Understanding Lasers: Chapter 2 Quiz

Source: Jeff Hecht, Understanding Lasers: An Entry-Level Guide, fourth edition (2019), Chapter 2 quiz, printed pages 55–57. The questions are paraphrased.

Quick answers

Question

Answer

1

b, \(2.83\times10^{13}\ \mathrm{Hz}\)

2

c, \(6.63\times10^{-20}\ \mathrm{J}\)

3

b, \(3.00\ \mathrm{\mu m}\)

4

a, destructive

5

c, \(656\ \mathrm{nm}\)

6

d, \(333\ \mathrm{nm}\)

7

d, \(17.5^\circ\)

8

e, none; \(T=0.0625\)

9

d, \(20\ \mathrm{cm}\)

10

c, magnitude \(1\)

Worked reasoning

  1. Frequency from wavelength: b. Use \(c=f\lambda\):

    \[f=\frac{2.998\times10^8\ \mathrm{m/s}} {10.6\times10^{-6}\ \mathrm m} =2.83\times10^{13}\ \mathrm{Hz}.\]

  2. Photon energy: c. Planck’s relation gives

    \[E=hf=(6.626\times10^{-34}\ \mathrm{J\,s})(10^{14}\ \mathrm{Hz}) =6.63\times10^{-20}\ \mathrm J.\]

  3. Wavelength from frequency: b.

    \[\lambda=\frac{c}{f}=\frac{2.998\times10^8}{10^{14}} =2.998\times10^{-6}\ \mathrm m\approx3\ \mathrm{\mu m}.\]

  4. Equal waves separated by 180 degrees: a. One field is the negative of the other at every instant, so their amplitudes cancel: destructive interference.

  5. Hydrogen transition: c. The Rydberg relation for the magnitude of the \(n=2\leftrightarrow3\) transition is

    \[\frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right) =R_H\frac{5}{36}, \qquad \lambda\approx656\ \mathrm{nm}.\]

  6. Two absorbed photons followed by one emitted photon: d. Energies add, and \(E=hc/\lambda\):

    \[\frac{1}{\lambda_e}=\frac{1}{500\ \mathrm{nm}} +\frac{1}{1000\ \mathrm{nm}}, \qquad \lambda_e=333\ \mathrm{nm}.\]

  7. Refraction: d. Snell’s law gives

    \[n_1\sin\theta_1=n_2\sin\theta_2,qquad \theta_2=\sin^{-1}\!\left(\frac{1.2\sin30^\circ}{2.0}\right) =17.46^\circ.\]

  8. Transmission through four half-transmission layers: e. A \(2\ \mathrm{cm}\) sample contains four \(0.5\ \mathrm{cm}\) layers, so Beer–Lambert multiplication gives

    \[T=(0.5)^4=0.0625=6.25\%.\]

    No listed numerical choice equals this result.

    Important

    Answer-key discrepancy

    The printed key selects c, \(0.018\). That would follow from treating \(0.5\ \mathrm{cm}\) as a \(1/e\) absorption length, not from the stated fact that it transmits one half. For the wording as printed, e (none of the above) is correct.

  9. Thin-lens image distance: d.

    \[\frac1f=\frac1{d_o}+\frac1{d_i},\qquad \frac1{d_i}=\frac1{10}-\frac1{20}=\frac1{20}\ \mathrm{cm^{-1}},\]

    hence \(d_i=20\ \mathrm{cm}\).

  10. Image-to-object size ratio: c. The transverse magnification is \(m=-d_i/d_o=-1\); the minus sign means inverted, while the requested size ratio is \(|m|=1\).