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Answers to Selected Exercises Answers to selected exercises CHAPTER 1 3. (a) 300 GHz; (b) 21 cm. 5. E 3:98  10À19 J 2.48 eV. 7. Number of electrons at dynodes 1; 2; 3 ÁÁÁ 3; 9; 27; ÁÁÁ: 31; 32; 33 ÁÁÁand after 10 dynodes it is 310. Thus, Q 9:46  10À15 C. 8. (a) Æ1,000 counts; (b) 0.1%; p (c) a factor of 4 2 (to 0.05%). 9. F 3 206,265/128  1.0 1,611.45; F 11.725. Lens 8 the mirror 800. Doublingfocal length implies aberrations are 2 3 8 times smaller; earliest tele- scopes had longfocal lengths. 10. Readout noise dominates for 1 s; background limits S/N for 100 s. CHAPTER 2 1. Diraction-limited angular resolution corresponds to D 85 m: 206; 2651:22=D Angular resolution is 0.003 seconds ofp arc per micron of wavelength. Light- gathering power is only equivalent to 2  10-m 14.1 m telescope (not 85 m). Interferometry gives huge gain in resolution (85/10 8.5). 2. Using1.22 =D, then 0.0500 at 2 mm. Linear dimension (0.0500/206,26500 per rad)  (4.2 lt-yr  6  1012 miles per light-year): linear size 6.15 million miles. 516 Answers to selected exercises Baseline of 100 m 10 better resolution: linear size 615,000 miles or about 2.5 the distance to the Moon. 3. Seeing 0=r0: À6 00 (a) r0 (0.5  10 m/0.75 )206,265 0.14 m; 6=5 (b) r=r0 (2.2/0.5) 5.9, thus at 2.2 mm, r 0.83 m. À6 00 6=5 5. r0 (0.5  10 m/0.5 )206,265 0.21 m; r=r0 (1.65/0.5) 4.2, r 0.86 m; number of sub-apertures D=r2 10=0:862 134. 6. From 5, r0 0.21 m; isoplanatic angle 0.314r0=H (0.314  0.21m/ 5,000 m) 13.2 mrad or 2.7 arcsec. The isokinetic angle is larger by 00 D=r0 47.6, thus isokinetic angle 129.5 . CHAPTER 3 3. (a) prime focus is faster by (17/5)2 11.56. (b) Cassegrain is better in practice because the larger plate scale reduces the level of sky background. 4. Tails toward the edges of the ®eld implies coma in the optics. For a simple telescope/camera this could mean a misaligned secondary mirror. 8. FOV 40.96  34.13 arcminutes or 0.39 square degrees. Number of frames for 100 square degrees is 258. At 20 minutes each this takes 86 hours of observ- ing. One hemisphere is 20,627 square degrees, so to cover this takes 17,720 hours of observing(over 2 years of time). This ®eld of view is too small or the integration time is too long to be useful. 00 0 10. Limit on astigmatism of <0.5 implies FOV 20 . Parameters are R1 35 m, R2 4.85 m, K1 À1.00404, K2 À1.64344, and d 15.204 m. Answers may vary slightly. CHAPTER 4 No entries. CHAPTER 5 00 6 2. The f /d (206,265)dpix=Dtelpix (206,265 /rad)  (24 mm)/(10 m  10 mm/m) (0.200) 2.48. Yes, this camera would be challenging. 3.40  3.40. 3. Dcoll RDtel '=206,265=2 tan B, R 20,000, Dtel 10 m, ' 0.5: (a) 2 tan B 0:63)Dcoll 77 cm; (b) 2 tan B 4)Dcoll 12 cm. The echelle is more practical. Focal length 15  Dcoll: (a) 11.55 m. (b) 1.8 m. Answers to selected exercises 517 4. n 2.4, A 30 , c 2.2 mm, and R 500 for 2 pixels. Pixel size 27 mm. Assuming m 1 (®rst order), T n À 1 sin A=c 318 lines per millimeter and EFL 2dpixR= n À 1 tan A 33.4 mm. 6. Q N 0ÀN 45=N 0N 452,000 À 1,000=3,000 0:33; U N 22:5ÀN 67:5=N 22:5N 67:5 1,800 À 1,200=3,000 0:20; p p 0:332 0:2020:39 39%; 1 À1 2 tan U=Q15:6 ; p p N=N 1= N 1= 3000 1:8%: 7. R 20,000 at 0.5 mm; an air-spaced etalon of ®nesse 40; ®nd gap d and free 2 spectral range DFSP? DFSP =2nd, but 2nd R=F 250 mm, giving DFSP 0.001 mm. 1 8. Scan length 4 R (10 mm)(100,000)/4 25 cm. CHAPTER 6 p p Ê 2. n ng 4 2. Thickness =4n 2.2 mm/8 0.28 mm (or 2,750 A). 3. Match the followingthree detectors to a 0.2 m telescope and then to an 8 m telescope: for the small telescope seeing 200 and 1 pixel 100, for the large telescope seeing 0.500 and we have 0.2500/pixel. Kodak KAF-4200 CCD with 9 mm pixels in a 2,048  2,048 format: (a) f /d 9.3, FOV 340 (b) fd 0.93, FOV 8.50. SITe CCD with 22 mm pixels in a 1,024  1,024 format: (a) fd 22.7, FOV 170 (b) f /d 2.27, FOV 4.270. Hughes-SBRC InSb array with 27 mm pixels in a 1,024  1,024 format: (a) fd 27.8, FOV 170 (b) f /d 2.78, FOV 4.270. The Kodak CCD cannot be matched to a large telescope unless the scale is reduced to less than 0.100/pixel (FOV 3.40). 4. Linear size f and f 50 mm given 1:5 4 1:5À1= 128 1:5 2 1:5 À 12 230:00837 radians Linear size 418.5 mm CCD pixel, therefore must use a multi-element lens system. 5. For a 1-second-of-arc diameter blur the focal ratio is given by F 3 206,265=128 1,611; hence F 11:724 518 Answers to selected exercises 6. Sagittal image blur due to coma 6000 o axis (2.9  10À4 rad) for an f /3 mirror is 6000/16(3)2 0.4200. The tail is 3 longer or 1.2500. For f /1.5, the blur is (3/1.5)2 4 times greater 1.6700. 7. (a) Diameter of diraction blur 2.44 (0.5 mm)(2) 2.44 mm for f /2 lens at 500 nm. (b) Depth of focus: Df Æ2 F2 Æ4 mm. 8. Assuming 24  10À6 KÀ1, Young's modulus E 10  106 psi and the yield strength of the aluminum strut 40,000 psi. Since F EA DL=L and DL=L DT, therefore F=A À EDT 50,400 psi. The strut will buckle because the stress exceeds the yield strength. 2 9. A 5m , " 5% 2Fhc, Tc 77 K: (a) for Th 300 K, QH 57 W (b) for Th 275 K, Qh 40 W. Add ¯oatingshields or multi-layer insulation. 10. Spectrometer: slit width w, slit height h 1 2 AO 4Dcoll w=fcoll h=fcoll Seeing-limited camera: seeing disk diameter 1 2 1 2 2 AO 4Dcoll w=fcoll 4 =f coll CHAPTER 7 1. See Figures 7.7 and 7.8. Draw the timing diagram very carefully. 4. Inverted operation attracts minority carriers from the channel stops which ®ll surface traps and eliminate dark current. 8. EMCCD gain: with a 40 V clock there is a 1% chance per transfer of creating a second electron by avalanche multiplication. For 600 elements in the register the average gain is G 1:01600 392. CHAPTER 8 3. Last pixel has n 2,048 2,048 4,096 transfers. CTE 0.99999, fraction of original charge (Q0) left after n transfers is 0 n 4096 Q CTE Q0 0:99999 Q0 0:9599 or 96%. One can also use 0 À5 Q f1 À n 1 À CTEgQ0 f1 À 4096 1  10 gQ0 f1 À 0:04096gQ0 0:9590 or 96%. Answers to selected exercises 519 p 5. ``kTC-noise'' 1=e kTC electrons. For T 150 K, C 0.5  10À12F, À noise 201e . p 6. Shot noise on ``pre-¯ash'' 400 20 electrons.p This adds in quadrature with readout noise of 15 electrons. Final noise f 202 152g25eÀ. 7. Pickingup 50-cycle or 60-cycle harmonics from mains; probably due to a ground- loop. CHAPTER 9 À 2. V 0:25S 18:75;p linear regression ®t. Gain factor g 1/0.25 4.0e /DN. Readout noise R g 18.75 17.3eÀ. 3. C 0.1 pF, Asfd 0.75, n 16, Vfs 10 volt, g 25 electrons/DN. n The total gain is given by Ag VfsC=2 ge, therefore À12 16 À19 Ag 10 0:1  10 =2 25 1:6  10 3:81 Dividing Ag by the source follower gain (0.75) gives the gain of the preamp/ postamp combination as 5.09. 5. Flat-®elding is a multiplicative or gain correction. Fringing is an additive term. 00 À 10. Given m 24, V band, Dtel 4m, 0.30, pix 0.3 , R 10e , and Id 0 F 3:92  10À12  10À0:4 24 WcmÀ2 mmÀ1 125,600 cm2  0:55 mm 0:09 mm 0:3 1:99  10À19 J mm 9:2eÀ=s Assuming1 00 seeingthen star ¯ux is diluted over n 9 pixels.
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