Chapter 26: Geometrical Optics

Transcription

1 Answers to Even-Numbered Conceptual Questons 2. Three mages are formed of object B. One extends from ( 3 m, 1 m) to ( 3 m, 2 m) to ( 4 m, 2 m). Another mage forms an L from (3 m, 1 m) to (3 m, 2 m) to (4 m, 2 m). Fnally, the thrd mage extends from ( 3 m, 1 m) to ( 3 m, 2 m) to ( 4 m, 2 m). 4. The concave sde of the dsh collects the parallel rays comng from a geosynchronous satellte and focuses them at the focal pont of the dsh. The convex sde of the dsh would send the parallel rays outward on dvergent paths. The stuaton s analogous to that of lght n an optcal telescope. 6. No. Lght bends toward the normal when t enters a medum n whch ts speed of propagaton s less than t was n the frst medum as when lght passes from ar to water. On the other hand, lght bends away from the normal f t enters a medum n whch ts speed s ncreased as when lght passes from water to ar. 8. You are actually seeng lght from the sky, whch has been bent upward by refracton n the low-densty ar near the hot ground. See fgure for a case where one would see a tree reflected n the "pool of water." 10. The Sun s already well below the horzon when you see t settng. The reason s that as the Sun s lght enters the atmosphere from the vacuum of space, t s bent toward the normal; that s, toward the surface of Earth. Therefore, the lght from the Sun can stll reach us even when a straght lne from our eyes to the Sun would go below the horzon. 12. The ol used n the bottle to the left has an ndex of refracton that s equal to the ndex of refracton of the glass n the eye dropper. Therefore, lght s undeflected when t passes from the ol to the glass or from the glass to the ol. Because lght propagates the same as f the eye dropper were not present, the dropper s nvsble. 14. Note that the word SECRET, whch s n red, s nverted. On the other hand, the word CODE, whch s n blue, appears not to be nverted. One mght thnk that the dfferent ndex of refracton for blue lght versus red lght s responsble for ths behavor, but recall that we sad the word CODE appears not to be nverted. In fact, t s nverted, just lke the word SECRET, but because all of ts letters have a vertcal symmetry, t looks the same when nverted. Solutons to Problems and Conceptual Exercses 2. Calculate the fnal angle between ncdent and reflected beams: r Insght: Because the ncdent and reflected angles are equal, ncreasng the ncdent angle by an angle wll ncrease the angle between ncdent and reflected rays by

2 2. Pcture the Problem: The mage shows two mrrors orented at 120 wth respect to each other. A lght ray strkes the frst mrror wth an ncdent angle of 55. The reflected lght then reflects off the second mrror. Strategy: Set the reflected angle from the frst mrror equal to the ncdent angle. The two mrrors and the ray form a trangle. The sum of the nteror angles of a trangle s 180. Two of the nteror angles are the complementary angles 90for the ncdent and reflected rays. Use ths relaton to calculate the angle of ncdence for mrror 2, and set angle of reflecton for mrror 2 equal to the angle of ncdence for mrror 2. Soluton: 1. Wrte the angle of reflecton for mrror 1: 1r Use the nteror angles of the trangle to calculate the ncdent angle of the second mrror: 90 1r r. 3. Set the reflected angle equal to the ncdent angle: 2r 2 65 Insght: For any ncdent angle, the sum of the ncdent angle and the reflected angle wll equal the angle between the two mrrors. 4. Pcture the Problem: The mage shows the reflecton of sunlght off of a small mrror located 1.40 m above the floor at two dfferent tmes. Strategy: Use the nverse tangent functon to calculate the angle of reflecton from the heght of the mrror and the dstance of the mage on the floor for each of the tmes. Set the angles of reflecton equal to the angles of ncdence and subtract the result to calculate the change n the Sun s angle of elevaton. Soluton: 1. (a) Because the dstance from the wall ncreased, the angle of ncdence decreased. Ths mples that the sun was gong down. So, the observatons occurred n the afternoon. 2. (b) Calculate the ntal angle of reflecton: h 1.40 m 1 1 tan tan 29.3 d m 1 h m 3. Calculate the fnal angle of reflecton: 2r tan tan 20.5 d 3.75 m 4. Subtract the two angles: 2 1 2r 1r r 2 The sun s angle of elevaton decreased by 8.8. Insght: The tme between observatons can be calculated by dvdng the angle by 360 and multplyng the result by

3 the tme n one day. Ths tme s approxmately 35 mnutes. 6. Pcture the Problem: The mage shows you observng the mage of your feet n a vertcal mrror mounted on a wall that s 1.50 meters away. Strategy: To observe your eyes the mrror should be tlted by an angle equal to the angle of reflecton of the lght from the shoes. Calculate the angle of reflecton from the nverse tangent of half your heght dvded by the dstance to the mrror. Soluton: Calculate r : 1 1 h m 2 r tan tan 31.7 d 1.50m Insght: Rotatng the mrror by any angle between 0 and 31.7 wll enable you to see dfferent parts of your body. 26 3

4 8. Pcture the Problem: The mage shows a lght beam that s orented 15.0 from the vertcal as t reflects back and forth between two mrrors. Strategy: Let d equal the horzontal dstance traveled by the lght between reflectons off ether mrror. Calculate the dstance d by multplyng the separaton dstance by the tangent of the beam angle. Dvde the total dstance (168 cm) by d to calculate the total number of reflectons. From ths result calculate the number of reflectons off each mrror, where the frst reflecton s off the top mrror and the reflectons alternate between the mrrors. Soluton: 1. (a) Calculate the dstance d: d 68.0 cm tan cm 2. Dvde the horzontal dstance by d: x 168 cm N reflectons d 18.2 cm 3. Because all of the odd-numbered reflectons are off the top mrror (1, 3, 5, 7, and 9) the lght wll reflect 5 tmes off of the top mrror. 4. (b) All of the even-numbered reflectons are off the bottom mrror (2, 4, 6, and 8), so the lght wll reflect 4 tmes off of the bottom mrror. Insght: Ether decreasng the separaton dstance of the mrrors or ncreasng the length of the mrrors wll ncrease the number of reflectons. 10. Pcture the Problem: The mage shows the 12.5-ft snake lyng perpendcular to a mrror wth ts head a dstance d from the mrror. Strategy: The dstance the lght from the snake s tal has to travel to reach the snake s eye s the dstance to the mrror (12.5 ft + d) plus the dstance back to the snake s eye (d). Set ths dstance equal to the greatest dstance the snake can see and solve for d. Soluton: 1. Set the dstance the lght travels equal to the maxmum dstance the snake can see: d 12.5 ft d 26.0 ft 2. Solve for the dstance d: 2d 26.0 ft 12.5 ft 13.5 ft d 6.75 ft Insght: If the snake were lyng parallel to the mrror, t could le 11.4 ft from the mrror and stll clearly see ts tal. 12. Pcture the Problem: The mage shows 1.9-m-tall person standng 3.2 m from a wall. A 0.80-m-tall table les 1.5 m from the wall. A mrror on the wall has the mnmum heght such that the person can see the top of the table. 26 4

5 Strategy: The ncdent angle of lght from the table equals the reflected angle to your eyes. Calculate the tangent of the ncdent angle from the rato of the heght from the top of the table to the top of the mrror and the dstance of the table to the mrror. Calculate the tangent of the reflected angle from the heght from the top of the mrror to your eyes and your dstance from the mrror. Set the tangents equal and solve for the heght of the mrror. Soluton: 1. Calculate the tangent of the ncdent ray: 2. Calculate the tangent of the reflected ray: tan h h h 0.80 m 1.5 m table dtable hyou h 1.9 m h tanr d 3.2 m you 3. Set the equatons equal and solve for h: h 0.80 m 1.9 m h 1.5 m 3.2 m 3.2 m 0.80 m 1.5 m 1.9 m h 3.2 m 1.5 m Insght: The mage of the bottom of the table legs wll reflect off the mrror at a heght of 0.61 m above the floor. 1.2 m 26 5

6 14. Pcture the Problem: The mage shows you lookng through a mrror of heght hmrror 0.32 m whle standng xy-b 95 m from a buldng. You are holdng the mrror xm-e 0.50 m n front of your eye. Strategy: Snce the ncdent angle and the reflected angle are equal, the lght rays from the ends of the wndow wll ntersect at the same dstance behnd the mrror as the eye s n front of the mrror. Usng ths relatonshp, create two smlar trangles to relate the rato of the heght of the mrror to the mrror-eye dstance and the rato of the vsble heght of the buldng to the dstance from the eye to the buldng. The dstance from the eye to the buldng s equal to the sum of the dstance from the buldng to you and twce the dstance from you to the mrror. Set the two ratos equal and solve for the vsble heght of the buldng. Soluton: 1. (a) Set the ratos of the heghts to dstances equal: 2. Solve for the vsble heght of the buldng: hmrror H x 2x x H m-e m-e y-b m20.50 m 95 m h x x mrror m-e m-b xm-e 0.50 m 3. (b) Decreasng x m-e ncreases the rato h mrror,/x m-e and thus ncreases H. Therefore, f the mrror s moved closer, the answer to part (a) wll ncrease. Insght: If the mrror s brought to a dstance of 0.25 m from the eye, 122 m of the buldng would be vsble. 61 m 16. Pcture the Problem: The mage shows an ncdent ray approachng a corner reflector at 45. The mrrors of the corner reflector form an angle of Strategy: The sum of the angles of the ncdent and reflected waves from ther respectve normal lnes wll equal the angle between the two reflectors (see problem 2). Sum the ncdent and reflected angles for each reflecton to calculate the change n path of the ray. Subtract 180 from the result to calculate the devaton of the ray from reflecton drectly back to the source. Multply the tangent of ths deflecton by the Earth-Moon dstance to calculate the devaton dstance from the orgn. Soluton: 1. Calculate the ncdent angle of the second mrror: 2 1r r Fnd the devaton of the ray from ts ntal path: 3. Calculate the devaton of the returnng ray from the drect path back to the orgn: Calculate the devaton dstance: d d 5 earth-moon 2 tan km tan km Insght: A slght devaton n the angle between the mrrors results n a large shft n the locaton of the return beam. 18. Pcture the Problem: A sphercal mrror s made from a secton of a sphere of radus 0.86 m. 26 6

7 Strategy: Calculate the focal pont for the convex sde usng equaton 26-2 and the focal pont for the concave sde usng equaton Soluton: 1. (a) Calculate the convex focal length: f R 0.86 m 43 cm (b) Calculate the concave focal length: f R 0.86 m 43 cm 2 2 Insght: The magntudes of the focal lengths are equal, but the convex mrror has a negatve focal length. 26 7

8 20. Pcture the Problem: The mage shows parallel rays from the Sun convergng at the focal pont of a concave pece of glass. Strategy: Solve equaton 26-3 for the radus of curvature, gven that the rays converge at the focal pont. Soluton: Calculate the radus: R f cm 30 cm Insght: If the glass were turned around so that the lght was reflectng off the convex porton, a vrtual mage of the Sun would appear to be 15 cm behnd the glass. 22. Pcture the Problem: You hold a shny tablespoon at arm s length and look at the front sde of the spoon. Strategy: Consder the mages formed by concave mrrors when answerng the conceptual questons. Soluton: 1. (a) Lookng at the front sde of a spoon means you are lookng at a concave mrror. In addton, holdng the spoon at arm s length means that you are outsde the focal pont of the mrror clearly, the focal length of the front sde of a spoon s only a few centmeters. The stuaton, then, s lke that llustrated n the fgure at the rght. It follows that your mage s nverted. 2. (b) Referrng to the fgure above, we see that your mage s reduced n sze. 3. (c) Lght rays pass through the mage, and t therefore s a real mage. Insght: Fgure and Example 26-3 llustrate how a sngle concave mrror can form ether real or vrtual mages. 24. Pcture the Problem: An object s placed to the left of a concave mrror, beyond ts focal pont. Strategy: Consder the mages formed by concave mrrors when answerng the conceptual questons. Soluton: Referrng to the fgure at the rght, we see that as the object s moved farther to the left, the mage moves toward the rght. Insght: The mage wll appear at the focal pont of the mrror when the object s at nfnty. 26 8

9 26. Pcture the Problem: An object s placed n front of a concave mrror. The mrror produces a real, magnfed mage. Strategy: Use equaton 26-3 to calculate the focal length from the radus of curvature. Then use equaton 26-6 to calculate the mage dstance. Soluton: 1. Calculate f : 1 1 f R 40.0 cm 20.0 cm d d f do 20.0 cm 30.0 cm 2. Solve Eq for : 60.0 cm Insght: As shown n the ray dagram, the mage s 60.0 cm n front of the mrror. 28. Pcture the Problem: A 46-cm-tall object s placed 2.4 m n front of a concave mrror that has a focal length of 0.50 m. The mrror produces a real, reduced mage. Strategy: Sketch the parallel, focal, and center rays for the object. Then sketch the mage where the rays cross. From the dagram estmate the mage dstance and the mage heght. Soluton: 1. (a) A sketch of the three smple rays s shown above. 2. From the dagram, estmate the mage dstance: The mage s located 63 cm n front of the mrror. 3. Estmate the mage heght: The mage s approxmately 12 cm tall. 4. (b) The dagram ndcates that the mage s nverted. Insght: The exact locaton and heght can be calculated from the mrror equatons (equatons 26-6 and 26-4). 26 9

10 30. Pcture the Problem: A 46-cm-tall object s placed 2.4 m n front of a convex mrror that has a focal length of 0.50 m. The mrror produces a vrtual, reduced mage. Strategy: Sketch the parallel ray and the focal-pont ray for the object. Extrapolate the rays back behnd the mrror to locate the pont where the rays appear to cross. Sketch the mage where the rays cross. From the dagram estmate the mage dstance and the mage heght. Soluton 1. (a) A sketch of the P ray and the F ray s shown at the rght. 2. Estmate the mage dstance: Accordng to the dagram, the mage s located 41 cm behnd the mrror. 3. Estmate the mage heght: The mage s approxmately 7.9 cm tall. 4. (b) The dagram ndcates that the mage s uprght. Insght: The exact locaton and heght can be calculated from the mrror equatons (equatons 26-6 and 26-4). A centerof-curvature ray can also be drawn, but t makes the drawng very crowded and doesn t mprove our estmaton of the mage locaton or heght

11 32. Pcture the Problem: The mage of the Sun forms behnd the surface of a football player s helmet. A ray dagram s shown at rght. Strategy: Because the Sun can be consdered nfntely far away, the rays from the Sun are approxmately parallel. The resultng mage of the Sun wll occur exactly at the focal pont of the helmet s surface. Use equaton 26-2 to calculate the radus of curvature. Soluton: Calculate the R 2f 2d radus of curvature: 24.8 cm 9.6 cm Insght: For objects that are closer than the Sun, the mage wll be closer to the surface of the helmet. 34. Pcture the Problem: The mage shows a 1.7-m-tall person standng 0.60 m from a reflectng globe wth dameter 0.16 m. Strategy: Use equaton 26-2 to calculate the focal length of the globe, where the radus s one half of the dameter. Then use equaton 26-6 to calculate the mage dstance from the focal length and object dstance. Fnally, use equaton 26-4 to calculate the heght of the mage. Soluton: 1. (a) Calculate the focal length: f R D m 2 f m 2 2. Use equaton 26-4 to calculate the mage dstance: d 4.2 cm f do m 0.66 m The mage s 4.2 cm behnd the surface of the globe. 3. (b) Calculate the mage heght: h h d m o 1.7 m 11 cm d 0.66 m o Insght: The mage sze ncreases as the person moves toward the globe. If he stood 0.40 m from the globe, hs mage heght would be 17 cm

12 36. Pcture the Problem: The sketch shows an astronomer standng 20.0 m n front of a mrror of focal length 16.9 m. Her mage s shown behnd her. Strategy: Use equaton 26-6 to calculate the mage dstance. Then calculate the magnfcaton usng equaton Soluton: 1. (a) Solve equaton 26-6 for the mage dstance: 2. (b) Her mage s real because lght passes through t. 3. (c) Calculate the magnfcaton: d 110 m f do 16.9 m 20.0 m The astronomer s mage s located 0.11 km n front of the mrror. m d 110 m do 20.0 m Insght: As the astronomer moves toward the focal pont, her mage moves away from the mrror and the magnfcaton ncreases. As she reaches the focal pont, the mage dstance and magnfcaton become nfntely large Pcture the Problem: A concave mrror produces a real, nverted, and magnfed mage of an object. Strategy: Use the magnfcaton equaton (equaton 26-8) to calculate the mage dstance from the mrror. Then solve equaton 26-6 for the focal length. Soluton: 1. (a) Calculate the mage dstance: d md 2. (b) Calculate the focal length from equaton 26-6: o 3 22 cm 66 cm f 17 cm do d 22 cm 66 cm Insght: A concave mrror can also produce a vrtual mage wth the same magnfcaton. In ths case, the mage dstance wll be 66 cm behnd the mrror and the focal length would be 33 cm. 40. Pcture the Problem: A convex mrror produces a vrtual, uprght, and reduced mage of a shopper. Strategy: Solve equaton 26-8 for the mage dstance. Then solve equaton 26-6 for the mrror s focal length. Fnally, use equaton 26-2 to calculate the radus of curvature. Soluton: 1. (a) Because convex mrrors always produce uprght and reduced mages, the shopper s mage s uprght. 2. (b) Calculate the mage dstance: d h 6.4 n 1 ft 17 ft = 19 n 1.59 ft do ho 5.7 ft 12 n 26 12

13 3. Calculate the focal length: f ft do d 17 ft 1.59 ft 4. Calculate the radus of curvature: R f ft = 3.5 ft do d Insght: If the mrror had been concave nstead of convex, the shopper would have seen a real, nverted mage of herself 1.59 feet n front of the mrror

14 42. Pcture the Problem: The fgure shows an object 25 cm n front of a concave shavng / makeup mrror. The vrtual mage s located behnd the mrror and s twce as tall as the object. Strategy: Use equaton 26-8 to calculate the mage dstance from the magnfcaton. Then use equaton 26-6 to calculate the focal length, and fnally equaton 26-3 to calculate the radus of curvature. Soluton: 1. Calculate the mage dstance: d md 2. Calculate the focal length: o cm = 55 cm 3. Calculate the radus of curvature: R 246 cm 92 cm f 46 cm do d 25 cm 55 cm Insght: Increasng the radus of curvature of ths concave mrror wll decrease the magnfcaton of the mage, as well as reduce the mage dstance behnd the mrror. 44. Pcture the Problem: When a ray of lght travelng through ar enters a glass lens, t slows down. After travelng a short dstance t leaves the glass and re-enters the ar. Strategy: Consder the ndces of refracton of ar and glass when answerng the conceptual queston. Soluton: 1. (a) When lght goes from ar (n = cv = 1.0) to glass (n = cv = 1.0) t slows down; when t goes from glass to ar t speeds up. In general, the speed of lght s determned solely by the medum n whch t propagates, rrespectve of ts past hstory. We conclude that ts speed wll ncrease when t leaves the glass and enters the ar. 2. (b) The best explanaton s I. Its speed ncreases because the ray s now propagatng n a medum wth a smaller ndex of refracton. Statements II and III are each false. Insght: In reference to statement III, the unversal constant s the speed of lght through vacuum, not through a transparent medum such as glass. 46. Pcture the Problem: When rays of color A and color B are sent through a prsm, color A s bent more than color B. Strategy: Recall that the amount that a ray refracts depends upon the ndex of refracton of the materal that t enters. Soluton: Color B travels more rapdly. The reason s that color B, whch s bent less than color A, must have the smaller ndex of refracton. Because the speed of lght n a materal s cn, the color wth the smaller ndex of refracton has the greater speed. Insght: The fact that dfferent colors have dfferent speeds (and thus dfferent ndces of refracton) n the glass allows the prsm to dsperse whte lght nto a ranbow of color that emerges at dfferent angles for dfferent colors. 48. Pcture the Problem: A ktchen has twn sde-by-sde snks. One snk s flled wth water, the other s empty. You observe the bottom of each snk and judge the depth of each. Strategy: Consder the refracton lght when t leaves the water to answer the conceptual queston. Soluton: 1. (a) Water causes the apparent depth to decrease (see Fgure 26-22). Therefore, the left snk (wth water) appears to be shallower than the snk on the rght (wthout water). 2. (b) The best explanaton s II. Water bends the lght, makng an object under the water appear to be closer to the surface. Thus the water-flled snk appears shallower. Statements I and III each gnore the effects of refracton. Insght: If one snk were flled wth ethanol (n = 1.36) t would appear even shallower than the snk flled wth water (n = 1.33)

15 50. Pcture the Problem: Lght travels through a medum at a speed slower than the speed of lght. Strategy: Solve equaton for the ndex of refracton, where the speed s the dstance traveled dvded by tme. Soluton: Calculate the ndex of refracton: Insght: Lght travels at m/s nsde ths medum m/s s c c ct n 1.25 v d t d m 52. Pcture the Problem: The mage shows two lght rays enterng water. One ray enters at 10.0 and refracts to The second ray enters at 20.0 and refracts to Strategy: Solve Snell s Law (equaton 26-11) for the ndex of refracton of water usng the gven ncdent and refracted angles and the ndex of refracton for ar (gven n Table 26-2). Then calculate the percent dfference between the measured values and the ndex of refracton of water gven n Table Soluton: 1. Solve Snell s Law for the ndex of refracton: 2. Calculate n w usng 10.0: 3. Calculate the percent error: 4. Calculate the n w usng 20.0: nar sn nar sn nw sn 2 nw sn sn10.0 nw 1.25 sn % 6% sn 20.0 nw 1.28 sn Calculate the percent error: % 4% 1.33 Insght: The refracted angles would have been 7.52 and 14.9 f Ptolemy had measured wth perfect accuracy

16 54. Pcture the Problem: A lght ray refracts as t travels from ar to water to ce. Strategy: Solve Snell s Law (equaton 26-11) for the angle of ncdence usng the ndex of refracton for ce gven n Table Soluton: Solve equaton nar sn nce sn2 for the angle of ncdence: 1n ce sn sn2sn sn nar 1.00 Insght: Because the ndex of refracton for ce s greater than the ndex for ar, the refracted angle s smaller than the ncdent angle. 56. Pcture the Problem: The fgure shows a scuba dver who sees the sun at an angle of 35 from the vertcal. The ncdent sunlght makes an angle of θ wth the vertcal. Strategy: Use Snell s Law (equaton 26-11) to calculate the ncdent angle of the sunlght. The angle of the sun from the horzontal s 90. Soluton: 1. Calculate the ncdent angle: 2. Fnd the Sun s angle above the horzon: nar sn nw sn2 1 n w sn sn2sn sn 3550 nar Insght: The scuba dver perceves the Sun to be much hgher n the sky than does hs frend on the shore

17 58. Pcture the Problem: The fgure shows lght travelng from pont A to pont B through the two regons wth dfferent ndces of refracton. Strategy: Calculate the tme for lght to travel through medum 1 by dvdng the dstance by the speed of lght n the medum (gven by equaton 26-10). Do the same for the tme n medum 2 and add the two tmes together. Soluton: 1. Calculate the tme n medum 1: 2. Calculate the tme n medum 2: c d vt t n nd m t ns c m/s t nd m 7.60 ns c m/s 3. Sum the two tmes: ttotal t1 t ns ns 22.3 ns Insght: Even though regon 1 s more than twce as long as regon 2, the tme n regon 1 s not qute twce the tme n regon 2 because the velocty of the lght s greater n regon 1 than t s n regon Pcture the Problem: The mage shows a person lookng nto an empty glass at an angle that allows her to barely see the bottom of the glass. When lookng at the same angle after fllng the glass wth water, she can see the center of the bottom of the glass. Strategy: Use the empty glass to calculate the sne of the angle of refracton θ 2 n terms of the heght H of the glass and the wdth of the bottom W. Wth the glass full, calculate sne of the angle of ncdence θ 1 n terms of H and W 2. Then use Snell s Law (equaton 26-11) to calculate the heght of the glass. Soluton: 1. Calculate sn 2 : 2. Calculate sn 1 : 3. Solve Snell s Law for H: sn 2 sn n ar 1 W W H W 2 2 W 2 2 W W 4H H nar sn nw snrefr W W n W H W 4H w n n H W 6.2 cm 3.6 cm 4 n n w ar ar w Insght: If the glass were flled wth benzene (n = 1.5) nstead of water, the heght H would have to be 5.2 cm. The greater the ndex of refracton of the flud, the taller the glass would have to be n order to reproduce ths experment. 62. Pcture the Problem: The mage shows a lght beam enterng a prsm perpendcular to the long sde. It undergoes total nternal reflecton on both short sdes before extng vertcally downward through the long sde. Strategy: The angles of ncdence for both nternal reflectons are exactly 45. Solve equaton for the ndex of refracton for the prsm, where the external ndex of refracton s that of ar

18 Soluton: Solve equaton for the mnmum ndex of refracton: n n ar snc sn Insght: Because the ndex of refracton for glass s greater than ths mnmum ndex of refracton, lght wll totally nternally reflect off of a glass prsm, as n the Porro prsms of bnoculars. 64. Pcture the Problem: The mage shows a beam ncdent upon a horzontal glass surface. Ths beam refracts nto the glass. When the refracted lght hts the vertcal surface t totally nternally reflects. Strategy: From the fgure, note that snc sn 902 cos 2. Use Snell s Law (equaton 26-11) to wrte an equaton relatng the ndex of refracton and the refracted angle. Use equaton to wrte a second equaton relatng the refracted angle and ndex of refracton, where the sne of the crtcal angle s the cosne of the refracted angle. Square both of these equatons and sum them to elmnate the angle and solve for the ndex of refracton. Soluton: 1. (a) Wrte Snell s Law for the frst refracton: nsn2 sn n 1 snc sn 902 cos2 n n ncos 1 ar 2. Wrte equaton n terms of the refracted angle: 2 3. Sum the squares of the two equatons and solve for n: n sn n cos sn n sn cos n sn 1 n 2 2 sn 1 sn (b) Because the mnmum ndex of refracton s related to the ncdent angle as n sn 1, decreasng θ wll cause the mnmum value of n to be decreased. Insght: As an example of decreasng the ncdent angle, set the ncdent angle equal to 50. Verfy for yourself that n ths case, the mnmum ndex of refracton drops to Pcture the Problem: The mage shows a lght beam ncdent on a semcrcular dsk wth ndex of refracton After passng through the dsk, the lght beam hts a wall 5.00 cm above the center of the dsk and at a dstance 20.0 cm behnd the dsk. Strategy: Because the lght enters the semcrcular dsk at the center, t wll ext perpendcular to the radus of curvature. It wll therefore not refract upon extng. Calculate the angle of refracton θ 2 for the lght enterng the glass usng the tangent of the rght trangle created by the dstance to the wall and the vertcal dsplacement on the wall. Then use Snell s Law (equaton 26-11) to calculate the angle of ncdence. Soluton: 1. (a) If n glass s ncreased, the ray wll refract more upon enterng the glass. Therefore, f 2 s held constant, the desred value of must ncrease. 2. (b) Calculate the refracted angle: tan2 y x 1 1 tan y x tan 5.00 cm 20.0 cm

19 3. Solve Snell s Law for θ: nar sn ng sn2 n 1 g sn sn2 sn sn n ar Insght: As predcted, the ncdent angle for flnt glass (23.7) s greater than the ncdent angle (21.6) for the glass wth smaller ndex of refracton (see problem 59)

20 68. Pcture the Problem: The fgure shows a horzontal beam refractng nto a rghtsosceles prsm. The beam makes a 4 34 angle wth the horzontal after extng the prsm. The angles θ 2 and θ 3 are labeled as 2 and 3 due to lmted space n the dagram. Strategy: Use Snell s Law (equaton 26-11) to wrte an equaton for the refracted angle θ 2 n terms of the ndex of refracton n of the glass. Use Snell s Law agan to wrte an equaton for the ncdent angle θ 3 n terms of n and θ 4. Set the sum of the angles n the hghlghted trangle equal to 180 to create a relatonshp between the θ 2 and θ 3, and then use the relatons to calculate θ 2. Fnally, use θ 2 to calculate n. Soluton: 1. Wrte an expresson for n n terms of θ 2 : nar sn1 nsn2 n sn 45 sn 2. Wrte an expresson for n n terms of θ 3 : nar sn4 nsn3 n sn 34 sn 3. Sum the nteror angles of the trangle and solve for θ 3 : 4. Use a trgonometrc dentty to wrte sn : sn sn 45 sn 45cos cos 45 sn Set the two equatons from steps 1 and 2 sn 45 sn 34 sn 34 n equal and wrte sn 3 n terms of θ 2 : sn2 sn3 sn 45cos2 cos 45sn2 6. Multply both sdes by sn 2 and dvde by sn 34 : 7. Solve for θ 2 : sn 45 1 sn 34 sn 45cot cos 45 1 cos 45sn34 sn 45 2 cot 25 sn Calculate the ndex of refracton: n sn 45 sn Insght: It would requre an ndex of refracton greater than 1.7 n order to have a beam ext at an angle greater than 34. For example, f the ndex of refracton were 2.1, the beam would ext at θ 4 =

21 70. Pcture the Problem: The fgure shows an object located a dstance 1 f from a concave lens. 2 Strategy: Sketch the P ray, F ray, and M ray for the object. Extrapolate the rays back to fnd the poston where the rays cross. From the graph determne the approxmate mage poston, mage orentaton, and mage type. Soluton 1. (a) Sketch the three rays on the dagram: 2. Note the locaton of the mage: The mage s located between the object and the lens, at about 1 3 f 3. (b) Note the mage orentaton: The mage s uprght. 4. (c) Note the mage type: The mage s vrtual, snce t s on the same sde of the lens as the object. Insght: Concave lenses produce uprght, vrtual mages for all real objects

22 72. Pcture the Problem: The fgure shows an object located a dstance 1 f from a convex lens. 2 Strategy: Sketch the P ray, F ray, and M ray for the object. Extrapolate the rays back to fnd the poston where the rays cross. Use the dagram to determne the approxmate mage poston, mage orentaton, and mage type. Soluton 1. (a) Sketch the three rays on the dagram: 2. Note the locaton of the mage: The mage s located just to the left of f. 3. (b) Note the mage orentaton: The mage s uprght. 4. (c) Note the mage type: The mage s vrtual, snce t s on the same sde of the lens as the object. Insght: When the object dstance s less than the focal length of a convex lens, the mage s vrtual and uprght. 74. Pcture the Problem: The fgure shows a convex lens ( f = 14 cm) and a concave lens ( f = 7.0 cm) separated by 35 cm. An object s located 24 cm n front of the convex lens. Strategy: Sketch the P ray, F ray, and M ray for lght from the object as t passes through the convex lens. Use the rays to determne the locaton of the mage. Then treat the mage from the frst lens as the object for the second lens and sketch the P ray, F ray, and M ray for lght as t passes through the concave lens. Use the resultng dagram to determne the approxmate mage poston, mage orentaton, and mage type. Soluton 1. (a) Sketch the three rays on the dagram: 2. Note the locaton of the fnal mage: The mage s located just to the left of Lens (b) Note the fnal mage orentaton: The mage s nverted. 4. (c) Note the fnal mage type: Snce the fnal mage s on the same sde of Lens 2 as ts object (the orgnal mage, whch s real), t s vrtual. Insght: For any combnaton of lenses, the mage from each successve lens becomes the object for the next lens

23 76. Pcture the Problem: Sunlght passes through a convex lens to a sheet of paper 26 cm below the lens. Strategy: Use equaton to calculate the focal length of the lens. Because the object (the Sun) s so far away, 1 d 0. o Soluton: 1. Insert the object dstance nto equaton 26-16: d d f d o 2. Solve for the focal length: f d 26 cm Insght: Ths would not be possble wth a concave lens because the concave lens spreads the lght beams out

24 78. Pcture the Problem: The dagram (not to scale) shows an object 42 cm n front of a lens that produces an mage 17 cm behnd the lens. Strategy: Solve equaton for the focal length. Soluton: Fnd f : f 12 cm d do 17 cm 46 cm Insght: The mage s reduced because the object s farther from the lens than the mage. Its magnfcaton s Pcture the Problem: An object s a dstance d o meters n front of a camera lens that has a focal length of f. The mage appears on the flm a dstance d behnd the lens. Strategy: Solve equaton for the mage dstance. Then use equaton to calculate the magnfcaton. 1 1 Soluton: 1. (a) Calculate the mage dstance: d f do m 5.0 m 56 mm 2. (b) Calculate the magnfcaton: d m m do 5.0 m Insght: If the object were a 1.6-m-tall person, her mage on the flm would be nverted and 1.8 cm tall. 82. Pcture the Problem: A real mage wth magnfcaton 2.0 s produced on a wall a dstance d from a lens. Strategy: Solve equaton for the mage dstance, where the object dstance has been wrtten n terms of the magnfcaton usng equaton Soluton: 1. (a) To project a real mage onto the wall, a convex lens should be used. Because convex lenses have postve focal lengths, the lens wth focal length f 1 should be used. 2. (b) Replace d o n equaton wth the magnfcaton from equaton 26-18: m m f d d d d d o 3. Solve for the mage dstance: d f m m m Insght: Usng ether the lens equaton or the magnfcaton equaton, we see that the object should be placed 0.60 m n front of the lens

25 84. Pcture the Problem: The fgure shows a concave lens ( f = 7.0 cm) and a convex lens ( f = 14 cm) separated by 35 cm. An object s located 24 cm n front of the convex lens. Strategy: Use equaton to calculate the locaton of the mage from the frst lens. Subtract ths dstance from the dstance between the lenses to calculate the dstance of the mage from the second lens. Let ths dstance be the object dstance for the second lens and use equaton to calculate the locaton of the fnal mage. Multply the magnfcatons from each lens (from equaton 26-18) together to calculate the total magnfcaton. Soluton: 1. (a) Calculate the ntal mage dstance: d cm f1 do1 7.0 cm 24 cm do2 d d1 35 cm 5.42 cm cm 2. Calculate the second object dstance: 3. Calculate the fnal mage dstance: 4. Calculate the dstance from the frst lens to the fnal mage: 5. (b) Multply the magnfcatons from each lens together: d cm f2 d o2 14 cm cm 35 cm d 35 cm cm 56 cm 2 d 1 d cm 21.4 cm mm 1 2 do1 do2 24 cm cm 0.12 Insght: The fnal mage s real, nverted, and has been reduced to about 1/8 the sze of the object. 86. Pcture the Problem: The fgure shows a glass lens whch creates a vrtual, uprght mage of a dstant object. Strategy: Use equaton to calculate the focal length, where the object dstance s nfnte (so 1 d o = 0). Soluton: 1. (a) A concave lens can produce an mage of a dstant object wthn Albert s far pont, whch allows hm to focus upon t. So, Albert s eyeglasses are concave. 2. (b) Calculate the focal length: f 0 d 2.2 m do d d Insght: For these glasses, the mages of objects at fnte dstances wll be located at dstances less than 2.2 m. For example, the mage of an object at 10.0 m wll be located at a dstance of 1.8 meters from the lens. 88. Pcture the Problem: The fgure shows a lens held 23 cm above a page. When lookng through the lens you see a reduced, uprght mage. Strategy: Use equaton to calculate the mage dstance. Then use the mage and object dstances n equaton to calculate the focal length. Soluton: 1. (a) Because the magnfcaton s postve (uprght mage) and the object dstance s postve, equaton tells us that the mage dstance must be negatve, so ths s a vrtual mage

26 2. (b) Calculate the mage dstance: d md 3. Calculate the focal length: 4. (c) Because f < 0, the lenses are concave. o cm 15.4 cm f 47 cm d do 15.4 cm 23 cm Insght: Your frend s nearsghted (see Chapter 27). Ths concave lens wll take dstant objects and produce mages that are wthn your frend s far pont (the farthest dstance your frend can see clearly). If hs glasses are held 2 cm from hs eye, hs far pont must be 47 cm + 2 cm = 49 cm

27 90. Pcture the Problem: You take a pcture of a ranbow wth an nfrared camera, and your frend takes a pcture at the same tme wth vsble lght. Strategy: Consder the prncples nvolved n the producton of a ranbow (see Fgure 26-38) to answer the queston. Soluton: 1. (a) Referrng to Fgure 26-38, we see that the red part of the ranbow s at the top of the bow. Clearly, the nfrared part of the ranbow wll be adjacent to the red lght n the bow, whch means t wll be hgher n the sky. Therefore, the heght of the ranbow n the nfrared pcture s greater than the heght of the ranbow n the vsble-lght pcture. 2. (b) The best explanaton s I. The heght wll be greater because the top of a ranbow s red, and so nfrared lght would be even hgher. Statements II and III are each false. Insght: Lkewse, f a photo were taken n the ultravolet we expect the heght of the ranbow to be less than the heght of the ranbow n the vsble-lght pcture. 92. Pcture the Problem: The mage shows a horzontal whte beam enterng an equlateral prsm. As the lght refracts through the two surfaces, the volet lght (whch has a hgher ndex of refracton) bends more than the red lght. Strategy: The four pertnent angles for the volet beam only are dentfed n the fgure. The angle of ncdence θ 1 s 30 for both beams, but the red beam has a dfferent set of angles θ 2, θ 3, and θ 4 than the volet beam. Use Snell s Law (equaton 26-11) to calculate the angles of refracton for volet and red lght at each surface. Note that θ 2 and θ 3 form two angles of a trangle, where the thrd angle s 120. Use ths trangle to calculate θ 3 from θ 2 before applyng Snell s Law to the second surface. Subtract the fnal refracted angles θ 4 to calculate the dsperson. Soluton: 1. (a) Calculate the ntal refracted angles for both colors: 2. Calculate the ncdent angles for the second surface: 3. Calculate the refracted angles through the second surface: 2, v 2, r n sn 1.00sn 30 1 ar 1 1 sn sn nglass, v n sn 1.00sn 30 1 ar 1 1 sn sn nglass, r , v 602, v , r 2, r 4, v 4, r ar nglass, v sn 1 3, v sn sn sn nar 1.00 n 1 glass, r sn 3, r sn sn sn n Calculate the dsperson: 4, v 4, r Insght: When whte lght strkes ths prsm, t wll create a ranbow that has an angular wdth of Pcture the Problem: Near the bottom of the rearvew mrror of a car there s wrtten a message: OBJECTS IN THE MIRROR ARE CLOSER THAN THEY APPEAR. Strategy: Recall the types of mages formed by concave and convex mrrors when answerng the conceptual queston. Soluton: We note that the rearvew mrror n a car always shows an mage that s both erect and reduced. Referrng to Fgure and Table 26-1, we see that ths means the mrror s convex. If you examne such a mrror closely you wll 26 27

28 notce that t bulges outward. Insght: A concave mrror wll produce a reduced, nverted, real mage for any object that s more dstant than ts focal length. Ths would not be an approprate mage for the rearvew mrror of a car! 96. Pcture the Problem: A mrror s mmersed n water and ts focal length s measured. Strategy: Consder how the law of reflecton mght change n the presence of water to determne the water s effect on the focal length of a sphercal mrror. Soluton: 1. (a) The focal length of a mrror wll stay the same f t s mmersed n water. The dfference between ths scenaro and that of Conceptual Checkpont 26-5 s that the focal length of a lens depends on how much t bends lght, and the amount of bendng, n turn, depends on the dfference n ndex of refracton between the lens and ts surroundngs. For the mrror, the focal length s determned solely by ts radus of curvature, and the fact that the angle of ncdence and the angle of reflecton are the same whch s true regardless of the medum n whch the mrror s mmersed. 2. (b) The best explanaton s III. The focal length stays the same because t depends on the fact that the angle of ncdence s equal to the angle of reflecton for a mrror. Ths s unaffected by the presence of the water. Statements I and II each erroneously assume refracton wll take place. Only reflecton occurs at the mrror surface. Insght: A mrror also has the advantage over a lens n that t does not exhbt chromatc aberraton. If a lens s made of a dspersve materal t wll have a slghtly dfferent focal length for dfferent colors of lght, resultng n a red mage that appears at one locaton and a blue mage that appears at another. The law of reflecton s ndependent of wavelength (or color) so mrrors do not suffer from ths defect. 98. Pcture the Problem: The porton of a glass eyedropper that s mmersed n ol s nvsble. Strategy: Recall that lenses are dependent upon refracton to bend rays and form mages. Soluton: The answer s f. We can see from Conceptual Queston 12 that the ndex of refracton of the eyedropper glass perfectly matches the ndex of refracton of the ol. That s, lght s not deflected as t passes from the glass to the ol or from the ol to the glass. If a lens doesn t bend lght, t follows that ts focal pont s at nfnty. Insght: The focal length of a lens s defned n terms of the pont where parallel rays wll cross after passng through the lens. In ths case the rays reman parallel and wll never ntersect Pcture the Problem: You see an mage of yourself n a three-dmensonal corner reflector. Strategy: Consder the formaton of mages by plane mrrors when answerng the conceptual queston. Soluton: A three-dmensonal corner reflector produces an mage that s nverted. To see ths, magne a corner reflector at about wast level. Lght from your head approaches the reflector movng downward. After reflectng, the lght from your head moves on a parallel path but n the opposte drecton; that s, t now moves upward. Smlarly, lght from your feet moves upward before reflecton, but downward after reflecton. Now, f the reflected lght from your head moves upward, and the reflected lght from your feet moves downward, t follows that the mage of your head s below the mage of your feet your mage s nverted. Insght: In the case of an ordnary plane mrror a ray from your feet strkes the mrror at wast level and contnues movng upward toward your eye. Your bran follows the ray backwards to floor level, where the mage of your foot appears

29 102. Pcture the Problem: The mage shows a lght beam that s tlted 15.0 from the vertcal as t reflects back and forth between two mrrors. Strategy: Let d equal the horzontal dstance traveled by the lght between reflectons off ether mrror. Calculate the dstance d by multplyng the separaton dstance by the tangent of the angle. Dvde the total dstance (168 cm) by d to calculate the total number of reflectons. From ths result calculate the number of reflectons off each mrror, where the frst reflecton s off the top mrror and the reflectons alternate between the mrrors. Soluton: 1. (a) Increasng the separaton dstance between the two mrrors wll decrease the number of reflectons, because the beam wll travel farther to the rght between successve reflectons. 2. (b) Calculate the horzontal dstance between reflectons: d 145 cm tan cm 3. Dvde the total horzontal dstance by d: x 168 cm N reflectons d 38.9 cm Insght: When the mrrors were 68 cm apart the lght reflected nne tmes (see problem 8). Therefore, ncreasng the separaton dstance decreased the number of reflectons, as predcted

30 104. Pcture the Problem: The mage shows a glass float near the front surface of a cylndrcal contaner flled wth lqud. The refracton of lght as t leaves the cylnder makes the float appear larger than t s. Strategy: Sketch a ray dagram for the float at the center of the flud. Draw a horzontal ray through the top of the float, whch wll refract at the same angle as the ray from the float near the front (they are both horzontal rays at the same elevaton). Draw a second ray from the top of the float through the center of the front. Ths ray has a smaller ncdent angle than the ray from the ntal dagram, so ts refracted ray wll be smaller than the refracted ray from the front float. Extrapolate these two rays backwards to see where the mage of the float s formed. Soluton: 1. (a) A float near the front surface wll appear smaller than a float near the back. 2. (b) Sketch the two rays (parallel and through the center) for the glass float that s closer to the center of the cylnder than prevously. Note that both P rays follow the same path, but the M rays follow dfferent paths. Extrapolate the two new rays backwards to see where the new mage s formed. The dagram at rght ndcates that a float near the front surface of the cylnder (closest to you) appears smaller than a float near the back surface. The ray angles n the fgure have been exaggerated n order to emphasze the change n the mage sze and poston. Insght: As the float moves toward the back, the curvature of the cylnder makes the floats appear larger Pcture the Problem: The fgure shows a 1.7-m-tall person standng 2.2 m n front of a convex mrror of focal length 85 cm. Strategy: Use equaton 26-6 to calculate the mage dstance. Then use equaton 26-4 to calculate the mage heght. Soluton: 1. (a) Calculate d : 2. (b) We can see that the mage s uprght by examnng the fgure. 3. (c) Calculate the mage sze: d f do m 2.2 m d 61 cm The mage s located 61 cm behnd the mrror. d m 47 cm ho do 2.2 h Insght: If the person walks toward the mrror hs mage dstance wll decrease and the mage sze wll ncrease Pcture the Problem: Because the speed of lght n a substance s determned by ts ndex of refracton, the rato of speeds of lght n two substances s related to ther rato of ndces of refracton. Strategy: Wrte the speed n substance A as x tmes the speed n substance B. Then wrte the speeds usng equaton and solve for the rato of ndces of refracton

31 Soluton: 1. Wrte the speeds usng equaton 26-10: v A c n A xv B c x n B 2. Solve for the rato of ndces of refracton: n n A B Insght: If the speed of lght n one substance s one-thrd the speed of lght n another substance, the ndex of refracton s three tmes greater for the frst substance than t s for the second substance. 1 x 26 31

32 110. Pcture the Problem: The fgure shows a beam of lght ncdent upon a flm of ol at an angle θ 1 that vares from 0 to 90 from the vertcal. The lght s refracted n the ol and then refracted agan as t enters the water. Strategy: Use Snell s Law (equaton 26-11) to calculate the maxmum angle of refracton n the water. The maxmum angle wll occur when the angle of ncdence at the ar/ol surface s 90. Because the ar/ol and ol/water nterfaces are parallel, the angle of refracton at the ar/ol nterface wll equal the angle of ncdence at the ol/water nterface. Soluton: 1. (a) Wrte Snell s nar sn1 nol sn2 Law at the ar/ol nterface: 2. Wrte Snell s Law at the ol/water nterface: nol sn2 nwater sn 3. Combne the two equatons and solve for the maxmum angle of refracton n water: nar sn1 nwater sn 1 n ar sn sn1 sn sn nwater (b) The answer to part (a) wll nether ncrease nor decrease f an ol wth a larger ndex of refracton s used, because depends only on the orgnal angle of ncdence and the ndces of refracton of ar and water. w Insght: The angle of refracton n the water wll vary between 0 (when the ncdent ray s vertcal) to a maxmum of 48.8 when the ncdent ray s horzontal

33 112. Pcture the Problem: The fgure shows lght enterng an optcal fber at an ncdent angle = The optcal fber has an ndex of refracton n = Lght totally nternally reflects when t reaches the curved surface of the fber. Strategy: Use equaton to calculate the angle of refracton θ 2 as the lght enters the fber. Because the axs of the fber s perpendcular to the surface, the angle of ncdence on the curved surface and the angle of refracton sum to 90. From ths relatonshp we can calculate the angle of reflecton. Compare ths angle wth the crtcal angle gven by equaton Soluton: 1. (a) Calculate the angle of refracton: 1n ar sn sn sn sn n Calculate the angle θ on the curved sde: (b) Calculate the crtcal angle: 1n ar c sn sn 38.1 n 1.62 Snce 61.8 > 38.1, the lght totally nternally reflects. Insght: Lght enterng the face at any angle (0 to 90) wll totally nternally reflect off the curved surfaces. Ths s true for any substance wth an ndex of refracton that s greater than

34 114. Pcture the Problem: Fgure (a) shows a vertcal object 2.0 cm tall located 75.0 cm from a convex lens ( f = 30.0 cm). The real mage s located to the rght of the lens. Fgure (b) shows the same object lyng horzontally along wth ts horzontal mage. Strategy: Use equaton to locate the mage of the vertcal object. Then use equaton to calculate ts magnfcaton. For the horzontal object, use equaton to fnd the locaton of the mage of the tp and tal of the arrow. Set the dfference n these mage dstances equal to the sze of the mage. Then dvde the mage sze by the object sze to calculate the longtudnal magnfcaton. Soluton: 1. (a) Calculate d : 2. Calculate the magnfcaton: d f do cm 75.0 cm d 50.0 cm d 50.0 cm m 0.67 d 75.0 cm o 3. (b) Fnd the mage dstance of the tp: d, tp cm f do 30.0 cm 74.0 cm 4. Fnd the mage dstance of the tal: d, tal cm f do 30.0 cm 76.0 cm 5. Calculate the mage length: L d, tp d, tal cm cm 0.89 cm 6. Fnd the longtudnal magnfcaton: m L 0.89 cm 0.45 L o 2.00 cm Insght: The mage of a three-dmensonal object s compressed more n the longtudnal drecton by ths lens than t s reduced n ether of the perpendcular drectons Pcture the Problem: The fgure shows a convex lens and a concave lens. An object 30 cm n front of the convex lens produces a real mage 60 cm to the rght of the concave lens. Strategy: Use equaton to calculate the mage dstance from the frst lens. Subtract the mage dstance from the separaton dstance between the two lenses to calculate the object dstance for the second lens. Then nsert the object and mage dstances from the second lens to calculate the focal length of the second lens. Soluton: 1. Fnd the mage dstance from the frst lens: 2. Subtract from the separaton dstance to d cm f1 do cm 30.0 cm do2 d d cm 60.0 cm 20.0 cm 26 34

35 calculate the object dstance for the second lens: 3. Calculate the focal length of the second lens: f do2 d cm 60.0 cm 30.0 cm Insght: Because the object dstance for the second lens s negatve, t represents a vrtual object. The lght from the frst lens that would have created that mage s refracted by the second lens, whch ncreases the dstance at whch the real mage s formed. The lens equaton works equally well for both real and vrtual objects

36 118. Pcture the Problem: The fgure shows an object of heght h o located a dstance d o from a curved mrror of focal length f that produces an mage of heght h at a dstance d from the mrror. Strategy: Solve equaton 26-8 for the mage dstance n terms of the magnfcaton. Then replace the mage dstance n equaton 26-6 and solve for the focal length. Soluton: 1. Solve equaton 26-8 for d o : d mdo 2. Insert the mage dstance nto the mrror equaton and solve for the focal length: m1 m f d do d do mdo mdo m1 Insght: Ths equaton for the focal length s vald for both concave and convex mrrors. o 26 36

37 120. Pcture the Problem: A lght ncdent on the flat end of a glass cylnder refracts nto the glass and then totally nternally reflects off the curved surfaces of the glass. Strategy: Use the extreme case of 90 n Snell s Law (equaton 26-11) to calculate the maxmum refracted angle the lght can have n the glass. Then use the rght trangle made up of the refracted angle and the ncdent angle to the curved surface to calculate the mnmum angle of ncdence. Set ths angle equal to the crtcal angle for total nternal reflecton (equaton 26-12) and solve for the mnmum ndex of refracton. Soluton: 1. Calculate the maxmum 1 1 nsnr 1.00sn 90 r sn refracted angle from Snell s Law: n 2. Calculate the mnmum angle θ by r 90sn summng the angles of the rght trangle: n Set θ equal to the crtcal angle n equaton 26-12: sn sn 90sn n n 4. Take the nverse sne of both sdes of the equaton and solve for the ndex of refracton: sn 90sn n n 1 n sn n 2 2 n 2 1 2sn 90 Insght: If the fber s submerged n a flud wth ndex of refracton n, ext the mnmum ndex of refracton that ensures total nternal reflecton wthn the fber becomes n 2 n. ext 26 37

38 122. Pcture the Problem: The mage shows lght passng through two nterfaces between three meda. The two nterfaces are parallel. As lght passes through each nterface t refracts nto the next medum. Strategy: Use Snell s Law (equaton 26-11) at each nterface. Snce the nterfaces are parallel the refracted angle at the frst nterface s equal to the ncdent angle at the second nterface. Soluton: 1. Wrte Snell s Law at the frst nterface: n1sn1 n2sn 2 2. Wrte Snell s Law at the second nterface: n2sn2 n3sn3 3. Combne the two equatons: n1sn1 n3sn3 Insght: The fnal equaton demonstrates that Snell s Law apples to meda 1 and 3 wthout any reference to medum