1. (a) From Fig we find the smaller wavelength in question to be about 515 nm.

Transcription

1 Chapter (a) From Fig - we fid the smaller wavelegth i questio to be about 55 m (b) Similarly, the larger wavelegth is approximately 6 m () From Fig - the wavelegth at whih the eye is most sesitive is about 555 m (d) Usig the result i (), we have f m/s λ 555 m 4 54 Hz (e) The period is T /f (54 4 Hz) 5 5 s air, light travels at roughly m/s Therefore, for t s, we have a distae of 9 d t ( m / s) ( s) m Sie λ << λ, we fid f is equal to F λ HG λk J λ 9 ( m / s)( m) 9 ( 6 m) Hz 4 (a) The frequey of the radiatio is f m/s λ ( )( m) 47 Hz (b) The period of the radiatio is T f 47 s mi s Hz 5 f f is the frequey ad λ is the wavelegth of a eletromageti wave, the fλ The frequey is the same as the frequey of osillatio of the urret i the LC iruit of the geerator That is, f / π LC, where C is the apaitae ad L is the idutae Thus

2 CHAPTER The solutio for L is λ π LC h h m/sh 9 λ 55 m L 4π C 4π 7 F 99 This is exeedigly small 6 The emitted wavelegth is 5 6 λ π LC π ( 99 m/s) ( 5 H)( 5 F) 474m f 7 f P is the power ad t is the time iterval of oe pulse, the the eergy i a pulse is h h E P t W 9 s 5 J The amplitude of the mageti field i the wave is H B m 4 Em V/m 7 99 m/s T 9 (a) The amplitude of the mageti field is B m Em V/m 99 m/s T 67 T (b) Sie the E r -wave osillates i the z diretio ad travels i the x diretio, we have B x B z So, the osillatio of the mageti field is parallel to the y axis r r () The diretio (+x) of the eletromageti wave propagatio is determied by E B f the eletri field poits i +z, the the mageti field must poit i the y diretio With S uits uderstood, we may write 5 By Bmos t ( ) 9 5 π ( t x ) 5 x os π / π 67 os t x

3 (a) The amplitude of the mageti field i the wave is B m Em 5 V/m m/s (b) The itesity is the average of the Poytig vetor: E 5 V/m m Savg 7 µ 4π Tm/A 99 m/s b g h h The itesity is the average of the Poytig vetor: h 4 Th B S m m/s avg µ 6 6 H/m The itesity of the sigal at Proxima Cetauri is h T W/m 6 W/m 6 P W 4πr 4π 4 ly 9 46 b g 5 m / ly h 9 4 W/m (a) The mageti field amplitude of the wave is (b) The itesity is B m Em V/m m/s b g 7 E V/m m µ 4π Tm/A 99 m/s () The power of the soure is 4 (a) The power reeived is h h ( ) ( ) avg 9 T 5 W/m P 4π r 4π m 5 W/m 67 W (b) The power of the soure would be P r π ( m ) /4 ( W) 4 W 4π 67 m 6 ( )

4 4 CHAPTER L M N O P Q 4 5 W P 4πr 4π lyh946 m / lyhm P 6 4π 6 7 m h 5 (a) We use E m E m /µ to alulate E m : 7 µ 4π Tm/A 4 W/m 99 m/s V/m (b) The mageti field amplitude is therefore h h h 4 Em V/m B m 4 99 m/s 6 (a) The expressio E y E m si(kx ωt) it fits the requiremet at poit P [it] is dereasig with time if we imagie P is just to the right (x > ) of the oordiate origi (but at a value of x less tha π/k λ/4 whih is where there would be a maximum, at t ) t is importat to bear i mid, i this desriptio, that the wave is movig to the right Speifially, xp (/ k) si (/ 4) so that E y (/4) E m at t, there Also, E y with our hoie of expressio for E y Therefore, part (a) is aswered simply by solvig for x P Sie k πf/ we fid xp si m π f 4 (b) f we proeed to the right o the x axis (still studyig this sapshot of the wave at t ) we fid aother poit where E y at a distae of oe-half wavelegth from the previous poit where E y Thus (sie λ /f ) the ext poit is at x λ /f ad is osequetly a distae /f x P 45 m to the right of P 7 (a) The average rate of eergy flow per uit area, or itesity, is related to the eletri field amplitude E m by E / µ, so m 7 6 h h h Em µ 4π H/m 99 m/s W/m 7 V/m (b) The amplitude of the mageti field is give by B m Em 7 V/m 9 99 m/s 6 T T 5 W

5 5 () At a distae r from the trasmitter, the itesity is P/ π r, where P is the power of the trasmitter over the hemisphere havig a surfae area π r Thus ( ) ( ) P π π 6 r m W/m 6 W From the equatio immediately preedig Eq -, we see that the maximum value of B/ t is ωb m We a relate B m to the itesity: B m E µ m, ad relate the itesity to the power P (ad distae r) usig Eq -7 Fially, we relate ω to wavelegth λ usig ω k π/λ Puttig all this together, we obtai B µ P π t 4π λr max 6 44 T/s 9 Sie the surfae is perfetly absorbig, the radiatio pressure is give by p r /, where is the itesity Sie the bulb radiates uiformly i all diretios, the itesity a distae r from it is give by P/4πr, where P is the power of the bulb Thus The radiatio pressure is P p r r 5W 4 99 π 4π5m m/s b g h W/m pr 99 m/s 59 The plasma ompletely reflets all the eergy iidet o it, so the radiatio pressure is give by p r /, where is the itesity The itesity is P/A, where P is the power ad A is the area iterepted by the radiatio Thus Pa Pa p r 9 5 ( W) 6 ( m ) ( 99 m/s) P A 7 Pa (a) The radiatio pressure produes a fore equal to

6 6 CHAPTER 6 ( W/m ) ( 67 m) π4 Fr pr( π Re ) ( π Re ) 6 N 99 m/s (b) The gravitatioal pull of the Su o Earth is F GM M 4 ( 667 N m / kg ) ( kg) ( 59 kg) ( 5 m) s e grav des 6 N, whih is muh greater tha F r Let f be the fratio of the iidet beam itesity that is refleted The fratio absorbed is f The refleted portio exerts a radiatio pressure of p r f ad the absorbed portio exerts a radiatio pressure of p a ( f ) where is the iidet itesity The fator eters the first expressio beause the mometum of the refleted portio is reversed The total radiatio pressure is the sum of the two otributios: f + ( f) ( + f) ptotal pr + pa To relate the itesity ad eergy desity, we osider a tube with legth λ ad rosssetioal area A, lyig with its axis alog the propagatio diretio of a eletromageti wave The eletromageti eergy iside is U uaλ, where u is the eergy desity All this eergy passes through the ed i time t λ/, so the itesity is, U ua u At l Al Thus u / The itesity ad eergy desity are positive, regardless of the propagatio diretio For the partially refleted ad partially absorbed wave, the itesity just outside the surfae is + f ( + f ), where the first term is assoiated with the iidet beam ad the seod is assoiated with the refleted beam Cosequetly, the eergy desity is

7 7 the same as radiatio pressure ( + f ) u, 4 (a) We ote that the ross setio area of the beam is πd /4, where d is the diameter of the spot (d λ) The beam itesity is P πd / 4 π 5 W 9 97 W/m 9 6 m / 4 b g h (b) The radiatio pressure is 9 97 W/m pr 99 m/s Pa () omputig the orrespodig fore, we a use the power ad itesity to elimiate the area (metioed i part (a)) We obtai F F H G K J F H G K J hb g πd P 5 W Pa p p W / m r r r (d) The aeleratio of the sphere is 67 Fr Fr 6( 67 N) a m ρ( πd / 6) π(5 kg / m )[( )( 6 4 m/s m)] N 9 5 (a) Sie λ f, where λ is the wavelegth ad f is the frequey of the wave, f 99 λ m m/s Hz (b) The agular frequey is ω πf π( Hz) 6 rad / s () The agular wave umber is π π k rad / m λ m

8 CHAPTER (d) The mageti field amplitude is B m Em V/m 99 m/s 6 T (e) B ρ must be i the positive z diretio whe E ρ ρ ρ is i the positive y diretio i order for E B to be i the positive x diretio (the diretio of propagatio) (f) The itesity of the wave is E m (V/m) 7 µ (4π H/m)(99 m/s) 9W/m W/m (g) Sie the sheet is perfetly absorbig, the rate per uit area with whih mometum is delivered to it is /, so dp dt A ( 9 W/m )( m ) 99 m/s 7 N (h) The radiatio pressure is dp / dt pr A m 7 N 4 7 Pa 6 We require F grav F r or ad solve for the area A: G mm d es s A, A GmM s ( 6 67 N m / kg )( 5 kg)(99 kg)(99 m / s) des ( 4 W/m )( 5 m) 5 95 m 95 km 7 f the beam arries eergy U away from the spaeship, the it also arries mometum p U/ away Sie the total mometum of the spaeship ad light is oserved, this is the magitude of the mometum aquired by the spaeship f P is the power of the laser, the the eergy arried away i time t is U Pt We ote that there are 64 seods i a day Thus, p Pt/ ad, if m is mass of the spaeship, its speed is p Pt v m m ( W)(64 s) 9 ( 5 kg)(99 m / s) m/s

9 9 The mass of the ylider is m ρπ ( D / 4) H, where D is the diameter of the ylider Sie it is i equilibrium We solve for H: πhd gρ πd Fet mg Fr 4 4 P H gρ D π /4 gρ (46W) [π( m m) / 4](9m/s )( m/s)( kg/m ) 9 (a) The upward fore supplied by radiatio pressure i this ase (Eq -) must be equal to the magitude of the pull of gravity (mg) For a sphere, the projeted area (whih is a fator i Eq -) is that of a irle A πr (ot the etire surfae area of the sphere) ad the volume (eeded beause the mass is give by the desity multiplied by the volume: m ρv) is V 4 π r / Fially, the itesity is related to the power P of the light soure ad aother area fator 4πR, give by Eq -7 this way, with 4 ρ 9 kg/m, equatig the fores leads to 4π rg P 4πR ρ 46 W π r (b) Ay hae disturbae ould move the sphere from beig diretly above the soure, ad the the two fore vetors would o loger be alog the same axis Eq -7 suggests that the slope i a itesity versus iverse-square-distae graph ( plotted versus r ) is P/4π We estimate the slope to be about (i S uits) whih meas the power is P 4π() 5 W We shall assume that the Su is far eough from the partile to at as a isotropi poit soure of light (a) The fores that at o the dust partile are the radially outward radiatio fore F r r ad the radially iward (toward the Su) gravitatioal fore F r g Usig Eqs - ad -7, the radiatio fore a be writte as A PS π R PSR Fr, 4π r 4r

10 9 CHAPTER where R is the radius of the partile, ad A π R is the ross-setioal area O the other had, the gravitatioal fore o the partile is give by Newto s law of gravitatio (Eq -): GM Sm GM Sρ(4 πr /) 4πGM SρR Fg, r r r where m ρ(4 πr /) is the mass of the partile Whe the two fores balae, the partile travels i a straight path The oditio that Fr Fg implies whih a be solved to give PR S 4π GMSρR, 4r r 6 PS (9 W) GMS π 7 R 6π ρ 6 ( m/s)(5 kg/m )(667 m /kg s )(99 kg) 7 m (b) Sie F g varies with R ad F r varies with R, if the radius R is larger, the Fg > Fr, ad the path will be urved toward the Su (like path ) this ase, we replae os 7 by as the itesity of the light after passig through the first polarizer Therefore, f os ( 9 7 ) ( 4 W/m )(os ) 9 W/m The agle betwee the diretio of polarizatio of the light iidet o the first polarizig sheet ad the polarizig diretio of that sheet is θ 7 f is the itesity of the iidet light, the the itesity of the light trasmitted through the first sheet is os θ ( 4 W/m ) os 7 5 W/m The diretio of polarizatio of the trasmitted light makes a agle of 7 with the vertial ad a agle of θ with the horizotal θ is the agle it makes with the polarizig diretio of the seod polarizig sheet Cosequetly, the trasmitted itesity is os θ ( 5 W/m ) os 44 W/m 4 After passig through the first polarizer the iitial itesity redues by a fator of / After passig through the seod oe it is further redued by a fator of os (π θ θ ) os (θ + θ ) Fially, after passig through the third oe it is agai redued by a fator of os (π θ θ ) os (θ + θ ) Therefore,

11 9 f os ( θ+ θ)os ( θ+ θ) os (5 + 5 )os (5 + 5 ) 4 45 Thus, 45% of the light s iitial itesity is trasmitted 5 Let be the itesity of the upolarized light that is iidet o the first polarizig sheet The trasmitted itesity is, ad the diretio of polarizatio of the trasmitted light is θ 4 outerlokwise from the y axis i the diagram The polarizig diretio of the seod sheet is θ lokwise from the y axis, so the agle betwee the diretio of polarizatio that is iidet o that sheet ad the polarizig diretio of the sheet is The trasmitted itesity is os 6 os 6, ad the diretio of polarizatio of the trasmitted light is lokwise from the y axis The polarizig diretio of the third sheet is θ 4 outerlokwise from the y axis Cosequetly, the agle betwee the diretio of polarizatio of the light iidet o that sheet ad the polarizig diretio of the sheet is The trasmitted itesity is 4 os 6 os 6 Thus, % of the light s iitial itesity is trasmitted 6 We examie the poit where the graph reahes zero: θ 6º Sie the polarizers must be rossed for the itesity to vaish, the θ 6º 9º 7º Now we osider the ase θ 9º (whih is hard to judge from the graph) Sie θ is still equal to 7º, the the agle betwee the polarizers is ow θ º Aoutig for the automati redutio (by a fator of oe-half) wheever upolarized light passes through ay polarizig sheet, the our result is os ( θ) 44 44% 7 (a) Sie the iidet light is upolarized, half the itesity is trasmitted ad half is absorbed Thus the trasmitted itesity is 5 mw/m The itesity ad the eletri field amplitude are related by E / µ, so m Em µ ( 4π 7 H/m)( m/s)(5 W/m ) 9 V/m

12 9 CHAPTER (b) The radiatio pressure is p r a /, where a is the absorbed itesity Thus 5 W/m p r m/s 7 Pa (a) The fratio of light whih is trasmitted by the glasses is E E E E + E f f v v h E Ev + ( E ) v v 6 (b) Sie ow the horizotal ompoet of ρ E will pass through the glasses, f Eh E + E v h ( Ev ) E + ( E ) v v 4 9 As the polarized beam of itesity passes the first polarizer, its itesity is redued to os θ After passig through the seod polarizer whih makes a 9 agle with the first filter, the itesity is ( os θ)si θ / whih implies si θ os θ /, or siθ osθ siθ / / This leads to θ 7 or 4 We ote the poits at whih the urve is zero (θ ad 9 ) i Fig -45(b) We ifer that sheet is perpediular to oe of the other sheets at θ, ad that it is perpediular to the other of the other sheets whe θ 9 Without loss of geerality, we hoose θ, θ 9 Now, whe θ, it will be θ relative to sheet ad θ 6 relative to sheet Therefore, f i os ( )os θ ( θ ) 94% 4 Let be the itesity of the iidet beam ad f be the fratio that is polarized Thus, the itesity of the polarized portio is f After trasmissio, this portio otributes f os θ to the itesity of the trasmitted beam Here θ is the agle betwee the diretio of polarizatio of the radiatio ad the polarizig diretio of the filter The itesity of the upolarized portio of the iidet beam is ( f ) ad after trasmissio, this portio otributes ( f ) / to the trasmitted itesity Cosequetly, the trasmitted itesity is fos θ + ( f)

13 9 As the filter is rotated, os θ varies from a miimum of to a maximum of, so the trasmitted itesity varies from a miimum of to a maximum of mi ( f ) max f + ( f) ( + f) The ratio of max to mi is max mi + f f Settig the ratio equal to 5 ad solvig for f, we get f 67 4 We apply Eq -4 (oe) ad Eq -4 (twie) to obtai os θos (9 θ) Usig trig idetities, we rewrite this as si ( θ ) (a) Therefore we fid θ si 4 96 (b) Sie the first expressio we wrote is symmetri uder the exhage: θ 9 θ, the we see that the agle's omplemet, 74, is also a solutio 4 (a) The rotatio aot be doe with a sigle sheet f a sheet is plaed with its polarizig diretio at a agle of 9 to the diretio of polarizatio of the iidet radiatio, o radiatio is trasmitted t a be doe with two sheets We plae the first sheet with its polarizig diretio at some agle θ, betwee ad 9, to the diretio of polarizatio of the iidet radiatio Plae the seod sheet with its polarizig diretio at 9 to the polarizatio diretio of the iidet radiatio The trasmitted radiatio is the polarized at 9 to the iidet polarizatio diretio The itesity is os θ os (9 θ) os θsi θ, where is the iidet radiatio f θ is ot or 9, the trasmitted itesity is ot zero (b) Cosider sheets, with the polarizig diretio of the first sheet makig a agle of θ 9 / relative to the diretio of polarizatio of the iidet radiatio The polarizig diretio of eah suessive sheet is rotated 9 / i the same sese from the polarizig

14 94 CHAPTER diretio of the previous sheet The trasmitted radiatio is polarized, with its diretio of polarizatio makig a agle of 9 with the diretio of polarizatio of the iidet radiatio The itesity is os (9 / ) We wat the smallest iteger value of for whih this is greater tha 6 We start with ad alulate os (9 / ) f the result is greater tha 6, we have obtaied the solutio f it is less, irease by ad try agai We repeat this proess, ireasig by eah time, util we have a value for whih os (9 / ) is greater tha 6 The first oe will be 5 44 We ote the poits at whih the urve is zero (θ 6 ad 4 ) i Fig -45(b) We ifer that sheet is perpediular to oe of the other sheets at θ 6, ad that it is perpediular to the other of the other sheets whe θ 4 Without loss of geerality, we hoose θ 5, θ 5 Now, whe θ 9, it will be θ 6 relative to sheet ad θ 4 relative to sheet Therefore, f i os ( )os θ ( θ ) 7% 45 The law of refratio states siθ si θ We take medium to be the vauum, with ad θ Medium is the glass, with θ We solve for : siθ ( ) siθ Fsi HG K J si 4 46 (a) For the agles of iidee ad refratio to be equal, the graph i Fig -4(b) would osist of a y x lie at 45º i the plot stead, the urve for material falls uder suh a y x lie, whih tells us that all refratio agles are less tha iidet oes With θ < θ Sell s law implies > (b) Usig the same argumet as i (a), the value of for material is also greater tha that of water ( ) () t s easiest to examie the right ed-poit of eah urve With θ 9º ad θ ¾(9º), ad with (Table -) we fid, from Sell s law, 4 for material (d) Similarly, with θ 9º ad θ ½(9º), we obtai 9

15 95 47 The agle of iidee for the light ray o mirror B is 9 θ So the outgoig ray r' makes a agle 9 (9 θ) θ with the vertial diretio, ad is atiparallel to the iomig oe The agle betwee i ad r' is therefore 4 (a) For the agles of iidee ad refratio to be equal, the graph i Fig -5(b) would osist of a y x lie at 45º i the plot stead, the urve for material falls uder suh a y x lie, whih tells us that all refratio agles are less tha iidet oes With θ < θ Sell s law implies > (b) Usig the same argumet as i (a), the value of for material is also greater tha that of water ( ) () t s easiest to examie the topmost poit of eah urve With θ 9º ad θ ½(9º), ad with (Table -) we fid 9 from Sell s law (d) Similarly, with θ 9º ad θ ¾(9º), we obtai 4 49 Note that the ormal to the refratig surfae is vertial i the diagram The agle of refratio is θ 9 ad the agle of iidee is give by ta θ L/D, where D is the height of the tak ad L is its width Thus L m θ D 5 m ta ta 5 The law of refratio yields siθ ( ) siθ F 9 HG si 5 K J si where the idex of refratio of air was take to be uity 6, 5 (a) A simple impliatio of Sell s law is that θ θ whe Sie the agle of iidee is show i Fig -5(a) to be º, the we look for a poit i Fig -5(b) where θ º This seems to our whe 7 By iferee, the, 7 (b) From 7si(6º) 4si(θ ) we get θ 5 Cosider a ray that grazes the top of the pole, as show i the diagram that follows Here θ 9 θ 5, λ 5 m, ad λ 5 m The legth of the shadow is x + L x is give by x λ ta θ ( 5 m) ta 5 5 m

16 96 CHAPTER Aordig to the law of refratio, si θ si θ We take ad (from Table -) The, L is give by θ siθ F H G K J si F HG K J si si L λ ta θ ( 5 m) ta555 7 m The legth of the shadow is 5 m + 7 m 7 m 5 (a) Sell s law gives air si(5º) b si θ b ad air si(5º) r si θ r where we use subsripts b ad r for the blue ad red light rays Usig the ommo approximatio for air s idex ( air ) we fid the two agles of refratio to be 76 ad 57 Therefore, θ (b) Both of the refrated rays emerges from the other side with the same agle (5 ) with whih they were iidet o the first side (geerally speakig, light omes ito a blok at the same agle that it emerges with from the opposite parallel side) There is thus o differee (the differee is ) ad thus there is o dispersio i this ase 5 (a) Approximatig for air, we have si θ () siθ 569 θ 5 5 ad with the more aurate value for air i Table -, we obtai 56 (b) Eq -44 leads to so that siθ siθ siθ siθ 4 4

17 97 θ4 si siθ (a) From siθ siθ ad siθ siθ, we fid siθ siθ This has a simple impliatio: that θ θ whe Sie we are give θ 4º i Fig - 56(a) the we look for a poit i Fig -56(b) where θ 4º This seems to our at 6, so we ifer that 6 (b) Our first step i our solutio to part (a) shows that iformatio oerig disappears (aels) i the maipulatio Thus, we aot tell; we eed more iformatio () From 6si7 4siθ we obtai θ 9 55 We label the light ray s poit of etry A, the vertex of the prism B, ad the light ray s exit poit C Also, the poit i Fig -57 where ψ is defied (at the poit of itersetio of the extrapolatios of the iidet ad emerget rays) is deoted D The agle idiated by ADC is the supplemet of ψ, so we deote it ψ s ψ The agle of refratio i the glass is θ si θ The agles betwee the iterior ray ad the earby surfaes is the omplemet of θ, so we deote it θ 9 θ Now, the agles i the triagle ABC must add to : φ θ + φ θ Also, the agles i the triagle ADC must add to : ( θ θ) + ψs θ 9 + θ ψs whih simplifies to θ θ + ψ Combiig this with our previous result, we fid θ b φ + ψg Thus, the law of refratio yields bg bg b gh b g si θ si φ+ ψ si θ si φ 56 (a) We use subsripts b ad r for the blue ad red light rays Sell s law gives θ b si 4 si(7 ) 444 θ r si si(7 ) 449

18 9 CHAPTER for the refratio agles at the first surfae (where the ormal axis is vertial) These rays strike the seod surfae (where A is) at omplemetary agles to those just alulated (sie the ormal axis is horizotal for the seod surfae) Takig this ito osideratio, we agai use Sell s law to alulate the seod refratios (with whih the light re-eters the air): θ b si [4si(9 θ b )] 766 θ r si [si(9 θ r )] 7497 whih differ by (thus givig a raibow of agular width ) (b) Both of the refrated rays emerges from the bottom side with the same agle (7 ) with whih they were iidet o the topside (the ourree of a itermediate refletio [from side ] does ot alter this overall fat: light omes ito the blok at the same agle that it emerges with from the opposite parallel side) There is thus o differee (the differee is ) ad thus there is o raibow i this ase 57 Referee to Fig -4 may help i the visualizatio of why there appears to be a irle of light (osider revolvig that piture about a vertial axis) The depth ad the radius of that irle (whih is from poit a to poit f i that figure) is related to the taget of the agle of iidee Thus, the diameter D of the irle i questio is L F N M H G wk J O Q P F b g L NM H G KJ O QP D htaθ hta si m ta si m 5 The ritial agle is θ si si 4 F H G K J F H G K J 59 (a) We ote that the omplemet of the agle of refratio (i material ) is the ritial agle Thus, leads to θ 6 siθ osθ (b) reasig θ leads to a derease of the agle with whih the light strikes the iterfae betwee materials ad, so it beomes greater tha the ritial agle; therefore, there will be some trasmissio of light ito material 6 (a) Referee to Fig -4 may help i the visualizatio of why there appears to be a irle of light (osider revolvig that piture about a vertial axis) The depth ad the radius of that irle (whih is from poit a to poit f i that figure) is related to the

19 99 taget of the agle of iidee The diameter of the irle i questio is give by d h ta θ For water, so Eq -47 gives si θ /, or θ 475 Thus, d htaθ ( m)(ta 475 ) 456 m (b) The diameter d of the irle will irease if the fish deseds (ireasig h) 6 (a) the otatio of this problem, Eq -47 beomes θ si whih yields 9 for θ φ 6 (b) Applyig Eq -44 law to the iterfae betwee material ad material, we have whih yields θ si si θ () Dereasig θ will irease φ ad thus ause the ray to strike the iterfae (betwee materials ad ) at a agle larger tha θ Therefore, o trasmissio of light ito material a our 6 (a) The oditio (i Eq -44) required i the ritial agle alulatio is θ 9 Thus (with θ θ, whih we do t ompute here), leads to θ θ si / 54 siθ siθ siθ (b) Yes Reduig θ leads to a redutio of θ so that it beomes less tha the ritial agle; therefore, there will be some trasmissio of light ito material () We ote that the omplemet of the agle of refratio (i material ) is the ritial agle Thus, leads to θ 5 F si θ os θ H G K J (d) No Reduig θ leads to a irease of the agle with whih the light strikes the iterfae betwee materials ad, so it beomes greater tha the ritial agle Therefore, there will be o trasmissio of light ito material

20 CHAPTER 6 Whe examiig Fig -6, it is importat to ote that the agle (measured from the etral axis) for the light ray i air, θ, is ot the agle for the ray i the glass ore, whih we deote θ ' The law of refratio leads to siθ siθ assumig air The agle of iidee for the light ray strikig the oatig is the omplemet of θ ', whih we deote as θ' omp ad reall that siθ osθ si θ omp the ritial ase, θ' omp must equal θ speified by Eq -47 Therefore, F H G siθ si θ siθ K J omp whih leads to the result: si θ With 5 ad 5, we obtai h θ si (a) We ote that the upper-right orer is at a agle (measured from the poit where the light eters, ad measured relative to a ormal axis established at that poit [the ormal at that poit would be horizotal i Fig -6) is at ta (/) 7º The agle of refratio is give by air si 4º 56 si θ whih yields θ 4º if we use the ommo approximatio air, ad yields θ 44º if we use the more aurate value for air foud i Table - The value is less tha 7º whih meas that the light goes to side (b) The ray strikes a poit o side whih is 64 m below that upper-right orer, ad the (usig the fat that the agle is symmetrial upo refletio) strikes the top surfae (side ) at a poit 4 m to the left of that orer Sie 4 m is ertaily less tha m we have a self-osistey hek to the effet that the ray does ideed strike side as its seod refletio (if we had gotte 4 m istead of 4 m, the the situatio would be quite differet) () The ormal axes for sides ad are both horizotal, so the agle of iidee (i the plasti) at side is the same as the agle of refratio was at side Thus,

21 56 si 4º air si θ air θ air 4 (d) t strikes the top surfae (side ) at a agle (measured from the ormal axis there, whih i this ase would be a vertial axis) of 9º θ 66º whih is muh greater tha the ritial agle for total iteral refletio (si ( air /56 ) 99º) Therefore, o refratio ours whe the light strikes side (e) this ase, we have air si 7º 56 si θ whih yields θ 74º if we use the ommo approximatio air, ad yields θ 75º if we use the more aurate value for air foud i Table - This is greater tha the 7º metioed above (regardig the upper-right orer), so the ray strikes side istead of side (f) After bouig from side (at a poit fairly lose to that orer) to goes to side (g) Whe it boued from side, its agle of iidee (beause the ormal axis for side is orthogoal to that for side ) is 9º θ 5º whih is muh greater tha the ritial agle for total iteral refletio (whih, agai, is si ( air /56 ) 99º) Therefore, o refratio ours whe the light strikes side (h) For the same reasos impliit i the alulatio of part (), the refrated ray emerges from side with the same agle (7 ) that it etered side at (we see that the ourree of a itermediate refletio [from side ] does ot alter this overall fat: light omes ito the blok at the same agle that it emerges with from the opposite parallel side 65 (a) No refratio ours at the surfae ab, so the agle of iidee at surfae a is 9 φ For total iteral refletio at the seod surfae, g si (9 φ) must be greater tha a Here g is the idex of refratio for the glass ad a is the idex of refratio for air Sie si (9 φ) os φ, we wat the largest value of φ for whih g os φ a Reall that os φ dereases as φ ireases from zero Whe φ has the largest value for whih total iteral refletio ours, the g os φ a, or φ os F HG F K J H G K J a os 4 9 g 5 The idex of refratio for air is take to be uity (b) We ow replae the air with water f w is the idex of refratio for water, the the largest value of φ for whih total iteral refletio ours is w φ os os 5 F HG F K J H G K J 9 g

22 CHAPTER 66 (a) We refer to the etry poit for the origial iidet ray as poit A (whih we take to be o the left side of the prism, as i Fig -57), the prism vertex as poit B, ad the poit where the iterior ray strikes the right surfae of the prism as poit C The agle betwee lie AB ad the iterior ray is β (the omplemet of the agle of refratio at the first surfae), ad the agle betwee the lie BC ad the iterior ray is α (the omplemet of its agle of iidee whe it strikes the seod surfae) Whe the iidet ray is at the miimum agle for whih light is able to exit the prism, the light exits alog the seod fae That is, the agle of refratio at the seod fae is 9, ad the agle of iidee there for the iterior ray is the ritial agle for total iteral refletio Let θ be the agle of iidee for the origial iidet ray ad θ be the agle of refratio at the first fae, ad let θ be the agle of iidee at the seod fae The law of refratio, applied to poit C, yields si θ, so si θ / /6 65 θ 6 The iterior agles of the triagle ABC must sum to, so α + β Now, α 9 θ 5, so β Thus, θ 9 β The law of refratio, applied to poit A, yields Thus θ 56 si θ si θ 6 si 57 (b) We apply the law of refratio to poit C Sie the agle of refratio there is the same as the agle of iidee at A, si θ si θ Now, α + β, α 9 θ, ad β 9 θ, as before This meas θ + θ 6 Thus, the law of refratio leads to ( ) siθ si 6 θ siθ si 6 osθ os6 siθ where the trigoometri idetity si(a B) si A os B os A si B is used Next, we apply the law of refratio to poit A: ( ) siθ siθ siθ / siθ whih yields osθ si θ / si θ or h Thus, siθ si 6 / si θ os6 siθ b g

23 b g θ θ + os6 si si6 si Squarig both sides ad solvig for si θ, we obtai si si 6 6 si6 b+ os6 g+ si 6 b+ os 6 g+ si 6 θ ad θ 5 67 (a) A ray diagram is show below Let θ be the agle of iidee ad θ be the agle of refratio at the first surfae Let θ be the agle of iidee at the seod surfae The agle of refratio there is θ 4 9 The law of refratio, applied to the seod surfae, yields si θ si θ 4 As show i the diagram, the ormals to the surfaes at P ad Q are perpediular to eah other The iterior agles of the triagle formed by the ray ad the two ormals must sum to, so θ 9 θ ad b g siθ si 9 θ osθ si θ Aordig to the law of refratio, applied at Q, si θ The law of refratio, applied to poit P, yields si θ si θ, so si θ (si θ )/ ad si θ Squarig both sides ad solvig for, we get + si θ (b) The greatest possible value of si θ is, so the greatest possible value of is max 4

24 4 CHAPTER () For a give value of, if the agle of iidee at the first surfae is greater tha θ, the agle of refratio there is greater tha θ ad the agle of iidee at the seod fae is less tha θ ( 9 θ ) That is, it is less tha the ritial agle for total iteral refletio, so light leaves the seod surfae ad emerges ito the air (d) f the agle of iidee at the first surfae is less tha θ, the agle of refratio there is less tha θ ad the agle of iidee at the seod surfae is greater tha θ This is greater tha the ritial agle for total iteral refletio, so all the light is refleted at Q 6 (a) We use Eq -49: θ B ta w ta ( ) 5 (b) Yes, sie w depeds o the wavelegth of the light 69 The agle of iidee θ B for whih refleted light is fully polarized is give by Eq -4 of the text f is the idex of refratio for the medium of iidee ad is the idex of refratio for the seod medium, the θ B ta ( / ) ta (5/) 49 7 Sie the layers are parallel, the agle of refratio regardig the first surfae is the same as the agle of iidee regardig the seod surfae (as is suggested by the otatio i Fig -66) We reall that as part of the derivatio of Eq -49 (Brewster s agle), the refrated agle is the omplemet of the iidet agle: θ ( θ ) 9 θ We apply Eq -49 to both refratios, settig up a produt: (ta θb )(ta θb ) (ta θ)(ta θ) Now, sie θ is the omplemet of θ we have taθ ta( θ ) taθ Therefore, the produt of tagets ael ad we obtai / Cosequetly, the third medium is air: 7 (a) The first otributio to the overall deviatio is at the first refratio: δθ θ i θ r The ext otributio to the overall deviatio is the refletio Notig that the agle betwee the ray right before refletio ad the axis ormal to the bak surfae of the sphere is equal to θ r, ad reallig the law of refletio, we olude that the agle

25 5 by whih the ray turs (omparig the diretio of propagatio before ad after the refletio) is δθ θ r The fial otributio is the refratio suffered by the ray upo leavig the sphere: δθ θ i θ r agai Therefore, θdev δθ + δθ + δθ + θ 4 θ (b) We substitute θr si ( si θi ) ito the expressio derived i part (a), usig the two give values for The higher urve is for the blue light i r () We a expad the graph ad try to estimate the miimum, or searh for it with a more sophistiated umerial proedure We fid that the θ dev miimum for red light is 76 76, ad this ours at θ i 595 (d) For blue light, we fid that the θ dev miimum is 95 94, ad this ours at θ i 595 (e) The differee i θ dev i the previous two parts is 7 7 (a) The first otributio to the overall deviatio is at the first refratio: δθ θ i θ r The ext otributio(s) to the overall deviatio is (are) the refletio(s) Notig that the agle betwee the ray right before refletio ad the axis ormal to the bak surfae of the sphere is equal to θ r, ad reallig the law of refletio, we olude that the agle by whih the ray turs (omparig the diretio of propagatio before ad after [eah] refletio) is δθr θr Thus, for k refletios, we have δθ kθ r to aout for these otributios The fial otributio is the refratio suffered by the ray upo leavig the sphere: δθ θ i θ r agai Therefore, θdev δθ + δθ + δθ ( θi θr) + k( θr) k( ) + θ i ( k+ ) θr (b) For k ad (give i Problem -7), we searh for the seod-order raibow agle umerially We fid that the θ dev miimum for red light is 7 4, ad this ours at θ i 79

26 6 CHAPTER () Similarly, we fid that the seod-order θ dev miimum for blue light (for whih 4) is 4 5, ad this ours at θ i 75 (d) The differee i θ dev i the previous two parts is approximately (e) Settig k, we searh for the third-order raibow agle umerially We fid that the θ dev miimum for red light is 75, ad this ours at θ i 76 (f) Similarly, we fid that the third-order θ dev miimum for blue light is 9, ad this ours at θ i 766 (g) The differee i θ dev i the previous two parts is 44 7 Let θ 45 be the agle of iidee at the first surfae ad θ be the agle of refratio there Let θ be the agle of iidee at the seod surfae The oditio for total iteral refletio at the seod surfae is si θ We wat to fid the smallest value of the idex of refratio for whih this iequality holds The law of refratio, applied to the first surfae, yields si θ si θ Cosideratio of the triagle formed by the surfae of the slab ad the ray i the slab tells us that θ 9 θ Thus, the oditio for total iteral refletio beomes si(9 θ ) os θ Squarig this equatio ad usig si θ + os θ, we obtai ( si θ ) Substitutig si θ (/) si θ ow leads to F HG K J si θ si θ The largest value of for whih this equatio is true is give by si θ We solve for : + si θ + si Sie some of the agles i Fig -7 are measured from vertial axes ad some are measured from horizotal axes, we must be very areful i takig differees For istae, the agle differee betwee the first polarizer struk by the light ad the seod is º (or 7º depedig o how we measure it; it does ot matter i the fial result whether we put θ 7º or put θ º) Similarly, the agle differee betwee the seod ad the third is θ 4º, ad betwee the third ad the fourth is θ 4º, also Aoutig for the automati redutio (by a fator of oe-half) wheever upolarized light passes through ay polarizig sheet, the our result is the iidet itesity multiplied by

27 7 os( θ )os( )os( ) θ θ Thus, the light that emerges from the system has itesity equal to 5 W/m 75 Let θ be the agle of iidee ad θ be the agle of refratio at the left fae of the plate Let be the idex of refratio of the glass The, the law of refratio yields si θ si θ The agle of iidee at the right fae is also θ f θ is the agle of emergee there, the si θ si θ Thus si θ si θ ad θ θ The emergig ray is parallel to the iidet ray We wish to derive a expressio for x i terms of θ f D is the legth of the ray i the glass, the D os θ t ad D t/os θ The agle α i the diagram equals θ θ ad Thus, x D si α D si (θ θ ) t si ( θ θ ) x osθ f all the agles θ, θ, θ, ad θ θ are small ad measured i radias, the si θ θ, si θ θ, si(θ θ ) θ θ, ad os θ Thus x t(θ θ ) The law of refratio applied to the poit of iidee at the left fae of the plate is ow θ θ, so θ θ/ ad F t x t θ θ b g θ HG K J 76 (a) Suppose there are a total of N trasparet layers (N 5 i our ase) We label these layers from left to right with idies,,, N Let the idex of refratio of the air be We deote the iitial agle of iidee of the light ray upo the air-layer boudary as θ i ad the agle of the emergig light ray as θ f We ote that, sie all the boudaries are parallel to eah other, the agle of iidee θ j at the boudary betwee the j-th ad the (j + )-th layers is the same as the agle betwee the trasmitted light ray ad the

28 CHAPTER ormal i the j-th layer Thus, for the first boudary (the oe betwee the air ad the first layer) i si θ, siθ for the seod boudary si θ, siθ ad so o Fially, for the last boudary Multiplyig these equatios, we obtai F N si θ, siθ N i N HG K JF H G K JF H G K J F H G N K JF H G K JF H G K JF H G K J F siθ siθ siθ Λ Λ H G siθ siθ siθ siθ siθ f We see that the LHS of the equatio above a be redued to / while the RHS is equal to siθ i /siθ f Equatig these two expressios, we fid F siθ f siθi si θi, H G K J whih gives θ i θ f So for the two light rays i the problem statemet, the agle of the emergig light rays are both the same as their respetive iidet agles Thus, θ f for ray a, (b) ad θ f for ray b () this ase, all we eed to do is to hage the value of from (for air) to 5 (for glass) This does ot hage the result above That is, we still have θ f for ray a, (d) ad θ f for ray b Note that the result of this problem is fairly geeral t is idepedet of the umber of layers ad the thikess ad idex of refratio of eah layer 77 The time for light to travel a distae d i free spae is t d/, where is the speed of light ( m/s) (a) We take d to be 5 km 5 m The, f K J

29 9 t d 5 m 5 m/s (b) At full moo, the Moo ad Su are o opposite sides of Earth, so the distae traveled by the light is d (5 km) + ( 5 km) 5 km 5 m The time take by light to travel this distae is 4 s d t m/s 5 m 5 s 4 mi () We take d to be ( 9 km) 6 m The, t d 6 m 7 s 4 h m / s (d) We take d to be 65 ly ad the speed of light to be ly/y The, d 65 ly t 65 y ly / y The explosio took plae i the year or 5446 b 7 The law of refratio requires that si θ /si θ water ost We a hek that this is ideed valid for ay give pair of θ ad θ For example si / si, ad si / si 5 ', et Therefore, the idex of refratio of water is water 79 (a) From k ω where k 6 m, we obtai ω 4 rad/s The mageti field amplitude is, from Eq -5, B E/ (5 V/m)/ 67 T From the fat that ˆk (the diretio of propagatio), E ρ ρ E y j, ad B are mutually perpediular, we olude that the oly o-zero ompoet of B ρ is B x, so that we have B z+ t x 6 4 (67 T)si[( / m) ( / s) ]

30 CHAPTER (b) The wavelegth is λ π/k 6 6 m () The period is T π/ω 9 4 s (d) The itesity is 5 Vm µ F H G K J Wm (e) As oted i part (a), the oly ozero ompoet of ρ B is B x The mageti field osillates alog the x axis (f) The wavelegth foud i part (b) plaes this i the ifrared portio of the spetrum (a) Settig v i the wave relatio kv ω πf, we fid f 9 Hz (b) E rms E m / B m / V/m () (E rms ) /µ o 7 W/m (a) At r 4 m, the itesity is P P 4( W) πd π( θr 4 ) 4 π ( 7 rad)( 4 m) W m (b) P 4πr 4π(4m) ( W m ) 7 6 W (a) Assumig omplete absorptio, the radiatio pressure is p r 4 Wm ms 6 47 Nm (b) We ompare values by settig up a ratio: p p r Nm Nm 47 (a) The area of a hemisphere is A πr, ad we get P/A 5 µw/m (b) Our part (a) result multiplied by m gives 7 µw () The part (b) aswer divided by the A of part (a) leads to5 7 W/m

31 (d) The Eq -6 gives E rms 76 V/m E max E rms 7 V/m (e) B rms E rms / 5 6 T 5 ft 4 Aoutig for the automati redutio (by a fator of oe-half) wheever upolarized light passes through ay polarizig sheet, the our result is (os (º)) 5 We remid ourselves that whe the upolarized light passes through the first sheet, its itesity is redued by a fator of Thus, to ed up with a overall redutio of oethird, the seod sheet must ause a further derease by a fator of two-thirds (sie (/)(/) /) Thus, os θ / θ 5 6 (a) The magitude of the mageti field is B E Vm ms (b) With E ρ B ρ µ S ρ r r, where E Ek ˆ ad S S( ˆj), oe a verify easily that sie k ˆ ( ˆi) ˆj,B r has to be i the x diretio 7 From Fig -9 we fid max 47 for λ 4 m ad mi 456 for λ 7 m (a) The orrespodig Brewster s agles are (b) ad θ B,mi ta (456) T θ B,max ta max ta (47) 55, (a) Let r be the radius ad ρ be the desity of the partile Sie its volume is (4π/)r, its mass is m (4π/)ρr Let R be the distae from the Su to the partile ad let M be the mass of the Su The, the gravitatioal fore of attratio of the Su o the partile has magitude GMm 4πGMρ r Fg R R f P is the power output of the Su, the at the positio of the partile, the radiatio itesity is P/4πR, ad sie the partile is perfetly absorbig, the radiatio pressure o it is P pr 4 πr

32 CHAPTER All of the radiatio that passes through a irle of radius r ad area Aπr, perpediular to the diretio of propagatio, is absorbed by the partile, so the fore of the radiatio o the partile has magitude F r πpr Pr pra 4πR 4R The fore is radially outward from the Su Notie that both the fore of gravity ad the fore of the radiatio are iversely proportioal to R f oe of these fores is larger tha the other at some distae from the Su, the that fore is larger at all distaes The two fores deped o the partile radius r differetly: F g is proportioal to r ad F r is proportioal to r We expet a small radius partile to be blow away by the radiatio pressure ad a large radius partile with the same desity to be pulled iward toward the Su The ritial value for the radius is the value for whih the two fores are equal Equatig the expressios for F g ad F r, we solve for r: P r 6πGMρ (b) Aordig to Appedix C, M 99 kg ad P 9 6 W Thus, 6 ( 9 W) r 6π(667 N m / kg )( 99 kg)( kg / m )( m / s) 7 5 m 9 (a) The polarizatio diretio is defied by the eletri field (whih is perpediular to the mageti field i the wave, ad also perpediular to the diretio of wave travel) The give futio idiates the mageti field is alog the x axis (by the subsript o B) ad the wave motio is alog y axis (see the argumet of the sie futio) Thus, the eletri field diretio must be parallel to the z axis (b) Sie k is give as 57 7 /m, the λ π/k 4 7 m, whih meas f /λ 75 4 Hz () The mageti field amplitude is give as B m 4 6 T The eletri field amplitude E m is equal to B m divided by the speed of light The rms value of the eletri field is the E m divided by Eq -6 the gives 9 kw/m 9 Usig Eqs -4 ad -4, we obtai fial ( )( ) ( ) / os 45 os 45 5

33 9 With the idex of refratio 456 at the red ed, sie si θ /, the ritial agle is θ 4 for red (a) At a agle of iidee of θ 4 < θ, the refrated light is white (b) At a agle of iidee of θ 4 whih is slightly less tha θ, the refrated light is white but domiated by red ed () At a agle of iidee of θ 44 > θ, there is o refrated light 9 We apply Eq -4 (oe) ad Eq -4 (twie) to obtai ' ' os θos θ ' ' where θ 9 θ 6 ad θ 9 θ 6 This yields / 9 We write m ρς where ς 4πR is the volume Pluggig this ito F ma ad the ito Eq - (with A πr, assumig the light is i the form of plae waves), we fid ρ 4 πr πr a This simplifies to a 4ρR whih yields a 5 9 m/s 94 We apply Eq -4 (oe) ad Eq -4 (twie) to obtai ' ' os θos θ ' where θ (9 θ) + θ is the relative agle betwee the first ad the seod ' polarizig sheets, ad θ 9 θ 5 is the relative agle betwee the seod ad the third polarizig sheets Thus, we have / 4 95 We apply Eq -4 (oe) ad Eq -4 (twie) to obtai θ θ os os

34 4 CHAPTER With θ θ θ 6 4 ad θ θ + ( π / θ) 4 + 7, we get / 4 96 We use Eq - for the fore, where A is the area of the refletig surfae (4 m ) The itesity is gotte from Eq -7 where P P S is i Appedix C (see also Sample Problem -) ad r m (give i the problem statemet) Our result for the fore is 9 µn 97 Eq -5 gives B E/, whih relates the field values at ay istat ad so relates rms values to rms values, ad amplitude values to amplitude values, as the ase may be Thus, the rms value of the mageti field is B rms ( V/m)/( m/s) 667 T, whih (upo multipliatio by ) yields a amplitude value of mageti field equal to 94 T 9 (a) The Su is far eough away that we approximate its rays as parallel i this Figure That is, if the suray makes agle θ from horizotal whe the bird is i oe positio, the it makes the same agle θ whe the bird is ay other positio Therefore, its shadow o the groud moves as the bird moves: at 5 m/s (b) f the bird is i a positio, a distae x > from the wall, suh that its shadow is o the wall at a distae y h from the top of the wall, the it is lear from the Figure that taθ y/x Thus, dy dt dx ta θ ( 5 m/s)ta 7 dt m/s, whih meas that the distae y (whih was measured as a positive umber dowward from the top of the wall) is shrikig at the rate of 7 m/s () Sie taθ grows as θ < 9 ireases, the a larger value of dy/dt implies a larger value of θ The Su is higher i the sky whe the hawk glides by (d) With dy/dt 45 m/s, we fid v hawk dx dt dy / dt taθ so that we obtai θ 7 if we assume v hawk 5 m/s 99 (a) The wave is travelig i the y diretio (see 6-5 for the sigifiae of the relative sig betwee the spatial ad temporal argumets of the wave futio)

35 5 (b) Figure -5 may help i visualizig this The diretio of propagatio (alog the y axis) is perpediular to ρ B (presumably alog the x axis, sie the problem gives B x ad o other ompoet) ad both are perpediular to ρ E (whih determies the axis of polarizatio) Thus, the wave is z-polarized () Sie the mageti field amplitude is B m 4 µt, the (by Eq -5) E m 99 V/m V/m Dividig by yields E rms 4 V/m The, Eq -6 gives Erms 9 W/m µ (d) Sie k ω (equivalet to f λ), we have k m Summarizig the iformatio gathered so far, we have (with S uits uderstood) E y+ t z 6 5 ( V/m) si[(667 / m) ( / s) ] (e) λ π/k 94 m (f) This is a ifrared light (a) The agle of iidee θ B, at B is the omplemet of the ritial agle at A; its sie is B, θ siθ os so that the agle of refratio θ B, at B beomes θ B, si si 5 (b) From si θ si θ ( / ), we fid θ si 499 () The agle of iidee θ A, at A is the omplemet of the ritial agle at B; its sie is

36 6 CHAPTER siθ osθ A, so that the agle of refratio θ A, at A beomes (d) From θ A, si si 5 siθ siθ A,, we fid θ si 6 (e) The agle of iidee θ B, at B is the omplemet of the Brewster agle at A; its sie is siθ B, + so that the agle of refratio θ B, at B beomes (f) From we fid θ B, si 67 + siθ siθ Brewster + θ si 5 +, (a) ad (b) At the Brewster agle, θ iidet + θ refrated θ B + 9, so θ B 5 ad glass ta θ B ta 5 6 We take the derivative with respet to x of both sides of Eq -:

37 7 E E B B x x x x t x t Now we differetiate both sides of Eq - with respet to t: Substitutig get t F HG K J B B x xt t F HG K J E E εµ εµ t t E x B x t from the first equatio above ito the seod oe, we E E E E E εµ t x t εµ x x Similarly, we differetiate both sides of Eq - with respet to t E B x t t ad differetiate both sides of Eq - with respet to x B E εµ x x t Combiig these two equatios, we get, (a) From Eq -, B t εµ B x B x ad E t t E si( kx ωt) ω E si ( kx ω t), m m E x x E si( kx ωt) k si( kx ωt) ω E si ( kx ω t) m m Cosequetly, E t E x is satisfied Aalogously, oe a show that Eq - satisfies

38 CHAPTER B t B x (b) From E E f ( kx ± ω t), m E t E m f( kx± ωt) ω E d f m t du u kx± ωt ad E x E m f kx t Ek d ( ± ω ) f m t du u kx± ωt Sie ω k the right-had sides of these two equatios are equal Therefore, E t E x Chagig E to B ad repeatig the derivatio above shows that B Bm f ( kx ± ω t) satisfies B B t x 4 Sie itesity is power divided by area (ad the area is spherial i the isotropi ase), the the itesity at a distae of r m from the soure is P 4 Wm 4πr as illustrated i Sample Problem - Now, i Eq - for a totally absorbig area A, we ote that the exposed area of the small sphere is that o a flat irle A π( m) m Therefore, F A ( 4)( ) 7 N