PROBLEM 1-10 points. [ ] n 1 >n 2 >n 3 [ ] n 1 >n 3 >n 2 [ ] n 2 >n 1 >n 3 [ X ] n 2 >n 3 >n 1 [ ] n 3 >n 1 >n 2 [ ] n 3 >n 2 >n 1

Transcription

1 PROBLEM - 0 pints [5 pints] (a) Three media are placed n tp f ne anther. A ray f light starting in medium experiences ttal internal reflectin at the tp interface but sme f the light refracts int medium 3 when the ray reaches the bttm interface. If the tw interfaces are parallel, rank the media by their index f refractin, frm largest t smallest. [ ] n >n >n 3 [ ] n >n 3 >n [ ] n >n >n 3 [ X ] n >n 3 >n [ ] n 3 >n >n [ ] n 3 >n >n [ ] There is nt enugh infrmatin given abve t decide. [ ] Nne f the abve Briefly justify yur answer: Fr ttal internal reflectin t ccur at the interface between medium and medium, medium must have a larger index f refractin than medium, and the angle f incidence exceeds the critical angle. We have the same angle f incidence at the interface between medium and medium 3, where the beam in medium 3 bends away frm the nrmal. Medium 3 must have a smaller index f refractin than medium (s, medium has the largest index f refractin). We have the same angle f incidence at bth interfaces, but at the lwer interface the angle f incidence is less than the critical angle. Thus, the index f refractin f medium 3 must be clser t that f medium than is the index f refractin f medium. [5 pints] (b) Medium 3 is nw changed, and the rays fllw the paths shwn at right. Once again rank the media by their index f refractin, frm largest t smallest. [ ] n >n >n 3 [ ] n >n 3 >n [ ] n >n >n 3 [ ] n >n 3 >n [ ] n 3 >n >n [ X ] n 3 >n >n [ ] There is nt enugh infrmatin given abve t decide. [ ] Nne f the abve Briefly justify yur answer: In this case, the light bends tward the nrmal as it passes frm medium t medium 3, s medium 3 has the largest index f refractin.

2 PROBLEM 0 pints [ pint] (a) A beam f light traveling in air enters a rectangular glass blck with refractive index n. Assuming the light exits the blck alng the side ppsite t the side it entered, what path des the light fllw when it emerges frm the blck? [ ] The exact path it was fllwing when it entered the blck. [ X ] A path parallel t the riginal path, but displaced frm it. [ ] A path perpendicular t the riginal path. sinθ [ ] A path that makes an angle sin with the riginal path. n [ ] Nne f the abve. [ pints] (b) Briefly justify yur answer: The ray bends twards the nrmal at the first interface, and then exits the blck alng the parallel sides. Snell s law is applied exactly the same way at bth interfaces, s the angle at which the ray emerges frm the blck is θ. Thus, the ray fllws a path parallel t the riginal path, but displaces frm the riginal path because f the change in directin inside the blck. A laser beam is incident alng the nrmal t the side ac f a right-angled prism. The prism is in the shape f a triangle, with sides measuring 30 cm (bc), 40 cm (ac), and 50 cm (ab). The prism, which has an index f refractin f 4/3, is surrunded by air (n =.00). [ pint] (c) At what angle, measured frm the nrmal, des the laser beam emerge frm the side ab f the prism when the beam first encunters that glass-air interface? [ ] sin (3/ 5) = 37 [ ] sin (3/ 4) = 49 [ ] It desn t it experiences ttal internal reflectin [ X ] sin (4/5) = 53 [ ] 60 [ pints] (d) Briefly justify yur answer: The laser beam, traveling alng the nrmal t side ac, des nt change directin when it enters the prism. The angle f incidence at the side ab is then the same as the angle inside the triangle at vertex a, which has a sine f 3/5. Applying Snell s law, we get (4/3) (3/5) =.00sin(θ), which gives θ = sin - (4/5). [ pint] (e) Nw, yu can adjust the index f refractin f the prism. Given the gemetry abve, what is the critical index f refractin f the prism? Belw this value, the laser beam emerges frm the prism int air when it first encunters side ab f the prism, while abve this value the beam experiences ttal internal reflectin. [ ] [ ] 5/4 [ ] 4/3 [ ].5 [ X ] 5/3 [ ] [ ] There is n critical index f refractin the beam never experiences ttal internal reflectin

3 [3 pints] (f) Briefly justify yur answer: When the light is at the critical angle, inside the glass, the angle at which it emerges int the air is 90. Setting up Snell s law fr this situatin gives: n (3/5) =.00sin(90 ) =.00. Slving fr n gives 5/3.

4 PROBLEM 3 0 pints An bject is placed a certain distance frm a lens. The image created by the lens is exactly half as large as the bject. If the tw fcal pints f the lens are 0 cm frm the lens, where is the bject? Where is the image? [3 pints] (a) There are tw slutins t this prblem. Describe in wrds ne f the slutins, including such infrmatin as what kind f lens is being used, what side f the lens the image is n, and what the image characteristics are. Dn t draw it yet we will draw it in part (c). One slutin is when the lens is cnverging, and the bject is farther frm the lens than twice the fcal length. This gives a smaller, real, and inverted image, n the ppsite side f the lens than the bject. [3 pints] (b) Fr the slutin yu describe in (a), use equatins t calculate the bject distance and the image distance. Be careful with the signs. We have tw unknwns, but we can wrk with tw equatins, the magnificatin equatin and the thin-lens equatin. di Magnificatin: m = = = which gives d = di d Thin-lens equatin, with f = +0 cm: = + = + = 3. f d di di di di 3 This gives di = f = 30 cm, and d = di = 60 cm [3 pints] (c) Fr the slutin yu describe in (a), sketch a ray diagram n the axis belw. Hint: it s a gd idea t first draw the lens. The squares n the axis are 4 cm 4 cm.

5 [3 pints] (d) Describe in wrds the secnd slutin, including such infrmatin as what kind f lens is being used, what side f the lens the image is n, and what the image characteristics are. Dn t draw it yet we ll d that in (f). The secnd slutin is when the lens is diverging. A diverging lens prduces a smaller, virtual, and upright image, n the same side f the lens as the bject. [4 pints] (e) Fr the slutin yu describe in (d), use equatins t calculate the bject distance and the image distance. Be careful with the signs. Once again, we have tw unknwns, but we can wrk with tw equatins, the magnificatin equatin and the thin-lens equatin. di Magnificatin: m = =+ which gives d = di d Thin-lens equatin, with f = 0 cm: = + = + =. f d di di di di This gives di = f = 0 cm, and d = di = 0 cm [3 pints] (f) Fr the slutin yu describe in (d), sketch a ray diagram n the axis belw. Hint: it s a gd idea t first draw the lens. The squares n the axis are cm cm.