refracted and the wavelength of the light shortens. According to the equation! = w"y L

Transcription

1 Section 0.2: Single-Slit Diffraction Tutorial Practice, page 56. When light travels fro air to a ediu that is denser than air, such as ater, the light is refracted and the avelength of the light shortens. According to the equation! y L, λ is proportional to Δy, so hen λ is reduced, the central axiu ill also be reduced. So the central axiu ould be narroer if the equipent ere suberged in ater. 2. Given:! #7 ; 43 µ ; L 3.0 Required: Δy Analysis: Rearrange the equation! y L!y L!y L (3.0 )(7.328 # 0$7 # 0 $7 ) 4.3 # 0 $5 5.# 0 $2 to solve for the distance beteen adjacent inia;!y 5. c Stateent: The separation of adjacent inia is 5. c. 3. Given: 3.00! 0 6 ; # 25.0 Required: λ Analysis: The angle beteen the first dark fringes is equal to the angle for the idth of the central axiu, hich is tice the angle for the first dark fringe, given by sinθ λ. In this case, n and 2! 25.0 ;! sin n. 2θ 25.0 θ 2.5! sin n (3.0 # 0 $6 )sin2.5! 6.49 # 0 $7 Stateent: The avelength of the light is Given: θa 56 ; θb 34 a Required: b Analysis: The forula that relates the angles, the coon avelength, and the slit sizes is a sinθn λ; a sin! a and b sin! b. Divide the equations to deterine the ratio. n b Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.2-

2 a sin! a b sin! b a b sin! b sin! a sin34 sin56 a b 0.67 Stateent: The ratio of the slit idths, a b, is Section 0.2 Questions, page 59. Given: single-slit diffraction;! 794 n #7 ; L.0 n 9, y c Required: Analysis:!y y 9 n!y y 9 n!l y ;!y L L!y !y (7.94 #0$7 )(.0 ) 7.2 #0 3. #0 3 Stateent: The idth of the slit is.! Given: single-slit diffraction;! 600 n #7 ; $ 6.9 Required: Analysis: The first dark fringe is located here sinθ λ. Rearrange the equation sinθ λ to solve for slit idth; n! sin n. n Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.2-2

3 ! sin 6.00 #0$7 sin #0 $6 Stateent: The idth of the slit is Given: single-slit diffraction; λ 450 n ; L 0.0 ; 0.5.5! 0 4 Required: y, the distance beteen the first and third dark fringes λ Analysis: Δ y L ; y 2!y!y L (4.50 #0$7 )(0.0 ).50 #0 $4 3.0 #0 $2 y 2!y 6.0 #0 $2 y 6.0 c Stateent: The distance beteen the first and third dark fringes is 6.0 c. 4. Given: single-slit diffraction; λ 550 n ; L 2.0 ; y ! 0 3 Required: λ Analysis: Rearrange the equation y L!L to solve for slit idth; y!l y (5.50 0#7 )(2.0 ) # # Stateent: The idth of the slit is Given: single-slit diffraction; λ 630 n ; L 3.0 ; ! 0 4 Required: 2Δy, the idth of the central axiu Analysis: Multiply the equation!y L by 2 to obtain 2Δy. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.2-3

4 2!y 2L ( ) #0$ #0 $4.5#0 $2 ( ) 2!y.5 c Stateent: The idth of the central axiu is.5 c Given: Δ y 0.20 c.20 0 ; c ; L 60.0 c 0.60 Required: λ Analysis: Rearrange the equation!y L y to solve for avelength;! L! y L (2.95#0$4 )(.20 #0 $3 ) 6.0 #0 $! 5.90 #0 $7 7 Stateent: The avelength of the yello light is λ 7. (a) The distance beteen successive axia in single-slit diffraction is given by Δ y L. If I double the avelength, λ, then the distance Δy ill also double. The angles of the axia and the inia ould be approxiately doubled. λ (b) If I ultiplied both the avelength, λ, and the slit idth,, in the equation Δ y L by 2, the 2s ill cancel each other out. Therefore, there ill be no effect on Δy. The interference pattern ill be the sae. 8. Blue light has an average avelength of 475 n, and green light has an average avelength of 50 n. If I replaced the blue light ith the green light, then I ould be increasing the avelength. Therefore, spacing of the intensity axia ould be greater. 9. Given: single-slit diffraction Required: θ 0 Analysis: Assue the idth of a typical dooray is Assue the visible light has a avelength of 500 n. Rearrange the equation sin! n n to solve for the angle; $ n# '! n sin % & ( ) Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.2-4

5 $ n# '! n sin % & ( ) $ (0)( 5.00 *0 7 )' sin & ) % & 0.92 ( ) sin 5.43*0 6! 0 3.*0 4 ( ) Stateent: The angle of the tenth iniu for a dooray that is 0.92 ide is To iprove the resolution of a digital iage, I could use ore pixels per square centietre. Or, I could use a ider aperture (size of slit) to increase the resolution. Hoever, the ider aperture ould reduce the depth of field (range of the focus).. I ould be able to resolve the double stars in Mizar ith the telescope because, in addition to enlarging the iage, the telescope s aperture is ider than the aperture in y eye. The ider aperture increases the resolution, alloing e to see the to stars. 2. In a double-slit interference pattern, there are ore intensity axia than in a single-slit interference pattern. In the double-slit interference pattern, there is less space beteen fringes because the second slit causes additional destructive interference. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.2-5

6 Section 0.3: The Diffraction Grating Tutorial Practice, page 523. Slit separation and nuber of lines are related by the equation. As N increases, N decreases. The diffraction grating ith lines/c has ore lines per centietre than the second diffraction grating, so the separation beteen adjacent principal axia in the first grating ould have to be saller. 2. Given:! 660 n #7 ; N 8500 lines/c ; Required: θ, the angular separation beteen successive axia Analysis: Use the equation to calculate the slit separation. Then use the equation N λ sinθ to locate the axiu for ; sin!. N 8500 lines/ c! 00 c.76! 0 6 (to extra digits carried) sin! sin! ()(6.60 #0$7 ).76 #0 $6! 34 Stateent: The angular separation of successive axia is Given:! n #7 ; 3; $ Required: N Analysis: Use the equation λ sinθ to calculate the slit separation;!. Then use sin the equation to deterine the nuber of grating lines; N N.! sin (3)(6.943# 0$7 ) sin # 0 $6 (one extra digit carried) Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.3-

7 N ! 0 6! 00 c N 798 lines/c Stateent: The grating has 798 lines per centietre. Research This: Blu-ray Technology, page 524 A. Ansers ay vary. Saple anser: The technology is called Blu-ray because it uses a blue laser instead of the red laser used in DVDs. B. Ansers ay vary. Saple anser: In Blu-ray technology, the data are placed on the top of a disc coated ith a polycarbonate layer. There are pits in the disc. Each pit contains a signal that is interpreted as a zero or a one, uch like a coputer. The laser reads the pits. C. Ansers ay vary. Saple anser: Blu-ray technology can hold ore data than CDs and DVDs. The quantity of data is at least five ties greater than the quantity that a DVD can store, hich eans that iages can be uch ore detailed on Blu-ray discs. Blu-ray is also able to play uch faster than CDs and DVDs. D. Ansers ay vary. Saple anser: The iproveents are possible through the use of the blue laser, hich has a uch saller avelength than the red laser used ith CD and DVD players. Blu-ray technology can also store inforation on as any as 20 layers ithin the disc. E. Ansers ay vary. Saple anser: The iages are sharpest fro Blu-ray discs. Ansers ill vary based on presentation forat, but students should sho at least three iages to deonstrate the representative features of the three technologies. Hazards associated ith using high-poer lasers such as the lasers used in Blu-ray technology should be included. For exaple, if the laser is pointed at a person s eyes, that person s eyesight could be daaged. Section 0.3 Questions, page 525. The surface of a CD has any closely spaced parallel lines, like a diffraction grating. Consequently, hen hite light reflects fro the surface of a CD, e see a rainbo-like pattern because the surface acts like a diffraction grating. 2. Given: N 2800 lines/c Analysis: N N 2800 lines/c 3.6! 0 4 c 3.6! 0 6 Stateent: The distance beteen to lines in the diffraction grating is Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.3-2

8 3. Given: N lines/c; θ 3.2 ; θ ; θ Required: λ, λ2, λ 3 Analysis: Use the equation to calculate the slit separation. Then use the equation N λ sinθ to deterine the avelength for each value of ;! sin. N lines/c.0! 0 4 c.0! 0 6 For : For 3:! sin (.0 # 0$6 )sin # 0 $7! 58 n! sin (.0 # 0$6 )sin # 0 $7! 246 n For 2:! sin (.0 # 0 $6 )sin # 0 $7! 297 n Stateent: The avelengths that produce these axia are 58 n, 297 n, and 246 n. 4. Given: N 6000 lines/c; 2.0 c;! 450 n #7 ; Required:θ Analysis: Divide the nuber of slits, N, by the length of the slit: 6000 lines/c 3000 lines/c. Then use to calculate the slit separation, and rearrange 2 N the equation λ sinθ to calculate the angle; sin!. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.3-3

9 N 3000 lines/ c! 00 c 3.333! 0 6 (to extra digits carried) sin! ()(4.50 # 0$7 ) # 0 $6! 7.8 Stateent: Blue light produces the first intensity axiu at Given: λ n ; 25 µ ; Required:θ Analysis: λ sinθ ; sin! sin! ()(6.000 # 0$7 ) 2.5 # 0 $5!.4 Stateent: The first-order axiu in intensity is at the angle Given:! 780 n #7 ; L 0 ; $y 0.50 ; Required: Analysis: Use the sine ratio to calculate!. Then rearrange the equation calculate the slit separation;! sin. sin! sin! 0.05! sin ()(7.80 # 0$7 ) # 0 $5 Stateent: The spacing beteen the lines in the diffraction grating is λ sinθ to Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.3-4

10 7. Given: N 300 lines/c; L 0.84 ;!y 3.6 c ; 3 Required: λ Analysis: Use to calculate the slit separation. Use the tan ratio to calculate θ 3. Then N rearrange the equation λ sinθ to calculate the avelength;! sin. N 300 lines/ c! 00 c 3.333! 0 5 (to extra digits carried) tan! ! (to extra digits carried)! sin (3.333 # 0$5 )sin # 0 $7! 480 n Stateent: The avelength of the light is 480 n Given: N 3000 lines/c; λ Required: axiu value of Analysis: Use to calculate the slit separation. Then use the equation λ sinθ to N deterine the angle in ters of ; sin!. N 3000 lines/ c! 00 c 3.333! 0 6 (to extra digits carried) Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.3-5

11 For a axiu, sin! <.0 : sin! ()(5.4 # 0$7 ) 3.333# 0 $6 sin! We need <.0 < < Stateent: The axiu order nuber possible is the 6th order. 9. (a) Given: 0.50 n 5.0! 0 0 ; # n 5.0! 0 Required: θ, θ 2, θ 3 Analysis: Use the equation λ sinθ to calculate the angles; sin!. For : For 3: sin! sin! ()(5.0 #0$ ) 5.0 #0 $0! 5.7 For 2: sin! (3)(5.0 #0$ ) 5.0 #0 $0! 7 (2)(5.0 #0$ ) 5.0 #0 $0! 2 Stateent: The angles for the first three axia are 5.7, 2, and 7. (b) 5.0! 0 0 ; # 600 n 6.0! 0 7 ; Required: θ Analysis: λ sinθ ; sin! Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.3-6

12 sin! ()(6.0 #0$9 ) 5.0 #0 $0.2! no angle possible Stateent: The angle for the first bright fringe does not exist. (c) The avelength 600 n is ithin the range of visible light, but no fringe angle as possible in part (b). Visible light is not usually diffracted by crystal lattices. It ay be possible to get a fringe but only if the avelength of the light is sufficiently short. 0. Given:! A n #7 ; $ A 20.0 ; $ B 8.0 ; Required: n atosphere Analysis: Use the equation λ sinθ to calculate ;!. Then use the sae sin equation ith the value of and the ne angle, θ B, to calculate the avelength in the planet s atosphere. Take the ratio of the avelengths to deterine the index of refraction for the planet s atosphere, n atosphere! A! B.! sin! B sin ()(5.0 #0$7 ) sin #0 $6 (one extra digit carried) (.462 # 0$6 )sin8.0! B 4.58 # 0 $7 (one extra digit carried) n atosphere! A! B n atosphere # #7 Stateent: The index of refraction of the planet s atosphere is.. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.3-7

13 Section 0.4: Electroagnetic Radiation Tutorial Practice, page 530. Given: f 07. MHz.07! 0 8 Hz; c /s Required: λ Analysis: c! f ;! c f! c f / s Hz! 2.8 Stateent: The avelength of the signal is Given: f 3.0! 0 7 Hz; c /s Required: λ Analysis: c! f ;! c f! c f / s Hz.0 0 #9!.0 0 #7 c Stateent: The avelength of the X-rays is c. 3. Given:! 638 n #7 ; c /s Required: T Analysis:! f c f c! T f T! c T! c # /s T 2.0 #5 s Stateent: The period of the ave is 2.! 0 5 s. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.4-

14 4. Given: f 60 Hz; x k ; c /s Required: n, nuber of avelengths Analysis: Calculate the avelength of the electrical transission using the universal ave equation, c λf;! c f. Then divide the avelength by the distance across North Aerica; n! x.! c f n! x /s 60 Hz! n Stateent: The nuber of avelengths of the electrical transission required to cross North Aerica is. Section 0.4 Questions, page 53. Given: f 5.0! 0 4 Hz; c /s Required: λ Analysis: c! f ;! c f! c f / s Hz #7! n Stateent: The avelength of the light in the CD player is n. 2. Given:! 550 n #7 ; c /s Required: f Analysis:! f c f c! f c! /s #7 f Hz Stateent: The frequency of the light that is ost sensitive to the huan eye is Hz. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.4-2

15 3. Given: f 88 MHz 8.8! 0 7 Hz; f 2 08 MHz.08! 0 8 Hz; c /s Required: λ, λ 2 Analysis:! f c! c f! c f! 2 c f / s Hz! /s Hz! Stateent: The avelengths of the FM radio stations range fro 3.4 to Given:! 0.0 n.0 0 #0 ; c /s Required: f Analysis: c! f ; f c! f c! /s.0 0 #0 f Hz Stateent: The frequency of the X-ray is Hz. 5. (a) Given:! 2.24 c # ; c /s Required: f Analysis:! f c f c! f c! /s # f Hz Stateent: The frequency of the X-ray ave is Hz. (b) Most icroave ovens contain rotating carousels to heat the food evenly. By rotating the food, the nodes cannot heat the sae spot all the tie, so the food cooks ore evenly. With a icroave avelength of 2 c, the nodes are located ithin the oven itself. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.4-3

16 6. Given: f 2.4 GHz 2.4! 0 9 Hz; c /s Required: λ Analysis: c! f! c f! c f / s Hz.25 0 #! 2 c Stateent: The avelength of the radio aves in the cordless phone is 2 c. 7. Given: f 680 khz 6.80! 0 5 Hz; c /s Required: λ Analysis:! f c! c f! c f / s Hz ! 440 Stateent: The avelength of the broadcasting frequency used by 680 Nes is Given: 6.0 c 6.0! 0 2 ; f 7.5 GHz 7.5! 0 9 Hz; c /s; Required:θ Analysis: First calculate the avelength using the universal ave equation, c λf;! c. Then use the equation sin f λ θ to calculate the angle; sin!.! c f /s Hz! #2 Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.4-4

17 sin! sin! ()(4.0 #0$2 ) 6.0 #0 $2! 42 Stateent: The angle fro the central axiu to the first diffraction iniu is Soe of the radiation in the electroagnetic spectru is only detectable in deep space because Earth s atosphere absorbs the radiation at several different avelengths, so it does not pass through the atosphere to the surface. To be able to detect these portions of the electroagnetic spectru, the detectors have to be in space, above Earth s atosphere. Soe exaples of electroagnetic radiation fro space that does not reach Earth s surface are soe avelengths of infrared radiation fro distant objects, X-rays, and gaa rays. Additional inforation: All objects eit infrared radiation. To avoid interfering ith very faint astronoical objects eitting infrared radiation, the detectors (telescopes) need to be kept extreely cold. That is only possible in deep space. 0. Television correspondents in a distant part of the orld are so far aay that their responses are delayed due to the travel tie of the signal. There is also the tie required to process the signal.. (a) Given: f 75 MHz Hz; d 34 ; c /s Required: type of interference Analysis: Calculate the avelength of the signal, c λf;! c f. Divide the distance by avelength.! c f / s Hz! avelengths The interference is constructive interference because the path difference is a half-hole-nuber ultiple of the avelength, and there is a 80 phase change fro the reflection. Stateent: The kind of interference that results is constructive interference. (b) The signal is no being reflected fro , hich is exactly 23 avelengths. This is no destructive interference because the path difference is a hole-nuber ultiple of avelengths, and there is a 80 phase change fro the reflection. 2. Ansers ay vary. The report should explain ho an antenna converts electrical currents into electroagnetic radiation and that an antenna can be either a transitter or a receiver. The size of the antenna ill depend on the type of signals being transitted. For exaple, the avelengths of FM signals are 2 to 4. Generally, the length of the antenna should be about half the avelength of the radio aves you are trying to send or receive. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0.4-5

18 Chapter 0 Revie, pages Knoledge. (d) 2. (d) 3. (d) 4. (c) 5. (a) 6. (a) 7. (d) 8. (a) 9. (a) 0. (b). (c) 2. (c) 3. (d) 4. (b) 5. (a) 6. True 7. True 8. True 9. False. The ave theory of light best explains the bright and dark fringes surrounding the central axiu in a single-slit diffraction deonstration. 20. False. To achieve the best possible resolution of their iages, astronoers try to iniize the effects of diffraction fro distant stars. 2. False. Doubling the avelength ill increase the diaeter of the central axiu by to ties. 22. True 23. False. A CD is a good exaple of a reflection grating. 24. True 25. False. Diffraction gratings ill display interference patterns for all avelengths if the slit spacing is appropriate. 26. False. Gaa rays travel through a vacuu at the sae speed as icroaves. 27. False. Microave ovens ork by radiating the food ith aves of selected frequencies that interact strongly ith ater olecules. (Microaves have no effect on carbon dioxide.) 28. False. Light fro the Sun is not polarized before reflecting off particles in the atosphere. 29. True 30. False. Lidar technology relies on laser light. 3. True 32. True 33. False. All technology is not beneficial and has shortcoings. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-2

19 Understanding 34. (a), (b) Saple anser: In the diagra, rays 2 and 3 should both undergo a phase change, so rays 2 and 3 are 80 out of phase if t 2 λ. (a) Changing the avelength of the light ill result in varying aounts of reflection because each avelength of light has a different ideal thickness in the coating. (b) Changing the angle of incidence of the light ill result in varying aounts of reflection because the thickness of the coating is deterined ith the assuption that light hits the surface perpendicularly. 35. The path difference beteen these to reflected rays is 2t. 36. (a) Saple anser: A thin fil refers to a coating or fil applied to optical devices, such as lenses. The fil is so thin that it is not visible, coparable to the avelength of light or soe hole-nuber ultiple of the avelength of light. (b) To achieve optical interference, the thickness of the transparent fil should be approxiately zero, and the path length difference beteen rays reflected fro the top and botto surfaces is of the order of one avelength (or half of one avelength) of the ray of light. 37. Saple anser: In thick fils, the rays reflected fro both the botto and top surfaces of the fil ay not leave the surface at the sae location. This is due to the uch longer distance that the ray reflecting fro the botto surface has to travel, aking it ipossible to see interference effects. 38. Saple anser: As onochroatic light passes through a sall opening, the avelength of the light and the slit idth ill affect the angle of diffraction. A longer avelength of light ill diffract ore, and decreasing the size of the slit opening allos the light to diffract ore as ell. 39. No, the axia created by a diffraction grating do not all have the sae intensity. The central axiu is the brightest, and the axia becoe less intense as distance fro the centre increases. 40. Saple anser: FM radio aves have a shorter avelength than AM radio aves. Consequently, FM radio aves do not diffract as ell as the AM aves do around ountains or Earth s curvature, so FM aves lose reception before the AM aves. 4. Saple anser: Sound aves are copression longitudinal aves in a ediu. Radio aves are electroagnetic aves that do not need a ediu. Radio aves travel as electroagnetic aves at the speed of light through the air fro the copany s radio station to the personal radio device. The radio then plays a sound ave that reaches the listener s ear. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-3

20 42. Radio aves are transverse aves, ith the electric field in the sae plane, parallel to the antenna. If the receiving antenna is horizontal and the incident aves are vertically aligned, the charges in the antenna ill not be free to oscillate ith sufficient aplitude to hear a signal. 43. Saple anser: Optically active refers to transparent aterials that rotate the direction of polarized light. Very sall voltages can anipulate these aterials and create iages on a larger screen. Most digital atches, calculators, and cellphone displays ake use of polarizing filters and optically active aterials. 44. To tell hether or not a pair of sunglasses is polarized, I ould put on the glasses and tilt y head side to side. If the intensity of reflected glare changes as y head tilts, then the glasses are polarized. Alternatively, I could take to pairs of the sae glasses and hold the lenses perpendicular to each other. If all the light is blocked out, the glasses are polarized. Analysis and Application 45. (a) Given: n soap fil.35; t 2.50! 0 7 ; 0,, 2 Required: λ Analysis: Only one ave has a phase change on reflection. Use the forula for constructive interference of to aves hen phase change occurs in only one reflection;! + $ # 2% & ' 2t ;! 2tn soap fil n soap fil + % # $ 2& ' For 0:! 2tn soap fil + % # $ 2& ' 2(2.50 ( 0)7 )(.35) 0 + % # $ 2& '!.35 ( 0 )6 For : For 2:! 2(2.50 0#7 )(.35) +! 2(2.50 0#7 )(.35) $ ' $ % & 2( ) 2 + ' % & 2( )! #7! #7 Stateent: The three longest avelengths are , , and (b) Visible light falls ithin the range of to Therefore, the only visible ave in part (a) is the avelength , hen. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-4

21 46. Given: t 250 n 2.50!0 7 ; # 500 n 5.00!0 7 Required: n fil Analysis: Only one ave has a phase change on reflection. Use the forula for constructive interference of to aves hen phase change occurs in only one reflection;! + $ # 2% & '! + $ # 2% & ' 2t ; n n fil fil 2t Since the thickness of the fil is half the avelength of the light, use. ( 0 leads to an ipossible index of refraction, n.)! + $ # 2% & ' n fil 2t! + $ # 2% & ( 5.00 (0)7 ) (0 )7 ( ) n fil.50 Stateent: The index of refraction of the plastic fil is (a)! fil! air n 560 n.45! fil 390 n The avelength of the light in the fil is 390 n. (b) There is a phase change at both the air fil and fil glass surfaces because in each case the index of refraction of the reflecting surface is higher than the index of refraction of the incident ediu. (c) Given: n fil.45;! 560 n #7 Required: t Analysis: Use the forula for constructive interference of to aves hen phase change occurs at both reflections. Use n for iniu thickness. 2t n! n fil t n! 2n fil Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-5

22 t n! 2n fil ( ) #7 ( ) ( ) 2.45 t 90 n Stateent: The iniu thickness of fil that ill create constructive interference is 90 n. 48. Given: λ ; L 6.0 c ; there are 25 cycles of alternating light patterns along L Required: t Analysis:!x 6.0 0#2 25!x L 2t t L 2!x t L! 2x ( 6.0 #0$2 ) ( 6.5#0 $7 ) % #0$2 ( ' & 25 * ) t 8.#0 $6 Stateent: The hair is thick. 49. (a) Given: n fil.39;! #7 Required: t Analysis: Use the forula for destructive interference of to aves hen phase change occurs at both reflections. Use 0 for iniu thickness.! + $ # 2% & ' 2t n fil! + $ # 2% & ' t 2n fil Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-6

23 ! + $ # 2% & ' t 2n fil ( )! 0 + $ # 2% & 5.40 (0) ( ) t 9.7(0 )8 Stateent: The iniu thickness of fil that ill create destructive interference is (b) For the next to possible thicknesses, use and 2. For : For 2:! + $ # 2% & ( 5.40 ' 0(7 )! 2 + $ t # 2% & ( 2(.39) 5.40 ' 0(7 ) t 2(.39) t 2.9' 0 (7 t 4.85 ' 0 (7 Stateent: The next to sallest thicknesses of fil that ill create destructive interference are and Given: λ 630 n Analysis: Light rays reflecting off the CD surface ill change phase hether they are reflecting at the top of a pit or at the botto. The depth of the pit is t, so the one ray travels 2t farther. For destructive interference, this distance ust be 2 λ ; 2t 2! t 4! t 4! ( ) 630 n 4 t 60 n Stateent: The iniu pit depth is 60 n. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-7

24 5. Given: n.70; λ 630 n Required: t Analysis: Light rays reflecting off the CD surface ill change phase hether they are reflecting at the top of a pit or at the botto. The depth of the pit is t, so the one ray travels 2t farther. For destructive interference this distance ust be λ. Hoever, the 2 rays are travelling in the plastic ediu, so the avelength of the light is! plastic! noral n plastic ; 2t 2! t 4!! plastic! noral n plastic 630 n.70! plastic 37 n t 4! plastic (37 n) 4 t 93 n Stateent: The iniu pit depth needed to produce destructive interference is 93 n. 52. Given: λ 420 n ; there are 25 cycles of alternating light patterns in L Required: t Analysis: L 25Δx;!x L 2t 2t L!x 2t 25!x!x t 25 2 Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-8

25 t 25! 2 ( ) ( ) # t 5.30 #6 Stateent: The plate spacing at the right edge is Given: t 3.5 µ Required: λ Analysis: Since the aves reflect off the irrors an even nuber of ties, they ill eerge in phase and interfere constructively if 2t is a ultiple of λ; 2t n!! 2t n! 2t n! (2)(3.50#6 ) n We need 600 n < λ < 700 n. By trial and error, For n 0: For n 2:! (2)(3.5 0#6 ) 0! 700 n For n :! (2)(3.5 0#6 ) n! 580 n! (2)(3.5 0#6 ) 636 n! 640 n Stateent: The only possible avelength values are 640 n and 700 n. 54. (a) Given: λ 350 n ; n air.00 Required: t Analysis: For the glass to appear bright, there needs to be constructive interference. Phase change occurs at the air glass surface but not at the glass air surface. Use the forula for constructive interference and 0. Include a diagra.! + $ # 2% & '! + $ # 2% & ' 2t ; t n air 2n air Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-9

26 ! + $ # 2% & ' t 2n air ( ) ( )! 0 + $ # 2% & 3.5 ( 0) t 8.8 ( 0 )8 Stateent: For the glass to appear bright, the layer of air should be thick. (b) Given: λ ; n air.00 Required: t Analysis: For the glass to appear opaque, there needs to be destructive interference. Phase change occurs at the air glass interface but not at the glass air interface. Use the forula for destructive interference and n. 2t n! ; t n! n air 2n air t n! 2n air ()(3.5 0#7 ) 2(.00) t.8 0 #7 Stateent: For the glass to appear opaque, the layer of air should be thick. 55. Given: t c ; L ; there are 6 cycles of alternating patterns in Required: λ Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-0

27 Analysis:!x L 2t 2t!x L!x.23 #0$2 6!x 2.05#0 $3! 2tx L 2(.92 #0$5 )(2.05#0 $3 ) 9.8 #0 $2! 8.03#0 $7 Stateent: The avelength of the light is 8.03! Given: λ 560 n ; L 6.3 ; 2Δ y.3 c.3 0 Required: Analysis:! y L!L y!l y 2y.3#0 $2 y 6.5#0 $3 (5.60 #0$7 )(6.3 ) 6.5#0 $3 540 µ Stateent: The idth of the slit is 540 µ Given: n 3; ; θ3 5 Required: λ Analysis: sin! n n! sin n n sin! n n (8.2 #0$6 )(sin5 ) 3! 70 n Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-

28 Stateent: The avelength of the light is 70 n. 58. (a) Given: 5.5Δy 6. c Required: Δy, the distance beteen successive axia Analysis: The distance to the fifth axiu is 5.5Δy. 5.5!y 6. c 6. c 5.5. c (one extra digit carried)!y. c Stateent: The distance beteen successive axia is. c. (b) Given: L.5 ; ; Δy. c. 0 2 Required: λ Analysis: λl Δ y Δy λ L! y L (5.6 # 0$5 )(.# 0 $2 ).5! 40 n Stateent: The avelength of the violet light is 40 n. 59. Given: λ 50 n ; 7 µ ; Δy 2.6 c Required: L Analysis: λl Δ y Δy L λ L!y (.7 #0$5 )(2.6 #0 $2 ) 5.0 #0 $7 L 0.87 Stateent: The distance beteen the slit and the screen is Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-2

29 60. Given: N 8000 lines/c;! 660 n #7 ; Required: θ Analysis: Use to calculate the slit separation. Rearrange the equation N λ sinθ to calculate the angle; sin! N 8.0!0 3 lines/ c! 00 c.25!0 6 sin! ()(6.60 #0$7 ).25#0 $6! 32 Stateent: The red light produces a first-order axiu at (a) When the lines on one grating are perpendicular to the lines on the other grating, there ill be a rectangular array of dots that are brighter in the centre and fainter to the sides. (b) When the lines on one grating are parallel to the lines on the other grating, light ill only pass through slits that line up if the gratings are touching each other. Otherise, the light rays ill not pass through. The result ill be a diffraction pattern siilar to a pattern produced by a diffraction grating but ith spacing indicated by the slits that line up. 62. (a) Consider the case here. For diffraction to occur,! sin sin! ( 0 #8 )sin$ 0! sin! 0! (# 0 $8 )sin! # 0 $8 0! %! # 0 $8 Visible light has avelengths beteen 380 n and 740 n. Visible light does not fall in the diffraction range for λ above. Therefore, visible light cannot exhibit diffraction hen reflected off crystals that have a spacing of 0 8. (b) The aves that ill sho diffraction in this situation are those ith shorter avelengths, such as ultraviolet radiation, X-rays, and gaa rays. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-3

30 7 63. (a) Given: λ 430 n ; θ 6 ; Required: Analysis: Rearrange the equation λ sinθ to calculate the slit separation;! sin! sin ()(4.30 #0$7 ) sin6.56 #0 $6 (one extra digit carried) Stateent: The spacing beteen adjacent slits on the diffraction grating is (b) Given: Required: N Analysis: Rearrange the equation to deterine the nuber of lines per N centietre; N N.56! 0 6 N 6400 lines/c Stateent: The diffraction grating has 6400 lines/c. 64. (a) Given:! 650 n #7 ; θ a 34 ; θ b 3 ; Required: Analysis: Rearrange the equation λ sinθ to calculate the slit separation for each angle;!. Then average the to angles. sin a! sin ()(6.50 #0$7 ) a.62 #0 $6 (to extra digits carried) Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-4

31 b! sin ()(6.50 #0$7 ) b.262 #0 $6 (to extra digits carried) For the average: av a + b 2.62! ! ! 0 6 (to extra digits carried) av.2! 0 6 Stateent: The average slit spacing is.2! 0 6. (b) Given: Required: N Analysis: N N.22!0 6! 00 c N 8300 c Stateent: The diffraction grating has 8300 lines/c. (c) Since there are 8300 lines/c, there are 4 c 8300 lines/c lines on the CD. If there is one rotation per line, there ill be rotations, or rotations, to read the coplete CD. (d) The rotational speed is the nuber of rotations divided by the tie: rotations v 50 n v 660 rp The rotational speed is 660 rp. 65. Given: λ 400 n ; λ n ; L 4.0 ; ; N 2000 lines/c Required: Δy Analysis: Deterine the slit idth using N ; rearrange the equation! sin to calculate the angle, sin!. Then use trigonoetry to calculate the idth of the first-order rainbo spectru. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-5

32 N 2000 c! 5.0 0!4 c 5.0 0!6 sin# $ sin# ()(4.0 0!7 ) 5.0 0!6 # (to extra digits carried) ()(7.0 sin! 2 0#7 /) #6 /! (to extra digits carried) The rainbo spectru has an angular idth of (to extra digits carried). Using trigonoetry, the linear idth of the spectru is (4.0 )sin Stateent: The idth of the first-order rainbo spectru is Given: λ 660 n ; N 5000 lines/c Required:, nuber of axia Analysis: Calculate the slit idth, N ; rearrange the equation! sin to calculate the angle; sin!. The last observable axiu occurs at an angle of 90. N 5000 c! 2.0 0!6 sin! sin90 (6.60 # 0$7 ) 2.0 # 0 $ Stateent: There are three axia on each side of the central axiu. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-6

33 67. Given: f 2.0!0 9 Hz Required: λ Analysis: Rearrange the universal ave equation, c λf, to solve for avelength;! c f! c f / s Hz! 0.5 Stateent: The avelength of the signal is (a) Given: λ 940 n Required: f Analysis: Rearrange the universal ave equation, c λf, to solve for frequency; f c! f c! /s #7 f s # Stateent: The frequency of the radiation is Hz. (b) The radiation travels at the speed of light, /s; distance tie speed t d s !0 8 /s t 8.3!0 9 s It takes s for the signal to reach the television. 69. Given: f 89 GHz 8.9!0 0 Hz Required: λ Analysis: Rearrange the universal ave equation, c λf, to solve for avelength;! c f Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-7

34 ! c f / s Hz! #3 Stateent: The avelength of the signal is When unpolarized light passes through a polarizing filter, only half the light passes through. So, through the first filter, I out 2 I in, then I in 2 2 I in. I out 2 I in 2 cos 2! 2 2 I in cos I in % # $ 2& ' 2 I out 2 8 I in The intensity of the transitted light is 8 I in. 7. Given: I out 0 I in Required: θ, the rotation angle Analysis: When unpolarized light passes through a polarizing filter, only half the light passes through. So, through the first filter, I out 2 I in, then I in 2 2 I in. I out 2 I in 2 cos 2! 2 0 I in 2 I cos 2! in 2 5 cos2! 2 # & cos % (! $ 5 2 '! 2 63 Stateent: The rotation angle of the second polarizer relative to the first is approxiately 63. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-8

35 72. (a) The angle of reflection is 6. (b) According to Brester s la, the su of the angle of reflection and the angle of refraction is 90 because the reflected ray and the refracted ray are perpendicular to each other: The angle of refraction is 29. (c) tan! B n 2 n n 2 (tan! B )n (tan6 )(.00) n.8 The index of refraction of the aterial is approxiately.8. (d) tan! B n 2 n tan! B.33.00! B 53 The angle of incidence ould have to be approxiately 53 to yield copletely polarized light if the transparent aterial ere ater. 73. (a) The shado is green because the shado is actually light that gets through the filter. The green light ust be polarized in one direction, and the red and blue light ust be polarized in the perpendicular direction. (b) If the student rotated the polarizer by 90, the green light ould be blocked and the red and blue light ould transit, producing a agenta shado. 74. The polarizer ill block out the light fro the glare, but not fro the bulb because the light fro the bulb is linearly polarized in all directions. The polarizer blocks a different orientation of light at each angle. Hoever, hen the light is reflected off the table s surface, ost of the reflected light is polarized in a certain direction. Making the polarizer perpendicular to the linearly polarized reflected light ill cancel this light entirely. 75. When unpolarized light passes through a polarizing filter, only half the light passes through. So, through the first filter, I out I in 2 2 I 0. I out 2 I in 2 cos 2! 2 2 I 0 cos I 0 # $ % 2 & ' 2 I out 2 4 I 0 Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-9

36 Any light that passes through the third filter ill be unpolarized. When unpolarized light passes through a polarizing filter, only half the light passes through. So, through the third filter: I out 3 2 I out 2! 2 4 I $ 0 # % & I out 3 8 I 0 The intensity of the light that passes through all three filters is 8 I 0. Evaluation 76. Ansers ay vary. Saple anser: The sentence is incorrect. The light intensity is zero at the point of contact of the to plates, folloed by seven alternating bright and dark fringes. 77. Ansers ay vary. Saple anser: My anser ould have three significant digits because the value provided for the avelength has three significant digits. The value for the nuber of lines on the diffraction grating has to significant digits, but that nuber is counted and therefore does not affect ho any significant digits I need to sho in y anser. 78. Ansers ay vary. Saple anser: The analogy is not accurate. The fence blocks all polarizations of the rope aves except those along the picket direction. Malus s la, hoever, says that soe light ill alays pass through a filter, unless the incident light is polarized at exactly 90 to the filter polarization. 79. Ansers ay vary. Saple anser: Publishing aterials on the Internet uses no paper and ink, benefiting the environent. If Internet access is free, as in libraries, it can reach ore people, hich benefits society. Reflect on Your Learning 80. Ansers ay vary. Saple anser: What I found ost surprising in this chapter as learning hy butterfly ings shier and are iridescent. What I found ost interesting in this chapter as holography. I can learn ore about these topics ith further research using the Internet. 8. Ansers ay vary. Saple anser: I ould explain the concepts of thin-fil interference and polarization to a student ho has not taken physics by using exaples and explaining these concepts in the siplest ay that I could. If y explanation is too coplex, the student ill not understand hat I a saying. Exaples allo the student to visualize hat I a teaching and deonstrate soe applications of the concepts. 82. Ansers ay vary. Saple anser: I no see the physics concepts that ere explored in this chapter everyhere. Light is everyhere in our lives and I no have a basic understanding of the physics concepts of light. I also see these concepts hen I here y sunglasses, see butterflies, and use any ite that has a hologra that is used to prevent counterfeiting. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-20

37 Research 83. (a) Saple anser: The cuticle is the outer, non-cellular layer of a butterfly s exoskeleton. Scales are tiny overlapping pieces of chitin on a butterfly s ing, hich create the butterfly s colour. (b) Saple anser: Soe purposes of the colours of a butterfly are as follos: the colours function as a caouflage, absorb theral energy, help find a ate, and can act as a arning to potential predators. 84. (a) Saple anser: The advantage of binocular vision is that it allos the bald eagle to gauge distances precisely. (b) Saple anser: The fovea in eagles contain a uch higher concentration of sensory cells than the fovea in huans, giving eagles a ore detailed iage. Each eagle eye has to fovea, hereas each huan eye only has one. (c) Saple anser: Each eye of a bald eagle has to eyelids. The inner eyelid keeps the eye et and clear hile also shielding the eye. The eyebros of a bald eagle help protect the fro injury and help shield their eyes fro the Sun. The flexible lens of a bald eagle s eye is soft and can focus rapidly. This allos the eagle to quickly focus beteen near and far objects. 85. (a) The VLA stands at an elevation of 224. (b) There are 27 independent antennas present in the observatory. (c) The diaeter of one of the circular dishes is 25. (d) With the VLA, astronoers observe radio galaxies, quasars, pulsars, supernova renants, black holes, the Sun, radio-eitting stars, gaa ray bursts, other planets, astrophysical asers, and the hydrogen gas that akes up uch of our galaxy. (e) One advantage of receiving radio aves instead of other types of aves is that radio aves can penetrate clouds and are unaffected by rain, so it does not atter if it is cloudy or raining hen gathering radio signals. Radio aves are also unaffected by sunlight, so radio observatories can be used at night. 86. (a) QR stands for quick response. (b) QR codes originated in Japan, here they ere first used in the autootive industry. (c) Saple anser: QR codes can hold uch ore inforation than barcodes, such as ebsite links and text. They are also quick and easy to scan ith obile applications, such as cellphones. (d) Saple anser: A barcode is considered one-diensional because the bars are in a line and the laser scans the lines. The QR code is considered to-diensional and the sybols are not arranged in a line, so a laser cannot read the sybols. A digital caera ith the QR reader softare can be used to read the QR code. A laser is not needed to scan the code. 87. Ansers ay vary. Saple anser: Fessenden s ain ork as in the area called radiotelephony, in hich a changing sound signal is iposed, or odulated, onto an electroagnetic ave. The odern ireless telephone, or cellphone, is one of the ost recent and idespread exaples of radiotelephony. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-2

38 Chapter 0 Self-Quiz, page 549. (c) 2. (a) 3. (c) 4. (c) 5. (a) 6. (b) 7. (a) 8. False. A thin-fil anti-reflective coating of constant thickness can block out soe avelengths of light but not all. 9. True 0. True. False. A ajor difference beteen DVD technology and CD technology is that the light used to read DVDs has a shorter avelength than the light used to read CDs. 2. False. Earth s atosphere absorbs ultraviolet radiation ost strongly at higher frequencies. 3. True 4. False. If a aterial can reorient the direction of linearly polarized light, then the aterial is said to be optically active. 5. False. Photoelasticity is a easure of ho easily a aterial refracts light hen under stress. 6. True 7. False. GPS devices have positive and negative ipacts on society. Copyright 202 Nelson Education Ltd. Chapter 0: Applications of the Wave Nature of Light 0-