Physics Spring 2007 Final Exam Solution. This version incorporates minor typographic corrections in the exam problems.

Transcription

1 Physics 02- Spring 2007 Final Exam Solution This version incorporates minor typographic corrections in the exam problems. Grading note: Point values are specified for each problem. Within a problem, some parts may be weighted more heavily than others. The total is 00 points.. [0 points] Sketch electric field lines and equipotential lines for the following charge distribution. There is a 6 µc charge in the center and two +3 µc charges, one to the left and one to the right. Either make a single drawing, using one color for the electric field lines and another color for the equipotential lines, or make two separate drawings. In either case, clearly label which are field lines and which are equipotentials. +3 µ C 6 µ C +3 µ C In the equipotential plot below, the dashed lines are negative and the thin solid lines are positive. The bold solid lines are V = 0. There is a set of equipotentials around each of the three charges. Because the central charge has a larger magnitude than the outer charges, the equipotentials around it are bigger, which might be shown by having equipotentials of larger radii, or by having more equipotentials. You were not required to draw the subtle shapes of the V = 0 lines and of the negative equipotentials at voltages just below 0. Here are two views of the field lines. The left field line plot has the same scale as the equipotential plot above. The right plot shows a larger area. All field lines coming off the + charges eventually end at the charge. The field lines from the left + charge all end on the left side of the charge, while the field lines from the right + charge end on the right side of the + charge.

2 2. [0 points] (a) A proton travels into a capacitor which is charged up as shown in the figure. What is the direction of the force on the proton (if any) as the proton passes through the capacitor? The electric field within the capacitor plate goes from the + charges to the charges, so it is downward. The force due to an electric field is F = q E. Since q is positive, the force is parallel to the field, hence it is downward. (b) A proton travels along the central axis of a current-carrying solenoid, as shown in the figure. What is the direction of the force on the proton (if any) as the proton passes through the solenoid? I The magnetic field within the solenoid is parallel to the solenoid axis, so the proton is traveling parallel to the field. The force due to a magnetic field is F = q v B. In this case, v is parallel to B, hence v B = 0, and there is zero force.. 2

3 3. [0 points] (a) Red light with a wavelength of 750 nm falls on a single slit of width 750 µm Draw a graph of intensity versus position for light which falls on a screen.00 m away. Your drawing should clearly show any relevant dimensions, i.e., put numbers on the axes as needed. (Reminder: nm=0 9 m; µm=0 6 m; mm=0 3 m.) A single slit of width a produces a broad central bright, surrounded by a series of dimmer spots. The intensity goes to zero at the following positions: y = n λd a = n( m)( m) m = n(0 3 m) = n( mm.) y mm (b) Red light with a wavelength of 750 nm falls on a pair of slits. Each slit has width 750 µm, and the centers of the two slits are separated by 3.0 mm Draw a graph of intensity versus position for light which falls on a screen.00 m away. Again, make sure your drawing clearly includes any relevant dimensions. The outer envelope of the intensity pattern is the same as in part (a). Within this envelope, there are periodic bright and dark bands with spacing y = n λd d = n( m)( m) m = n( m) = n(0.25 mm.) y mm 3

4 4. [0 points] Three light bulbs are connected in series with a battery, as shown. Initially the switch is open. A B C (a) When the switch is closed, describe what happens to the brightness of each light: Does it stay the same? Get brighter? Get dimmer? Turn off entirely? Let R b be the resistance of one light bulb. Open switch. There are three bulbs in series, so the total resistance of the circuit is 3R b. The current is I = V R = V 3R b. The power consumed by each bulb is P = I 2 R b = V 2 9R b. The total power consumed by three bulbs, which equals the total power supplied by the battery, is three times this, or V 2 3R b. Closed switch. The closed switch is a short circuit across the third bulb. There cannot be a voltage difference across that bulb, so no current flows through it, and it can be ignored. The circuit, then, effectively has two bulbs in series. The total resistance is 2R b. The current is I = V 2R b. The power consumed by each of bulbs A and B is P = I 2 R b = V 2 4R b. The total power consumed by two bulbs, which equals the total power supplied by the battery, is two times this, or V 2 2R b. The answer. Bulbs A and B each consume more power when the switch is closed, hence bulbs A and B are brighter. Bulb C has no current, hence bulb C turns off entirely. (b) When the switch is closed, how much power does the battery have to supply? Your answer should be the ratio of power supplied with the switch closed to the power supplied with the switch open, P closed /P open. Based on the analysis in part (a), P closed P open = V 2 2R b V 2 3R b = 3 2 4

5 5. [5 points] The beginning of sunrise is the moment when the Sun s rays are first visible. See the illustration below. It shows the view looking down upon the Earth from above the north pole. The person in the figure is standing on the equator. He has been on the dark (night-time) side of the Earth, but the Earth s rotation is now bringing him onto the daytime side. The Sun s rays are just beginning to hit him. Rays of Light from the Sun Earth s Rotation The figure above does not account for the atmosphere. picture, but with an atmosphere surrounding the Earth. Consider the same Rays of Light from the Sun Atmosphere Earth s Earth s Rotation What is the effect of the atmosphere on the time at which sunrise occurs? Specifically, does the atmosphere cause the Sun s rays to hit someone on Earth earlier than they otherwise would, or later than they otherwise would, or what? Briefly justify your answer. (Answers given without explanations will not get credit.) Light rays are refracted by the atmosphere (see figure; the dashed line is the normal line). The sun s rays hit a person earlier than they would if there were no atmosphere. For this same reason, sunsets are later than they would be if there were no atmosphere. Earth s Atmosphere Earth s Rotation 5

6 6. [0 points] (a) Derive a formula for the temperature of the Earth s surface, T, in terms of the power output of the Sun, P, the distance from the Sun to the Earth, d, the radius of the Earth, r, and/or any physical constants. To simplify matters, ignore the effect of the Earth s atmosphere. Assume that the Earth absorbs all of the sunlight which falls on it, and that every point on the Earth s surface has the same temperature. This problem is continued on the next page... The intensity of sunlight at the distance of the Earth is I = P /(4πd 2 ). The power captured by the Earth is proportional to the cross-sectional area of the Earth, πr 2. The thermal emission by the Earth is σat 4 = σ4πr 2 T 4. The power going out of the Earth must equal the power coming into the earth, so... P out = P in σ4πr 2 T 4 = P πr 2 4πd 2 σ4t 4 = P 4πd 2 T 4 = T = P 6σπd 2 ( P 6σπd 2 ) /4 An interesting thing to notice is that the temperature does not depend on the size of the Earth. Any object at the same distance from the Sun will reach the same equilibrium temperature (assuming it has no internal sources of energy, that it is a blackbody, that there is no atmospheric warming effect, etc.) You weren t asked for a numerical answer to this problem, but if you plug in P = W and d =.50 0 m, given on the next page, and σ = J K 4 m 2 s, given on the equation sheet, you find a temperature of 278 K, which is very close to the average surface temperature of the Earth. As we discussed in class, it is somewhat lucky that we get the right answer with this simple calculation. In fact, some of the sunlight which hits the Earth is reflected away, so the total power coming into the earth is less than we assume in this calculation. On the other hand, the atmosphere absorbs radiation on its way out from the Earth, and it radiates energy back down upon us, heating us up. 6

7 (b) Here is an excerpt from an article titled Astronomers Find Planet Outside Solar System published in the New York Times on April 25, The most enticing property yet found outside our solar system is about 20 light years away in the constellation Libra, a team of European astronomers said today. The astronomers have discovered a planet... orbiting a dim red star known as Gliese 58. It is the smallest of the 200 or so planets that are now known to exist outside of our solar system, called extrasolar or exoplanets. Moreover, it orbits its home start within the so-called habitable zone where surface water, the staff of life, could exist if other conditions are right, said Stephane Udry of the Geneva Observatory. We are at the right place for that, said Dr. Udry, the lead author of a paper describing the discovery that has been submitted to the journal Astronomy & Astrophysics..... The planet, officially known as Gliese 58c circles the star every 3 days at a distance of about miles. According to model soft planet formation... such a planet should be composed of rock and water. The article goes on to say that, depending on the nature of this planet s atmosphere (which is unknown), it might have a surface temperature similar to that of Earth, around 280 K. Calculate the distance the new planet must be from the star it orbits, Gliese 58, assuming that the planet has the same surface temperature as the Earth. As in part (a), assume the planet has no atmosphere, and make any other assumptions you find necessary. Some of the following information may be useful: The Sun emits power at a rate P = W. The power output of the star Gliese 58 is 0.03P or W. The Earth is.50 0 m from the Sun and has radius m. Space for working this problem is given on the following page... 7

8 This page is extra space for your answer to problem 6b. Use the formula from part (a). Solve for d and plug in the power output of Gliese 58, P = W, the Earth s surface temperature of T = 280 K, and the Stefan-Boltzmann constant, σ = J K 4 m 2 s : T = T 4 = d 2 = d = ( P 6σπd 2 P 6σπd 2 P 6σπT 4 P 6σπT 4 ) /4 d = m You could also get this number by noting that d is proportional to P, so if P = 0.03P, the distance from the star to the planet must be 0.03 times the distance from the Sun to the Earth,: d = 0.03 (.5 0 m) = m. This answer is slightly different than the one given above because of rounding of the assumed temperature of 280 K in the above calculation. The New York Times article gave the distance as about 7 million miles, which is. 0 0 m. The smaller distance would give a higher temperature, 350 K, which is within the range of liquid water on Earth. Of course, there are many further complications to the story. We think of 0 C and 00 C as the freezing point and boiling point of water, so that temperatures within this range should allow liquid water to exist. However, the freezing and boiling points depend on atmospheric pressure. With a heavier atmosphere, the boiling point would be raised substantially. Once we allow for an atmosphere, we also need to consider the possibility of a greenhouse effect heating the surface. Further, we need to consider possible absorption of the star s light by the atmosphere, since a dim red star will have a much higher fraction of its output as infrared radiation than a yellower star like our Sun. 8

9 7. [0 points] For each of the two situations given below, give (i) the type of image formed (real or virtual), (ii) whether the image is upright or inverted, (iii) how tall the image is, and (iv) where the image is located (give a distance from the image to the lens, and say whether the image is on the same side of the lens as the object or the opposite side.) (a) A 20 cm tall flower is placed 30 cm to the left of a convex lens of focal length 0 cm. f = 0 cm, s 0 = 30 cm, y 0 = 20 cm. s o + s i = f 30 cm + = s i 0 cm s i = 5 cm M T = y i y o = s i s o y i cm = 5 20 cm 30 cm y i = 0 cm (i) Positive s i implies real image. (ii) Negative y i implies inverted image. (iii) Size of y i is 0 cm.. (iv) Positive s i gives location of image as 5 cm on the opposite side of the lens as the object. (b) A 20 cm tall flower is placed 30 cm to the left of a concave lens of focal length 0 cm. f = 0 cm, s 0 = 30 cm, y 0 = 20 cm. s o + s i = f 30 cm + = s i 0 cm s i = 7.5 cm M T = y i y o = s i s o y i cm = cm 30 cm y i = 5 cm (i) Negative s i implies virtual image. (ii) Positive y i implies upright image. (iii) Size of y i is 5 cm.. (iv) Negative s i gives location of image as 7.5 cm on the same side of the lens as the object. 9

10 8. [0 points] A radio speaker is on a shelf 0 m directly over your head. It puts out a continuous tone. You have various sound equipment, and you measure the frequency of the tone to be 500 Hz and the intensity of the tone to be 70 db. At time t = 0, the speaker suddenly falls off the shelf and is accelerated downward by gravity. This means that its height above you at time t is y = 0 m 4.9 m s 2 t2 and its speed at time t is v y = 4.9 m s 2 t. Note that v y is negative, since it is falling downward. The speed of sound is 343 m s. (a) What frequency do you measure for the tone as the speaker falls? Answer this in two ways: (i) give a general formula, which depends on t, and (ii) give a specific answer numerical answer for t =.0 s. (i) Use the Doppler formula. The observer speed is zero, v 0 = 0. The wave speed is v = 343 m s. The source speed is 4.9 m s 2 t. The upper sign (minus sign) is used in the denominator because the source is moving toward the observer. (Intuitively, you should expect the pitch to increase as the speakers falls toward you; this is accomplished by the minus sign.) f o = ` ± v ov ` v s v f s = 4.9 m s 2 t 343 m s «(500 Hz) = (ii) Plug t =.0 s into the equation to get f 0 = 507 Hz. 500 Hz s t (b) What intensity, in decibels, do you measure as the speaker falls? Again, answer this in two ways: (i) give a general formula, which depends on t, and (ii) give a specific answer numerical answer for t =.0 s. Let I 0 and β 0 be the intensity in Watts/m 2 and in decibels, respectively, at time t = 0. Let I and β be the intensity at time t. Let P be the power emitted by the radio. P P I 0 = 4π(0 m) 2 I = 4π(y(t)) 2 Depending on how you solve this problem, you might use the value of 70 db to calculate I 0 = W/m 2 and P = W. However, these values need not be explicitly calculated if you do the calculation as follows. Start with the definition of decibel: «I β β 0 = 0 db log I 0 2 P» 4π(y(t)) β 70 db = 0 db log 4 2 (0 m) 2 5=0 db log P (y(t)) 2 4π(0 m) " 2 # (0 m) 2 β = 70 db+0 db log (0 m 4.9 m s 2 t 2 ) 2 = 70 db+0 db log = 70 db 20 db log ` 0.49 s t 2 At t =.0 s, this gives β = 70 db + 0 db log(3.84) = 76 db. 3» ( 0.49 s t 2 ) 2 0

11 9. [0 points] Light from a laser with wavelength λ = 630 nm is split into two beams which are directed through a series of mirrors so that they follow the paths shown in the diagram. The upper and lower beam paths are identical except that the upper path goes through a glass block with index of refraction n =.500 and unknown length d. Calculate two possible values of d (other then zero) for which a bright spot (constructive interference) will be seen on the paper. d glass laser d piece of paper The key is to compare the light going through the glass block of length d on the upper path with light going through an equivalent length d of air on the lower path (drawn in the figure above, but not shown on the exam as distributed). The two paths are identical in length outside of these two regions, so the interference pattern can be worked out by comparing the difference in the light propagation within these two regions. Let λ a = 630 nm be the wavelength in air. The wavelength of the light in the glass is λ g = λ a /n = 420 nm. If the glass block has length d, there are d/λ g wavelengths within the glass. Similarly, in the length d of air, there are d/λ a wavelengths. In order to get constructive interference, the number of wavelengths must be equal or must differ by an integer. Let m be any integer. For constructive interference, d d = m λ g λ a ( d ) = m λ g λ a m d = λ g λ a d = m 420 nm 630 nm d = m(260 nm) Thus the answers are any numbers from the sequence 260 nm, 2520 nm, 3780 nm, etc. Another way to look at this is that the answer is any distance which is a multiple of both 420 nm and of 630 nm.

12 0. [5 points] Two resistors and two inductors are wired up as shown. Initially, both switches are open. switch 0V 0Ω switch 2 20Ω 0 H 0 H (a) At t = 0, switch is closed. Switch 2 remains open. What is the current through the 0 Ω resistor immediately after it is closed? The current immediately after the switch is closed is zero because the current was zero before the switch was closed, and the inductor prevents a sudden change in current. (b) If no other switches were opened or closed, what would the current through the 0 Ω resistor be after a long time, that is, as t? A long time after the switch is closed, the current is steady, so there is no voltage drop across the inductor. The entire 0 V is across the 0 Ω resistor, so the current is I = V/R = 0/0 = amp. (c) Write a formula for the current through the 0 Ω resistor as a function of time, I (t) assuming that no switches are opened or closed after time t = 0. The time constant is τ = L/R = 0/0 = s. The current starts at zero and rises to amp, so the formula is t/ I (t) = ( amp) ( e s) (d) Write a formula for the current through the 20 Ω resistor as a function of time, I 2 (t) assuming that no switches are opened or closed after time t = 0. Follow the same reasoning as parts (a) through (c), but note that the final current is I = V/R = 0/20 = 0.5 amp, and the time constant is τ = L/R = 0/20 = 0.5 s, so the current is t/0.5 I 2 (t) = (0.5 amp) ( e s) 2

13 (e) Guess what! We can t leave well enough alone. At time t = second, switch 2 is closed. What is the voltage difference across the inductors immediately after the switch is closed? (Hints: (i) consider what about the circuit behavior must be the same immediately before and immediately after switch 2 is closed; (ii) After the switch is closed, various components are parallel to each other. Look for a way to simplify the circuit.) switch 0V 0Ω switch 2 20Ω 0 H 0 H V=? This is tricky! From parts (c) and (d), we can calculate the currents through the inductors just before switch 2 is closed, at t = : ( I (t) = ( amp) e t/ s) ( = ( amp) e / s) = amp ( I 2 (t) = (0.5 amp) e t/0.5 s) ( = (0.5 amp) e /0.5 s) = amp The total current flowing through the circuit just before switch 2 is closed is I = amp amp =.064 amp. The same total current must flow immediately after the switch is closed. After the switch is closed, the two resistors are parallel to one another. Thus their total resistance is R R 2 /(R + R 2 ) = 6.67 Ω. The voltage across the resistors is, therefore, by V = IR = (.064)(6.67) = 7.0 V. The voltage across the inductors is the battery voltage minus the voltage across the resistors, 0.00 V 7.0 V = 2.90 V. 3