I. Multiple choice (55 points): 1. The neutrons in a parallel beam, each having kinetic energy of ½ eV, are directed through two slits 0.50 mm apart. What is the wavelength of the neutron? a. 0.0134 nm b. 0.0243 nm c. 0.0406 nm d. 0.134 nm e. 0.243 nm from l= h/p, p = sqrt(2mE) = 1.63 e ­23 kg­m/s, l = 4 e ­11 m or 0.04 nm 2. From problem 1, the wavelength is adjusted by changing the energy of the neutrons. It is now 2.0 nm. How far apart will the interference peaks on a screen 2.50 m away be? a. 0.010 mm b. 0.10 mm c. 1.0 mm d. 10 mm e. none of the above 2 e ­9 m/ 0.5 e ­3 m = x / 2.5 m, x = 10 e­6 m or 1e­5, or 0.010 mm 3. A farsighted person has a nearpoint of 41 cm from her eyes. She wants glasses that will let her see objects at a distance of only 25 cm from her eyes. Determine the focal length of the glasses needed if the glasses are 2 cm from her eyes. (a) 39 cm (b) ­39 cm (c) 41 cm (d) 56 cm (e) ­56 cm need to subtract 2 cm from nearpoint and 25 cm, 1/f = ­1/39 +1/23, f = 56 cm 4. A nearsighted person has a farpoint of 512 cm. He wants contact lenses that will let him see objects at a great distance. Determine the focal length of the contact lens needed. (a) ­51.2 cm (b) 256 cm (c) ­256 cm (d) 512 cm (e) ­512 cm no subtraction needed: 1/f = 1/infinity – 1/512, f=­512 cm
5. When an object is placed 15 cm from a lens, a virtual image is formed. Which one of the following conclusions is incorrect? A) The lens may be a convex or concave. B) If the image is upright, the lens must be a diverging lens. C) If the image is reduced, the lens must be a diverging lens. D) If the lens is a diverging lens, the image distance must be less than 15 cm. E) If the lens is a converging lens, the focal length must be greater than 15 cm. An object placed inside the focal length of a converging lens will create an upright virtual image. 6. A child sitting at the edge of a swimming pool sees a coin resting on the bottom of the pool. The coin appears to be 2.0 ft directly below the water's surface. How deep is the pool at the location of the coin? A) 1.5 ft B) 2.0 ft C) 2.7 ft D) 3.2 ft E) 4.0 ft d’ = d/n, 2=d/1.33, d = 2.7 ft 7. A beam of light passes from air into water. Which is necessarily true? A) The frequency is unchanged and the wavelength increases. B) The frequency is unchanged and the wavelength decreases. C) The wavelength is unchanged and the frequency decreases. D) Both the wavelength and frequency increase. E) Both the wavelength and frequency decrease. 8. Complete the following statement: The term dispersion refers to the fact that the index of refraction of materials a) depends on the radius of curvature of the medium . b) depends on the wavelength of light. c) depends on the angle of incidence. d) depends on the intensity of light. e) depends on the polarization of light.
9. A convex lens with two identical sides has a focal length of 28 cm and an index of refraction of 1.7, what is the radius of curvature of the spherical surfaces of the lens? (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm (E) 80 cm 1/f = (n­1)(1/R1 – 1/R2), 1/28 cm = 0.7 x 2/R 10. J.J. Thomson’s conclusion that cathode ray particles are fundamental constituents of atoms was based primarily on which observation? (A) They have a negative charge (B) They are the same for all cathode materials (C) The mass is much less than that of a hydrogen (D) They penetrate very thin metal foils (E) They exhibit wavelike properties 11. The figure above shows a collimated beam of X­rays striking a polycrystalline sample. The energy of the X rays is 154.0 keV. The diameter of the smallest diffraction circle observed on a photographic plate located L = 1.60 m from the sample is D = 9.00 cm. What is the spacing, d, between atomic planes in the sample? (a) 0.0143 nm (b) 0.0286 nm (c) 0.0805 nm (d) 0.143 nm (e) 0.286 nm
q = 0.045m/2 x 1.6m = l/2d, l = 1240 eV­nm/154 keV = 0.00805 nm, d= 0.00805 x 1.6/0.045 = 0.286 nm
II. 25 points – The picture above is a schematic of the lens system of an astronomical telescope, with eyepiece on the right (fe = 8.1 cm) and the objective lens on the left (fo = 52.0 cm) . (a) (5)If a 22.0 meter tall tree is 3.5 km away, how tall will the image be after the first lens? So di/do = ­hi/ho, di=0.52 m, do = 3500 m, ho = 22.0 m 0.52/3500 = x/22, x = ­3.3 mm, or just 3.3 mm tall. (b) (5)What is the total angular magnification produced by the two lenses? M = ­fo/fe = ­52/8.1 = ­6.4 x (c). (7) If the total telescope length is 60 cm, where does the final image appear? Is it real or virtual? Upside down or right side up? Final image is virtual and rightside up relative to second image, upside down relative to object. Final image appears at 1/f = 1/do2 + 1/di2, 1/8.1 = 1/8 + 1/di2, di2 = ­ 650 cm or 65 meters in front of eyepiece. (d) (8) If the opening on the telescope is 1.2 m in diameter, what is the farthest distance in km that this telescope can resolve the headlights of a car that are 1.6 meters apart? Assume the wavelength emitted by the lights is 550 nm. D sin q = 1.22 l gives us the distance between bright spot and first dark rim. The first dark rim must be at a maximum of ½ the distance between the lights, or 0.8 meters.
q = 1.22 * 5.5 e­7 m/1.2 m = 0.8 m/L, L = 0.14 e7, or 1.4 e6 m, or 1400 km.