Answers to the Conceptual Questions

Transcription

1 Chapter 18 Reractn Lght 219 Resurce CD. They are rganzed by textbk chapter, and each anmatn cmes wthn a shell that prvdes nrmatn n hw t use the anmatn, explratn actvtes, and a shrt quz. Answers t the Cnceptual Questns 1. Because the beam bends away rm the nrmal, the arrw labeled C s crrect. 2. Accuntng r the reractn n and ut the water and the relectn rm the mrrr, the ray labeled C s the mst lkely path. 3. The lght wuld llw the path labeled C. The amunt reractn at the bundary depends n the msmatch between the ndces reractn, whch s less r glass and water than r glass and ar. 4. The curve r damnd wuld be belw the curve r glass because damnd has a hgher ndex reractn. 5. The bendng lght at the surace the stream causes the llusn shallwness. 6. The ball wuld seem arther away; the eect s ppste that when the ball s submerged. 7. The graph n g shws that the crtcal angle s greater at the water-ar surace, 49 versus rm the graph n g yu can determne that ttal relectn ccurs when the ncdent angle s greater than The lght rays ht the sdes the lght ppe at angles greater than the crtcal angle and are therere ttally nternally relected. 10. Ths wuld be a greater prblem because the crtcal angle r ttal nternal relectn depends n the msmatch between the ndces reractn, whch s less r glass and water than r glass and ar. 11. Mars appears t twnkle mre when t appears pnt-lke, whch ccurs when t s arther away. 12. We see stars twnkle rm Earth because the starlght s bent by mmentary changes n the lcal densty the atmsphere. There s almst n atmsphere at the alttude the Space Shuttle. 13. The separatn wll appear t decrease near the hrzn because the lght s bent dwnward mre the clser t the hrzn t rgnates. 14. In the presence an atmsphere a star wuld appear t mve mre slwly when t s near ether hrzn because the dwnward bendng the lght n the atmsphere. 15. The Sun appears t rse bere t actually des. Therere, sunrse s earler. 16. The Sun appears t rse bere and set ater t actually des, yeldng a lnger day and a shrter nght. 17. Because the lght bends dwnward mre the clser t the hrzn t rgnates, the Sun and Mn appear t be shrtened n the vertcal drectn, gvng them val shapes. 18. Reractn n the Earth s atmsphere causes lght rm the Sun t bend arund the Earth and strke the Mn. 19. Assumng that there s n delectn the spear as t enters the water, yu wuld am lw because the sh appears t be hgher than t actually s. 20. Yu shuld am drectly at the Mn because yur laser beam wll retrace the path the mnlght cmng t yur eyes. 21. N, because the law relectn s ndependent the clr the lght. 22. The damnd has a hgher ndex reractn and, therere, mre ttal nternal relectn. Damnd als has a lt dspersn. 23. Blue lght has a greater ndex reractn than yellw lght and therere has a smaller crtcal angle. The bend wll nt be a prblem r the blue lght. Red lght has a smaller ndex reractn and wll leak ut. 24. O the clrs n the ranbw, red has the lwest ndex reractn and therere the largest crtcal angle. The red lght wll leak ut rst. 25. The bendng the lght rays near the ht sand causes the mages surrundng trees and rdges t be

2 220 Chapter 18 Reractn Lght nverted as they were beng relected rm the surace a pnd. Reer t gure B n the eature n "Mrages." 26. Warm ar near the cean s surace des nt allw lght rm the base the sland t reach the marner s eyes and the tp part the sland appears t lat n the ar. See the eature n "Mrages." 27. Yu wll see the center clrs the ranbw, yellw and green. 28. In a ranbw, water drplets bend blue lght mre than green lght. I the drplets near the kte are redrectng green lght tward the grund lr, they must be redrectng blue lght tward the hgher lrs. 29. The angle rmed by the lne rm the Sun t the randrp and the lne rm yur eye t the randrp must have a xed value r a gven clr. 30. The crcular ranbw wuld be belw the plane. The plane s shadw wuld be n the center the ranbw. 31. The tp the ranbw wll set n the west at abut md mrnng. 32. The tp the ranbw wll appear n the east at abut md aternn. They wll be separated by A dvergng lens wuld spread the lght ut. Yu culd als use a cnvergng lens wth a very shrt cal length. 34. A cnvergng lens wuld cus the Sun s energy t ts cal pnt. 35. Send the tw beams n parallel t the ptc axs and nd the pnt where they crss. Sme chalk dust wll help yu see the beams. 36. Usng the reversblty lght rays, any beam rgnatng at the cal pnt wll emerge parallel t the ptc axs. 37. I yu send the tw beams n parallel t the ptc axs, they wll dverge. I yu recrd the dvergng paths n a pece paper, the lnes can be extended backward t nd ther crssng pnt, whch s the cal pnt the lens. 38. Usng the reversblty lght rays, any beam that s amed at the cal pnt n the ther sde a dvergng lens wll emerge parallel t the ptc axs. 39. A cnvergng lens wll create a real mage n the screen. 40. N, because prsms prduce n cusng eect: ncmng parallel rays wuld emerge parallel t each ther. 41. It passes thrugh the prncpal cal pnt. 42. The ray emerges parallel t the ptc axs. 43. The nly change s that the mage becmes dmmer. 44. The prmary eect s a dmmng the mage. A secndary eect s a sharpenng the mage due t reduced sphercal aberratn. 45. The lens made rm the glass wth the larger ndex reractn wll have the shrter cal length because the lght s bent mre at the suraces. 46. T ncrease the cal length, yu want the lens t bend the lght less, whch requres a lwer ndex reractn. 47. An mage the tp the candle s rmed when all the lght rm the tp the candle that passes thrugh the lens cnverges at the mage lcatn and then agan dverges. Yu shuld be anywhere t the rght the mage lcatn t see the lght dvergng rm the mage. 48. Yu shuld be anywhere t the rght the mage lcatn t see the lght dvergng rm the mage. Only a pece lm culd recrd the mage at the mage lcatn. 49. The mage n the retna must be a real mage as lght s requred t actvate the sensrs n the retna. 50. Yu culd argue that the presence lght n the lm s requred t expse the lm. 51. The pupl regulates the amunt lght enterng the eye.

3 Chapter 18 Reractn Lght The daphragm cntrls the amunt lght that reaches the lm. 53. Dspersn ccurs when lght s reracted, nt relected. 54. Green lght s bent mre than red lght and thus wuld have a shrter cal length. 55. Cnvergng lenses prvde the extra cnvergence that the eyes can n lnger prvde. 56. All real mages rm sngle lenses are nverted s the mage rmed by glasses must be vrtual. 57. It shuld be a cnvergng lens because the eye rms real mages n the retna. 58. They are cnvergng because they act lke magners. 59. The rays ntersect at a pnt arther rm the lens and the mage gets larger. 60. The rays ntersect at a pnt arther rm the lens and the mage gets larger. 61. The vertcal symmetry the letters means that letters such as E, O, and X can be nverted and stll lk the same. 62. Derent clrs have derent cal lengths wth red beng lngest. Imagne mvng the screen away rm the lens. At the pnt where the blue lght s n cus the red lght rms the rnge. When the red lght s n cus, the blue lght s spreadng ut and rms the rnge. Answers t the Exercses 1. rm gure 18-2, we get 40 and 36, respectvely. 2. rm gure 18-2, we get 20 and 23, respectvely. 3. rm gure 18-2, the sh s dwn at 32 rm the nrmal, r 13 belw the mage the sh. 4. rm gure 18-2, we get Our estmate s Our estmate s water 22

4 222 Chapter 18 Reractn Lght 11. The mage s lcated at the ther cal pnt and s twce as bg. 12. The mage s lcated at 3/2 and s ne-hal the sze. 13. The mage s lcated 26.7 cm rm the lens n the near sde. 14. The mage s lcated 80 cm rm the lens n the ar sde. 15. When the mage s smaller than the bject, t must be clser t the lens than the bject. Ray dagrams shw that ths requres that the bject be placed urther than 2 rm the lens. 16. T be magned, the mage must be arther away than the bject. Ray dagrams shw that ths requres that the bject be placed between and 2 rm the lens.

5 Chapter 18 Reractn Lght The dvergng lens wll spread the prncpal rays slghtly, such that ther cnvergence wll be shted t the rght, as shwn. 18. Ths can be shwn by drawng a ew sample ray dagrams. 19. The mage s vrtual, erect, and reduced n sze. 20. The mage s lcated at /2 n the near sde. It s vrtual, erect, and reduced n sze. 21. d = 5 dpters = 0.2 m = 22. d = 8dpters = m = 23. de = d1 + d2 = + = 7dpters 0.2 m 0.5 m e = = = m d 7dpters e 24. d2 = d d1 = 5 dpters 4 dpters = 1 dpter 2 e = 1m d = 1dpter = 2

6 224 Chapter 18 Reractn Lght 25. de = d1 + d2 = = 2dpters 0.5 m 0.25 m e 1 = = 0.5 m dvergng d e 26. d2 = de d1 = dpter 1 dpter = dpter 2 2 = 2 2m d = dpter = Answers t the Prblems n Prblem Slvng d 3m 1. d = = = 2.25 m belw the surace n 4 3 d 15 cm 2. n = = = 1.67 d 9cm = = 3 3m = 4m d 3cm d = = = 2cm n d nd = 1 1 = 1 1 = 1 = 300 cm s 60 cm 75 cm 300 cm 00 cm m = = = 4 h = m h = 4( 15cm) = 60cm s 75 cm 6. = = = = 50 cm real s 25 cm 50 cm 50 cm 50 cm m = = = 1 ( nverted ) s 50 cm 7. 1 = = = 1 = 0 cm s 30 cm 15 cm 30 cm = = 3 8 cm = 24 cm 8. s ms 1 2 = + = + = = 12 cm s 8cm 24cm 24cm 9. 1 = 1 1 = 1 1 = 49 s = m = 10.2 cm s 10 cm 5 m 5 m = = = s = 15.8 cm s 15 cm 300 cm 300 cm = = = = 3.33 cm s 5cm 10cm 10cm

7 Chapter 18 Reractn Lght = = = = 17.1 cm s 30 cm 40 cm 120 cm ( 17.1cm) m = = = h = m h = ( 3 cm) = 1.28 cm s 40 cm 1 5 = = = = 12 cm vrtual s 30 cm 20 cm 60 cm ( 12 cm) m = = = 0.6 erect s 20 cm 1 4 = = = = 11.3 cm s 45 cm 15 cm 45 cm ( 11.3 cm) m = = = erect s 15 cm 1 2 = = = s = 15 cm s 30 cm 10 cm 30 cm 10 cm h 20 cm m = = = h = = = 30 cm s 15 cm m m ms 17. m h 2cm = = = = = cm = 15 cm h 6cm 1 2 = + = + = = 22.5 cm s 45 cm 15 cm 45 cm h 18. m h 19. m h 20. m h 50 mm 1 m = = s s 40 m = 1000 mm = m h = mm = 2.75 mm 200 mm 1 m = = 5 10 s s 40 m = 1000 mm = m h = mm = 11 mm 1.67 cm 1 m = = s 3m = 100cm = m h = cm = 1 cm 1.67 cm 1 m = = = s 40 m 100 cm 4 = m h = m = 5.02 mm = + = + = 1.67 cm s 1.67 cm 4

8 226 Chapter 18 Reractn Lght d = 59.9 dpters = m = = + = + = s 8cm 1.67cm cm cm d = = 72.4 dpters cm = 1 m 1m 1m 23. d = d1 + d2 = 2 6 = 4 = = = 25cm d 4 1m 1m 24. d = d1 + d2 = 8 4 = 4 = = = 25cm d 4 d 7cm 25. = n = = 2.33 cm m 3 d 200 cm 26. = n = = 66.7 cm m 3