Unit 14 Solutions

Unit 14 Solutions - Chapter 22 Electromagnetic waves...

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Unformatted text preview: Chapter 22 Electromagnetic waves CHAPTER 22 PROBLEMS 5, [6] 5. Strategy and Solution At a point due south of the transmitter, the EM waves are traveling due south. The electric field at this point is oriented vertically, like the antenna. The magnetic field is perpendicular to both the electric field and the direction the EM waves travel, so the magnetic field must be oriented east-west. 6. Strategy and Solution At a point due north of the transmitter, the EM waves are traveling due north. The electric field at this point is oriented vertically, like the antenna. A receiver should be oriented the same as the transmitter, so the orientation of the receiver should be vertical. CHAPTER 22 PROBLEMS: [12], [15], 17, 19, 27, 32 12. Strategy The frequency, wavelength, and speed of EM radiation are related by λ f = c. Solution Compute the wavelength of the radio waves. c 3.00 × 108 m s λ= = = 3.3 m f 90.9 × 106 Hz 15. Strategy The speed of light in matter is given by v = c n , where n is the index of refraction. Solution Compute the index of refraction of topaz. c 3.00 × 108 m s n= = = 1.62 v 1.85 × 108 m s 17. Strategy The time of travel is equal to the distance traveled divided by the rate of travel. Solution Compute the time it takes for light to travel 50.0 cm. d 50.0 × 10−2 m Δt = = = 1.67 × 10−9 s = 1.67 ns c 3.00 × 108 m s 19. (a) Strategy The wavelength is shorter in matter than it is in vacuum. Use Eq. (22-5). Solution Compute the wavelength of light inside the glass. λ 692 nm λg = v = = 455 nm n 1.52 (b) Strategy The frequency in glass is the same as the frequency in air. Use c = f λ . Solution Compute the frequency of light inside the glass. c 3.00 × 108 m s fg = fa = = = 4.34 × 1014 Hz −9 λa 692 × 10 m 27. Strategy The electric and magnetic fields oscillate with the same frequency, and their amplitudes are proportional to each other. Use Eq. (22-6). 887 Solution Compute the amplitude of the magnetic field. E 0.60 × 10−3 V m Bm = m = = 2.0 × 10−12 T 8 c 3.00 × 10 m s The frequency of the magnetic field is 30 GHz, the same as that of the electric field. 32. (a) Strategy and Solution Since the electric field depends on the value of y but not on the values of x or z, the wave moves parallel to the y-axis. The direction can be found by noting that as t increases in ky − ωt + π /6, y must increase to maintain the relative phase. So, the wave is moving in the +y-direction. (b) Strategy The amplitudes of the electric and magnetic fields are proportional to each other. Use Eq. (22-7). The direction of propagation of EM waves is determined by E × B. Solution Since the electric field is in the +z-direction when t = 0 and y = 0 and the wave is traveling in the +y-direction, by E × B and the RHR, the magnetic field at t = 0 and y = 0 must point in the +x-direction. The components are Bx = y v E B z x Em sin(ky − ω t + π /6), B y = Bz = 0 . c CHAPTER 22 PROBLEMS: 35, [43], [46], 47, [48], 56, 68 35. Strategy Use Eqs. (22-10), (22-12), and (22-13). Solution Find the intensity of the beam. 〈 P〉 10.0 × 10−3 W I= = = 180 W m 2 1 π (0.85 × 10−2 m) 2 A 4 Find the rms value of the electric field. I = 〈u 〉 c = (e0 Erms 2 )c, so Erms = I e0 c = 180 W m 2 [8.854 × 10−12 C2 (N ⋅ m 2 )](3.00 × 108 m s) = 260 V m . 43. Strategy Use Eq. (22-14). Solution (a) Compute the average power of the incident EM wave. 〈 P〉 = IA cos θ = (1.0 × 103 W m 2 )(4.0 m)(6.0 m) cos 0° = 24 × 103 W Since the panels are only 12% efficient, the power supplied is 0.12(24 × 103 W) = 2.9 kW . (b) Compute the average power supplied by the panels. 〈 P〉 = IA cos θ × efficiency = (0.80 × 103 W m 2 )(4.0 m)(6.0 m) cos 60.0°(0.12) = 1.2 kW (c) In part (a) the panels provide more than enough power on average, while in part (b) the panels provide less power than required. The excess energy produced at midday should be stored for use at night; in general, a supplemental energy source is necessary. 46. Strategy Use Eq. (22-16b). 888 Solution (a) The intensity of the light passing through the first polarizing sheet is I1 = I 0 cos 2 θ1 = I 0 cos 2 θ . The intensity of the light passing through the second sheet is I 2 = I1 cos 2 θ 2 = I1 cos 2 (90° − θ ) = I 0 cos 2 θ cos 2 (90° − θ ) = I 0 cos 2 θ sin 2 θ = 1 I 0 sin 2 2θ , 4 since cos(90° − θ ) = sin θ and 2 cos θ sin θ = sin 2θ . (b) The sine function reaches a maximum when its argument is 90°. 2θ = 90°, so θ = 45° . 47. Strategy Use Eqs. (22-15) and (22-16b). Solution Since the light is initially unpolarized, the intensity of the light after passing through the first polarizer is I1 = I 0 2. The intensity after passing through the second polarizer is I 2 = I1 cos 2 30.0° = ( I 0 2) cos 2 30.0°. The intensity after passing through the third polarizer is I3 = I 2 cos 2 (45.0° − 30.0°) = ( I 0 2) cos 2 30.0° cos 2 15.0°. Solve for the initial intensity, I 0 . I3 = I0 2I3 2(23.0 W m 2 ) cos2 30.0° cos2 15.0°, so I 0 = = = 65.7 W m2 . 2 cos 2 30.0° cos 2 15.0° cos2 30.0° cos 2 15.0° 48. Strategy Use Eqs. (22-15) and (22-16b). Solution Since the light is initially unpolarized, the intensity of the light after passing through the first polarizer is I1 = I 0 2. The intensity after passing through the second polarizer is I 2 = I1 cos 2 30.0° = ( I 0 2) cos 2 30.0°. The angular difference between the polarization axis and the transmission axis for the next two polarizers is 30.0°; we multiply I 2 by cos 2 30.0° twice to find the final intensity of the light, I 4 . Find the percent I 4 is of I 0 . I4 × 100% = I0 1 I cos6 30.0° 20 × 100% I0 = 1 cos6 30.0°× 100% = 21.1% 2 56. Strategy First try Eq. (22-18). If the resulting velocity is small compared to c, we are done. If the velocity is not small compared to c, we will have to use Eq. (22-17). Solution Find the speed of the star with respect to Earth. ⎛v⎞ f o ≈ fs ⎜1 + rel ⎟ c⎠ ⎝ vrel f o λs 1+ ≈ = c fs λo ⎛λ ⎞ ⎛ 659.6 nm ⎞ vrel ≈ c ⎜ s − 1⎟ = (3.00 × 108 m s) ⎜ − 1⎟ = −680 km s λo ⎠ ⎝ 661.1 nm ⎠ ⎝ This velocity is small compared to c, so the use of Eq. (22-18) was justified. vrel < 0, so the star is moving at 680 km/s away from the Earth. 68. Strategy (a) Refer to Figure 22.7. (b) The length of the pulse is equal to the distance EM radiation travels in 20.0 ps. (c) The speed of the EM wave is reduced to v = c n in water. (d) Divide the length of a pulse by the 889 wavelength to find the number that fit in one pulse. (e) The total EM energy in one pulse is equal to the power times the time of the pulse. Solution (a) According to Figure 22.7, the pulse is in the infrared part of the EM spectrum. (b) length = cΔt = (3.00 × 108 m s)(20.0 × 10−12 s) = 6.00 mm (c) length = (d) c 3.00 × 108 m s Δt = ( 20.0 × 10−12 s) = 4.51 mm n 1.33 length of pulse 6.00 × 10−3 m = = 5660 wavelength 1060 × 10−9 m (e) U = PΔt = (60.0 × 10−3 W)(20.0 × 10−12 s) = 1.20 pJ Chapter 23 REFLECTION AND REFRACTION OF LIGHT Chapter 23 Conceptual Questions 2,7,9,[13,14,22] 2. In a virtual image, light rays do not actually come from a point on an image but instead only appear to diverge from a single point. In a real image, the rays actually do pass through the image point. A real image may be detected using photographic film. 7. The wavelength changes, as do the speed and direction of propagation. The frequency of the radiation remains unchanged. 9. (a) The image is upright. (b) The image is inverted. (c) The real image is formed at a distance greater than 2f, and therefore, cannot be seen. 13. A virtual image would be formed on the left side of the lens, while a real image would be formed on the right side. 14. The larger index of refraction for the diamond (n ≈ 2.4) results in a smaller critical angle for total internal reflection. Thus, a greater portion of the light incident on the gemstone is reflected back out, giving it a greater sparkling brilliance. An artificial diamond made from glass (n ≈ 1.5) would be even less brilliant than the cubic zirconium (n ≈ 1.9), since the index of refraction for glass is smaller. 22. Yes, the reflected light is partially polarized perpendicular to the plane of incidence, which is vertical in this case. Chapter 23 Problems: 7, 10, [12], 15, 17 7. Strategy Redraw the figure, including the angle of incidence and the angle of reflection. Use the laws of reflection. 890 Solution From the figure, we see that the angle of incidence is θi = 90° − 50° = 40°. Using the laws of reflection, we find that the angle of reflection is θ r = θi = 40°. The angles θi , θ r , and δ must add to 180°. Solve for δ . θi θ r δ 50° θi + θ r + δ = 40° + 40° + δ = 180°, so δ = 100° . 10. Strategy Use Snell’s law, Eq. (23-4). Solution Find the angle with respect to the normal at which the fish sees the Sun. ⎛n ⎞ ⎛ 1.000 ⎞ ni sin θi = nt sin θ t , so θ t = sin −1 ⎜ i sin θi ⎟ = sin −1 ⎜ sin 30.0° ⎟ = 22.0° . ⎝ 1.333 ⎠ ⎝ nt ⎠ 12. Strategy Use Snell’s law, Eq. (23-4). Let the index of refraction for air be n, and let n1 , n2 , n3 , and n4 be 1.20, 1.40, 1.32, and 1.28, respectively. Let θ be the angle of emergence. Solution Find the angle the beam makes with the normal when it emerges into the air after passing through the entire stack of four flat transparent materials. n sin 60.0° = n1 sin θ1 = n2 sin θ 2 = n3 sin θ3 = n4 sin θ 4 = n sin θ , so θ = 60.0° . 15. Strategy Draw a diagram. Use Snell’s law, Eq. (23-4). Solution Find θ1. Relate θ1 and θ3 . n1 sin(90° − θ1 ) = n3 sin θ3 Relate θ 2 and θ3 . n2 sin(90° − θ 2 ) = n3 sin θ3 Eliminate n3 sin θ3 and solve for θ1. 90 – θ 2 ˚ θ3 θ2 θ3 n2 = 1.00 n3 = 1.20 90 – θ 1 ˚ n1 = 1.40 θ1 n1 sin(90° − θ1 ) = n2 sin(90° − θ 2 ) ⎡n ⎤ 90° − θ1 = sin −1 ⎢ 2 sin(90° − θ 2 ) ⎥ ⎣ n1 ⎦ ⎡1.00 ⎤ θ1 = 90° − sin −1 ⎢ sin(90° − 5.00°) ⎥ = 44.6° 1.40 ⎣ ⎦ 17. Strategy Draw a diagram. Use Snell’s law, Eq. (23-4). Solution Initially, the coin is just hidden from view, so the angle of the observer’s eye 6.5 . After the water is poured into the mug, the with respect to the vertical is θ1 = tan −1 8.9 observer can just see the near end of the coin. Find θ 2 . 6.5 ⎞ ⎛ ⎛ −1 6.5 ⎞ ⎤ −1 ⎡ n1 n1 sin θ1 = n1 sin ⎜ tan −1 ⎟ = n2 sin θ 2 , so θ 2 = sin ⎢ sin ⎜ tan ⎟⎥ . 8.9 ⎠ 8.9 ⎠ ⎦ ⎝ ⎝ ⎣ n2 From the diagram, we see that tan θ 2 = 6.5 cm − d , where d is the diameter of the coin. 8.9 cm Find d. 891 θ1 θ2 8.9 cm 6.5 cm ⎧ ⎡n 6.5 cm − d 6.5 ⎞ ⎤ ⎫ ⎪ ⎪ ⎛ = tan ⎨sin −1 ⎢ 1 sin ⎜ tan −1 ⎟ ⎥ ⎬ , so 8.9 cm 8.9 ⎠ ⎦ ⎪ n2 ⎝ ⎪ ⎣ ⎩ ⎭ ⎧ ⎡1.000 ⎛ −1 6.5 ⎞ ⎤ ⎫ sin ⎜ tan d = 6.5 cm − (8.9 cm) tan ⎨sin −1 ⎢ ⎟ ⎬ = 2.1 cm . 8.9 ⎠ ⎥ ⎭ ⎣1.333 ⎝ ⎦ ⎩ Chapter 23 Problems: [23], 25, 28, 32, [33] 23. Strategy From Table 23.1, the index of refraction for diamond is 2.419, for air it is 1.000, and for water it is 1.333. Total internal reflection occurs when the angle of incidence is greater than or equal to the critical angle. Use Eq. (23-5a). Solution (a) Calculate the critical angle for diamond surrounded by air. n 1.000 θc = sin −1 t = sin −1 = 24.42° ni 2.419 (b) Calculate the critical angle for diamond under water. n 1.333 θc = sin −1 t = sin −1 = 33.44° ni 2.419 (c) Under water, the larger critical angle means that fewer light rays are totally reflected at the bottom surfaces of the diamond. Thus, less light is reflected back toward the viewer. 25. Strategy Total internal reflection occurs when the angle of incidence is greater than or equal to the critical angle. Use Eqs. (23-5). Solution (a) Find the index of refraction of the glass. n n 1.000 θc = sin −1 t , so sin θc = t and ni = = 1.556 . ni ni sin 40.00° (b) No; rays from the defect could reach all points above the glass since for 0 ≤ θi ≤ θc , 0 ≤ θ t ≤ 90° . 28. Strategy Total internal reflection occurs when the angle of incidence is greater than or equal to the critical angle. Use Eqs. (23-5) and Snell’s law, Eq. (23-4). Solution When the light is incident on the Plexiglas tank, some is transmitted at angle θ1 , so n sin θi = n1 sin θ1 where n = 1.00 for air and n1 = 1.51 for Plexiglas. At the Plexiglas carbon disulfide interface, θ1 is the incident angle and θ 2 is the transmitted angle, so n2 sin θ 2 = n1 sin θ1 where n2 = 1.628 for carbon disulfide. The ray passes through the carbon disulfide and is incident on the bottom tank-liquid interface at angle θ 2 . Here the light must experience total internal reflection, so θ 2 = θ c = sin −1 n sin θi = n1 sin θ1 = n2 sin θ 2 = n2 n1 n , or 1 = sin θ 2 . Find θi . n2 n2 n n1 1.51 = n1 , so θi = sin −1 1 = sin −1 = sin −1 1.51, or sin θi = 1.51, n 1.00 n2 which is impossible since sin θ ≤ 1 for all θ . Thus, there is no angle θi for which light is transmitted into the carbon disulfide but not into the Plexiglas at the bottom of the tank. 892 32. (a) Strategy The reflected light is totally polarized when the angle of incidence equals Brewster’s angle. Use Eq. (23-6). Solution Compute Brewster’s angle. n 1.333 θ B = tan −1 t = tan −1 = 53.12° ni 1.000 The angle with respect to the horizontal is the complement of this angle. θB θB θ θ nt = 1.333 θt θ = 90° − 53.12° = 36.88° (b) Strategy and Solution For Brewster’s angle, the reflected light is polarized perpendicular to the plane of incidence. (c) Strategy When the angle of incidence is Brewster’s angle, the incident and transmitted rays are complementary. Solution Find the angle of transmission. θ t = 90° − θi = 90° − 53.12° = 36.88° The angle below the horizontal is the complement of this angle. 90° − 36.88° = 53.12° 33. (a) Strategy The reflected light is totally polarized when the angle of incidence equals Brewster’s angle. Use Eq. (23-6). Solution Compute Brewster’s angle. n 1.309 θ B = tan −1 t = tan −1 = 52.62° ni 1.000 The angle with respect to the horizontal is the complement of this angle. θB θB θ θ nt = 1.309 θt θ = 90° − 52.62° = 37.38° (b) Strategy and Solution For Brewster’s angle, the reflected light is polarized perpendicular to the plane of incidence. (c) Strategy When the angle of incidence is Brewster’s angle, the incident and transmitted rays are complementary. Solution Find the angle of transmission. θ t = 90° − θi = 90° − 52.62° = 37.38° The angle below the horizontal is the complement of this angle. 90° − 37.38° = 52.62° Chapter 23 Problems: 36, [38], 39, 46, [48], 49 36. Strategy The equation derived in Example 23.4 can be used for this problem (with namber = nw ) since namber > nair . Solution Find the depth of the insect. n apparent depth 1.000 = air = , so actual depth = apparent depth ⋅ 1.546 = 7.00 mm ⋅ 1.546 = 10.8 mm . actual depth namber 1.546 893 38. Strategy As found in Conceptual Example 23.5, the mirror must be at least half as tall as Norah. Solution Find the required height of the mirror. 1.64 m = 0.82 m 2 39. Strategy For a plane mirror, a point source and its image are at the same distance from the mirror (on opposite sides) and both lie on the same normal line. Draw a ray diagram. Use geometry and the laws of reflection. Solution Suppose the mirror is hung at the proper height (see the figure). The top of Daniel’s head is at point A, his eyes are at point B, and his shoes are at point D. Lines AD and CE are perpendicular, and θi = θ r , 1 so triangles BCE and DCE are congruent and BC = CD = BD . 2 Similarly, triangles CDE and FED are congruent, so 1 1 EF = CD = BD = ⋅1.82 m = 0.91 m . 2 2 A B mirror d θr C θi d D E d F 46. Strategy In a convex mirror, the image is virtual and is formed behind the mirror. To find the image distance q, use the mirror equation. Solution The object distance and focal length are p = 12.0 cm and f = − R 2 = − 4.00 cm, respectively. 111 1 1 + = , so q = 1 1 = = −3.00 cm. 1 1 pqf − p − 4.00 cm − 12.0 cm f The image is formed 3.00 cm behind the mirror. Object Image 2.00 cm 3.00 cm 12.0 cm F C 4.00 cm 8.00 cm 48. Strategy The object distance is p = 1.20 cm and the magnification is m = 3.00. Use the magnification and mirror equations. The focal length is half of the radius of curvature. Solution Use the magnification equation to find the image distance. q m = − , so q = −mp. p Find the focal length. ⎛ 1 1⎞ f =⎜ + ⎟ ⎝ p q⎠ −1 ⎛1 1⎞ =⎜ + ⎟ p −mp ⎠ ⎝ −1 1⎞ ⎛ = p ⎜1 − ⎟ ⎝ m⎠ −1 1⎞ ⎛ = (1.20 cm) ⎜1 − ⎟ ⎝ 3.00 ⎠ 894 −1 = 1.80 cm . Compute the radius of curvature. 1 f = R, so R = 2 f = 2(1.80 cm) = 3.60 cm . 2 49. Strategy Since the image is real, the image distance is positive and the image is inverted (negative). The image is twice the size of the object, so h′ = −2h. Use the magnification and mirror equations. Solution Relate the object and image locations. h′ −2h q = = − , so q = 2 p. h h p Find the distance of the object from the mirror. 111 1 3 1 3 3⎛ R⎞ 3 +=+ = = , so p = f = ⎜ ⎟ = (25.0 cm) = 18.8 cm. p q p 2p 2p f 2 2⎝ 2 ⎠ 4 The object is 18.8 cm in front of the mirror. Object Image C F 18.8 cm Chapter 23 Problems: 57, 63, 64, [65], [70], 73 57. (a) Strategy Use the thin lens equation to find the object distance. Then, draw the ray diagram. Solution Find the object distance. 111 qf (5.00 cm)(3.50 cm) + = , so p = = = 11.7 cm . 5.00 cm − 3.50 cm pqf q− f The diagram is shown. Converging lens 11.7 cm Image Object f f (b) Strategy and Solution Light rays actually pass through the image location, so the image is real . (c) Strategy Use the magnification equation. Solution Compute the magnification of the image. q 5.00 cm m=− =− = − 0.429 11.667 cm p 63. Strategy Draw a ray diagram using the three principal rays. Solution Find the height and position of the image. 895 3.00 cm Diverging Lens Object Virtual image F F 6.00 cm 12.0 cm 12.0 cm The image is located 6.00 cm from the lens on the same side as the object and has a height of 1.50 cm. 64. Strategy All images are virtual since a diverging lens is used. The magnification determines the size and orientation of the image. Use the thin lens and magnification equations. Solution (a) Solve the thin lens equation for q. 111 1 + = , so q = 1 1 . pqf −p f Substitute the given values. 1 1 q= = −3.08 cm for p = 5.00 cm. q = = − 4.00 cm for p = 8.00 cm. 1 1 1 1 − 8.00 cm − 5.00 cm − 8.00 cm − 8.00 cm For p-values of 14.0 cm, 16.0 cm, and 20.0 cm, the corresponding q values are −5.09 cm, −5.33 cm, and −5.71 cm, respectively. The results are summarized in the table. p (cm) q (cm) m = −q p Real or virtual Orientation Relative size 5.00 –3.08 0.615 virtual upright diminished 8.00 – 4.00 0.500 virtual upright diminished 14.0 –5.09 0.364 virtual upright diminished 16.0 –5.33 0.333 virtual upright diminished 20.0 –5.71 0.286 virtual upright diminished (b) Solving the magnification equation for the image height yields h′ = mh. For p = 5.00 cm , h = 4.00 cm , and m = 0.6153 , we have h′ = 0.6153 ⋅ 4.00 cm = 2.46 cm . For p = 20.0 cm , h = 4.00 cm , and m = 0.286 , we have h′ = 0.286 ⋅ 4.00 cm = 1.14 cm . 65. Strategy Use the thin lens and magnification equations. Solution (a) Solve the thin lens equation for q. 111 pf + = , so q = . pqf p− f For f = 8.00 cm and p = 5.00 cm , we have 896 (5.00 cm)(8.00 cm) = −13.3 cm. 5.00 cm − 8.00 cm Since q is negative, the image is virtual and on the same side of the lens as the object. The magnification is q −13.33 cm m=− =− = 2.67. p 5.00 cm Since m is positive and greater than 1, the image is upright and enlarged. The same process is used for the other object distances, and the results are summarized in the table on the following page. q= p (cm) q (cm) m Real or virtual Orientation Relative size 5.00 –13.3 2.67 virtual upright enlarged 14.0 18.7 –1.33 real inverted enlarged 16.0 16.0 –1.00 real inverted same 20.0 13.3 – 0.667 real inverted diminished (b) The image height can be found from the magnification equation. h′ = mh For p = 5.00 cm , h = 4.00 cm , and m = 2.67 we have h′ = 2.67 ⋅ 4.00 cm = 10.7 cm . For p = 20.0 cm , h = 4.00 cm , and m = − 0.667 we have h′ = − 0.667 ⋅ 4.00 cm = −2.67 cm . 70. Strategy Use the thin lens and magnification equations. Since the lens is diverging, the focal length is negative, so f = −20.0 cm. p = 40.0 cm and h = 6.00 cm are given. Solution Find the image position q. −1 −1 ⎛1 1⎞ 111 1 1 ⎛ ⎞ + = , so q = ⎜ − ⎟ = ⎜ − ⎟ = −13.3 cm. pqf f p⎠ −20.0 cm 40.0 cm ⎠ ⎝ ⎝ Find the image size. h′ q q −13.3 cm m = = − , so h′ = − h = − (6.00 cm) = 2.00 cm. h p p 40.0 cm q < 0 and h′ > 0, so the image is virtual and upright; it is located 13.3 cm in front of the lens and is 2.00 cm tall. 73. (a) Strategy The image is virtual, so the image distance is negative. Use the magnification equation. Solution Find the object distance. h′ q h 8.0 cm m = = − , so p = − q = − (− 4.0 cm) = 9.1 cm . h p h′ 3.5 cm (b) Strategy and Solution The image is upright, virtual, smaller than the object, and closer to the mirror than the object. The mirror is convex . (c) Strategy The radius of curvature is twice the absolute value of the focal length. Use the mirror equation. Solution Find the focal length of the mirror. ⎛ 1 1⎞ 111 + = , so f = ⎜ + ⎟ pqf ⎝ p q⎠ −1 ⎛ h′ 1 ⎞ = ⎜− + ⎟ ⎝ hq q ⎠ −1 ⎛ h′ ⎞ = q ⎜1 − ⎟ h⎠ ⎝ 897 −1 ⎛ 3.5 ⎞ = (− 4.0 cm) ⎜1 − ⎟ ⎝ 8.0 ⎠ −1 = −7.1 cm . Compute the radius of curvature. R = 2 f = 2(7.1 cm) = 14 cm 898 ...
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This note was uploaded on 09/30/2011 for the course PHYS 1101 at Cornell.

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