Tài liệu Physics exercises_solution: Chapter 34 pptx

34.1: If up is the y  -direction and right is the x  -direction, then the object is at ),,(atis),,( 00200 yxPyx    and mirror 1 flips the y -values, so the image is at ),( 00 yx which is . 3 P  34.2: Using similar triangles, m.3.24 m0.350 m0.350m28.0 m040.0 mirror tree mirrortree mirror tree mirror tree    d d hh d d h h 34.3: A plane mirror does not change the height of the object in the image, nor does the distance from the mirror change. So, the image is cm2.39 to the right of the mirror, and its height is cm.85.4 34.4: a) cm.0.17 2 cm0.34 2  R f b) If the spherical mirror is immersed in water, its focal length is unchanged—it just depends upon the physical geometry of the mirror. 34.5: a) b) cm,0.33 cm5.16 1 cm0.22 21111        s sfss to the left of the mirror. cm,20.1 cm16.5 cm33.0 cm)600.0(     s s yy and the image is inverted and real. 34.6: a) b) cm,60.6 cm5.16 1 cm0.22 21111        s sfss to the right of the mirror. cm,240.0 cm16.5 cm).606( cm)600.0(       s s yy and the image is upright and virtual. 34.7: m.75.1 m1058.5 1 m75.1 11111 10          s sfss m.1013.2)m106794)(1014.3( 1014.3 1058.5 75.1 4311 11 10         myym 34.8: cm40.1 cm0.21 1 cm00.3 21111 ,cm00.3        s sfss R (in the ball). The magnification is .0667.0 cm0.21 cm40.1      s s m 34.9: a) .Also. 111111 sf f s s m fs sf s fs fs sfsfss               b) For ,0,0      sfsf so the image is always on the outgoing side and is real. The magnification is ,0   sf f m since .sf  c) For ,12    f f mfs which means the image is always smaller and inverted since the magnification is negative. For .102  f f mffsfsf d) Concave mirror: ,00      sfs and we have a virtual image to the right of the mirror. ,1 f f m so the image is upright and larger than the object. 34.10: For a convex mirror, .00        fs fs fs sf sf Therefore the image is always virtual. Also ,0        sf f sf f sf f m so the image is erect, and ,since1 fsfm  so the image is smaller. 34.11: a) b) .0,for0     sfss c) .0for0 fss     d) If the object is just outside the focal point, then the image position approaches positive infinity. e) If the object is just inside the focal point, the image is at negative infinity, “behind” the mirror. f) If the object is at infinity, then the image is at the focal point. g) If the object is next to the mirror, then the image is also at the mirror. h) i) The image is erect if .fs  j) The image is inverted if .fs  k) The image is larger if .20 fs   l) The image is smaller if .0or2   sfs m) As the object is moved closer and closer to the focal point, the magnification INCREASES to infinite values. 34.12: a) a) 0.for0   sfs b) .0andsfor0   sfs c) If the object is at infinity, the image is at the outward going focal point. d) If the object is next to the mirror, then the image is also at the mirror. For the answers to (e), (f), (g), and (h), refer to the graph on the next page. e) The image is erect (magnification greater than zero) for .fs  f) The image is inverted (magnification less than zero) for .fs  g) The image is larger than the object (magnification greater than one) for .02  sf h) The image is smaller than the object (magnification less than one) for .2and0 fss  34.13: a) b) cm,45.5 cm0.12 1 cm0.20 21111        s sfss to the right of the mirror. cm,409.0 cm12.0 cm5.45 cm)9.0(       s s yy and the image is upright and virtual. 34.14: a) ,00.4 0 . 12 0.48      s s m where s  comes from part (b). b) cm.0.48 cm0.12 1 cm0.32 21111        s sfss Since s  is negative, the image is virtual. c) 34.15: cm.67.20 00.1 cm50.3 309.1 0        s ss n s n ba 34.16: a) cm,26.50 00.1 cm00.7 33.1 0        s ss n s n ba so the fish appears cm26.5 below the surface. b) cm,8.240 00.1 cm0.33 33.1 0        s ss n s n ba so the image of the fish appears cm8.24 below the surface. 34.17: a) For ,ndwith,and0             baba annR we have: .)()(  abba b a bbabb nnnn n n nn  But ,and,, R h s h s h      so subbing them in one finds: . )( R nn s n s n abba     Also, the magnification calculation yields: .tanandtan sn sn y y m s yn s yn s y s y b aba ba               b) For                 abbaba nnnnR :havewe,andwith,and0 , )( and,,But.)( R nn s n s n R h s h s h nn abba ab          so subbing them in one finds: . )( R nn s n s n abba     Also, the magnification calculation yields: .tantan sn sn y y m s yn s yn nn a aba bbaa         34.18: a) cm.00.8 cm00.3 60.060.11            s sR nn s n s n abba b) cm.7.13 cm00.3 60.060.1 cm0.12 1          s sR nn s n s n abba c) cm.33.5 cm00.3 60.060.1 cm00.2 1          s sR nn s n s n abba 34.19: . )( )()( )( b n Rs Rs s s n Rs Rs Rs sR n R nn s n s n ba abba               .52.1)60.1( cm)3.00cm0.90( cm)3.00cm601( cm160 cm.090     a n 34.20: cm.8.14 cm00.4 60.060.1 cm0.24 1          s sR nn s n s n abba mm,0.578mm50.1 )cm0.24)(60.1( cm8.14                        y sn sn y b a so the image height is mm,578.0 and is inverted. 34.21: cm.35.8 cm00.4 60.060.1 cm0.24 1            s sR nn s n s n abba mm,0.326mm50.1 )cm0.24)(60.1( cm)35.8(                        y sn sn y b a so the image height is mm,326.0 and is erect. 34.22: a) cm,0.14 cm0.14 33.000.1 cm0.14 33.1             s sR nn s n s n abba so the fish appears to be at the center of the bowl. .33.1 cm)0.17)(00.1( cm)0.17)(33.1(                       sn sn m b a b) cm,4.56 cm0.14 33.033.100.1              s sR nn s n s n abba which is outside the bowl. 34.23: For :cm18  s a) cm.0.63 cm0.18 1 cm0.14 11111        s sfss b) .50.3 0 . 18 0.63    s s m c) and d) From the magnification, we see that the image is real and inverted. For :cm00.7  s a) cm.0.14 cm00.7 1 cm0.14 11111        s sfss b) .00.2 00 . 7 0.14      s s m c) and d) From the magnification, we see that the image is virtual and erect. 34.24: a) cm,0.48 cm0.12 1 cm0.16 11111      f ffss and the lens is diverging. b) cm,638.0 16.0 12.0)( cm)850.0(                   s s yy and is erect. c) 34.25: .25.3 400.0 30.1 s s y y m     Also: cm75.1525.31 cm00.7 1 cm00.7 11111            s s s s ssfss (to the left). cm,85.4 cm75.15 1 cm0.7 11    s s and the image is virtual ).0since(   s 34.26: :Also.711.0 20.3 50.4 ss s s y y m       cm217 0.90 11 711.0 1111          s ssfss (to the right). cm,154cm)217(711.0    s and the image is real (since ).0   s 34.27:                       cm50.3 1 cm00.5 1 )48.0( 1 cm0.18 111 )1( 11 21 sRR n ss cm3.10     s (to the left of the lens). [...]... (3000)  0.127 rad 34. 55: a) M   34. 56: f 1  f 2  d ss  f1  d ss  f 2  1.80 m  0.0900 m  1.71 m 171 f M  1   19.0 9.00 f2 34. 57: f  19.0  f  (19.0) D  (19.0)(1.02 m)  19.4 m D 34. 58: y  y s f    3  y  θf  (0.014) (18 m)  4.40  10 m s s  180  R  0.650 m  d  f 1  f 2  0.661 m 2 f 0.650 m b) M  1   59.1 f 2 0.011 m 34. 59: a) f 1  1 1 1 1 1 1  34. 60:   ... fit on the 24-mm  s 3.90 36-mm film 34. 36: 34. 37: 1 1 1 1 1 1       s  1020 cm s s f s 20.4 cm 20.0 cm 34. 38: y    5.00 m s f y y (70.7 m)  0.0372 m  37.2 mm s s 9.50  10 3 m 28 mm s f  m  1.4  10 4 s s 200,000 mm 105 mm s f b) m    m   5.3  10  4 s s 200,000 mm 300 mm s f c) m    m   1.5  10 3 s s 200,000 mm 34. 39: a) m  34. 40: a) s1    s  f1  12 cm 1... Again, for Ex (34. 10) and (34. 12), the change from a convex mirror to a diverging lens changes nothing in the exercises, except for the interpretation of the location of the images, as explained in part (b) above 1 1 1 1 1 1       s  7.0 cm s s f s 12.0 cm  17.0 cm s (17.0) y  0.800 cm m    2.4  y   0 .34 cm, so the object is  2 4 s 7.2 m 0 .34 cm tall, erect, same side 34. 32: 1 1... “expected.” 34. 66: 34. 67: a) R  0 and s  0, so a real image ( s   0) is produced for virtual object positions between the focal point and vertex of the mirror So for a 24.0 cm radius mirror, the virtual object positions must be between the vertex and 12.0 cm to the right of the s s mirror b) The image orientation is erect, since m      0 s s c) 34. 68: The derivations of Eqs (34. 6) and (34. 7)... cm 17.0 cm s  17.0 y  0.800 cm m   0.646  y    1.24 cm tall, erect, same side  26.3 s m 0.646 34. 33: 1 1 1 1 1 1       f  11.1 cm, converging s s f f 16.0 cm 36.0 cm  s   36  b) y  y      (0.80 cm)     1.8 cm, so the image is inverted  s  16  c) 34. 34: a) 34. 35: a) m   0.024 m  y s y   s  s  600 m   240 m   0.0600 m  60 mm  y s y   1 1 1 1 1... b)  power   1.67 diopters f s s    0.600 m f 34. 47: a) 25.0 cm 25.0 cm   4.17 6.00 cm f 1 1 1 1 1 1 b)       s  4.84 cm s s f s  25.0 cm 6.00 cm 34. 48: a) Angular magnification M  1 1 1 1 1 1       s  6.06 cm s s f s  25.0 m 8.00 cm s  25.0 cm  4.13  y   ym  (1.00 mm)(4.13)  4.13 mm b) m   s 6.06 cm 34. 49: a) 34. 50:   y y 2.00 mm  f    80.0 mm  8.00 cm f...  f  nliq   n  nliq       34. 99: The image formed by the converging lens is 30.0 cm from the converging lens, and becomes a virtual object for the diverging lens at a position 15.0 cm to the right of the diverging lens The final image is projected 15  19.2  34. 2 cm from the diverging lens 1 1 1 1 1 1       f  26.7 cm s s f  15.0 cm 34. 2 cm f 34. 100: The first image formed by the... s   6.50s      s s s  s  6.50s 4.00 cm 1 1  1  1   s  3.38 cm, s   6.50s  22.0 cm  s  6.50  4.00 34. 51: m    (250 mm)s1 (250 mm)(160 mm  5.0 mm)   317 (5.00 mm)(26.0 mm) f1 f 2 y y  0.10 mm   3.15  10  4 mm b) m   y  y 317 m 34. 52: a) M  34. 53: a) The image from the objective is at the focal point of the eyepiece, so  s1  d oe  f 2  19.7 cm  1.80 cm  17.9... light entered the aperture, and the film must be exposed four times as long for the correct exposure 34. 41: a) f 4  4  34. 42: The square of the aperture diameter (~ the area) is proportional to the length of the 2  1   8 mm   1  exposure time required  s    23.1 mm    250 s    30      34. 43: a) A real image is formed at the film, so the lens must be convex 1 1 1 1 s f sf and s ... length) and have a radius of curvature equal to 3.0 m 34. 61: If you move away from the mirror at 2.40 m s, then your image moves away from the mirror at the same speed, but in the opposite direction Therefore you see the image receding at 4.80 m s, the sum of your speed and that of the image in the mirror 34. 62: a) There are three images formed 34. 63: The minimum length mirror for a woman to see her . in Fig. (34. 29b). 34. 31: a) The lens equation is the same for both thin lenses and spherical mirrors, so the derivation of the equations in Ex. (34. 9) is. the lens. c) Again, for Ex. (34. 10) and (34. 12), the change from a convex mirror to a diverging lens changes nothing in the exercises, except for the interpretation