CHAPTER 37. Answer to Checkpoint Questions

Transcription

1 1010 CHAPTER 37 DIFFRACTION CHAPTER 37 Answer to Checkpoint Questions 1. (a) expan; (b) expan. (a) secon sie axiu; (b) :5 3. (a) re; (b) violet 4. iinish 5. (a) increase; (b) sae 6. (a) left; (b) less Answer to Questions 1. (a) contract; (b) contract. (a) the 5 iniu; (b) (approxiately) the axiu between the 4 an 5 inia 3. with egaphone (larger opening, less iraction) 4. (a) a an c tie, then b an tie; (b) a an b tie, then c an tie 5. four 6. re 7. (a) larger; (b) re 8. (a) less; (b) greater; (c) greater 9. (a) ecrease; (b) sae; (c) in place 10. (a) ecrease; (b) ecrease; (c) to the right 11. (a) A; (b) left; (c) left; () right

2 CHAPTER 37 DIFFRACTION (a) increase; (b) rst orer Solutions to Exercises & Probles 1E The conition for a iniu in a single-slit iraction pattern is given by Eq. 37-3: a sin, where a is the slit with, is the wavelength, an is an integer. Solve for : a sin (0:0 )(sin 1:8 )1 6: n: E (a) sin 1 (1:50 c:00 ) 0:430 : (b) For the th iraction iniu a sin : Solve for a: a sin (441 n) 0:118 : sin 0:430 3E Use a sin. The angle is easure fro the forwar irection, so for the situation escribe in the proble it is 0:60, for 1. Thus a sin sin 0:60 6: : 4E (a) Use a sin. For a an 1 the angle is the sae as for b an. Thus a b. (b) Let a be the integer associate with a iniu in the pattern prouce by light with wavelength a an let b be the integer associate with a iniu in the pattern prouce by light with wavelength b. A iniu in one pattern coincies with a iniu in the other if they occur at the sae angle. This eans a a b b. Since a b, the inia coincie if a b. Thus every other iniu of the b pattern coincies with a iniu of the a pattern. 5E (a) Use Eq to calculate the separation between the rst ( 1 1) an fth ( 5) inia: y D sin D D D a a a ( 1 ) :

3 101 CHAPTER 37 DIFFRACTION solve for a: a D( 1 ) y (400 )( )(5 1) 0:35 :5 : (b) For 1 sin a (1)( ) :5 The angle is sin 1 (: 10 4 ) : 10 4 ra: : 10 4 : 6E Fro Eq a sin 1 sin 45:0 1:41 : 7E (a) A plane wave is incient on the lens so it is brought to focus in the focal plane of the lens, a istance of 70 c fro the lens. (b) Waves leaving the lens at an angle to the forwar irection interfere to prouce an intensity iniu if a sin, where a is the slit with, is the wavelength, an is an integer. The istance on the screen fro the center of the pattern to the iniu is given by y D tan, where D is the istance fro the lens to the screen. For the conitions of this proble sin a (1)( ) 0: : : This eans 1: ra an y (7010 ) tan(1: ra) 1: P The conition for a iniu of intensity in a single-slit iraction pattern is a sin, where a is the slit with, is the wavelength, an is an integer. To n the angular position of the rst iniu to one sie of the central axiu set 1: 1 sin 1 a sin 1 5:89 1: ra : If D is the istance fro the slit to the screen, the istance on the screen fro the center of the pattern to the iniu is y 1 D tan 1 (3:00 ) tan(5: ra) 1:

4 CHAPTER 37 DIFFRACTION 1013 To n the secon iniu set : ( sin 1 ) 1:178 1: ra : The istance fro the pattern center to the iniu is y D tan (3:00 ) tan(1: ra) 3: The separation of the two inia is y y y 1 3:534 1:767 1:77. 9P Let the rst iiu be a istance y fro the central axis which is perpenicular to the speaker. Then sin y(d + y ) 1 a a (for 1). Slove for y: y D p (a) 1 D p (afvs ) p [(0:300 )(3000 Hz)(343 /s)] 1 41: : 10P Fro y Da we get D y D a a (63:8 n)(:60) [10 ( 10)] 4:0 : 1:37 11E Fro Eq :10 (x sin ) (sin 30 ) 67 ra : 589 n This is equivalent to 67 ra 84 :79 ra 160 : 1E (a) sin 1 (1:1 c3:5 ) 0:18 : (b) Use Eq. 37-6: a sin (0:05 )(sin 0:18 ) 0:46 ra : 538 n

5 1014 CHAPTER 37 DIFFRACTION 13P If you ivie the original slit into N strips an represent the light fro each strip, when it reaches the screen, by a phasor, then at the central axiu in the iraction pattern you a N phasors, all in the sae irection an each with the sae aplitue. The intensity there is proportional to N. If you ouble the slit with you nee N phasors if they are each to have the aplitue of the phasors you use for the narrow slit. The intensity at the central axiu is proportional to (N) an is therefore 4 ties the intensity for the narrow slit. The energy reaching the screen per unit tie, however, is only twice the energy reaching it per unit tie when the narrow slit is in place. The energy is siply reistribute. For exaple, the central peak is now half as wie an the integral of the intensity over the peak is only twice the analogous integral for the narrow slit. 14P Think of the Huygens' explanation of iraction phenoenon. When A is in place only the Huygens' wavelets that pass through the hole get to point P. Suppose they prouce a resultant electric el E A. When B is in place the light that was blocke by A gets to P an the light that passe through the hole in A is blocke. Suppose the electric el at P is now E B. The su E A + E B is the resultant of all waves that get to P when neither A nor B are present. Since P is in the geoetric shaow this is zero. Thus E A E B an since the intensity is proportional to the square of the electric el, the intensity at P is the sae when A is present as when B is present. 15P (a) The intensity for a single-slit iraction pattern is given by I I sin ; where (a) sin, a is the slit with an is the wavelength. The angle is easure fro the forwar irection. You want I I, so sin 1 : (b) Evaluate sin an for 1:39 ra an copare the results. To be sure that 1:39 ra is closer to the correct value for than any other value with 3 signicant igits, you shoul also try 1:385 ra an 1:395 ra. (c) Since (a) sin, Now 1:39 0:44, so sin 1 : a 0:44 sin 1 : a

6 CHAPTER 37 DIFFRACTION 1015 The angular separation of the two points of half intensity, one on either sie of the center of the iraction pattern, is 0:44 sin 1 : a () For a 1:0, for a 5:0, an for a 10, sin 1 (0:441:0) 0:916 ra 53 ; sin 1 (0:445:0) 0:177 ra 10 ; sin 1 (0:4410) 0:0884 ra 5:1 : 16P (a) The intensity for a single-slit iraction pattern is given by I I sin ; where (a) sin. Here a is the slit with an is the wavelength. To n the axia an inia, set the erivative of I with respect to equal to zero an solve for. The erivative is I I sin ( cos sin ) : 3 The erivative vanishes if 6 0 but sin 0. This yiels, where is an integer. Except for 0 these are the intensity inia: I 0 for. The erivative also vanishes for cos sin 0. This conition can be written tan. These are the axia. (b) The values of that satisfy tan can be foun by trial an error on a pocket calculator or coputer. Each of the is slightly less than one of the values ( + 1) ra, so start with these values. The rst few are 0, 4:4934, 7:75, 10:9041, 14:066, an 17:07. They can also be foun graphically. As in the iagra to the right, plot y tan an y on the sae graph. The intersections of the line with the tan curves are the solutions. The rst two solutions liste above are shown on the iagra. y 0 ytan α yα π / π 3π / α (ra)

7 1016 CHAPTER 37 DIFFRACTION (c) Write ( + 1 ) for the axia. For the central axiu, 0 an 1. For the next, 4:4934 an 0:930. For the next 7:75 an 1: P Since the slit with is uch less than the wavelength of the light, the central peak of the single-slit iraction pattern is sprea across the screen an the iraction envelope can be ignore. Consier 3 waves, one fro each slit. Since the slits are evenly space the phase ierence for waves fro the rst an secon slits is the sae as the phase ierence for waves fro the secon an thir slits. The electric els of the waves at the screen can be written E 1 E 0 sin(!t), E E 0 sin(!t + ), an E 3 E 0 sin(!t + ), where () sin. Here is the separation of ajacent slits an is the wavelength. The phasor iagra is shown to the right. It yiels E E 0 cos + E 0 + E 0 cos E 0 (1 + cos ) E 3 φ φ for the aplitue of the resultant wave. Since the intensity of a wave is proportional to the square of the electric el, we ay write I AE 0(1+ cos ), where A is a constant of proportionality. If I is the intensity at the center of the pattern, for which 0, then I 9AE 0. Take A to be I 9E 0 an obtain E φ ω t E 1 E φ I I 9 (1 + cos ) I cos + 4 cos : 18E Use Eq. 37-1: sin 1:. In our case :5 1:5, so 1: sin 1:(550 n) sin 1:5 31 : 19E (a) Use the Rayleigh criteria. To resolve two point sources the central axiu of the iraction pattern of one ust lie at or beyon the rst iniu of the iraction pattern of the other. This eans the angular separation of the sources ust be at least R 1:, where is the wavelength an is the iaeter of the aperture. For the healights of this proble R 1:( ) 1:34 5: ra : (b) If L is the istance fro the healights to the eye when the healights are just resolvable an D is the separation of the healights, then D L tan R L R, where the sall angle

8 CHAPTER 37 DIFFRACTION 1017 approxiation tan R R was ae. This is vali if R is easure in raians. Thus L D R (1:4 )(1: ra) 1: k. 0E (a) Use Eq : R 1: (1:)( ) 5:0 1: ra : (b) The linear separation is D L R ( )(1: ra) 1 : 1E The iniu separation is D in e R e 1: (1:)(3:8 108 )( ) 5:1 50 : E L ax D R D 1: (5: )(4: ) 1:( ) 30 : 3E 1: D in L R L (1:)(50 )( ) 5:00 30:5 : 4E (a) Use Rayleigh's criterion: two objects can be resolve if their angular separation at the observer is greater than R 1:, where is the wavelength of the light an is the iaeter of the aperture (the eye or irror). If L is the istance fro the observer to the objects then the sallest separation D they can have an still be resolvable is D L tan R L R, where R is easure in raians. The sall angle approxiation tan R R was ae. Thus D 1:L 1:(8: )( ) 5: : : k : This istance is greater than the iaeter of Mars. One part of the planet's surface cannot be resolve fro another part.

9 1018 CHAPTER 37 DIFFRACTION (b) Now 5:1 an D 1:(8: )( ) 5:1 1: k : 5E L ax D R D 1: (5:0 10 )(4: ) 1:(0: ) 1: k : 6E 1: D in L R L (6: 103 )(1:)(1:6 10 ) :3 53 : 7P (a) The iaeter is 1: D L R L ( )(1:)(1: ) 0: :1 : (b) II (D) (0: :71 ) 1: : 8P (a) L ax D 1: ( )(1: ) 1:( ) 0:19 : (b) The wavelength of the blue light is shorter so L ax / 1 will be larger. 9P Accoring to Rayleigh's criterion (Eq ) R 1:. In our case R DL, where D 60 is the size of the object your eyes ust resolve, an L is the liiting viewing istance in question. Also 3:00 is the iaeter of your pupil. Solve for L: L D 1: (60 )(3:00 ) 1:(550 n) 7 c :

10 CHAPTER 37 DIFFRACTION P (a) Use Eq. 37-1: 1: sin 1 sin 1 1:(vs f) sin 1 (1:)(1450 /s) 6:8 : ( Hz)(0:60 ) (b) Now f 1: Hz so 1: (1:)(1450 /s) (1: Hz)(0:60 ) :9 > 1 : Since sin cannot excee 1 there is no iniu. 31P Use 1: DL: 1:L D (1:)(0 i)(1610 /i)( ) (30 ft)(0:305 /ft) 4:7 c : 3P Fro R 1: DL we get 1:L D (1:)( )( ) 0:30 0:36 : 33P (a) The rst iniu in the iraction pattern is at an angular position, easure fro the center of the pattern, such that sin 1:, where is the wavelength an is the iaeter of the antenna. If f is the frequency then the wavelength is cf (3: /s)( Hz) 1: Thus 1: sin 1 1:(1: sin 1 ) 3:0 55: ra : The angular with of the central axiu is twice this, or 6: ra (0:346 ).

11 100 CHAPTER 37 DIFFRACTION (b) Now 1:6 c an :3, so sin 1 1:(1:6 10 ) :3 8: ra : The angular with of the central axiu is 1:7 10 ra (0:97 ). 34P (a) The angular separation is (b) The istance is R 1: (1:)( ) 0:76 8: ra 0:18 00 : D L R (10 ly)(9: k/ly)(0:18) (3600)(180) 8: k : (c) The iaeter is R L (0:18)()(14 ) (3600)(180) : :05 : 35P (a) Since 1:, the larger the wavelength the larger the raius of the rst iniu (an secon axiu, ect). Therefore the white pattern is outline by re lights (with longer wavelength than blue lights). (b) 1: 1:( ) 1:5(0:50 )(180 ) 1: : 36P The energy of the bea of light which is projecte onto the oon is concentrate is a circular spot of iaeter 1, where 1 e (1: 0 ), with 0 the iaeter of the irror on Earth an e the Earth-Moon separation. The fraction of energy picke up by the reector of iaeter on the Moon is then 0 ( 1 ). This reecte light, upon reaching the Earth, has a circular cross section of iaeter 3 satisfying 3 e (1: ). The fraction of the reecte energy that is picke up by the telescope is then 00 ( 0 3 ) : Thus the fraction of the original energy picke up by the etector is (:44 e 0 )(:44 e ) :44 e 4 (:6 )(0:10 ) :44(0: )(3: : )

12 CHAPTER 37 DIFFRACTION E Bright interference fringes occur at angles given by sin, where is the slit separation, is the wavelength, an is an integer. For the slits of this proble 11a, so a sin 11. The rst iniu of the iraction pattern occurs at the angle 1 given by a sin 1 an the secon occurs at the angle given by a sin, where a is the slit with. You want to count the values of for which 1 < <, or what is the sae, the values of for which sin 1 < sin < sin. This eans 1 < (11) <. The values are 6, 7, 8, 9, an 10. There are ve bright fringes in all. 38E The nuber is (a) 1 (aa) 1 3: 39E It is clear fro Eq that for a single slit of with a the iraction pattern is given by sin() I I ; where a sin. Now, if we put a in Eq , then a sin, an Eq reuces to I I (cos ) sin sin cos sin() I I ; where the trigonoetric ientity sin() sin cos was use. Thus Eq inee reuces to the iraction pattern for a single slit of with a. 40P (a) Let the location of the fourth bright fringe coincie with the rst iniu of iraction pattern: sin 4 a, or 4a. (b) Any bright fringe which happens to be at the sae location with a iraction iniu will vanish. So let sin 1 a 1 4a a, or 1 4 where 1; ; 3; : The fringes issing are thus the 4th, the 8th, the 1th,, i.e., every fourth fringe is issing. 41P The angular location of the th bright fringe is given by sin so the linear separation between two ajacent fringe is D y (D sin ) D D :

13 10 CHAPTER 37 DIFFRACTION 4P (a) The angular positions of the bright interference fringes are given by sin, where is the slit separation, is the wavelength, an is an integer. The rst iraction iniu occurs at the angle 1 given by a sin 1, where a is the slit with. The iraction peak extens fro 1 to + 1, so you want to count the nuber of values of for which 1 < < + 1, or what is the sae, the nuber of values of for which sin 1 < sin < + sin 1. This eans 1a < < 1a or a < < +a. Now a (0: )(30: ) 5:00, so the values of are 4, 3,, 1, 0, +1, +, +3, an +4. There are nine fringes. (b) The intensity at the screen is given by I I cos sin ; where (a) sin, () sin, an I is the intensity at the center of the pattern. For the thir bright interference fringe sin 3, so 3 ra an cos 1. Siilarly, 3a 35:00 0:600 ra an (sin ) (sin 0:600) (0:600) 0:55. The intensity ratio is II 0:55. 43P (a) The rst iniu of the iraction pattern is at 5:00 so a sin 0:440 sin 5:00 5:05. (b) Since the fourth bright fringe is issing 4a 4(5:05 ) 0: : (c) For the 1 bright fringe a sin (5:05 ) sin 1:5 0:440 0:787 ra ; so the intensity of the 1 fringe is sin sin 0:787 ra I I (7:0 W/c ) 5:7 W/c ; 0:787 which agrees with what Fig inicates. Siilarly for in agreeent with Fig I :9 W/c, also 44P As the phase ierence is varie fro zero to, both the intensity prole of the iraction an the location of the interference axiu change. At, the original central iraction envelop is now a iniu, an the two axia of the iraction intensity prole are now centere where the rst inia were. The locations of the intensity axia/inia ue to interference exchange, with the original locations of the axia now those of inia, an vice versa.

14 CHAPTER 37 DIFFRACTION 103 As is further varie fro to, the intensity pattern is graually change back, resuing the original pattern at. 45E (a) 0: : :33 : (b) Let sin ( 0; 1; ; ), we n 0 for 0, sin 1 () sin 1 (0:589 3:30 ) 10: for 1, an siilarly 0:7 for ; 3: for 3; 45 for 4, an 6: for 5. Since jj > 1 for jj 6 these are all the axia. 46E (a) Let sin an solve for : sin (1:0 00)(sin 30 ) 500 n ; where 1; ; 3. In the visible light range can assue the following values: 1 4; 5 an 3 6. The corresponing wavelengths are n4 65 n; 500 n5 500 n, an n6 416 n. (b) The colors are orange (for 1 65 n), blue-green (for 500 n), an violet (for n). 47E The angular location of the th orer iraction axiu is given by sin. To be able to observe the fth-orer one we ust let sin j 5 5 < 1, or < 5 1:00 n n : So all wavelengths shorter than 635 n can be use. 48E The ruling separation is 1(400 1 ) : Diraction lines occur at angles such that sin, where is the wavelength an is an integer. Notice that for a given orer the line associate with a long wavelength is prouce at a greater angle than the line associate with a shorter wavelength. Take to be the longest wavelength in the visible spectru (700 n) an n the greatest integer value of such that is less than 90. That is, n the greatest integer value of for which <. Since (: )( ) 3:57 that value is 3. There are 3 coplete orers on each sie of the 0 orer. The secon an thir orers overlap (see 59P).

15 104 CHAPTER 37 DIFFRACTION 49E Let the total nuber of lines on the grating be N, then LN where L 3:00 c. For the secon orer iraction axiu sin (LN) sin, so N L sin (3:00 10 )(sin 33 ) ( ) 13; 600 : 50E Use Eq for iraction axia: sin. In our case since the angle between the 1 an 1 axia is 6 the angle corresponing to 1 is Solve for : sin (1)(550 n) sin 13 :4 : 51P Let sin (LN) sin, we get (LN) sin (1:0 107 n)(sin 30 ) (1)(10; 000) 500 n : 5P (a) Maxia of a two-slit interference pattern occur at angles given by sin, where is the slit separation, is the wavelength, an is an integer. The two lines are ajacent so their orer nubers ier by unity. Let be the orer nuber for the line with sin 0: an + 1 be the orer nuber for the line with sin 0:3. Then 0: an 0:3 ( + 1). Subtract the rst equation fro the secon to obtain 0:1, or 0:1 ( )0:1 6: (b) Minia of the single-slit iraction pattern occur at angles given by a sin, where a is the slit with. Since the fourth orer interference axiu is issing it ust fall at one of these angles. If a is the sallest slit with for which this orer is issing the angle ust be given by a sin. It is also given by sin 4, so a 4 (6: )4 1: (c) First set 90 an n the largest value of for which < sin. This is the highest orer that is iracte towar the screen. The conition is the sae as < an since (6: )( ) 10:0, the highest orer seen is the 9 orer. The fourth an eighth orers are issing so the observable orers are 0, 1,, 3, 5, 6, 7, an 9. 53P (a) For the axiu with the greatest value of ( M) we have M a sin <, so M < 900 n600 n 1:5, or M 1. Thus three axia can be seen, with 0; 1:

16 CHAPTER 37 DIFFRACTION 105 (b) Fro Eq hw N cos tan sin N cos tan N sin n 900 n 1 N 0:051 : sin 1 54P The angular positions of the rst-orer iraction lines are given by sin, where is the slit separation an is the wavelength. Let 1 be the shorter wavelength (430 n) an be the angular position of the line associate with it. Let be the longer wavelength (680 n) an let + be the angular position of the line associate with it. Here 0. Then sin 1 an sin( + ). Use a trigonoetric ientity to replace sin(+) with sin cos +cos sin, then use the equation for the rst line to replace sin with 1 an cos with p 1 1. After ultiplying by you shoul obtain 1 cos + p 1 sin. Rearrange to get p 1 sin 1 cos. Square both sies an solve for. You shoul get r ( 1 cos ) + ( 1 sin ) sin s [(680 n) (430 n) cos 0 ] + [(430 n) sin 0 ] sin n 9: : There are 1 1(9: ) 1090 rulings per. 55P Use Eq. 37-5: sin. For 1 an for sin (1:73) sin(17:6 ) 1 (1:73) sin(37:3 ) 53 n ; 54 n : Siilarly we ay copute the values of corresponing to the angles for 3. The average value of these 0 s is 53 n.

17 106 CHAPTER 37 DIFFRACTION 56P The ierence in path lengths between the two ajacent light rays shown to the right is x jabj + jbcj sin + sin. The conition for bright fringes to occur is thus x (sin + sin ) ; where 0; 1; ; : 1 A ψ θ ψ B θ C 57P Fro the gure to the right we see that the angular eviation of the rst-orer axiu fro the incient irection is + 1. Here sin 1 sin (600 n1:50 ) sin 0:400 sin : Thus sin 1 sin incient wave ψ ψ iffracte wave θ 1 + sin 1 (0:400 sin ) : The plot is as follows. The angles are given in raians δ (ra) ψ (ra) 58P Fro Eq we get (sin ) ;

18 CHAPTER 37 DIFFRACTION 107 but for sall ; sin ( sin ) cos so cos p 1 sin p 1 () p () : 59P Fro sin we see that if two spectral lines (labele 1 an, respectively) overlap, eaning they share the sae value of, then 1 1. For 1 an 3 this becoes : Since 400 n < < 1 < 700 n we can always n suitable values of 1 an which satisfy this conition. For exaple n, an 400 n, or n an 440 n, etc. So these two spectra always overlap, regarless of the value of. 60P In this case a, an the forula for the locations of the th iraction iniu, a sin () sin, ay be re-written as () sin, which we recognize as the forula for the location of the ()-th axiu of interference. Thus all the ()-th (i.e., even) orers of axia will be eliinate (exept 0). 61P At the location of the hole sin c 0:164, an fro sin we n sin 5:0 c p (30 c) + (5:0 c) 0:164; so sin (1: n350)(0:164) 470 n : Since for white light > 400 n the only integer allowe here is 1. At one ege of the hole 477 n, while at the other ege " # 1:00 sin n p 560 n : 350 (30 c) + (6:0 c) So the range of wavelength is fro 470 to 560 n. 6P The erivation is siilar to that use to obtain Eq At the rst iniu beyon the th principal axiu two waves fro ajacent slits have a phase ierence of + (N), where N is the nuber of slits. This iplies a ierence in path length

19 108 CHAPTER 37 DIFFRACTION of L () + (N). If is the angular position of the th axiu then the ierence in path length is also given by L sin( + ). Thus sin( + hw ) + (N). Use the trigonoetric ientity sin( + hw ) sin cos hw + cos sin hw. Since hw is sall, we ay approxiate sin hw by hw in raians an cos hw by unity. Thus sin + hw cos + (N). Use the conition sin to obtain hw cos N an hw N cos : 63E Let R N an solve for N: N (589:6 n + 589:0 n) (589:6 n 589:0 n) 491 : 64E (a) Solve fro R N: N (b) Since sin ax < 1, 500 n (600 )(5:0 )(3) 0:056 n56 p : ax < 1 (600 )( ) 3:3 ; thus ax 3. No higher orers of axia can be seen. 65E If a grating just resolves two wavelengths whose ean is an whose separation is then its resolving power is ene by R. The text shows this is N, where N is the nuber of rulings in the grating an is the orer of the lines. Thus N an N 656:3 n (1)(0:18 n) 3650 rulings : 66E (a) Fro R N we n N (415:496 n + 415:487 n) (415:96 n 415:487 n) 3; 100 :

20 CHAPTER 37 DIFFRACTION 109 (b) The axia are foun at sin 1 sin 1 ()(415:5 n) 4: n3; 100 8:7 : 67E (a) Fro sin we n sin (b) The total with of the ruling is 3(589:3 n) sin 10 1: n 10 : L N R N (589:3 n)(10 ) 3(589:59 n 589:00 n) 3: :3 : 68E The ispersion of a grating is given by D, where is the angular position of a line associate with wavelength. The angular position an wavelength are relate by sin, where is the slit separation an is an integer. Dierentiate this with respect to to obtain () cos or Now () sin, so D cos : D sin cos tan : The trigonoetric ientity tan sin cos was use. 69E (a) For the rst orer axia sin, which gives sin 1 sin n 18 ; n40; 000 so fro 68E D tan tan n 0:03 n: Siilarly for an 3 we have 38 an 68, an the corresponing values of ispersion are 0:076 n an 0:4 n, respectively.

21 1030 CHAPTER 37 DIFFRACTION (b) R N ; 000 for ( 1); 80; 000 (for ); an 10; 000 (for 3). 70P (a) We require that sin 1; sin 30, where 1; an n. This gives s (600 n) 400 n : sin 30 sin 30 For a grating of given totla with L we have N L / 1, so we nee to iniize to axiize R N / 1. Thus we choose 400 n. (b) Let the thir-orer axiu for 600 n be the rst iniu for the single-slit iraction prole. This requires that sin 3 a sin, or a n3 800 n: (c) Let sin ax 1 to obtain ax 400 n 800 n 3 : Since the thir orer is issing the only axia present are the ones with 0; 1 an. 71P (a) Fro the expression for the half-with hw (given by Eq. 37-8) an that for the resolving power R (given by Eq. 37-3), we n the prouct of hw an R to be hw R N N cos cos sin cos tan ; where we use sin (see Eq. 37-5). (b) For rst orer 1, so the corresponing angle 1 satises sin 1. Thus the prouct in question is given by tan 1 sin 1 cos 1 sin 1 p 1 sin 1 1 p (1 sin 1 ) 1 1 p () 1 1 p (900 n600 n) 1 0:89 : 7P (a) Since the resolving power of a grating is given by R an by N, the range of wavelengths that can just be resolve in orer is N. Here N is the nuber

22 CHAPTER 37 DIFFRACTION 1031 of rulings in the grating. The frequency f is relate to the wavelength by f c, where c is the spee of light. This eans f + f 0, so (f)f ( c)f, where f c was use. The negative sign eans siply that an increase in frequency correspons to a ecrease in wavelength. If we interpret f as the range of frequencies that can be resolve we ay take it to be positive. Then an c f f N c N : (b) The ierence in travel tie for waves traveling along the two extree rays is t Lc, where L is the ierence in path length. The waves originate at slits that are separate by (N 1), where is the slit separation an N is the nuber of slits, so the path ierence is L (N 1) sin an the tie ierence is t (N 1) sin If N is large this ay be approxiate by t (Nc) sin. The lens oes not aect the travel tie. (c) Substitute the expressions you erive for t an f to obtain c N sin f t N c c : sin 1 : The conition sin for a iraction line was use to obtain the last result. 73E Bragg's law gives the conition for a iraction axiu: sin ; where is the spacing of the crystal planes an is the wavelength. The angle is easure fro the noral to the planes. For a secon orer reection, so sin (0: ) :6 sin :6 n : 74E Use Eq For sallest value of let 1: in sin 1 sin 1 (1)(30 p) :9 : (0: p)

23 103 CHAPTER 37 DIFFRACTION 75E For rst orer reection sin 1 an for the secon orer one sin. Solve for : sin 1 ( sin 1 ) sin 1 ( sin 3:4 ) 6:8 : 76E Use Eq Fro the peak on the left at angle 1 sin 1 1, or 1 sin 1 (0:94 n) sin(0:75 ) 0:05 n 5 p. Fro the next peak sin (0:94 n) sin(1:15 ) 0:038 n 38 p. You can check that the thir peak fro left is just the secon-orer one for 1. 77E For the rst bea sin 1 A an for the secon one sin 3 B. The values of an A can then be solve. (a) (b) 3 B sin 3(97 p) sin 60 1:7 10 p : A sin 1 (1:7 10 p)(sin 3 ) 1:3 10 p : 78E sin (39:8 p)(sin 30:0 ) 39:8 p : 79P There are two unknowns, the X-ray wavelength an the plane separation, so ata for scattering at two angles fro the sae planes shoul suce. The observations obey Bragg's law, so sin 1 1 an sin : However, these cannot be solve for the unknowns. For exaple, use rst equation to eliinate fro the secon. You obtain sin 1 1 sin, an equation that oes not contain either of the unknowns. 80P The wavelengths satisfy sin (75 p)(sin 45 ) 389 p. In the range of wavelengths given, the allowe values of are 3; 4, with corresponing wavelengths being 389 p3 130 p an 389 p4 97: p, respectively.

24 CHAPTER 37 DIFFRACTION P The angle of incience on the reection planes is 63:8 45:0 18:8, an the plane-plane separation is a 0 p. Thus fro sin we get a 0 p p sin 0:60 n p sin 18:8 0:570 n : 8P (a) The sets of planes with the next ve saller interplanar spacings (after a 0 ) are shown in the iagra to the right. In ters of a 0 the spacings are: (i): a 0 p 0:7071a 0 ; (ii): a 0 p 5 0:447a 0 ; (iii): a 0 p 10 0:316a 0 ; (iv): a 0 p 13 0:774a 0 ; (v): a 0 p 17 0:45a 0 : (i) (ii) (iii) (iv) (v) (b) Since any crystal plane passes through lattice points its slope can be written as the ratio of two integers. Consier a set planes with slope n, as shown in the iagra to the right. The rst an last planes shown pass through ajacent lattice points along a horizontal line an there are 1 planes between. If h is the separation of the rst an last planes, then the interplanar spacing is h. If the planes ake the angle with the horizontal, then the noral to planes (shown otte) akes the angle 90. The istance h is given by h a 0 cos an the interplanar spacing is h (a 0 ) cos. Since p tan n, tan n an cos tan p n +. Thus a 0 na 0 θ φ h a 0 cos a 0 p n + :

25 1034 CHAPTER 37 DIFFRACTION 83P The angles of incience which correspon to intensity axia in reecte bea of light satisfy sin, or sin (0:15 n) (0:5 n) 4:03 : Since j sin j < 1 the allowe values for are 1; ; 3; 4: Corresponingly the values of are 14:4 ; 9:7 ; 48:1 ; an 8:8. Therefore the crystal shoul be rotate counterclockwise by 48:1 45:0 3:1 or 8:8 45:0 37:8, or clockwise by 45:0 14:4 30:6 or 45:0 9:7 15:3 : 84 0:143 ra, II 4:7 10 ; 0:47 ra, II 1:65 10 ; 0:353 ra, II 8: