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 ------------- .. csv-table:: :header: "Question", "Answer" "1", "**b**" "2", "**c**" "3", "**a**" "4", "**b**, :math:`1.008\ \mathrm{GB}`" "5", "**d**" "6", "**a**, :math:`5.0\times10^5` points/s" "7", "**d** in the printed key; see convention note" "8", "**c**" "9", "No listed answer; :math:`156{,}250` channels" "10", "**e**, :math:`2.56\ \mathrm s`" Worked reasoning ---------------- #. **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. #. **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. #. **Why Blu-ray uses violet: a.** Diffraction-limited spot size scales with wavelength, so a shorter wavelength reads smaller marks and closer tracks. #. **Capacity from wavelength alone: b.** Linear feature size scales as :math:`\lambda`, so areal density scales approximately as :math:`1/\lambda^2`: .. math:: C_{DVD}=700\ \mathrm{MB}\left(\frac{780}{650}\right)^2 =1008\ \mathrm{MB}=1.008\ \mathrm{GB}. #. **Other DVD improvements: d.** Higher-numerical-aperture optics reduce the spot further, while improved coding and compression store useful content more efficiently. #. **Maximum lidar point rate: a.** The farthest target requires a :math:`600\ \mathrm{m}` round trip: .. math:: t_{rt}=\frac{2R}{c}=\frac{600}{3.00\times10^8} =2.00\ \mathrm{\mu s}, .. math:: f_{\max}=\frac1{t_{rt}}=5.00\times10^5\ \mathrm{s^{-1}}. The 1-ns pulse duration is negligible compared with this wait time. #. **Distance scale of a 1-ns pulse: d in the key.** Its free-space spatial length is .. math:: \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, :math:`R=ct/2`, so the pulse-duration-limited *range resolution* is often quoted as :math:`c\tau/2=0.15\ \mathrm m`. The choices do not include that value. #. **Single-drum colour printing: c.** The photoconductor is written and developed successively with different toner colours, transferring the colour separations during multiple passes. #. **Voice channels in 10 Gbit/s: no listed answer.** Direct division gives .. math:: 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**, :math:`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. #. **Earth--Moon round trip: e.** .. math:: t=\frac{2R}{c} =\frac{2(384{,}000\ \mathrm{km})}{299{,}792\ \mathrm{km/s}} =2.56\ \mathrm s.