Pearson Physics Level 30 Unit VII Electromagnetic Radiation: Unit VII Review Solutions

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1 Pearson Pysics Level 30 Unit VII Electromagnetic Radiation: Unit VII Review Solutions Student Book pages Vocabulary 1. angle of diffraction: te angle formed between te perpendicular bisector and te straigt line to a nodal or antinodal point on te interference pattern antinode: a point of interaction between waves, at wic only constructive interference occurs blackbody: an object tat completely absorbs any ligt energy tat falls on it blackbody radiation curve: a grap of te intensity of ligt emitted versus wavelengt for an object of a given temperature Compton effect: te cange in wavelengt of te scattered X ray poton Compton scattering: te scattering of an X ray by an electron converging: te process wereby refracted or reflected rays converge or meet at a single point critical angle: for any two media, te size of te incident angle for wic te angle of refraction is 90 diffraction: te cange in sape and direction of a wave front as a result of encountering a small opening or aperture in a barrier, or a corner diffraction grating: a seet of glass or plastic etced wit a large number of parallel lines; wen ligt is incident on te grating, eac line or slit acts as one individual ligt source dispersion: separation of wite ligt into its components diverging: te process wereby refracted or reflected rays are spread out suc tat tey never meet at a single point electromagnetic wave (radiation): periodic variation in perpendicular electric and magnetic fields, propagating at rigt angles to bot fields focal point (F): te point were ligt rays tat are parallel to and close to te principal axis converge, or appear to diverge from, after being reflected or refracted frequency (f): number of cycles per unit of time Heisenberg s uncertainty principle: It is impossible to know bot te position and momentum of a particle wit unlimited precision at te same time. Huygens Principle: a model of wave teory, wic predicted te motion of a wave front as being many small point sources propagating outward in a concentric circle at te same speed as te wave itself image attitude: te orientation caracteristic of an image, weter erect or inverted incandescent: glowing wit eat interference: an effect resulting from te passage of two like waves troug eac oter law of reflection: te angle of reflection is equal to te angle of incidence magnification (m): te relationsip of te size of te image to te size of te object 1

2 node: a point of interaction between waves, at wic only destructive interference occurs particle model: describes electromagnetic radiation as a stream of tiny particles radiating out from a source pat lengt: te distance between a point source and a cosen point in space period (T): time required to complete one cycle potoelectric effect: te emission of electrons wen a metal is illuminated by sort wavelengts of ligt potoelectrons: electrons emitted from a metal because of te potoelectric effect poton: a quantum of ligt Planck s formula: Ligt comes in quanta of energy tat can be calculated using te formula E nf, were is a constant of proportionality aving te value of J s. polarization: production of a state in wic te plane of te electric field for eac electromagnetic wave occurs in only one direction quantized: limited to wole multiples of a basic amount, te quantum quantum: te smallest amount or bundle of energy tat a wavelengt of ligt can possess quantum indeterminacy: te probability of finding a particle at a particular location in a double-slit interference pattern refraction: a cange in te direction of ligt as it passes at an angle from one medium to anoter refractive index (n): a ratio comparing te speed of ligt in a vacuum to te measured speed of ligt in a medium Snell s Law: for any angle of incidence greater tan zero, te ratio sinθ i /sinθ r is a constant for any ligt ray passing troug te boundary between two media spectrum: te bands of colours making up wite ligt; in order: red, orange, yellow, green, blue, and violet stopping potential: te potential difference for wic te kinetic energy of a potoelectron equals te work needed to move troug a potential difference, V tresold frequency: te minimum frequency tat a poton can ave to cause potoemission from a metal total internal reflection: reflection of all incident ligt back into te medium tat as te iger refractive index wave model: describes electromagnetic radiation as a stream of transverse waves radiating out from a source wave-particle duality: ligt as bot wave-like and particle-like properties wavelengt (λ): te distance between adjacent points tat vibrate in pase wit one anoter in a wave work function: te minimum energy tat a poton can ave to cause potoemission from a metal. Te work function is specific for every metal. Knowledge Capter 13. Te currently accepted quantum model describes ligt and all oter radiation as discrete bundles or packets of energy. Eac packet, or poton, is a particle tat as

3 wave caracteristics. In te quantum model, EMR as two aspects of beaviour, one being wavelike and te oter being particle-like. 3. According to Ampere s Law, if te magnetic field canges over time, ten tere must be a current flowing. But according to te classical laws of pysics, current flow was not possible in empty space, in te absence of a conductor. However, Maxwell observed tat te electric field produced by a capacitor could actually ave te same effect as a moving carge; terefore, a conductor is not necessary to produce an electromagnetic wave. 4. Hertz s experiment provided evidence tat supported Maxwell s predictions about electromagnetic radiation. In tis experiment, an induction coil rapidly canged te electric field across a spark gap (te radiator). Wen te electric field between te two electrodes forming te spark gap reaced a sufficiently ig value, te electrons jumped from one electrode to te oter. As te carge was rapidly transferred, te electric field underwent a rapid cange tat caused a canging magnetic field, wic ten caused a canging electric field, and so on. An electromagnetic wave was produced and it radiated outward in all directions. Wen te canging electromagnetic field encountered te metal antenna, it induced a current flow, wic was measured. Te induced current flow in te antenna oscillated at a frequency identical to tat of te canging electric field in te radiator. Tis was conclusive evidence tat Hertz s device ad indeed produced te electromagnetic radiation tat was being observed at te antenna. 5. A canging magnetic field will produce a canging electric field (and vice versa). If te magnetic field lines eld witin a closed pat are constant and uncanging, tere will not be an electric field along te closed pat. Similarly, if te electric field is constant and uncanging, it will not produce a magnetic field. 6. A spark produces EMR since tere will be a rapidly canging electric field as te electrons accelerate from one electrode to anoter. Te rapidly canging electric field induces a canging magnetic field, and so on, wit bot fields varying over time and travelling outward in all directions from te spark location. 7. Te microwaves in a microwave oven consist of varying electric and magnetic fields. Wen tey are incident on a metal conductor, suc as a spoon, te varying electric field will induce current flow and carge separation in te conductor. Te carge separation and current flow in te spoon can cause sparks to be produced.. Converging mirror 3

4 Diverging mirror Incident Ray 1 is parallel to te principal axis. Any ray tat is parallel to te principal axis will reflect troug te focal point on a converging mirror, or appear to ave originated from te focal point on a diverging mirror. Incident Ray travels troug or from or toward te focal point. Any ray tat passes troug te focal point on a converging mirror, or is directed at te focal point on a diverging mirror, will be reflected back parallel to te principal axis. Incident Ray 3 travels troug or from or toward te centre of curvature. Any ray tat passes troug te centre of curvature on a converging mirror, or is directed at te centre of curvature on a diverging mirror, will be reflected directly back along te incident pat. 9. Te radius of curvature is twice as long as te focal lengt (f) of te mirror. 10. A virtual focal point represents te position were several reflected rays appear to ave converged or originated. For example, any ray tat is directed at te virtual focal point on a diverging mirror will be reflected back parallel to te principal axis. And any ray tat travels parallel to te principal axis will be reflected suc tat it appears to ave originated or passed troug from te virtual focal point. 11. Two rays from te object are incident on concave mirror 1 and are reflected parallel to te principal axis. Tese two rays are reflected from concave mirror and converge at te focal point. A small ole is cut in mirror 1 so tat te real image forms above te top of mirror Wen you pull te concave side of a polised spoon away from your nose, your image disappears wen te distance between te spoon and your nose is equal to te focal lengt of te reflector. At te focal lengt, te reflected rays travel outward, parallel to one anoter, never converging or appearing to ave converged at any point. Terefore, no image is formed. 13. A paddle appears bent wen you put it in water because of te visual effects created by refraction. As te ligt rays from te submerged section of te paddle exit te water and enter te air, tey are refracted, canging te direction of travel and producing te visual effect of a bent object. 4

5 14. According to Snell s Law, te wavelengt of EMR canges wen it enters a new medium. Snell s Law supports te wave model of ligt, since it relates wavelengt to te angles of incidence and refraction and to te velocity of te wave. sinθ1 v1 λ1 n sinθ v λ n1 15. Wen ligt passes from air (n 1.00) into water (n 1.33), Snell s law predicts tat te wavelengt decreases by a factor of 1.33 according to: λ1 n λ n1 λ λ λ1 1.33λ λ1 1.33λ 16. As eac person enters te water, is or er speed of travel is reduced compared to tose wo are still running on te land. Te difference in speed between tose in te water and tose wo are still travelling on land causes te line (or wave front) to cange directions. 17. Newton set up two prisms. Te first one was exposed to wite ligt wic produced a full spectrum of colours. Te second prism was exposed only to red (monocromatic) ligt coming from te first prism. Te second prism did not produce any more colours; only te red ligt emerged. As a second test, Newton inverted te second prism and allowed te spectrum of colours from te first prism to enter it. Te spectrum disappeared and only wite ligt emerged from te second prism. Based on tese observations, Newton concluded tat wite ligt is made up of all te colours in te spectrum and te prism was separating or dispersing all te colours. 1. Huygens Principle describes a wave front as consisting of many small point sources of tiny secondary waves, called wavelets, wic propagate outward in a concentric circle at te same speed as te wave itself. Te line tangent to all te wavelets constitutes te wave front. 19. As te secondary wavelets emerge from te opening, tey form a new wave front, travelling outward wit curved edges. 5

6 0. Young conducted is experiment using a pinole in a window sutter and a card e described as a slip of card, about one-tirtiet of an inc in breadt (tickness). Te card was positioned edgewise in te centre of a sunbeam suc tat it split te sunbeam into two coerent beams separated by a very small distance. Ligt coming from bot sides of te card was in pase and te wave fronts created a fixed interference pattern of ligt and dark bands, called interference fringes, on a nearby screen. 1. Diffraction occurs wen a wave front encounters a small opening or aperture in a barrier. Wen passing troug te opening, te wave front canges bot sape and direction according to Huygens Principle, wic describes EMR as a wave. In addition, te wavelets tat constitute te wave front may also interfere, a process wic can only be explained by assuming EMR as wave properties.. In 11, Dominique Arago experimentally verified Poissson s Brigt Spot, providing strong evidence for te wave model of ligt. Poissson s brigt spot is explained by constructive interference occurring at te center of a sadow wen monocromatic ligt diffracts around te edge of a small disc. Te prediction of tis pattern, based on wave interference, and te verification of te pattern was crucial to te acceptance of te wave model of ligt. 3. Wen unpolarized ligt is incident on a polarizing filter, only one plane of te electric field is allowed to pass troug, causing plane polarized ligt to emerge. If a second 6

7 polarizing filter is eld at rigt angles to te plane polarized ligt, ten te plane polarized ligt also is absorbed fully blocking all of te electromagnetic radiation from passing troug te filters. 4. Electromagnetic waves are defined by varying electric and magnetic fields travelling troug space. Te periodic variations in te transverse electric and magnetic fields occur at rigt angles to one anoter and to te direction of propagation. Terefore, an electromagnetic wave is tree dimensional. Capter No. Te energy of a quantum of ligt (poton) is given by E nf, were n is te number of potons, is Planck's constant, and f is frequency. A quantum of blue ligt as a iger frequency and terefore iger energy tan a quantum of red ligt. 6. Given λ 550 nm poton energy (E) Use te equation E nf, were n 1. E f c λ 34 m ( J s ) s J Paraprase 19 Te energy of a 550-nm poton is J. 7. Temperature affects te sape of te blackbody curve, and especially te colour at wic maximum intensity of emission occurs. For example, a bluis-wite-coloured star is otter tan a reddis-coloured star. If you compare te blackbody curve for incandescent objects of known temperatures wit te blackbody curves produced by te spectra of stars, you can estimate te temperature of te star. Tis temperature is only approximate, owever, because stars are not perfect blackbody radiators.. gamma ray poton, 10-nm poton, ultraviolet poton, 600-nm poton, infrared poton, microwave poton 9. Given λ 500 nm frequency (f) c fλ c f λ 7

8 m f s Hz Paraprase 14 Te frequency of blue ligt of wavelengt 500 nm is Hz. 30. Ultraviolet potons are energetic enoug (ave a ig enoug frequency, from te equation E f) to cause cell damage in te skin (sunburn), wereas visible ligt potons are not sufficiently energetic. 31. Tresold frequency is te minimum frequency of illumination tat will cause potoemission to occur from a metal surface. Ligt below tis frequency, regardless of its intensity, will not cause emission of electrons from te metal surface. 3. Te energy of potoelectrons does not depend on te intensity of te incident ligt. Te number of electrons emitted increases wit increasing intensity, owever. 33. Given W 0 3. ev minimum wavelengt for potoemission ( λ ) Ligt tat as a frequency equal to te tresold frequency will just cause potoemission. Te kinetic energy of te emitted electrons will be zero. Convert 3. ev into joules and use te equation Ek f W0 0 f0 W0 f0 W0 W0 f0 c λ f c W0 c W0 m J s ( s ) λ 19 J ( 3. ev ) ev 7 3. Paraprase Te minimum wavelengt of ligt tat will cause potoemission is 3 nm. 34. Te Compton effect is te elastic scattering of potons from matter often X rays scattering off electrons or oter small, subatomic particles. Te potoelectric effect is te absorption of te energy of a poton, wic leads to te emission of an electron

9 from a metal surface. Even toug tey are distinct penomena, tey bot exibit te particle nature of ligt. 35. Te term wave particle duality refers to te unusual situation tat bot matter and energy, under appropriate conditions, exibit eiter wave-like or particle-like properties. Waves and particles are opposing concepts, and tis dual nature of ligt is referred to as te wave particle duality. 36. Given λ 300 nm poton momentum (p) De Broglie s relation also applies to potons, so use te equation p. λ J s p N s Paraprase 7 Te momentum of te 300-nm poton is.1 10 N s. 37. Given v e 3000 km/s de Broglie wavelengt ( λ ) p λ λ p mv J s λ kg /s ( )( ) Paraprase Te electron as a de Broglie wavelengt of 0.43 nm. 3. Te de Broglie wavelengt is inversely related to momentum. Te neutron is sligtly more massive tan te proton, so te neutron as greater momentum, and ence a smaller de Broglie wavelengt, tan te proton. 39. If a particle is confined to a fixed region in space, ten te uncertainty in its position is, at most, equal to te size of te region in wic te particle is located. According to Heisenberg s uncertainty principle, te smaller te uncertainty in position, te greater is te uncertainty in momentum of te particle. So, if you know rougly were te particle is, tere must also be an uncertainty in its momentum. Tis means tat te particle cannot ave zero momentum. Because momentum and energy are related 9

10 by te equation energy. E k p, te particle cannot be at rest and ence must ave some m Applications 40. If visible ligt (incident on two small oles) is a particle, we would expect to observe a large brigt region were te ligt from eac slit overlaps. Or, if te oles were sufficiently far apart, you would observe two brigt regions, one for eac opening. Wat is actually observed wen ligt passes troug two small oles is an interference pattern. Tis pattern supports te wave model of ligt since only waves would be expected to sow te constructive and destructive interference required to produce te observed pattern. 10

11 41. Given f 90.9 MHz P 50 kw te number of potons per second (n) E Use te equations E nf and P. Te radio station emits 50 kw or 50 kj/s, so E Δ t 50 kj. Rearrange Planck s equation to solve for te number of potons. E n f 50 kj 34 ( J s)(90.9 MHz) J 34 6 ( J s )( Hz ) Paraprase Te radio station emits radio potons eac second. 4. An antenna is a conductor wic is able to sense electromagnetic radiation by tuning into te induced current flow in te antenna. Te rate of cange of te induced current in te antenna corresponds to te frequency of te varying electric field in te electromagnetic wave. 43. Given Δ t.56 s c 3.00 /s distance (Δd) Te distance from te surface of Eart to te surface of te moon is equal to one alf te round trip distance tat te ligt travels in.56 s at te speed of ligt in a vacuum (c). Δd c Δ t Δ d cδt Δ d cδt roundtrip Δ d (3.00 /s)(.56 s) roundtrip Δ d 7.6 roundtrip 7.6 Δ doneway Δ d 3.4 oneway Δ d km oneway 5 11

12 Paraprase Te surface of te moon is km from te surface of Eart. 44. No, increasing te intensity of te ligt does not cange te energy of eac poton. You can only cange te energy of te potons by canging te frequency of te potons. Increasing te intensity increases te number of potons. 45. Given 1 rotation 16 sides v.97 /s rot f Hz 60 s distance (d) Unit analysis can be used to determine te distance tat te ligt can travel during 1/16t of a rotation. Te given information must be arranged as follows rot m s d side s rot 1 rot.97 1 s dtotal 16 side 1 s rot dtotal m m d m Paraprase Te fixed mirror sould be place m from te 16-sided rotating mirror. 46. Given 1 rotation sides c 3.00 /s 3 4 d (5.00 )() 1.00 frequency of rotation (rot/s Hz) Unit analysis can be used to determine te frequency of rotation in Hertz. Te given information must be arranged as follows rot m 1 rot f Hz side s m s Solution 1 rot f rot f 3750 s 4 side 1s 1.00 f Hz 3 1

13 Paraprase Te -sided mirror was rotating at a frequency of Hz. 47. Given A sixteen-sided mirror 1 rotation 16 sides rot f Hz s time to make one sixteent of a rotation (t) Unit analysis can be used to determine te time. Te given information must be arranged as follows: rot s s t side rot side 1 rot 1 s t 16 side rot 4 s t side Paraprase Te 16-sided rotating mirror takes s to make 1/16t of a rotation. 4. At nigt, visible ligt is not intense enoug for police searc and rescue personnel to see people. Since uman bodies release infrared radiation wic as an intensity greater tan tat of te background, tey can be sensed by observing te infrared radiation, wic emanates from all warm objects, suc as people and animals. 49. Given v material m/s c in air m/s Te refractive index of te unknown material (n) Te index of refraction is a ratio of te speed of ligt in te medium to te speed of ligt in a vacuum. Te index will be used to identify te unknown material. c n v c 3.00 /s nunknown v 1.4 /s nunknown.4 Paraprase and Verify Te refractive index of te unknown material is.4, wic is equal to te refractive index of diamond. Terefore, te unknown material is diamond. 50. Given λ 50 nm W 0.3 ev to determine weter te metal surface will emit electrons 13

14 Calculate te tresold frequency for te metal and te frequency of a 50-nm poton. Potoemission will occur if te frequency of te 50-nm poton is greater tan te tresold frequency. Convert te work function units from electron volts into joules. Use te equations f0 W0 W0 f0 and c f λ For tresold frequency, 19 J (.3 ev ) ev f J s Hz Te frequency of te 50-nm poton is m f s Hz Te poton s frequency is greater tan te tresold frequency. Paraprase Te metal surface will emit electrons wen illuminated by 50-nm ligt. 51. Given An eigt-sided mirror 1 rot side f Hz 4 4 d speed of ligt (c) percent error Te ligt travels te given, round-trip distance in te time it takes te rotating mirror to make one eigt of a rotation. 1 Δd T c f Δ t 14

15 1 T Hz 1 Δ t T 1 3 t ( Δ s ) Δ t s s c s c.91 /s (experimental value accepted value) % error 100 accepted value (.91 /s /s) % error /s % error 3.00% Paraprase Te experimentally determined speed of ligt is m/s. Tis differs by 3% from te currently accepted value for te speed of ligt in air. 5. Above Eart s atmospere, te aluminium will be illuminated by te strong UV rays and soft X rays emitted by te Sun. Tese frequencies will be well above te tresold frequency for aluminium. As a result, te aluminium metal will emit potoelectrons. Since te aluminium is losing electrons, it will become positively carged. 53. Given f 5.0 cm d i 3.0 cm i 1.0 cm object distance (d o ) object eigt and attitude Te object position can be found using te mirror equation as follows: f do di do f di do 5.0 cm 3.0 cm d cm o Te object eigt and attitude can be found using te ratio of te image and object eigt to te image and object distance. Te object attitude is determined by te sign of te object eigt. Te relative size of te object is determined by comparing te image and object eigts. i di d o o 15

16 o d i d o i (1.0 cm)( cm) o ( 3.0 cm) o +.5cm Paraprase and Verify Te object is located 7.5 cm from te mirror. It is erect (indicated by te positive object eigt), and.5 cm ig (larger tan te virtual image). 54. Use te following expressions: c E f and p λ λ Poton A as four times more energy tan poton B, so it must ave four times te frequency and ence one-quarter of te wavelengt. Set up te following ratios: c EA E λa B c λb λb λa 4 λ 4λ B A and pa p λ A B λ B λb λ A 4 pa 4 pb So, it follows tat poton A as four times te momentum of poton B. 55. Given θ 1 10 From Table 13.4: n n 1.54 angle of refraction (θ ) Te angle of refraction is calculated using Snell s Law. 16

17 sinθ n 1 sinθ n1 1 n1sinθ θ 1 sin n 1 θ sin (1.33)(sin10 ) 1.54 θ.6 Paraprase Te angle of refraction is Given λ 0.05 nm i θ 90 m e kg te cange in energy of te electron and poton ( Δ E) Te X ray will lose some energy to te electron. Use te Compton equation to predict te cange in wavelengt of te X ray. Δ λ λf λi ( 1 cosθ) mc λf λi + ( 1 cosθ) mc J s cos kg 3.00 /s 11 ( )( ) ( ).74 From tis cange, determine te energy cange. 1 1 Δ E ( ff fi) c λ f λ i 1 1 Δ E c λ f λ i 34 m 1 1 ( J s ) s m m J According to te law of conservation of energy, tis energy, lost by te X ray, is gained by te electron. Paraprase During te scattering event, te X-ray poton s wavelengt increased to 0.07 nm. Te electron gained J of energy. 57. Given o 3.0 cm d o 10.0 cm f 5.0 cm 17

18 image distance (d i ) image eigt ( i ) image attributes (location, type, attitude, and magnification) Te image distance and eigt can be calculated by using te tin lens and magnification equations. Te image attributes can be determined based on te sign and/or magnitude of te image distance and eigt do di f di f do di 5.0 cm 10.0 cm di 10 cm i di m d o d i o i d o o (10cm)(3.0 cm) i 10.0 cm i 3.0 cm Paraprase and Verify Te image forms 10 cm from te lens wit a eigt of 3.0 cm. Te image is real (indicated by te positive image distance), inverted (indicated by te negative eigt), and te same size as te object (indicated by comparing te eigt of te image and object). 5. (a) Te underlying principle of tis form of propulsion is te Compton effect, tat is, tat potons ave momentum. (b) Given P 1 kw λ 500 nm m 1000 kg speed of spacecraft after one year (v f ) Use te energy output of te laser and te energy of eac 500-nm poton to determine ow many 500-nm potons are emitted by te laser drive eac second. c n ΔE P λ 1000 J/s 1 s 1 s ΔEλ n c 1

19 n ( ) 9 ( 1000 J ) m 34 ( J s m ) s Te total momentum carried by all te potons eac second is p n λ J s ( ) N s According to te law of conservation of momentum, te spacecraft must acquire Δp an equal and opposite momentum eac second. Use te basic expressions F Δ t and vf vi + aδ t, were Δ t 1 year Δp s. Since F, te force Δ t exerted by te potons as tey carry momentum away from te rocket is N. Te magnitude of te rocket s acceleration is F a m N 1000 kg /s Te final speed of te rocket is vf vi + aδt 9 m s s ( ) m/s Paraprase In principle, te propulsion system works, but it is not very efficient it only acieves a speed of 10.5 cm/s after one year! 59. Te visible spectrum can be recomposed by refraction in a prism or by mixing all te colours using a ig speed rotating disc. 60. Given V 75 kv electron wavelengt ( λ ) Use te basic relation λ. Te electrons acquire kinetic energy, according to te p p equation Ek. m 19

20 Terefore, p me k mqv Substitute tis equation for momentum into te wave equation: λ mqv J s λ kg C V ( )( )( ) Paraprase Te wavelengt of te electrons in te electron microscope is only nm, wic is muc smaller tan te wavelengt of ligt used in a conventional microscope. 61. If two waves arrive at te screen one alf wavelengt out of pase, ten destructive interference will occur in tis region. A region of destructive interference is observed as a dark fringe (or node) on te screen. 6. Given L 1.0 nm te rest energy of te electron (E k ) For te electron to be contained in te box, its wavelengt must be tat of a standing wave. For minimum energy, te wavelengt of te electron must be equal to twice te lengt of te box, or nm. Use te equations p and E λ J s p N s Use tis value for momentum to calculate rest energy. E 5 ( N s) 9.11 ( 10 kg) k 31 0 k p. m J 0.3 ev Paraprase Te rest or minimum energy of te electron confined to a 1.0-nm box is 0.3 ev. 0

21 63. Given 1 cm 4 d cm lines 6 d 1.00 n 1 7 λviolet λred 6.50 angle of diffraction ( θ ) Te first antinode is a region of constructive interference, were te pat difference must be one wole wavelengt. Terefore, te generalization for te difference in pat lengt for an antinode can be used to calculate te angle of diffraction for eac wavelengt. d sinθ λ n 1 λn θ sin d (4.0 )(1) 1 (6.50 )(1) θviolet sin θ 6 red sin θviolet 4. θred 40.5 Paraprase Te angle of diffraction to te first antinode for violet ligt is 4. ; for red ligt it is Given 1 cm 3 d cm slits 5 d λ 7.00 n 1 L.50 m te distance between adjacent antinodes (x) Taking te inverse of te lines/cm gives te slit separation (d). We assume tat n 1 and tat te distance between te central antinode and te first brigt fringe is te same as te distance between any two antinodes in te same interference pattern. Te first brigt region of constructive interference occurs wen te pat difference is one wole wavelengt. Terefore, te generalization for te difference in pat lengt for an antinode can be used to calculate te distance between any two antinodes (x). xd λ nl λnl x d 1

22 7 (7.00 )(1)(.50 m ) x x.63 Paraprase Adjacent antinodes are separated by a distance of.6 10 m. 65. Given m 000 kg v 60 km/ d m Δ t 10 s te uncertainty in te teacer s speed ( Δ v) Use Heisenberg s uncertainty principle, ΔΔ x p, to determine Δp, te uncertainty 4π in momentum. You can estimate Δx by arguing tat it cannot be greater tan m, te space over wic te spotting plane saw te car. Δp 4πΔx J s Δp 4 π ( m) N s Use tis answer for Δp to calculate Δv from te equation Δp mδv. Δp Δ v m N s 000 kg /s Paraprase Your pysics teacer s excuse is pretty lame! Quantum effects cannot be invoked ere. Tese effects are undetectable for a macroscopic object suc as an automobile. Extensions 66. Te first layer is te unique digital identifier of te cell pone, used to verify te user, log activity, and locate te user relative to te cell tower infrastructure. Te second layer is te digital stream of data wic corresponds to te audio signals coming to and from te user s cell pone. 67. Given Δx 0.0 nm te approximate momentum of an electron in a ydrogen atom (p) te approximate energy of an electron in a ydrogen atom ( E k )

23 Use Heisenberg s uncertainty principle, Δp 4πΔx J s 9 4 π (0.0 ) N s p Ten apply Ek. m E 5 (.6 10 N s) 9.11 ( 10 kg) k 31 0 ΔΔ x p 4π, to estimate Δp J Paraprase Te energy of an electron confined to a ydrogen atom sould be no less tan ev, wit a momentum no less tan.6 10 N s. 6. Given 7 Δ d.00 c 3.00 /s time (Δt) Te time it takes te signal to travel from te satellite to te receiver depends on te satellite s altitude and te speed at wic te signal travels. All EMR is assumed to travel at te speed of ligt in air or in a vacuum. Δd c Δ t 7 Δ d.00 Δ t / c 3.00 /s / Δ t s Paraprase Te signal travels from te satellite to te receiver in s. 69. Tere are many problems wit a large value. You definitely would not want to go sunbating! A 500-nm poton would carry a great deal of energy. If 1 J s, ten te energy of a single 500-nm poton would be E f c λ m s (1 J s ) J 3

24 Tis amount of energy is about te same as te energy released by illion kilograms of gasoline! If 1 J s, ten te de Broglie wavelengt of a typical student moving at 1 m/s would be about λ p mv 1 J s (50 kg)(1 m/s) cm Walking troug a doorway could be dangerous because you may start to diffract! Recall from Capter 13 tat, wen te size of an opening troug wic waves pass is comparable to te size of te waves temselves, diffraction effects become significant. 70. A fibre-optic tube transmits ligt using total internal reflection. Te fibre itself is composed of a material wit a ig index of refraction (wic as a relatively small critical angle). Ligt wic is reflected troug te fibre is always incident to te boundary at an angle larger tan te critical angle. Terefore, none of te ligt is lost or absorbed outside of te fibre, ensuring tat te signal intensity is preserved over long distances. 71. Given P W n potons/s te distance from wic you could see te Sun (d) Tis problem needs to be solved in several steps. 1. First estimate ow many visible ligt potons leave te surface of te Sun eac second. To do tis, estimate tat alf of te energy output of te Sun ( J/s) leaves as blue-green potons of average wavelengt 500 nm. c Use te equation E nf n and solve for n. λ Te number of blue-green potons leaving te Sun eac second is approximately c E nf n λ Eλ n c 6 9 (.0 10 J )( m ) 34 ( J s m ) s Use te value for n to determine ow many potons per square metre leave te surface of te Sun (te surface density of potons per square metre). Divide te number of potons leaving te Sun by te surface area of te Sun: 4

25 n σ potons, were σ potons is te poton density. 4π R number of potons A σ eye ( π r ) potons n pupil 4π R Solve for R (te Sun s radius). n R ( π rpupil ) 4π (10 000) n R ( rpupil ) 4(10 000) 3. Write an expression tat sows ow te surface density of potons per square metre canges as you move away from te Sun. As you move outward from te Sun, te density drops wit te square of te distance. n Use te equation σ potons. 4π R Substitute r pupil m and n potons/s: ( m) ( ) R Determine te distance from te Sun at wic te total number of potons entering your eye is One ligt year (te distance ligt travels in one year) is a more convenient unit to use. Since one ligt year equals m, te distance from wic you can see te Sun is d 15 m ca 41 ca Paraprase and Verify Wit a dark adapted eye, you sould be able to see te Sun from a distance of about 41 ligt years. Tis estimate is actually very good: Simple astronomy using te distance modulus formula suggests tat a person wit very good vision under dark skies could see a star similar to our Sun from a distance of 5 ca (ligt years). 7. Given E 00 ev d 50 nm L 1 m to predict wat you will see wen te electrons reac te pospor screen Tis problem involves two-slit interference. Determine te wavelengt of te electrons using te equation 5

26 λ p me k were E k is te 00-eV energy of te electrons. Te de Broglie wavelengt for te 00-eV electrons is λ mek J s 31 J 9.11 ( 10 kg) ( 00 ev) ev nm Te two-slit interference formula is nλ dsinθ, were n is te spectral order, d is te slit spacing, and θ is te angle at wic antinodes form. You sould expect to see a series of antinodes and nodes on te pospor screen, wic will appear as alternating brigt and dark bands. Te brigt bands will be located at a distance given x by te expression tanθ, were x is te distance of te brigt bands on te L pospor screen and L is te 1-m separation between te screen and te two slits. Using te two-slit formula gives te following values for θ for te location of antinodes: nλ sinθ d n 0.06 nm ( ) 50 nm n( ) Te first tree antinodes occur at angles of 0.099, 0.199, and 0.9, on eiter side of te central antinode, wic is directly beind te midpoint of te two slits. Te brigt bands appear at locations given by and m, on eiter side of te central maximum. Paraprase A series of brigt and dark lines will appear on te pospor screen. x Ltanθ, were x , , Skills Practice 73. Reasoning and Solution Incident Ray 1 is parallel to te principal axis. Any ray tat is parallel to te principal axis will appear to ave originated from te focal point on a diverging mirror. Incident Ray travels toward te centre of curvature. Any ray tat is directed at te centre of curvature on a diverging mirror, will be reflected directly back along te incident pat. Incident Ray 3 travels towards te vertex. Te vertex acts as a plane reflector, were te angle of incidence is equal to te angle of reflection. 6

27 Ray Diagram Te image is located approx. 7.0 cm from te mirror. It is virtual, erect, and magnified by approximately 0.5. Verification di f do cm 5 cm 0.1 cm 0.04 cm d 7.14 cm i di m do 7.14 cm 5.0 cm 0.6 Te image distance is 7.14 cm, and te magnification is 0.6. Te scale drawing gives a good approximation to te calculated values of image location and magnification. c 74. Use te formula E f to create a column of poton energies. Convert tese λ energies into units of electron volts. Plot poton energy versus maximum kinetic energy and note were te x-intercept occurs. Te x-intercept represents a poton energy tat is just equal to te work function for te metal, Ek f W max 0, because E k max is zero wen te incident poton frequency equals te tresold frequency. By inspecting te grap, you can see tat te work function, W 0, is.5 ev. 7

28 Energy of potons (ev) Wavelengt (nm) Maximum Kinetic Energy (ev) 75. A converging lens refracts all te rays tat travel parallel to te principal axis troug a common point, called te focal point. Terefore, te term converging is used to identify te lens. A diverging lens refracts all of te rays tat travel parallel to te

29 principal axis outwards; suc tat tey will never converge at a real point in space. Terefore, te term diverging is used to identify te lens. 76. Given E k 50 kev te wavelengt of te electron ( λ ) te momentum of te electron (p) 31 Te mass of an electron is kg. First calculate momentum using te equation p Ek m p me k ( )( )( ) p kg ev J/eV kg m/s Use tis answer to find wavelengt using te equation λ p J s λ kg m/s Paraprase Te electron as a momentum of kg m/s and a wavelengt of m. 77. Reasoning and Solution Incident Ray 1 is parallel to te principal axis. Any ray tat is parallel to te principal axis will be refracted troug te focal point on a converging lens. Incident Ray travels towards te optical centre of te lens. Any ray tat passes troug te optical centre is laterally sifted, but not refracted. Wen te object is located at te focal point of te lens, te refracted rays emerge parallel and never converge to form a real or virtual image. 7. Given λ 10 nm 9

30 79. momentum (p) p λ J s p N s Paraprase 6 Te X-ray poton as a momentum of N s. Property of Ligt straigt-line propagation reflection refraction polarization dispersion interference and diffraction potoelectric effect, potoemission Compton scattering Best explained by wave or particle model wave or particle model wave or particle model wave model wave model wave model particle model particle model Te best way to answer te question, Is ligt a wave or a particle? is to state tat it is neiter! Te quantum world sows you tat te beaviour of ligt depends on wat property of ligt you are testing: Sometimes ligt exibits wave properties and sometimes it exibits particle properties. Self-assessment Students' answers in tis section will vary greatly depending on teir experiences, dept of understanding, and te dept of te researc tat tey do into eac of te topics. 30