Unformatted text preview: PHY2054 Spring 2007 Prof. Eugene Dunnam Prof. Paul Avery Apr. 11, 2007 Exam 3 Solutions
1. An 8.0 mH inductor and a 2.5 resistor are wired in series to a 23.6 V ideal battery. A switch in the circuit is closed at time 0, at which time the current is 0. At a time 5.4 msec after the switch is thrown the potential difference in volts across the inductor is: (1) 4.4 (2) 19.2 (3) 13.0 (4) 15.2 (5) 7.7 The current in an RL circuit, starting at 0, is i = imax 1 - e -t / , where = L / R and imax = V / R and V is the voltage. Thus VL = V - iR = V - V 1 - e -t / = Ve-t / We calculate = L / R = 3.2 msec. Plugging in yields VL = 4.4 V. 2. Refer to the previous problem. How much energy in joules is stored in the inductor after the switch has been closed a long time? (1) 0.36 (2) 0 (3) 0.24 (4) 0.012 (5) 0.17 The energy stored in the inductor is U L = 1 Li 2 . To find the current, we see that after a long time 2 ( ( ) ) the current reaches its full value of i = 23.6 / 2.5 = 9.44 A. Thus U L = 0.36 J. 1 PHY2054 Spring 2007 3. An LC circuit has an oscillation frequency that corresponds to a wavelength of 3 km in free space. If C = 0.1 F, then L must be about: (1) 25 H (2) 10 mH (3) 1 mH (4) 2.5 H (5) 1 pH The frequency is c / = 105 Hz. Using f = 1/ 2 LC yields L = 25 H. 4. The primary of a 3:1 step-up transformer is connected to a source and the secondary is connected to a resistor R. The power dissipated by R in this situation is P. If R is connected directly to the source it will dissipate a power of: (1) P / 9 (2) P / 3 (3) P (4) 3P (5) 9P When the resistor is connected to the secondary, the voltage is 3V and the resistance is R, thus the power is 9V2 / R. If you now hook the resistor to the battery alone, the voltage is V and resistance is R, so the power is V2 / R. Thus the power is 1 / 9 as large. 5. Which of the following choices shows forms of electromagnetic radiation strictly increasing in frequency? (1) radio, visible, ultraviolet, X-ray (2) infrared, X-ray, ultraviolet, gamma ray (3) infrared, television, visible, ultraviolet (4) FM radio, AM radio, infrared, X-ray (5) ultraviolet, visible, infrared, television This was described in the chapter on E&M radiation. 2 PHY2054 Spring 2007 6. A laser beam of wavelength 632.8 nm has intensity 13.3 W/m2. The amplitudes of the associated electric and magnetic fields, in V/m and T, respectively are: (1) 100, 0.33 (2) 70.7, 0.24 (3) 10,000, 33.4 (4) 56.5, 0.19 (5) none of these
2 The intensity is relate to the peak E and B fields by I = Emax Bmax / 2 0 = Emax / 20c . We solve this for Emax yielding Emax = 100 V/m. Using B = E / c yields Bmax = 0.33 T. 7. The laser beam in the above question has an area of 0.2 cm2. What force (in pico N) will this beam exert on a 1 cm2 mirror from which it is totally reflected? (1) 1.77 (2) 8.86 (3) 0.89 (4) 0.77 (5) none of these The total force for a beam of intensity I is F = 2 IA / c , where A is the area of the beam, c is the speed of light and the factor of 2 accounts for reflection. The total force is then calculated to be 1.77 pN. 8. A band of frequencies for many cellular telephones is centered near 800 MHz. At this frequency, what is the photon energy in eV (not joules)? (1) 3.3 10-6 (2) 4.2 10-5 (3) 8.9 10-3 (4) 8.0 108 (5) 3.8 10-1 The energy of each photon of frequency f is hf, where h = Planck's constant = 6.6 10-34 J-sec. This relation yields a photon energy of 5.26 10-25 J. Converting to eV yields 3.3 10-6 eV. 3 PHY2054 Spring 2007 9. A 60 resistor, a 4.0 F capacitor and a 0.60 H inductor are connected in series to a 240 Hz sinusoidal voltage source with Vmax = 160 V. What is the maximum current in amps delivered to the circuit? (1) 0.22 (2) 0.18 (3) 0.68 (4) 2.67 (5) none of these This is a forced oscillation situation, so the current amplitude is given by imax = Vmax / Z , where Z = R2 + ( X L - X C )
2 is the impedance, Vmax = 160, R is the resistance, and the reactances are X L = 2 fL and X C = 1/ 2 fC . Using a frequency of f = 240 Hz, we obtain X L = 905 and X C = 166 , from which we obtain Z = 741 and imax = 0.22 A. 10. As shown in the figure, a beam of monochromatic light is incident at an angle 1 = 60 upon a transparent slab of glass (n = 1.47) with parallel sides, 2.5 cm thick. What is the angle of emergence 2 and displacement d of the beam (measured as shown along the bottom of the slab) when it emerges? (1) 60, 2.51 cm (2) 60, 1.82 cm (3) 36, 4.33 cm (4) 36, 1.52 cm (5) none of these Since the beam enters and emerges from air, 2 = 1 = 60 . The refracted angle in the slab is found from Snell's law and is 36. The distance d can then be computed from 2.5 tan 60 - 2.5 tan 36 = 4.33 - 1.82 = 2.51 cm. 11. Which of the following improves the performance of optical fibers? (1) make ncore and ncladding as different as possible (2) introduce sharp bends (3) increase the light absorption of the core (4) increase the light absorption of the cladding (5) increase the roughness of the boundary between core and cladding As discussed in the chapter, a larger separation of the indices of refraction of the core and cladding allows the laser beam to be totally internally reflected over a larger entry angle. The other choices hurt the ability of the fiber to conduct light. 4 PHY2054 Spring 2007 12. Which optical conditions do not play a role in the formation of rainbows? (A) Dispersion (B) Reflection (C) Refraction (D) Polarization (E) Sun behind observer (1) D only (2) E only (3) A, E (4) B, E (5) C only This was discussed in detail in class and in the book. Polarization is not involved. 13. An RLC series circuit, connected to an applied sinusoidal voltage source of amplitude E, is at resonance. Thus: (1) the voltage amplitude across R is equal to E (2) the voltage amplitude across R is zero (3) the voltage amplitude across C is zero (4) the voltage amplitude across L equals E (5) the applied voltage and current are 90 out of phase with each other At resonance, the effects of the inductor and capacitor on the current cancel, so the net voltage difference of E appears across the resistor. 14. You stand 2 m in front of a plane mirror and take a picture of your image in the mirror with a camera. For what distance (in m) must the camera lens be focused and is the image formed on the camera film real, or virtual? (1) 4, real (2) 4, virtual (3) 2, real (4) 2, virtual (5) 6, real The image is 2 m behind the mirror, so it is 4 m from the camera. The image on the film must be real to make an image. 5 PHY2054 Spring 2007 15. A woman stands in front of and facing a plane mirror and holds a large rectangular card of area A just below her chin. In order for her to see the entire image of the card in the mirror, the least mirror area needed is: (1) A / 4 (2) A (3) that of the pupil of her eye (4) A / 2 (5) an amount which decreases with her distance from the mirror This is similar to a WebAssign problem. The person can see her entire length in a mirror 1 / 2 her height. The same reasoning applies to left right. Thus the area of the mirror is A / 4. 16. An erect object is located in front of a convex mirror a distance greater than the focal length. The image formed by the mirror is: (1) virtual, erect, and smaller than the object (2) real, inverted, and smaller than the object (3) virtual, inverted, and larger than the object (4) real, inverted, and larger than the object (5) real, erect, and larger than the object A convex mirror has f < 0. The relation
1 q = 1 f - 1 , shows that |q| < p and is negative. The p magnification is m = -q / p , which is positive and less than 1. Thus the image is virtual, erect and smaller. 17. Light of frequency 6.5 1014 Hz travels through vacuum and passes through a 1 mm - thick sheet of polystyrene (n = 1.59). Approximately how many complete wavelengths are inside the polystyrene? (1) 3450 (2) 2170 (3) 1360 (4) 0.00029 (5) none of these The wavelength in free space is = c / f = 462 nm. However, the wavelength in the material is shortened by the index of refraction. So = 290 nm and number of wavelengths in the material is approximately 1 mm / = 3450. 6 PHY2054 Spring 2007 18. A dentist's inspection mirror is designed to produce a non-inverted image that is twice the size of the object when the mirror is held 1.5 cm from a tooth. What is the focal length of the mirror (in cm), and is the image real, or virtual? (1) +3.0, virtual (2) +1.0, real (3) -3.0, real (4) -3.0, virtual (5) +1.0, virtual The image magnification is m = +2 = -q / p. Thus q = -2p and the image is virtual. Plugging p and q into the lens equation yields f = +3.0 cm. 19. What is the focal length (in cm) of a lens that casts an image that is enlarged 3.4 times, on a screen 2.3 m from the lens? (1) +52 (2) +68 (3) -96 (4) +230 (5) none of these The image magnification is m = -3.4 = -q / p (it is negative because the image is real, q > 0). Since q = 2.3m, we see that p = 68 cm and from the lens equation, f = +52 cm. 20. An erect object is placed at a distance 2f in front of a double-concave lens of focal length |f|. The resultant image is: (1) none of these (2) real, inverted, enlarged (3) virtual, erect, enlarged (4) virtual, erect, same size (5) virtual, inverted, reduced A double concave lens is a diverging lens (f < 0), so as seen by the simple calculation in problem 16 the image is always virtual and smaller. 7 ...
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