.2}{2}=10\%,\quad x_{\mathrm{As}}=\frac{1}{2}=50\%.\]

  4. Brief solution

    1. Reasoning and answer.

    Exciton: d. It is a bound electron–hole pair: the electron is excited relative to the filled valence band but has not recombined with the hole.

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    Exciton: d. It is a bound electron–hole pair: the electron is excited relative to the filled valence band but has not recombined with the hole.

  5. Brief solution

    1. Reasoning and answer.

    Double-heterostructure advantage: b. Higher-bandgap layers confine injected carriers to the thin active layer, increasing the probability of radiative recombination. They also help confine the optical mode.

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    Double-heterostructure advantage: b. Higher-bandgap layers confine injected carriers to the thin active layer, increasing the probability of radiative recombination. They also help confine the optical mode.

  6. Brief solution

    1. Reasoning and answer.

    Structures possible in GaAlAs: e. The material system supports Fabry–Perot and distributed-feedback edge emitters, VCSELs, and gain chips used in external cavities.

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    Structures possible in GaAlAs: e. The material system supports Fabry–Perot and distributed-feedback edge emitters, VCSELs, and gain chips used in external cavities.

  7. Brief solution

    1. Reasoning and answer.

    Shortest cavity: d. A VCSEL cavity runs vertically through only a few micrometres of epitaxial material, much shorter than edge-emitter or free-space cavities.

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    Shortest cavity: d. A VCSEL cavity runs vertically through only a few micrometres of epitaxial material, much shorter than edge-emitter or free-space cavities.

  8. Brief solution

    1. Reasoning and answer.

    High-definition optical-disc source: a. InGaN diodes emit violet-blue light, whose shorter wavelength focuses to the small spot required for high-density Blu-ray data.

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    High-definition optical-disc source: a. InGaN diodes emit violet-blue light, whose shorter wavelength focuses to the small spot required for high-density Blu-ray data.

  9. Brief solution

    1. Reasoning and answer.

    Bandgap wavelength: c. Using \(E(\mathrm{eV})\lambda(\mathrm{nm})\approx1240\),

    2. Key calculation.

    \[\lambda=\frac{1240\ \mathrm{eV\,nm}}{1.5\ \mathrm{eV}} =827\ \mathrm{nm}.\]
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    Bandgap wavelength: c. Using \(E(\mathrm{eV})\lambda(\mathrm{nm})\approx1240\),

    \[\lambda=\frac{1240\ \mathrm{eV\,nm}}{1.5\ \mathrm{eV}} =827\ \mathrm{nm}.\]

  10. Brief solution

    1. Reasoning and answer.

    Single-longitudinal-mode diode: d. A distributed-feedback grating selects one cavity mode across the gain region.

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    Single-longitudinal-mode diode: d. A distributed-feedback grating selects one cavity mode across the gain region.

  11. Brief solution

    1. Reasoning and answer.

    Shortest-wavelength family: e. Wide-bandgap AlGaN reaches farther into the ultraviolet than GaInN, AlGaInP, GaAlAs, or InGaAsP.

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    Shortest-wavelength family: e. Wide-bandgap AlGaN reaches farther into the ultraviolet than GaInN, AlGaInP, GaAlAs, or InGaAsP.

  12. Brief solution

    1. Reasoning and answer.

    Red pointer diode: b. AlGaInP is the standard material family for efficient visible-red diode emission.

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    Red pointer diode: b. AlGaInP is the standard material family for efficient visible-red diode emission.

  13. Brief solution

    1. Reasoning and answer.

    Low-threshold, efficient, good-beam source: e. VCSELs combine a tiny active volume with strong mirrors and a circular, low-divergence output mode.

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    Low-threshold, efficient, good-beam source: e. VCSELs combine a tiny active volume with strong mirrors and a circular, low-divergence output mode.

  14. Brief solution

    1. Reasoning and answer.

    Maximum efficient power without beam-quality priority: d. Stacking multiple diode arrays combines many broad emitting stripes and scales total power, at the cost of poorer spatial quality.

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    Maximum efficient power without beam-quality priority: d. Stacking multiple diode arrays combines many broad emitting stripes and scales total power, at the cost of poorer spatial quality.

  15. Brief solution

    1. Reasoning and answer.

    Energy of a 400-nm photon: e.

    2. Key calculation.

    \[E=\frac{1240\ \mathrm{eV\,nm}}{400\ \mathrm{nm}} =3.10\ \mathrm{eV}.\]
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    Energy of a 400-nm photon: e.

    \[E=\frac{1240\ \mathrm{eV\,nm}}{400\ \mathrm{nm}} =3.10\ \mathrm{eV}.\]

    Check: a shorter wavelength than \(1240\ \mathrm{nm}\) must have more than \(1\ \mathrm{eV}\) of energy.