Understanding Lasers: Chapter 5 Quiz

Source: Jeff Hecht, Understanding Lasers: An Entry-Level Guide, fourth edition (2019), Chapter 5 quiz, printed pages 165–167. The questions are paraphrased.

Quick answers

Question

Answer

1

a, blue focal length is \(3.33\ \mathrm{cm}\) shorter

2

d, about \(30\%\)

3

d, about \(17.2\%\)

4

c, magnesium fluoride

5

b, silicon

6

b, interference filter

7

e, \(173.5\ \mathrm{nm}\)

8

e, about \(1996\ \mathrm{nm}\)

9

c

10

a, semiconductor diode

11

a, silicon

12

b

Worked reasoning

  1. Chromatic focal shift: a. For a symmetric thin biconvex lens,

    \[\frac1f=(n-1)\left(\frac1R-\frac1{-R}\right) =\frac{2(n-1)}R.\]

    Thus \(f_{400}=20/[2(0.60)]=16.67\ \mathrm{cm}\) and \(f_{700}=20/[2(0.50)]=20.00\ \mathrm{cm}\). The 400-nm focus is \(3.33\ \mathrm{cm}\) shorter.

  2. Bare silicon reflection: d. Normal-incidence power reflectance is

    \[R=\left(\frac{n_2-n_1}{n_2+n_1}\right)^2 =\left(\frac{3.42-1}{3.42+1}\right)^2=0.300.\]

  3. Reflection with an index-2 coating: d. Ignoring interference and multiplying interface transmissions,

    \[R_{12}=\left(\frac{2-1}{2+1}\right)^2=0.1111, \qquad R_{23}=\left(\frac{3.42-2}{3.42+2}\right)^2=0.0686,\]
    \[R_{\mathrm{total}}=1-(1-R_{12})(1-R_{23}) =1-(0.8889)(0.9314)=0.172.\]

  4. Visible-window material: c. Magnesium fluoride transmits throughout the 0.4–0.7 micrometre band; the semiconductor choices have absorption edges that exclude part or all of it.

  5. Unsuitable 0.9–1.0 micrometre material: b. Silicon absorbs below its roughly \(1.1\ \mathrm{\mu m}\) band-edge wavelength. The other listed optical materials transmit in this band.

  6. Reject one narrow laser line: b. A narrow notch interference filter can reject the laser wavelength while passing nearby wavelengths. A neutral-density filter would attenuate the whole band.

  7. Fourth harmonic: e. Harmonic frequency is multiplied by four, so wavelength is divided by four:

    \[\lambda_4=\frac{694\ \mathrm{nm}}4=173.5\ \mathrm{nm}.\]

  8. Difference-frequency wavelength: e.

    \[\frac1{\lambda_d}=\left|\frac1{694\ \mathrm{nm}} -\frac1{1064\ \mathrm{nm}}\right|,qquad \lambda_d=1995.9\ \mathrm{nm}.\]

  9. Raman shifting: c. Raman interaction exchanges a modest vibrational energy with the medium, shifting the input frequency and wavelength rather than simply doubling or intensity-modulating it.

  10. Direct current modulation: a. A diode laser’s carrier population and optical output respond directly and rapidly to drive current.

  11. Green detector: a. A silicon photodiode responds well at \(525\ \mathrm{nm}\); the other listed compound-semiconductor detectors are aimed mainly at longer wavelengths or are unsuitable absorbers there.

  12. Decibels: b. A decibel expresses a logarithmic power ratio:

    \[L_{\mathrm{dB}}=10\log_{10}\left(\frac{P_2}{P_1}\right).\]