Física para ciencias e ingenieria serway 7ed vol 2 (solucionario)

http://www.elsolucionario.blogspot.com LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS Chapter 15 Electric Forces and Electric Fields Quick Quizzes (b) Object A must have a net charge because two neutral objects not attract each other Since object A is attracted to positively-charged object B, the net charge on A must be negative (b) By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other (c) The electric field at point P is due to charges other than the test charge Thus, it is unchanged when the test charge is altered However, the direction of the force this field exerts on the test change is reversed when the sign of the test charge is changed (a) If a test charge is at the center of the ring, the force exerted on the test charge by charge on any small segment of the ring will be balanced by the force exerted by charge on the diametrically opposite segment of the ring The net force on the test charge, and hence the electric field at this location, must then be zero (c) and (d) The electron and the proton have equal magnitude charges of opposite signs The forces exerted on these particles by the electric field have equal magnitude and opposite directions The electron experiences an acceleration of greater magnitude than does the proton because the electron’s mass is much smaller than that of the proton (a) The field is greatest at point A because this is where the field lines are closest together The absence of lines at point C indicates that the electric field there is zero (c) When a plane area A is in a uniform electric field E, the flux through that area is Φ E = EA cosθ where θ is the angle the electric field makes with the line normal to the plane of A If A lies in the xy-plane and E is in the z-direction, then θ = 0° and Φ E = EA = ( 5.00 N C ) ( 4.00 m ) = 20.0 N ⋅ m C (b) If θ = 60° in Quick Quiz 15.7 above, then Φ E = EA cosθ = ( 5.00 N C ) ( 4.00 m ) cos ( 60° ) = 10.0 N ⋅ m C (d) Gauss’s law states that the electric flux through any closed surface is equal to the net enclosed charge divided by the permittivity of free space For the surface shown in Figure 15.28, the net enclosed charge is Q = −6 C which gives Φ E = Q ∈0 = − ( C ) ∈0 10 CHAPTER 15 (b) and (d) Since the net flux through the surface is zero, Gauss’s law says that the net change enclosed by that surface must be zero as stated in (b) Statement (d) must be true because there would be a net flux through the surface if more lines entered the surface than left it (or vise-versa) Electric Forces and Electric Fields Answers to Even Numbered Conceptual Questions Conducting shoes are worn to avoid the build up of a static charge on them as the wearer walks Rubber-soled shoes acquire a charge by friction with the floor and could discharge with a spark, possibly causing an explosive burning situation, where the burning is enhanced by the oxygen Electrons are more mobile than protons and are more easily freed from atoms than are protons No Object A might have a charge opposite in sign to that of B, but it also might be neutral In this latter case, object B causes object A to be polarized, pulling charge of one sign to the near face of A and pushing an equal amount of charge of the opposite sign to the far face Then the force of attraction exerted by B on the induced charge on the near side of A is slightly larger than the force of repulsion exerted by B on the induced charge on the far side of A Therefore, the net force on A is toward B +++ + + B + + +++ –+ – +A –+ If the test charge was large, its presence would tend to move the charges creating the field you are investigating and, thus, alter the field you wish to investigate 10 She is not shocked She becomes part of the dome of the Van de Graaff, and charges flow onto her body They not jump to her body via a spark, however, so she is not shocked 12 An electric field once established by a positive or negative charge extends in all directions from the charge Thus, it can exist in empty space if that is what surrounds the charge 14 No Life would be no different if electrons were positively charged and protons were negatively charged Opposite charges would still attract, and like charges would still repel The designation of charges as positive and negative is merely a definition 16 The antenna is similar to a lightning rod and can induce a bolt to strike it A wire from the antenna to the ground provides a pathway for the charges to move away from the house in case a lightning strike does occur 18 (a) If the charge is tripled, the flux through the surface is also tripled, because the net flux is proportional to the charge inside the surface (b) The flux remains constant when the volume changes because the surface surrounds the same amount of charge, regardless of its volume (c) The flux does not change when the shape of the closed surface changes (d) The flux through the closed surface remains unchanged as the charge inside the surface is moved to another location inside that surface (e) The flux is zero because the charge inside the surface is zero All of these conclusions are arrived at through an understanding of Gauss’s law 20 (a) –Q (b) +Q (c) (d) (e) +Q (See the discussion of Faraday’s ice-pail experiment in the textbook.) 22 CHAPTER 15 The magnitude of the electric force on the electron of charge e due to a uniform electric G field E is F = eE Thus, the force is constant Compare this to the force on a projectile of mass m moving in the gravitational field of the Earth The magnitude of the gravitational force is mg In both cases, the particle is subject to a constant force in the vertical direction and has an initial velocity in the horizontal direction Thus, the path will be the same in each case—the electron will move as a projectile with an acceleration in the vertical direction and constant velocity in the horizontal direction Once the electron leaves the region between the plates, the electric field disappears, and the electron continues moving in a straight line according to Newton’s first law Electric Forces and Electric Fields Answers to Even Numbered Problems 5.71 × 1013 C F = 1.91( ke q2 a2 ) along the diagonal toward the negative charge 2.25 × 10 −9 N m 5.08 m 10 F6 = 46.7 N ( left ) , F1.5 = 157 N ( right ) , F−2 = 111 N ( left ) 12 3.89 × 10 −7 N at 11.3° below + x axis 14 x = 0.634 d , stable if third bead has positive charge 16 1.45 m beyond the –3.00 nC charge 18 (a) 2.00 × 107 N C to the right 20 (a) 5.27 × 1013 m s 22 1.63 × 10 N C directed opposite to the proton’s velocity 24 1.88 × 10 N C at 4.40° below + x axis 26 at y = + 0.85 m 28 (a) 34 (a) zero 36 ~ µm 38 (a) 40 (a) –55.7 nC (b) negative, with a spherically symmetric distribution 42 5.65 × 10 N ⋅ m C 48 (a) 8.2 ì 10 N 50 5.25 C q1 q2 = − 2.0 × 10 N ⋅ m C (b) (b) 40.0 N to the left 5.27 × 10 m s (b) q2 > 0, q1 < (b) 1.8 × 10 N C (b) (b) 2.2 × 10 m s (c) 1.1 × 10 N C CHAPTER 15 3.43 µ C 52 (a) downward 54 2.51 × 10 −10 56 (a) 0.307 s (b) Yes, the absence of gravity produces a 2.28% difference 60 (d) 62 2.0 µ C 64 (a) (b) JG E = at x = +9.47 m, y = 37.0° or 53.0° (b) 1.66 × 10 −7 s and 2.21 × 10 −7 s Electric Forces and Electric Fields Problem Solutions 15.1 Since the charges have opposite signs, the force is one of attraction Its magnitude is F= 15.2 k e q1 q2 r2 −9 −9  N ⋅ m  ( 4.5 × 10 C )( 2.8 × 10 C ) =  8.99 × 109 = 1.1 × 10 −8 N  2 C  ( 3.2 m )  The electrical force would need to have the same magnitude as the current gravitational force, or ke q2 M m = G E moon r r q= giving GME mmoon ke This yields q= 15.3 F= ( 6.67 × 10 −11 N ⋅ m kg )( 5.98 × 10 24 kg )( 7.36 × 10 22 kg ) 8.99 × 10 N ⋅ m C 2 k e ( e )( 79 e ) r2 ( 158 ) ( 1.60 × 10 −19 C )  N⋅m  =  8.99 × 10 = 91 N  C2   ( 2.0 × 10−14 m ) ( repulsion ) = 5.71 × 1013 C 15.4 CHAPTER 15 F1 = F2 = and and F3 = ke q (a ) = ur F1 ke q 2a2 k q  ke q kq + e ( 0.707 ) = 1.35  e  2a a  a  ΣFy = F1 + F3 sin 45° =  ke q2  ke q2 ke q2 + = 0.707 1.35 ( )   a2 2a2  a  ( ΣFx ) ( ) + ΣFy 2 = 1.91 2 +q a ur F3 45° a ur F2  ΣFy  ke q2 −1 and θ = tan −1   = tan ( 1) = 45° a F Σ  x G  k q2  F R = 1.91  e  along the diagonal toward the negative charge  a  so (a) ke q a2 –q a r ΣFx = F2 + F3 cos 45° = FR = 15.5 –q The attractive forces exerted on the positive charge by the negative charges are shown in the sketch and have magnitudes F= k 2e  e( ) r2 N ⋅ m2 =  8.99 × 10 C2  −19     ( 1.60 × 10 )  = 36.8 N   ( 5.00 × 10 −15 m ) (b) The mass of an alpha particle is m = 4.002 u , where u = 1.66 × 10 −27 kg is the unified mass unit The acceleration of either alpha particle is then a= F 36.8 N = = 5.54 × 10 27 m s m 4.002 ( 1.66 × 10 −27 kg ) –q Electric Forces and Electric Fields 15.6 The attractive force between the charged ends tends to compress the molecule Its magnitude is −19 k ( 1e )  N ⋅ m  ( 1.60 × 10 C ) F = e =  8.99 × 109 = 4.89 × 10 −17 N  2 − r C  ( 2.17 × 10 m )  The compression of the “spring” is x = ( 0.010 ) r = ( 0.010 ) ( 2.17 × 10 −6 m ) = 2.17 × 10 −8 m , so the spring constant is k = 15.7 F 4.89 × 10 −17 N = = 2.25 × 10 −9 N m x 2.17 × 10 −8 m 1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton and one electron Thus, each charge has magnitude q = N A e The distance separating these charges is r = RE , where RE is Earth’s radius Thus, F= ( ke N A e ( RE ) ) 2  N ⋅ m2 =  8.99 × 10 C2  15.8 The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus, k e e r = me g or 15.9 −19 23  ( 6.02 × 10 )( 1.60 × 10 C )  = 5.12 × 10 N   ( 6.38 × 10 m ) k e2 r= e = me g ( 8.99 × 10 N ⋅ m C )(1.60 × 10 C ) ( 9.11 × 10 kg )( 9.80 m s ) −31 −19 2 = 5.08 m (a) The spherically symmetric charge distributions behave as if all charge was located at the centers of the spheres Therefore, the magnitude of the attractive force is F= k e q1 q2 r2 −9 −9  N ⋅ m  ( 12 × 10 C )( 18 × 10 C ) =  8.99 × 109 = 2.2 × 10 −5 N  2 C ( 0.30 m )   348 CHAPTER 25 25.22 (a) The lateral magnification produced by the objective lens of a good compound microscope is closely approximated by M1 ≈ − L fO , where L is the length of the microscope tube and fO is the focal length of this lens Thus, if L = 20.0 cm and M1 = −50.0 (inverted image), the focal length of the objective lens is fO ≈ − L 20.0 cm =− = + 0.400 cm M1 −50.0 (b) When the compound microscope is adjusted for most comfortable viewing (with parallel rays entering the relaxed eye), the angular magnification produced by the eyepiece lens is me = 25 cm f e If me = 20.0 , the focal length of the eyepiece is fe = 25.0 cm 25.0 cm = = + 1.25 cm me 20.0 (c) The overall magnification is 25.23 m = M1 me = ( −50.0 )( 20.0 ) = − 000  25 cm  The overall magnification is m = M1 me = M1    fe  where M1 is the magnification produced by the objective lens Therefore, the required focal length for the eye piece is fe = 25.24 M1 ( 25 cm ) ( − 12 ) ( 25 cm ) = = 2.1 cm m − 140 Note: Here, we need to determine the overall lateral magnification of the microscope, M = he′ h1 where he′ is the size of the image formed by the eyepiece, and h1 is the size of the object for the objective lens The lateral magnification of the objective lens is M1 = h1′ h1 = − q1 p1 and that of the eyepiece is Me = he′ he = − qe pe Since the object of the eyepiece is the image formed by the objective lens, he = h1′ , and the overall lateral magnification is M = M1 Me Using the thin lens equation, the object distance for the eyepiece is found to be pe = qe f e ( − 29.0 cm ) ( 0.950 cm ) = 0.920 cm = qe − f e − 29.0 cm − 0.950 cm and the magnification produced by the eyepiece is Me = − qe ( − 29.0 cm ) = + 31.5 =− 0.920 cm pe Optical Instruments 349 The image distance for the objective lens is then q1 = L − pe = 29.0 cm − 0.920 cm = 28.1 cm and the object distance for this lens is p1 = q1 f o ( 28.1 cm )( 1.622 cm ) = = 1.72 cm 28.1 cm − 1.622 cm q1 − f o The magnification by the objective lens is given by M1 = − q1 ( 28.1 cm ) =− = − 16.3 p1 1.72 cm and the overall lateral magnification is M = M1 Me = ( − 16.3 )( + 31.5 ) = − 514 The lateral size of the final image is he′ = qe ⋅ θ = ( 29.0 cm ) ( 1.43 × 10 −3 rad ) = 4.15 × 10 −2 cm and the size of the red blood cell serving as the original object is h1 = 25.25 he′ 4.15 × 10 −4 m = = 8.06 × 10 −7 m = 0.806 µ m M 514 Some of the approximations made in the textbook while deriving the overall magnification of a compound microscope are not valid in this case Therefore, we start with the eyepiece and work backwards to determine the overall magnification If the eye is relaxed, the eyepiece image is at infinity ( qe → − ∞ ) , so the object distance is pe = f e = 2.50 cm , and the angular magnification by the eyepiece is me = 25.0 cm 25.0 cm = = 10.0 2.50 cm fe The image distance for the objective lens is then, q1 = L − pe = 15.0 cm − 2.50 cm=12.5 cm and the object distance is p1 = q1 f o (12.5 cm ) (1.00 cm ) = 1.09 cm = 12.5 cm − 1.00 cm q1 − f o 350 CHAPTER 25 The magnification by the objective lens is M1 = − q1 ( 12.5 cm ) = − 11.5 , and the =− 1.09 cm p1 overall magnification of the microscope is m = M1 me = ( − 11.5 ) ( 10.0 ) = − 115 25.26 The moon may be considered an infinitely distant object ( p → ∞ ) when object w viewed with this lens, so the image distance will be q = f o = 500 cm Considering the rays that pass undeviated through the center of this lens as shown in the sketch, observe that the angular widths of the image and the object are equal Thus, if w is the linear width of an object forming a 1.00 cm wide image, then θ= or 25.27 image q 1.0 cm q 3.8 ´ 108 m fo w 1.0 cm 1.0 cm = = 3.8 × 10 m 500 cm fo  1.0 cm  mi  w = ( 3.8 × 108 m )    = 1.6 × 10 mi  500 cm  609 m  The length of the telescope is L = f o + f e = 92 cm and the angular magnification is m= fo = 45 fe Therefore, f o = 45 f e and L = f o + f e = 45 f e + f e = 46 f e = 92 cm , giving f e = 2.0 cm 25.28 and f o = 92 cm − f e or f o = 90 cm Use the larger focal length (lowest power) lens as the objective element and the shorter focal length (largest power) lens for the eye piece The focal lengths are fo = 1 = + 0.833 m , and f e = = + 0.111 m + 1.20 diopters + 9.00 diopters 351 Optical Instruments (a) The angular magnification (or magnifying power) of the telescope is then m= f o + 0.833 m = = 7.50 f e + 0.111 m (b) The length of the telescope is L = f o + f e = 0.833 m + 0.111 m = 0.944 m 25.29 (a) From the thin lens equation, q = lens is M = h′ h = − q p = − f h′ = M h = − pf , so the lateral magnification by the objective p− f ( p − f ) Therefore, the image size will be fh fh = p− f f −p (b) If p >> f , then f − p ≈ − p and h′ ≈ − fh p (c) Suppose the telescope observes the space station at the zenith Then, 25.30 h′ ≈ − fh ( 4.00 m )( 108.6 m ) =− = − 1.07 × 10 -3 m = − 1.07 mm 407 × 10 m p (b) The objective forms a real, diminished, inverted image of a very distant object at q1 = f o This image is a virtual object for the eyepiece at pe = − f e , q0 q0 F0 F0 h¢ I giving 1 1 = − = + =0 qe p e f e − f e fe and qe → ∞ (a) Parallel rays emerge from the eyepiece, so the eye observes a virtual image L1 Fe q Fe O L1 352 CHAPTER 25 (c) The angular magnification is m = fo = 3.00 , giving fe f o = 3.00 f e Also, the length of the telescope is L = f o + f e = 3.00 f e − f e = 10.0 cm , giving fe = − fe = − 25.31 10.0 cm = − 5.00 cm and f o = 3.00 f e = 15.0 cm 2.00 The lens for the left eye forms an upright, virtual image at qL = − 50.0 cm when the object distance is pL = 25.0 cm , so the thin lens equation gives its focal length as fL = ( 25.0 cm ) ( − 50.0 cm ) pL qL = = 50.0 cm 25.0 cm − 50.0 cm pL + qL Similarly for the other lens, qR = − 100 cm when pR = 25.0 cm , and f R = 33.3 cm (a) Using the lens for the left eye as the objective, m= f o f L 50.0 cm = = = 1.50 f e f R 33.3 cm (b) Using the lens for the right eye as the eyepiece and, for maximum magnification, requiring that the final image be formed at the normal near point ( qe = − 25.0 cm ) gives pe = qe f e ( − 25.0 cm ) ( 33.3 cm ) = + 14.3 cm = qe − f e − 25.0 cm − 33.3 cm The maximum magnification by the eyepiece is then me = + 25.0 cm 25.0 cm = 1+ = + 1.75 33.3 cm fe and the image distance for the objective is q1 = L − pe = 10.0 cm − 14.3 cm = − 4.30 cm Optical Instruments The thin lens equation then gives the object distance for the objective as p1 = q1 f1 ( − 4.30 cm ) ( 50.0 cm ) = + 3.94 cm = q1 − f1 − 4.30 cm − 50.0 cm The magnification by the objective is then M1 = − q1 ( − 4.30 cm ) = + 1.09 =− 3.94 cm p1 and the overall magnification is m = M1 me = ( + 1.09 )( + 1.75 ) = 1.90 25.32 The angular resolution needed is s r 300 m = 7.9 × 10 −7 rad 3.8 × 10 m θ = = For a circular aperture θ = 1.22 so 25.33 D = 1.22 λ θ D  500 × 10 −9 m  = 1.22   = 0.77 m (about 30 inches) −7  7.9 × 10 rad  If just resolved, the angular separation is θ = θ = 1.22 Thus, the altitude is 25.34 λ λ  500 × 10 −9 m  −6 = 1.22   = 2.03 × 10 rad D 0.300 m   h= d θ = 1.00 m = 4.92 × 10 m = 492 km −6 2.03 × 10 rad For a narrow slit, Rayleigh’s criterion gives θ = λ a = 500 × 10 −9 m = 1.00 × 10 −3 = 1.00 mrad −3 0.500 × 10 m 353 354 CHAPTER 25 25.35 The limit of resolution in air is θ air = 1.22 λ D = 0.60 µ rad In oil, the limiting angle of resolution will be or 25.36 θ oil θ oil = 1.22 = λoil θ noil D air = 1.22 = (λ noil ) D λ  =  1.22  D  noil  0.60 µ rad = 0.40 µ rad 1.5 (a) The wavelength of the light within the eye is λn = λ n Thus, the limiting angle of resolution for light passing through the pupil (a circular aperture with diameter D = 2.00 mm ), is θ = 1.22 λn D = 1.22 λ nD ( 500 × 10 m ) = ( 1.33 ) ( 2.00 × 10 m ) −9 = 1.22 −3 2.29 × 10 −4 rad (b) From s = rθ , the distance from the eye that two points separated by a distance s = 1.00 cm will intercept this minimum angle of resolution is r= 25.37 s θ = 1.00 cm = 4.36 × 10 cm = 43.6 m -4 2.29 × 10 rad The minimum angle of resolution when light of 500 nm wavelength passes through a 20inch diameter circular aperture is θ = 1.22 λ D ( 500 × 10 = 1.22 m )  39.37 inches  −6   = 1.2 × 10 rad 20 inches  1m  −9 If two stars, 8.0 lightyears away, are just resolved by a telescope of 20-in diameter, their separation from each other is   9.461 × 1015 m   −6 10 s = rθ =  8.0 ly    ( 1.2 × 10 rad ) = 9.1 × 10 m = 9.1 × 10 km ly     25.38 If just resolved, the angular separation of the objects is θ = θ = 1.22   550 × 10 −9 m   and s = r θ = ( 200 × 10 m ) 1.22    = 0.38 m = 38 cm  0.35 m    λ D Optical Instruments 25.39 If just resolved, the angular separation of the objects is θ = θ = 1.22 355 λ D   500 × 10 −9 m   and s = r θ = ( 8.0 × 107 km ) 1.22    = 9.8 km  5.00 m    25.40 The resolving power of a diffraction grating is R= λ = Nm ∆λ (a) The number of lines the grating must have to resolve the Hα line in the first order is N= R λ ∆λ 656.2 nm = = = 3.6 × 10 lines 0.18 nm m ( 1) (b) In the second order ( m = ) , N = 25.41 R 656.2 nm = = 1.8 × 10 lines 2 ( 0.18 nm ) cm = 1.67 × 10 −4 cm = 1.67 × 10 −6 m , and the highest order of 000 600 nm light that can be observed is The grating spacing is d = mmax = d sin 90° λ (1.67 × 10 = −6 600 × 10 m ) ( 1) −9 m = 2.78 → orders The total number of slits is N = ( 15.0 cm ) ( 000 slits cm ) = 9.00 × 10 , and the resolving power of the grating in the second order is Ravailable = Nm = ( 9.00 × 10 ) = 1.80 × 10 The resolving power required to separate the given spectral lines is Rneeded = λ 600.000 nm = = 2.0 × 10 ∆λ 0.003 nm These lines cannot be separated with this grating 25.42 A fringe shift occurs when the mirror moves distance λ Thus, if the mirror moves distance ∆L = 0.180 mm , the number of fringe shifts observed is N shifts −3 ∆L ( ∆L ) ( 0.180 × 10 m ) = = = = 1.31 × 10 fringe shifts 550 × 10 −9 m λ λ 356 CHAPTER 25 25.43 A fringe shift occurs when the mirror moves distance λ Thus, the distance moved (length of the bacterium) as 310 shifts occur is  650 × 10 −9 m  λ −5 ∆L = N shifts   = 310   = 5.04 × 10 m = 50.4 µ m 4     25.44 A fringe shift occurs when the mirror moves distance λ Thus, the distance the mirror moves as 250 fringe shifts are counted is  632.8 × 10 −9 m  λ −5 ∆L = N shifts   = 250   = 3.96 × 10 m = 39.6 µ m 4   25.45 When the optical path length that light must travel as it goes down one arm of a Michelson’s interferometer changes by one wavelength, four fringe shifts will occur (one shift for every quarter-wavelength change in path length) The number of wavelengths (in a vacuum) that fit in a distance equal to a thickness t is N vac = t λ The number of wavelengths that fit in this thickness while traveling through the transparent material is N n = t λn = t ( λ n ) = nt λ Thus, the change number of wavelengths that fit in the path down this arm of the interferometer is ∆N = N n − N vac = ( n − 1) t λ and the number of fringe shifts that will occur as the sheet is inserted will be # fringe shifts = ( ∆N ) = ( n − 1) 25.46  15.0 × 10 −6 m  = ( 1.40 − 1)   = 40 −9 λ  600 × 10 m  t A fringe shift will occur each time the effective length of the tube changes by a quarter of a wavelength (that is, for each additional wavelength fitted into the length of the tube, fringe shifts occur) If L is the length of the tube, the number of fringe shifts observed as the tube is filled with gas is  L  L L L  4L − = N shifts =  −  =  ngas −  λ ngas λ  λ  λn λ  ( )  600 × 10 −9 m   λ   ( 160 ) = 1.000 Hence, ngas = +   N shifts = +  −2  4L   ( 5.00 × 10 m )  Optical Instruments 25.47 357 Removing air from the cell alters the wavelength of the light passing through the cell Four fringe shifts will occur for each additional wavelength fitted into the length of the cell Therefore, the number of fringe shifts that occur as the cell is evacuated will be  L  L L L  4L − = N shifts =  −  =  ( nair − 1)  λn λ   λ nair λ  λ or 25.48 N shifts = ( 5.00 × 10 −2 m ) 590 × 10 −9 m (1.000 29 − 1) = 98.3 98 complete shifts (a) Since this eye can already focus on objects located at the near point of a normal eye (25 cm), no correction is needed for near objects To correct the distant vision, a corrective lens (located 2.0 cm from the eye) should form virtual images of very distant objects at 23 cm in front of the lens (or at the far point of the eye) Thus, we must require that q = −23 cm when p → ∞ This gives P= 1 1 = + =0+ = − 4.3 diopters f p q − 0.23 m (b) A corrective lens in contact with the cornea should form virtual images of very distant objects at the far point of the eye Therefore, we require that that q = −25 cm when p → ∞ , giving P= 1 1 = + =0+ = − 4.0 diopters f p q − 0.25 m   When the contact lens  f = = − 25 cm  is in place, the object distance which yields P   a virtual image at the near point of the eye (that is, q = −16 cm ) is given by p= 25.49 qf ( −16 cm )( −25 cm ) = = 44 cm q − f −16 cm − ( −25 cm ) (a) The lens should form an upright, virtual image at the near point of the eye q = −75.0 cm when the object distance is p = 25.0 cm The thin lens equation then gives f = pq ( 25.0 cm )( −75.0 cm ) = = 37.5 cm = 0.375 m p+q 25.0 cm − 75.0 cm so the needed power is P = 1 = = + 2.67 diopters f 0.375 m 358 CHAPTER 25 (b) If the object distance must be p = 26.0 cm to position the image at q = −75.0 cm , the actual focal length is f= pq ( 26.0 cm )( −75.0 cm ) = = 0.398 m p+q 26.0 cm − 75.0 cm and P = 1 = = + 2.51 diopters f 0.398 m The error in the power is ∆ P = ( 2.67 − 2.51) diopters = 0.16 diopters too low 25.50 (a) If q = 2.00 cm when p = 1.00 m = 100 cm , the thin lens equation gives the focal length as f= pq ( 100 cm )( 2.00 cm ) = = 1.96 cm p + q 100 cm + 2.00 cm (b) The f-number of a lens aperture is the focal length of the lens divided by the diameter of the aperture Thus, the smallest f-number occurs with the largest diameter of the aperture For the typical eyeball focused on objects 1.00 m away, this is ( f -number )min = f Dmax = 1.96 cm = 3.27 0.600 cm (c) The largest f-number of the typical eyeball focused on a 1.00-m-distance object is ( f -number )max = 25.51 f Dmin = 1.96 cm = 9.80 0.200 cm (a) The implanted lens should give an image distance of q = 22.4 mm for distant ( p → ∞ ) objects The thin lens equation then gives the focal length as f = q = 22 mm , so the power of the implanted lens should be Pimplant = 1 = = + 44.6 diopters f 22 × 10 −3 m Optical Instruments 359 (b) When the object distance is p = 33.0 cm , the corrective lens should produce parallel rays ( q → ∞ ) Then the implanted lens will focus the final image on the retina From the thin lens equation, the required focal length is f = p = 33.0 cm , and the power of this lens should be Pcorrective = 25.52 1 = = + 3.03 diopters f 0.330 m When viewed from a distance of 50 meters, the angular length of a mouse (assumed to have an actual length of ≈ 10 cm ) is s r θ= = 0.10 m = 2.0 × 10 −3 radians 50 m Thus, the limiting angle of resolution of the eye of the hawk must be θ ≤ θ = 2.0 × 10 −3 rad 25.53 The resolving power of the grating is R = λ ∆λ = Nm Thus, the total number of lines needed on the grating to resolve the wavelengths in order m is N= R λ = m m ( ∆λ ) (a) For the sodium doublet in the first order, N= 589.30 nm = 1.0 × 10 ( 1)( 0.59 nm ) (b) In the third order, we need N = 589.30 nm = 3.3 × 10 ( )( 0.59 nm ) 360 CHAPTER 25 25.54 (a) The image distance for the objective lens is ( 40.0 m ) ( 8.00 × 10 -2 m ) p1 f1 q1 = = = 8.02 × 10 −2 m = 8.02 cm -2 p1 − f1 40.0 m − 8.00 × 10 m The magnification by the objective is M1 = h′ h = − q1 p1 , so the size of the image formed by this lens is q   8.02 × 10 −2 m  h′ = h M1 = h   = ( 30.0 cm )   = 0.060 cm 40.0 m    p1  (b) To have parallel rays emerge from the eyepiece, its virtual object must be at its focal point, or pe = f e = − 2.00 cm (c) The distance between the lenses is L = q1 + pe = 8.02 cm − 2.00 cm = 6.02 cm (d) The overall angular magnification is m = 25.55 f1 8.00 cm = = 4.00 fe − 2.00 cm The angular magnification is m = θ θo , where θ is the angle subtended by the final image, and θo is the angle subtended by the object as shown in the figure When the telescope is adjusted for minimum eyestrain, the rays entering the eye are parallel Thus, the objective lens must form its image at the focal point of the eyepiece Objective Eyepiece Object fe qo A qo B E h¢ D C q1 fe q F parallel rays emerge From triangle ABC, θo ≈ tan θ o = h′ q1 and from triangle DEF, θ ≈ tan θ = h′ f e The θ h′ f e q1 angular magnification is then m = = = θo h′ q1 f e Optical Instruments 361 From the thin lens equation, the image distance of the objective lens in this case is q1 = p1 f1 ( 300 cm )( 20.0 cm ) = = 21.4 cm p1 − f1 300 cm − 20.0 cm With an eyepiece of focal length f e = 2.00 cm , the angular magnification for this telescope is m= 25.56 We use q1 21.4 cm = = 10.7 f e 2.00 cm n1 n2 n2 − n1 + = , with p → ∞ and q equal to the cornea to retina distance Then, p q R  n − n1   1.34 − 1.00  R = q  = ( 2.00 cm )   = 0.507 cm = 5.07 mm 1.34    n2  25.57 When a converging lens forms a real image of a very distant object, the image distance equals the focal length of the lens Thus, if the scout started a fire by focusing sunlight on kindling 5.00 cm from the lens, f = q = 5.00 cm (a) When the lens is used as a simple magnifier, maximum magnification is produced when the upright, virtual image is formed at the near point of the eye ( q = −15 cm in this case) The object distance required to form an image at this location is p= qf ( −15 cm )( 5.0 cm ) 15 cm = = −15 cm − 5.0 cm q− f 4.0 and the lateral magnification produced is M=− q −15 cm =− = + 4.0 p 15 cm 4.0 (b) When the object is viewed directly while positioned at the near point of the eye, its angular size is θ = h 15 cm When the object is viewed by the relaxed eye while using the lens as a simple magnifier (with the object at the focal point so parallel rays enter the eye), the angular size of the upright, virtual image is θ = h f Thus, the angular magnification gained by using the lens is m= h f θ 15 cm 15 cm = = = = 3.0 f 5.0 cm θ h 15 cm 362 CHAPTER 25 ... lowered inside the sphere neutralizes the inner surface, leaving zero charge on the inside This leaves − 5µ C on the outside surface of the sphere (d) When the object is removed, the sphere is... Because ∆V = Ed and E does not change, ∆V increases as d increases Because the same charge is stored at a higher potential difference, the capacitance has decreased Because Energy stored = Q 2C... magnitude of q2 is three times the magnitude of q1 because times as many lines emerge from q2 as enter q1 (a) Then, q1 q2 = − q2 = q1 Electric Forces and Electric Fields (b) 19 q2 > because lines emerge