1. An incident ray from the object to the mirror, parallel to the principal axis and then reflected through the focal point F.
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1 Hmewrk- Capter REASONING Te bject stance ( = cm) s srter tan te cal lengt ( = 8 cm) te mrrr, s we expect te mage t be vrtual, appearng ben te mrrr. Takng Fgure 25.8a as ur mel, we wll trace ut: tree rays rm te tp te bject t te surace te mrrr, ten tree relecte rays, an nally tree vrtual rays extenng ben te mrrr an meetng at te tp te mage. Te scale te ray tracng wll etermne te lcatn an egt te mage. Te tree sets rays are:. An ncent ray rm te bject t te mrrr, parallel t te prncpal axs an ten relecte trug te cal pnt F. 2. An ncent ray rm te bject t te mrrr, rectly away rm te cal pnt F an ten relecte parallel t te prncpal axs. (Te ncent ray cannt pass rm te bject trug te cal pnt, as ts wul take t away rm te mrrr, an t wul nt be relecte.) 3. An ncent ray rm te bject t te mrrr, rectly away rm te center curvature C, ten relecte back trug C. SOLUTION C Image (vrtual) 7.6 cm tall F cm 28 cm Object 3.0 cm tall Scale: 6.0 cm
2 a. Te ray agram ncates tat te mage s 28 cm ben te mrrr. b. We see rm te ray agram tat te mage s 7.6 cm tall. 6. REASONING We wll start by rawng te tw stuatns n wc te bject s 25 cm an 5 cm rm te mrrr, makng sure tat all stances (nclung te raus curvature te mrrr) an egts are t scale. Fr eac lcatn te bject, we wll raw several rays t lcate te mage (see te Reasnng Strategy r cnvex mrrrs n Sectn 25.5). Once te mages ave been lcate, we can realy answer te questns regarng ter pstns an egts. SOLUTION Te llwng tw ray agrams llustrate te stuatns were te bjects are at erent stances rm te cnvex mrrr.
3 Image Object 3 25 cm F C Image Object 3 F C 5 cm
4 a. As te bject mves clser t te mrrr, t can be seen tat te magntue te mage stance becmes smaller. b. As te bject mves clser t te mrrr, te magntue te mage egt becmes larger. c. By measurng te mage egts, we n tat te rat te mage egt wen te bject stance s 5 cm t tat wen te bject stance s 25 cm s REASONING Fr an mage tat s n rnt a mrrr, te mage stance s pstve. Snce te mage s nverte, te mage egt s negatve. Gven te mage stance, te mrrr equatn can be use t etermne te cal lengt, but t s a value r te bject stance s als neee. Te bject an mage egts, tgeter wt te knwlege tat te mage s nverte, allws us t calculate te magncatn m. Te magncatn m s gven by m = / (Equatn 25.4), were an are te mage an bject stances, respectvely. SOLUTION Accrng t Equatn 25.4, te magncatn s m r Substtutng ts result nt te mrrr equatn, we btan F HG O Q F 5 H G I 3 cmk J L cm N M b. g b3.5 cmg I K J P 0. cm r 9. cm
5 23. REASONING Snce te mage s ben te mrrr, te mage s vrtual, an te mage stance s negatve, s tat 34.0 cm. Te bject stance s gven as 7.50 cm. Te mrrr equatn relates tese stances t te cal lengt te mrrr. I te cal lengt s pstve, te mrrr s cncave. I te cal lengt s negatve, te mrrr s cnvex. SOLUTION Accrng t te mrrr equatn (Equatn 25.3), we ave r 7.50 cm 34.0 cm 9.62 cm Snce te cal lengt s pstve, te mrrr s cncave. 26. REASONING Te magncatn m s gven by m = / (Equatn 25.4), were an are te mage an bject stances, respectvely. Te bject stance s knwn, an we can btan te mage stance rm te mrrr equatn: (25.3) SOLUTION Slvng te mrrr equatn (Equatn 25.3) r te mage stance gves r = r Substtutng ts result nt te magncatn equatn (Equatn 25.4) gves
6 m /c Usng ts result wt te gven values r te cal lengt an bject stances, we n Smaller bject stance m cm cm 9. 0 cm b gb g Greater bject stance m cm cm 8. 0 cm b gb g REASONING Te cal lengt te water rp s gven by te mrrr equatn (Equatn 25.3), were an are, respectvely, te bject stance an te mage stance. Te bject stance ( = 3.0 cm) s gven, an we wll etermne te mage stance rm te magncatn equatn m (Equatn 25.4), were s te ameter te lwer an s te ameter ts mage. Te water rp acts as a cnvex spercal mrrr, s te mage s uprgt. Terere, te mage egt s pstve, an we expect te cal lengt t be negatve. SOLUTION Slvng (Equatn 25.4) r an takng te recprcal, we btan r () Substtutng Equatn () nt (Equatn 25.3) yels
7 (2) Takng te recprcal Equatn (2), we n tat te cal lengt te water rp s 3.0 cm 0.6 cm 2.0 cm 0.0 cm Capter REASONING a. Te reracte ray s swn crrectly. Wen lgt ges rm a meum lwer nex reractn (n =.4) t ne ger nex reractn (n =.6), te reracte ray s bent twar te nrmal, as t es n part (a). b. Te reracte ray s swn ncrrectly. Wen lgt ges rm a meum lwer nex reractn (n =.5) t ne ger nex reractn (n =.6), te reracte ray must ben twar te nrmal, nt away rm t, as part (b) te rawng sws. c. Te reracte ray s swn crrectly. Wen lgt ges rm a meum ger nex reractn (n =.6) t ne lwer nex reractn (n =.4), te reracte ray bens away rm te nrmal, as t es part (c) te rawng.. Te reracte ray s swn ncrrectly. Wen te angle ncence s 0, te angle reractn s als 0, regarless te nces reractn. SOLUTION a. Te angle reractn 2 s gven by Snell s law, Equatn 26.2, as
8 n sn.4 sn 55 sn sn n 2 b. Te actual angle reractn s n sn.5 sn 55 sn sn n 2 c. Te angle reractn s n sn.6 sn 55 sn sn n 2. Te actual angle reractn s nsn.6 sn 0 2 sn sn 0 n REASONING AND SOLUTION Te angle ncence s un rm te rawng t be 8.0 m = tan = m Snell's law gves te angle reractn t be sn 2 = (n /n 2 ) sn = (.000/.333) sn 73 = 0.72 r 2 = 46 Te stance s un rm te rawng t be
9 = 8.0 m + (4.0 m) tan 2 = 2. m
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