Tài liệu Physics exercises_solution: Chapter 35 docx

35.1: Measuring with a ruler from both 21 and SS to there different points in the antinodal line labeled m = 3, we find that the difference in path length is three times the wavelength of the wave, as measured from one crest to the next on the diagram. 35.2: a) At ,λ4, 121   rrS and this path difference stays the same all along the axis,-y so ,rrSm λ4,At.4 122  and the path difference below this point, along the negative y -axis, stays the same, so .4   m b) c) The maximum and minimum m-values are determined by the largest integer less than or equal to . λ d d) If ,77λ 2 1 7  md so there will be a total of 15 antinodes between the sources. (Another antinode cannot be squeezed in until the separation becomes six times the wavelength.) 35.3: a) For constructive interference the path diference is ,2,1,0,λ    nm The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. Thus only the path difference of zero is possible. This occurs midway between the two sources, 2.50 m from A. b) For destructive interference the path difference is ,2,1,0,λ)( 2 1  mm A path difference of 00.32λ   / m is possible but a path difference as large as 00.92/λ3  m is not possible. For a point a distance x from A and Bx from00.5  the path difference is m00.1givesm00.3)m00.5( m00.4givesm00.3)m00.5( ).m00.5(     xxx xxx xx 35.4: a) The path difference is 120 m, so for destructive interference: .m024λm120 2 λ  b) The longest wavelength for constructive interference is .m120λ  35.5: For constructive interference, we need λ)m00.9(λ 12 mxxmrr       ,10,1,2,3,Form.8.25m,7.00m,5.75m,4.50m,3.25m,2.00m,75.0 ).m25.1(m5.4 Hz)102(120 )sm1000.3( m5.4 2 m5.4 2 λ m5.4 6 8      mx m m f mcm x .3,2   (Don’t confuse this m with the unit meters, also represented by an “m”). 35.6: a) The brightest wavelengths are when constructive interference occurs: nm.408 5 nm2040 λ andnm510 4 nm2040 λ,nm680 3 nm2040 λλλ 43   s m d md b) The path-length difference is the same, so the wavelengths are the same as part (a). 35.7: Destructive interference occurs for: .nm453 5.4 nm2040 λandnm358 5.3 nm2040 λ 21 λ 43    m d 35.8: a) For the number of antinodes we have: ,90settingso,,2317.0 Hz)10(1.079m)(12.0 )sm10(3.00 λ sin 8 8      m m df mc d m the maximum integer value is four. The angles are         9.67and,0.44,6.27,4.13 for .4,3,2,1,0      m b) The nodes are given by sin ).21(2317.0 λ)21(    m d m  So the angles are .3,2,1,0for2.54,4.35,3.20,65.6            m 35.9: .m105.90 m2.20 m)10(2.82m)10(4.60 λ λ 7 34        R yd d R y 35.10: For bright fringes: .mm1.14m101.14 m0.0106 m)10(5.02(20)m)(1.20 λ 3 7      m y Rm d 35.11: Recall m104.50 m)10(5.00)m(0.750)23( λ 4 7 2323        d R λ yyy d Rm y m .mm0.833m1033.8 4 23   y 35.12: The width of a bright fringe can be defined to be the distance between its two adjacent destructive minima. Assuming the small angle formula for destructive interference , λ)( 2 1 d m Ry m   the distance between any two successive minima is mm.8.00 m)10(0.200 m)10(400 m)00.4( λ 3 9 1        d Ryy nn Thus, the answer to both part (a) and part (b) is that the width is 8.00 mm. 35.13: Use the information given about the bright fringe to find the distance d between the two slits: .m103.72 m104.84 m)10(600m)(3.00 λ so),6.35.Eq( λ 4 3 9 1 11 1        y R d d R y (R is much greater than d, so Eq.35.6 is valid.) The dark fringes are located by ,2,1,0,λ)(sin 2 1  mmd  The first order dark fringe is located by sin 22 λ where,2λ d  is the wavelength we are seeking. d R RRy 2 λ sintan 2   We want 2 λ such that . 1 yy  This gives 1200λ2λand 2 λλ 12 21  d R d R nm. 35.14: Using Eq.35.6 for small angles, , λ d m Ry m  we see that the distance between corresponding bright fringes is .mm3.17m)(10470)(660 m)10(0.300 (1)m)(5.00 λ 9 3      d Rm y 35.15: We need to find the positions of the first and second dark lines: .m0.0541)tan(8.79m)0.350(tan 79.8 m102(1.80 m105.50 arcsin 2 λ arcsin 11 6 7 1                         Ry d Also                      3.27 m)102(1.80 m)103(5.50 arcsin 2 λ3 arcsin 6 7 2 d  m.0.1805tan(27.30)m)035.0(tan 22    θRy The fringe separation is then .m0.1264m0.0541m1805.0 12  yyy 35.16: (a) Dark fringe implies destructive interference. λ 2 1 sin   d m1064.1 0.11sin2 m10624 sin2 λ 6 9        d (b) Bright fringes: λsin maxmax mθd  The largest that θ can be is mdm Since6.2λ/so,90 m10624 m101.64 max 9 6      is an integer, its maximum value is 2. There are 5 bright fringes, the central spot and 2 on each side of it. Dark fringes:   .λsin 2 1  mθd This equation has solutions for ;9.34;0.11      θ and .6.72   Therefore, there are 6 dark fringes. 35.17: Bright fringes for wavelength λ are located by .λsin md   First-order )1(  m is closest to the central bright line, so ./λsin d       0.401andm)10m)/(0.10010(700singivesnm700λ 0.229andm)10m)/(0.10010(400singivesnm400λ 39 39   The angular width of the visible spectrum is thus .0.1720.2290.401      35.18: m.0.193m101.93 m104.20 m)10(4.50m)1.80( λλ 4 3 7        y R d d R y 35.19: The phase difference  is given by )Eq.35.13(sin)λ/2(    d  rad16700.23sin]m)10(500m)10340.0(2[ 93    35.20: λ differencePath 2     radians119 cm2 cm486cm524 2             35.21: a) Eq.(35.10): 0 2 0 2 0 750.0)0.30(cos)2(cos IIII   b) rad)3/(0.60    Eq. so),)(λ/2(:)11.35( 12 rr     nm806/λλ2/)3/(λ)2/()( 12   rr 35.22: a) The source separation is 9.00 m, and the wavelength of the wave is m.0.20 Hz1050.1 m/s1000.3 λ 7 8     f c So there is only one antinode between the sources ),0(  m and it is a perpendicular bisector of the line connecting the sources. b)                            sin m)(20.0 m)00.9( cossincos 2 cos 2 0 2 0 2 0 I λ d III )sin)41.1((cos 2 0  I ;295045580030forSo, 00 I.I,;θI.I,θ  .026090117060 00 I.I,;θI.I,θ  35.23: a) The distance from the central maximum to the first minimum is half the distance to the first maximum, so: m.1088.8 m)102(2.60 m)10(6.60m)700.0( 2 λ 4 4 7        d R y b) The intensity is half that of the maximum intensity when you are halfway to the first minimum, which is m.1044.4 4  Remember, all angles are .small 35.24: a) m,50.2 Hz1020.1 m/s1000.3 λ 8 8     f c and we have: rad.4.52m)8.1( m50.2 2 )( λ 2 21     rr b) .404.0 2 rad52.4 cos 2 cos 0 2 0 2 0 IIII                 35.25: a) To the first maximum: m.1081.3 m101.30 m)10(5.50m)900.0( λ 3 4 7 1        d R y So the distance to the first minimum is one half this, 1.91 mm. b) The first maximum and minimum are where the waves have phase differences of zero and pi, respectively. Halfway between these poi nts, the phase difference between the waves is :So. 2  .W/m1000.2 24 cos 2 cos 26 0 2 0 2 0                 I III  35.26: From Eq. (35.14), .sin λ cos 2 0          d II So the intensity goes to zero when the cosine’s argument becomes an odd integer of     )2/1(sin λ :isThat. 2 m d ),2/1(λsin   md  which is Eq. (35.5). 35.27: By placing the paper between the pieces of glass, the space forms a cavity whose height varies along the length. If twice the height at any given point is one wavelength (recall it has to make a return trip), constructive interference occurs. The distance between the maxima (i.e., the # of meters per fringe) will be 0.0235 rad10095.4 m))1500/1((2 m105.46 arctan 2 λ arctan tan2 λ 2 λ 4 7                      xh l x   35.28: The distance between maxima is cm.0369.0 m)102(8.00 cm)(9.00m)1056.6( 2 λ 5 7       h l x So the number of fringes per centimeter is 1.27 1   x fringes/cm. 35.29: Both parts of the light undergo half-cycle phase shifts when they reflect, so for destructive interference nm.114m1014.1 )42.1(4 m1050.6 4 λ 4 λ 7 7 0      n t 35.30: There is a half-cycle phase shift at both interfaces, so for destructive interference: nm.5.80 4(1.49) nm480 4 λ 4 λ 0  n t 35.31: Destructive interference for 800λ 1  nm incident light. Let n be the refractive index of the oil. There is a 2/λ phase shift for the reflection at the air- oil interface but no phase shift for the reflection at the oil-water interface. Therefore, there is a net 2/λ phase difference due to the reflections, and the condition for destructive interference is )./λ(2 nmt  Smallest nonzero thickness means .λ2so,1 1  tnm The condition for constructive interference with incident wavelength λ is onsoandnm,320λ2,for nm533 λ,1for nm1600 λ,0for nm.800λwhere),/(λλso,λ2But . λ)(2and)/λ)((2 1 2 1 11 2 1 2 1      m m m mtn mtnnmt The visible wavelength for which there is constructive interference is 533 nm. 35.32: a) The number of wavelengths is given by the total extra distance traveled, divided by the wavelength, so the number is .5.36 m106.48 (1.35)m)1076.8(2 λ 2 λ 7 6 0       tnx b) The phase difference for the two parts of the light is zero because the path difference is a half- integer multiple of the wavelength and the top surface reflection has a half-cycle phase shift, while the bottom surface does not. 35.33: Both rays, the one reflected from the pit and the one reflected from the flat region between the pits, undergo the same phase change due to reflection. The condition for destructive interference is ),/λ)((2 2 1 nmt  where n is the refractive index of the plastic substrate. The minimum thickness is for ,0  m and equals m.0.11nm1108)]nm/[(4)(1.790)4/(λ      nt 35.34: A half-cycle phase change occurs, so for destructive interference nm.180 2(1.33) nm480 2 λ 2 λ 0  n t 35.35: a) To have a strong reflection, constructive interference is desired. One part of the light undergoes a half-cycle phase shift, so: . 2 1 nm771 2 1 (1.33)nm)290(2 2 1 2 λ λ 2 1 2                                mmm dn n md For an integer value of zero, the wavelength is not visible (infrared) but for 1  m , the wavelength is 514 nm, which is green. b) When the wall thickness is 340 nm, the first visible constructive interference occurs again for         2 1 nm904 λyieldsand1 m m = 603 nm, which is orange. 35.36: a) Since there is a half-cycle phase shift at just one of the interfaces, the minimum thickness for constructive interference is: nm.3.74 4(1.85) nm550 4 λ 4 λ 0  n t b) The next smallest thickness for constructive interference is with another half wavelength thickness added:   nm.223 )85.1(4 nm5503 4 λ3 4 λ3 0  n t 35.37: mm.0.570m1070.5 2 )m1033.6(1800 2 λ 4 7      m x 35.38: a) For Jan, the total shift was m.1048.2 2 m)1006.6(818 2 1λ 4 7 1      m x For Linda, the total shift was m.1005.2 2 m)1002.5(818 2 λ 4 72 2      m x b) The net displacement of the mirror is the difference of the above values: mm.0.043mm0.205mm248.0 21  xxx 35.39: Immersion in water just changes the wavelength of the light from Exercise 35.11, so: mm,626.0 1.33 mm833.0λ vacuum  n y dn R y using the solution from Exercise 35.11. 35.40: Destructive interference occurs 1.7 m from the centerline. 22 1 m)2.6(m)0.12( r =13.51 m 22 2 m)8.2(m)0.12( r =12.32 m For destructive interference, m19.12/λ 21  rr and m.4.2λ  The wavelength we have calculated is the distance between the wave crests. Note: The distance of the person from the gaps is not large compared to the separation of the gaps, so the path length is not accurately given by .sin  d 35.41: a) Hearing minimum intensity sound means that the path lengths from the individual speakers to you differ by a half-cycle, and are hence out of phase by  180 at that position. b) By moving the speakers toward you by 0.398 m, a ma ximum is heard, which means that you moved the speakers one-half wavelength from the min and the signals are back in phase. Therefore the wavelength of the signals is 0.796 m, and the frequency is m0.796 m/s340 λ  v f =427 Hz. c) To reach the next maximum, one must move an additional distance of one wavelength, a distance of 0.796 m. 35.42: To find destructive interference, λ 2 1 )200( 22 12        mxxmrrd .λ 2 1 2 1 λ 2 1 m000,20 λ 2 1 2 λ 2 1 )m200( 2 2 222                                    m m x mxmxx The wavelength is calculated by .m7.51 Hz1080.5 sm1000.3 λ 6 8     f c . m0.20;3andm,1.90:2andm,219:1andm,761:0          xmxmxmxm 35.43: At points on the same side of the centerline as point ,A the path from B is longer than the path from ,A and the path difference d sin θ puts speaker A ahead of speaker B in phase. Constructive interference occurs when     ,2,1,0,2381.0 3 2 λ 3 2 sin ,2,1,0, λ 2 1 6 λsin                        mmdm mmd    ,4;8.60,3;4.39,2;4.23,1;13.9,0          mmmmm no solution At points on the other side of the centerline, the path from A is longer than the path from B , and the path difference d sin θ puts speaker A behind speaker B in phase. Constructive interference occurs when     , 2,1,0,2381.0 3 1 λ 3 2 sin ,2,1,0, λ 2 1 6 λsin                        mmdm mmd   ,4;5.52,3;7.33,2;5.18,1;55.4,0          mmmmm no solution [...]... The path difference must now retard A’s sound by 5 1 4 λ, 4 λ,  1 5  d sin   λ, λ,  gives  7.2,  38.5 4 4 sin  2  35. 45: a) If the two sources are out of phase by one half-cycle, we must add an extra half a wavelength to the path difference equations Eq (35. 1) and Eq (35. 2) This exactly changes one for the other, for m  m  1 and m  1  m, since m 2 2 in any integer b) If one source leads...  0.910 mm 2 So the diameter of the third bright ring is 1.82 mm 35. 58: As found in Problem (35. 51), the radius of the mth bright ring is in general: (2m  1)λR r , 2 for R  λ Introducing a liquid between the lens and the plate just changes the λ wavelength from λ  n So: 0.850 mm (2m  1)λR r r ( n)     0.737 mm 2n 1.33 n 35. 59: a) Adding glass over the top slit increases the effective path... is the same for both types of light 35. 51: First we need to find the angles at which the intensity drops by one-half from the value of the m th bright fringe d m d   d  I I  I 0 cos 2  sin    0  sin    (m  1 2) 2 λ λ  λ  2 λ 3λ λ   ; m  1:  m   m  0 :  m    m  4d 4d 2d  so there is no dependence on the m - value of the fringe 35. 52: There is just one half-cycle... 2 n  35. 53: a) There is a half-cycle phase change at the glass, so for constructive interference: 2 1  x  2 d  x  2 h     x   m  λ 2 2  2 1   x 2  4h 2  x   m  λ 2  Similarly for destructive interference: x 2  4h 2  x  mλ b) The longest wavelength for constructive interference is when m  0 : λ x 2  4h 2  x  m 1 2 (14 cm) 2  4(24 cm)2  14 cm  72 cm 12 35. 54:...  ( y  d ) 2  x 2  ( y  d ) 2  (m  1 )λ 2 2 2 35. 48: a) E p  E12  E 2  2 E1 E 2 cos(   )  E 2  4 E 2  4 E 2 cos   5E 2  4 E 2 cos  I  5   4   1  0 cE p 2   0 c  E 2    E 2  cos   2  2   2   9   0  I 0   0 cE 2 2 5 4  So I  I 0   cos   9 9  1 b) I min  I 0 which occurs when   n (n odd) 9 35. 49: For this film on this glass, there is a net... occur at (d sin   L(n  1))  m  d sin   mλ  L(n  1) λ 35. 60: The passage of fringes indicates an effective change in path length, since the wavelength of the light is getting shorter as more gas enters the tube 2L 2L 2L mλ m    (n  1)  (n  1)  λn λ λ 2L So here: 48(5.46  10 7 m) (n  1)   2.62  10 4 2(0.0500 m) 35. 61: There are two effects to be considered: first, the expansion... be adjusted for both 2 destructive and constructive interference, by this amount So for constructive inference: r1  r2  (m   2 )λ, and for destructive interference, r1  r2  (m  1 2   2 )λ 35. 46: a) The electric field is the sum of the two wave functions, and can be written: E p (t)  E2 (t )  E1 (t )  E cos(t )  E cos(t   )  E p (t )  2 Ecos( / 2) cos(ωt   / 2) b) E p (t ) ... cycle from both of the original waves e) The instantaneous magnitude of the Poynting vector is: | S |  0 cE p (t )   0 c(4 E 2 cos 2 ( 2) cos 2 (t   2)) 2 For a time average, cos 2 (t   2)  35. 47: a) 1 , so | S av | 2 0 cE 2 cos 2 ( 2) 2 r  mλ r1  x 2  ( y  d ) 2 r2  x 2  ( y  d ) 2 So r  x 2  ( y  d ) 2  x 2  ( y  d ) 2  mλ b) The definition of hyperbola is the locus... the path length in the layers change (always to a larger value than the normal incidence case) If the path length changes, then so do the wavelengths that will interfere constructively upon reflection 35. 55: a) Intensified reflected light means we have constructive interference There is one half-cycle phase shift, so: 2tn 2(485 nm) (1.53) 1484 nm 1 λ  2t   m    λ    1 (m  2 ) (m  1 ) (m... nm (m  3) is the only wavelength of visible light that is intensified We could also think of this as the result of internal reflections interfering with the outgoing ray without any extra phase shifts 35. 56: a) There is one half-cycle phase shift, so for constructive interference: 2tn 2(380 nm) (1.45) 1102 nm 1 λ  2t   m   0  λ    1 (m  2 ) (m  1 ) (m  1 ) 2 n  2 2 Therefore, we have . the same as part (a). 35. 7: Destructive interference occurs for: .nm453 5.4 nm2040 λandnm358 5.3 nm2040 λ 21 λ 43    m d 35. 8: a) For the number. are .3,2,1,0for2.54,4 .35, 3.20,65.6            m 35. 9: .m105.90 m2.20 m)10(2.82m)10(4.60 λ λ 7 34        R yd d R y 35. 10: For bright