Understanding Lasers: Chapter 12 Quiz
Source: Jeff Hecht, Understanding Lasers: An Entry-Level Guide, fourth edition (2019), Chapter 12 quiz, printed pages 470–473. The questions are paraphrased.
Quick answers
Question |
Answer |
|---|---|
1 |
b |
2 |
c |
3 |
a |
4 |
b, \(1.008\ \mathrm{GB}\) |
5 |
d |
6 |
a, \(5.0\times10^5\) points/s |
7 |
d in the printed key; see convention note |
8 |
c |
9 |
No listed answer; \(156{,}250\) channels |
10 |
e, \(2.56\ \mathrm s\) |
Worked reasoning
Brief solution
1. Reasoning and answer.
Long-life vent monitor: b. A rarely serviced diode laser can last much longer than an incandescent bulb. Its directionality is also useful, but the reliability advantage is the book’s intended reason.
Show detailed steps
Long-life vent monitor: b. A rarely serviced diode laser can last much longer than an incandescent bulb. Its directionality is also useful, but the reliability advantage is the book’s intended reason.
Brief solution
1. Reasoning and answer.
Scanner rejection of room light: c. A narrow optical filter passes the scanner’s laser line while rejecting most broadband fluorescent light, greatly improving signal-to-background ratio.
Show detailed steps
Scanner rejection of room light: c. A narrow optical filter passes the scanner’s laser line while rejecting most broadband fluorescent light, greatly improving signal-to-background ratio.
Brief solution
1. Reasoning and answer.
Why Blu-ray uses violet: a. Diffraction-limited spot size scales with wavelength, so a shorter wavelength reads smaller marks and closer tracks.
Show detailed steps
Why Blu-ray uses violet: a. Diffraction-limited spot size scales with wavelength, so a shorter wavelength reads smaller marks and closer tracks.
Brief solution
1. Reasoning and answer.
Capacity from wavelength alone: b. Linear feature size scales as \(\lambda\), so areal density scales approximately as \(1/\lambda^2\):
2. Key calculation.
\[C_{DVD}=700\ \mathrm{MB}\left(\frac{780}{650}\right)^2 =1008\ \mathrm{MB}=1.008\ \mathrm{GB}.\]Show detailed steps
Capacity from wavelength alone: b. Linear feature size scales as \(\lambda\), so areal density scales approximately as \(1/\lambda^2\):
\[C_{DVD}=700\ \mathrm{MB}\left(\frac{780}{650}\right)^2 =1008\ \mathrm{MB}=1.008\ \mathrm{GB}.\]Brief solution
1. Reasoning and answer.
Other DVD improvements: d. Higher-numerical-aperture optics reduce the spot further, while improved coding and compression store useful content more efficiently.
Show detailed steps
Other DVD improvements: d. Higher-numerical-aperture optics reduce the spot further, while improved coding and compression store useful content more efficiently.
Brief solution
1. Reasoning and answer.
Maximum lidar point rate: a. The farthest target requires a \(600\ \mathrm{m}\) round trip:
2. Key calculation.
\[f_{\max}=\frac1{t_{rt}}=5.00\times10^5\ \mathrm{s^{-1}}.\]Show detailed steps
Maximum lidar point rate: a. The farthest target requires a \(600\ \mathrm{m}\) round trip:
\[t_{rt}=\frac{2R}{c}=\frac{600}{3.00\times10^8} =2.00\ \mathrm{\mu s},\]\[f_{\max}=\frac1{t_{rt}}=5.00\times10^5\ \mathrm{s^{-1}}.\]The 1-ns pulse duration is negligible compared with this wait time.
Brief solution
1. Reasoning and answer.
Distance scale of a 1-ns pulse: d in the key. Its free-space spatial length is
2. Key calculation.
\[\ell=c\tau=(3.00\times10^8)(10^{-9})=0.30\ \mathrm m.\]Show detailed steps
Distance scale of a 1-ns pulse: d in the key. Its free-space spatial length is
\[\ell=c\tau=(3.00\times10^8)(10^{-9})=0.30\ \mathrm m.\]This matches choice d and the printed key. In a two-way time-of-flight range calculation, however, \(R=ct/2\), so the pulse-duration-limited range resolution is often quoted as \(c\tau/2=0.15\ \mathrm m\). The choices do not include that value.
Brief solution
1. Reasoning and answer.
Single-drum colour printing: c. The photoconductor is written and developed successively with different toner colours, transferring the colour separations during multiple passes.
Show detailed steps
Single-drum colour printing: c. The photoconductor is written and developed successively with different toner colours, transferring the colour separations during multiple passes.
Brief solution
1. Reasoning and answer.
Voice channels in 10 Gbit/s: no listed answer. Direct division gives
2. Key calculation.
\[N=\frac{10\times10^9\ \mathrm{bit/s}} {64\times10^3\ \mathrm{bit/s}} =156{,}250.\]Show detailed steps
Voice channels in 10 Gbit/s: no listed answer. Direct division gives
\[N=\frac{10\times10^9\ \mathrm{bit/s}} {64\times10^3\ \mathrm{bit/s}} =156{,}250.\]Important
Answer-key discrepancy
The printed key selects d, \(178{,}000\), but that value does not follow from the two rates stated in the question. Protocol overhead would reduce, not increase, the number of payload channels.
Brief solution
1. Reasoning and answer.
Earth–Moon round trip: e.
2. Key calculation.
\[t=\frac{2R}{c} =\frac{2(384{,}000\ \mathrm{km})}{299{,}792\ \mathrm{km/s}} =2.56\ \mathrm s.\]Show detailed steps
Earth–Moon round trip: e.
\[t=\frac{2R}{c} =\frac{2(384{,}000\ \mathrm{km})}{299{,}792\ \mathrm{km/s}} =2.56\ \mathrm s.\]