Physics 202 Problems  Week 8 Worked Problems Chapter 25: 7, 23, 36, 62, 72


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1 Physics 202 Problems  Week 8 Worked Problems Chapter 25: 7, 23, 36, 62, 72 Problem 25.7) A light beam traveling in the negative z direction has a magnetic field B = ( T )ˆx + ( T )ŷ at a given instant of time. The electric field in the beam has a magnitude of 1.40 N/C at the same time. (a) Does the electric field at this time have a z component that is positive, negative, or zero? Explain. (b) Write E in terms of unit vectors. (a) The electric field must be perpendicular to both B and the direction of propagation, ˆk = ẑ. Therefore, the electric field can have no component along the zdirection. So, Ez = 0. (b) There are two methods to solving this problem. The first is a geometric approach and the second uses the crossproduct. Both methods yield the same result, so it s really just a matter of personal preference (or your familiarity with the method). I don t recommend trying to do it with a method you are not familiar with on an exam. Method 1: Geometric From part (a), we know that the electric field must be in the xy plane and that it must be perpendicular to the magnetic field vector. We also know that E, B and ˆk obey the righthandrule. That is to say, if we take our right hand and points our fingers in the directions of E and then curl our fingers toward B, then our thumb will point in the direction of propagation, ˆk, These things are combined to give us the picture below. By geometry, the angle that E makes with the xaxis is the same as the angle that B makes with the yaxis. We can exploit this in the following way. Using trigonometry, we find that tan θ = B x B y
2 This leads us to θ = We can then use this angle to break E into its x and y components. E x = E cos θ = 1.22 N/C E y = E sin θ = 0.69 N/C E z = 0 We can then write E = (1.22 N/C)ˆx + (0.69 N/C)ŷ. Method 2: CrossProduct Maxwell s Equations, the theory behind electromagnetic waves, gives us a relationship between E, B and ˆk for plane waves. E = cˆk B OR B = ˆk E c where c = m/s is the speed of light in vacuum (or air). Since we know both ˆk and B, we can calculate E. E = c( ˆk) B = cẑ B We can simply write B out in component form We can distribute ẑ as follows E = cẑ (B xˆx + B y ŷ) E = cb x (ẑ ˆx) + cb y (ẑ ŷ) Note: We can find the crossproduct of unit vectors by simply applying the righthandrule. If we draw out the unit vectors for our coordinate system as shown below. We can simply take our right hand and point our fingers in the direction of the first vector. Then we curl our fingers toward the second vector. The direction our thumb points is the result. Since we are doing crossproducts of unit vectors, the result should also be a unit vector, which is just a direction. For example, ˆx ŷ = ẑ. This is actually how you define a righthanded coordinate system like the one drawn above. For this problem, we will need ẑ ˆx = ŷ and ẑ ŷ = ˆx
3 Finally, we find E = cb x ŷ cb y ˆx. If we evaluate this expression using the values for B x and B y we were given, we arrive at the same result as in the first method. E = (1.22 N/C)ˆx + (0.69 N/C)ŷ Problem 25.23) Consider a spiral galaxy that is moving directly away from Earth with a speed of V = m/s at its center, as shown in the figure below The galaxy is also rotating about its center, so that points in the spiral arms are moving with speed v = m/s relative to the center. If light is emitted in both arms with a frequency of Hz, what frequency is detected by astronomers observing the arms that is moving (a) toward (b) away from Earth? (Measurements of this type are used to map out the speed of various regions in distant, rotating galaxies.) We know that the observed frequency, f will be doppler shifted due to the relative motion between Earth and the arms of the spiral galaxy. The observed frequency is given by ( f = f 1 ± u ) c where, we take + when the motion is together and when the motion is apart. (a) For the arm moving toward the Earth u = V v = m/s. Therefore, f = f ( 1 + u ) f = Hz c (b) For the arm moving away from the Earth u = V + v = 10 6 m/s. Therefore, f = f ( 1 u ) f = Hz c
4 Problem 25.36) A television is tuned to a station broadcasting at a frequency of Hz. For best reception, the rabbitear antenna used by the TV should be adjusted to have a tiptotip length equal to half a wavelength of the broadcast signal. Find the optimum length of the antenna. We are trying to find the length of the antenna, L, such that it is half a wavelength long, i.e. L = λ 2 We know that wavelength and frequency are related Therefore, λf = c in vacuum (or air) Which leads us to an antenna length of λ = c f L = c 2f Finally, the optimal antenna length is L = 2.27 m. Problem 25.62) Laser Surgery: Each pulse produced by an argonfluoride eximer laser used in PRK and LASIK opthalmic surgery lasts only 10 ns but delivers an energy of 2.50 mj. (a) What is the power delivered in each pulse? (b) If the beam has a diameter of mm, what is the average intensity of the beam during each pulse? (c) If the laser emits 55 pulses per second, what is the average power it generates? (a) The power in a pulse from the laser is given by the energy delivered divided by the the time in which the energy is delivered, i.e. P pulse = U pulse t For this laser, the power delivered by a single pulse is P pulse = W. (b) The average intensity of the pulse is given by I av = P pulse A where A is the area into which the energy is flowing. cylindrical beam of light, Since the laser beam is basically a
5 Combining these we find A = πr 2 = πd2 4 I av = 4P pulse πd 2 = W m 2 (c) The average power of the laser is given by the total energy delivered per unit time. So the average power of the laser is 0.14 W. P av = U t = 55U pulse 1 s Problem 25.72) A heliumneon laser emits a beam of unpolarized light that passes through three Polaroid filters, as shown in the figure below. The intensity of the laser beam is I 0. (a) What is the intensity of the beam at point A? (b) What is the intensity of the beam at point B? (c) What is the intensity of the beam at point C? (d) If the middle filter is removed, what is the intensity of the beam at point C? (a) Since the light from the laser is unpolarized and it passes through one polarizer before point A, the intensity at point A will be half the incident intensity. I A = 1 2 I 0
6 (b) After the light passes through the first polarizer, the polarization is along the direction of the first polarizer s transmitting axis. When the light goes through the second polarizer, the transmitted intensity, or the intensity at point B, will be given by I B = I A cos 2 θ where θ is the angle between the second polarizer s transmitting axis and the polarization of the incident light. Therefore, I B = I A cos 2 (30 o ) We find that the intensity at point B is I B = 3 4 I A = 0.375I 0 (c) After the light passes through the second polarizer, the polarization is along the direction of the second polarizer s transmitting axis. When the light goes through the third polarizer, the transmitted intensity, or the intensity at point C, will be given by I C = I B cos 2 θ where θ is now the angle between the third polarizer s transmitting axis and the polarization of the incident light. Therefore, I C = I B cos 2 (60 o ) We find that the intensity at point C is I C = 1 4 I B = 0.094I 0 (d) If the second polarizer is removed, then the intensity at point C goes to zero, I C = 0. This occurs because the polarization of the light after the first polarizer is perpendicular to the transmitting axis of the third polarizer. I C = I A cos 2 θ = I A cos 2 (90 o ) = 0
7 Physics 202 Problems  Week 8 Worked Problems Chapter 26: 8, 13, 32, 84 Problem 26.8) How many times does the light beam shown in the figure below reflect from (a) the top and (b) the bottom mirror? The first thing that we will want to do is figure out how many total times the beam of light will reflect. To do this we can find the distance along the length of the mirrors between bounces, l. We can create the triangle shown above to do this. Clearly, l = d tan θ Then the total number of reflections should be the total length of the mirrors, L = 168 cm, divided by l. N tot = L l = L d tan θ = 9.22 The decimal tells us that there is a fraction of a bounce. This does not really help us so we can round down to 9. In general, for a problem like this one you should always round down to the nearest whole number to avoid over counting. So let s leave it as N tot = 9. (a) Looking back it the first drawing we can see that the bounces on the top are the odd bounces. Knowing this we can write { 2Ntop 1 if N N tot = tot is even if N tot is odd 2N top Since we have N tot is an odd number, we arrive at
8 N top = N tot + 1 = 5 2 (b) Looking back it the first drawing we can see that the bounces on the bottom are the even bounces. Knowing this we can write { 2Nbot + 1 if N N tot = tot is even if N tot is odd 2N bot Since we have N tot is an odd number, we arrive at N bot = N tot 1 2 = 4 Problem 26.13) You hold a small mirror 0.5 m in front of your eyes, as shown below The mirror is 0.32 m high, and in it you see the image of a tall building behind you. (a) If the building is 95 m behind you, what vertical height of the building, H, can be seen in the mirror at any one time? (b) If you move the answer closer to your eyes, does your answer to part (a) increase, decrease, or stay the same? Explain. (a) The furthest distance that you could see on the building is going to occur when you look at the edges of the mirror. We know that if we trace a ray from your eye to the mirror, it will reflect at the same angle that it hits the mirror as shown below.
9 We can create two triangles from the above diagram and use them to find H. From the two triangles we see that tan θ = h 2d = We can solve this for H to get our final answer. H h 2(D + d) H = h + h ( d (D + d) = h 2 + D ) d We find that the vertical height of the building that can be seen in the mirror at any one time is H = 61 m. (b) If you move the mirror closer to your eyes, then you will be able to more of the building. This is due to the fact that the height of the building that you can see is inversely proportional to d.
10 Problem 26.32) The vertical image produced by a convex mirror is onequarter the size of the object. (a) If the object is 32 cm in front of the mirror, what is the image distance? (b) What is the focal length of the mirror? (a) A convex mirror will always create a virtual (upright) image as shown below. Using the sign convention of the book, a virtual image has positive magnification. Therefore, the image distance is given by m = h i h o = d i d o = 1 4 d i = 1 4 d o and the image is formed at d i = 8 cm, so 8 cm behind the mirror. (b) Now that we have both the object and image distances, we can easily find the focal length using the mirror (thin lens) equation: Therefore, the focal length is given by 1 d o + 1 d i = 1 f ( 1 f = + 1 d o d i ) 1 We find that the focal length of the convex mirror is f = 6 cm. Note: The focal length of a convex mirror is negative, f = R 2, and the focal length of a concave mirror is positive, f = R 2, under the sign convention of the text.
11 Problem 26.84) (a) Find the two locations of where an object can be placed in front of a concave mirror with a radius of curvature of 39 cm such that its image is twice its size. (b) In each of these cases, state whether the image is real or virtual, upright or inverted. (a) For a concave mirror, both a real and virtual image can be produced that is twice the size of the object. We can say that the magnification is m = h i h o = d i d o = ±2 The positive value corresponds to the virtual (upright) image and the negative value corresponds to the real (inverted) image. From this, we find that d i = md o = ±2d o We also know that a concave mirror with a radius of curvature, R, has a focal length given by f = R 2. We also know that the mirror equation applies to the locations of the object and 1 image. d o + 1 d i = 1 f = 2 R. We can use the fact that d i = md o, to solve for d o. 1 1 = 2 d o md o R m 1 md o = 2 R d o = (m 1)R 2m We can then insert our two values of m m, ±2. We find that the object should be placed at { 9.8 cm m = 2 d o = 29 cm m = 2 This result is shown in the diagram below. (b) For the answer to part (a), m = 2 corresponds to an upright, virtual image, and m = 2 corresponds to an inverted, real image.
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