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Get Full Access to Contemporary Abstract Algebra - 8 Edition - Chapter 12 - Problem 32e
Get Full Access to Contemporary Abstract Algebra - 8 Edition - Chapter 12 - Problem 32e

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# Let n be an integer greater than 1. In a ring in which xn

ISBN: 9781133599708 52

## Solution for problem 32E Chapter 12

Contemporary Abstract Algebra | 8th Edition

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Contemporary Abstract Algebra | 8th Edition

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Problem 32E

Let n be an integer greater than 1. In a ring in which xn = x for all x, show that ab = 0 implies ba = 0.

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Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 problems will be included in Exam 3 ) Due: 9:00pm on Friday, March 18, 2016 To understand how points are awarded, read therading Policy for this assignment. Locate the Object Part A You stand near the edge of a swimming pool and observe through the water an object lying on the bottom of the pool. Which of the following statements correctly describes what you see Hint 1. How to approach the problem When you look at the object, light rays coming from the object travel through the water, cross the water-air interface, and reach your eyes. Because of the difference in index of refraction between water and air, a ray coming from below the water surface changes in direction at the water-air interface. Your eyes, however, can see only the direction of the ray emerging from the water surface, and nothing about that ray tells you that it bent as it crossed the water-air interface. Use the law of refraction to determine how the direction of propagation of the ray crossing the water-air interface changes. Finally, to determine the apparent position of the object, you may find it useful to draw a picture that includes a light ray traveling from the object, through the water-air interface, to your eyes. Hint 2. Find the change in direction of a ray passing through the water-air interface Consider a light ray traveling from water (n=1.33) into air (n=1.00). How does its direction of propagation change as it passes through the water-air interface Hint 1. Law of refraction The law of refraction gives us a relation between the indexes of refraction of the materials and the sines of the angles the light rays make with the normal to the interface. In particular, if light propagates from a material with index of refraction \texttip{n_{\rm a}}{n_a} into a material with index of refraction \texttip{n_{\rm b}}{n_b}, then \large{\frac{\sin\theta_{\rm a}}{\sin\theta_{\rm b}}=\frac{n_{\rm b}}{n_{\rm a}}}, where \texttip{\theta _{\rm a}}{theta_a} is the angle that the incident ray makes with the normal to the interface between the two materials and \texttip{\theta _{\rm b}}{theta_b} is the angle between the refracted ray and the same normal. ANSWER: The refracted ray bends away from the normal to the interface. The refracted ray bends toward the normal to the interface. The refracted ray travels along the same direction as the incident ray. Hint 3. A sketch of the situation As you look through the water, the apparent position of the object, Q', is different from its actual position Q. Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... ANSWER: The apparent depth of the object is less than the real depth. The apparent depth of the object is greater than the real depth. There is no difference between the apparent depth and the actual depth of the object. Correct Because of the difference in index of refraction between water and air, the actual position of the object is quite different from its apparent position to you. Since your eyes can see only the direction of a light ray emerging from the water surface, and nothing about that ray tells you that it bent as it crossed the water-air interface, the object will appear closer to the surface. This phenomenon makes a swimming pool always look shallower than it really is! Part B In the evening you go back to the swimming pool with your friends. The object is still lying on the bottom of the pool in the same location as before, but since it's dark now, you use a flashlight to show it to your friends. Remembering the apparent location of the object in daylight, where should you aim to illuminate the object with the flashlight Assume you hold the flashlight at eye level. Hint 1. How to approach the problem Recall that the path of a refracted ray is reversible. That is, light follows the same path when going from water into air as when going from air into water. Therefore, you have to aim at a point so that light rays coming from the flashlight would refract through the same angle as the rays that reached your eyes when you observed the object in daylight. ANSWER: Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... You should aim the flashlight at the same location as that of the apparent image seen in daylight. You should aim the flashlight at a point below (closer to the bottom of the pool than) the location of the apparent image seen in daylight. You should aim the flashlight at a point above (closer to the surface of the water than) the location of the apparent image seen in daylight. Correct Refracted Waves in Unknown Materials Part A When monochromatic light passes through the interface between two unknown materials at an angle \texttip{\theta }{theta} where 0^\circ< \theta < 90^\circ, no changes in the direction of propagation of light are observed. What can be said about the two materials Check all that apply. Hint 1. How to approach the problem As described by the law of refraction, when light propagates through the interface between two materials with different indexes of refraction, the direction of propagation changes. If no such phenomenon is observed, it means that the two materials have very similar optical properties. Are they necessarily exactly the same material ANSWER: The two materials have matching indexes of refraction. The second material through which light propagates has a lower index of refraction. The second material through which light propagates has a higher index of refraction. The two materials are identical. Correct By simply observing no change in the direction of propagation of light at the interface between two materials, you cannot be certain that the two materials are identical. In fact, different materials with matching indexes of refraction would produce a similar effect. For example, at the interface between glass Pyrex (n=1.474) and glycerin (n=1.473), no relevant change in the direction of propagation of light can be observed. Part B The same monochromatic light passes through the interface between two other unknown materials. This time the transmitted wave is observed to be farther from the normal to the interface than the incident wave. What can be said about these two materials and the light traveling through them Check all that apply. Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Hint 1. How to approach the problem If the transmitted wave is observed to be farther from the normal of the interface than the incident wave, you can determine how the angle between the transmitted ray and the normal to the interface relates to the angle between the incident ray and the normal. Then, you can use the law of refraction to derive a relation between the indexes of refraction of the two materials. Once you have information on the indexes of refraction of the materials, you can determine how the speed of light changes as it propagates through the interface. Hint 2. Law of refraction The law of refraction gives us a relation between the indexes of refraction of the materials and the sines of the angles that the ray of light makes with the normal to the interface. In particular, if light propagates from a material with index of refraction \texttip{n_{\rm a}}{n_a} into a material with index of refraction \texttip{n_{\rm b}}{n_b}, then \large{\frac{\sin\theta_{\rm a}}{\sin\theta_{\rm b}}=\frac{n_{\rm b}}{n_{\rm a}}}, where \texttip{\theta _{\rm a}}{theta_a} is the angle that the incident ray makes with the normal to the interface between the two materials and \texttip{\theta _{\rm b}}{theta_b} is the angle between the refracted ray and the same normal. Hint 3. Find how the sines of the angles relate Let \texttip{\theta _{\rm a}}{theta_a} be the angle between the incident ray and the normal to the interface, and let \texttip{\theta _{\rm b}}{theta_b} be the angle between the transmitted ray and the same normal. If the transmitted ray is farther from the normal to the interface than the incident ray, which of the following relations is correct Hint 1. How to approach the problem You may find it helpful to draw a sketch of the incident and transmitted rays, the interface, and the normal to the interface. Note that if the transmitted ray is farther from the normal to the interface than the incident ray, it means that it bends away from the normal. Hint 2. A sketch of the incident and transmitted rays ANSWER: Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... \sin\theta_{\rm a} = \sin\theta_{\rm b} \sin\theta_{\rm a} > \sin\theta_{\rm b} \sin\theta_{\rm a} < \sin\theta_{\rm b} Hint 4. Find in which material light travels faster If the same light propagates through two distinct materials with different indexes of refraction, in which material will light travel fastest Hint 1. Index of refraction The index of refraction \texttip{n}{n} of a material is the ratio of the speed of light $$\texttip{c}{c}$$ in vacuum to the speed of light $$\texttip{v}{v}$$ in the material, or $$\large{n=\frac{c}{v}}$$. In other words, the speed of light in a material is inversely proportional to the index of refraction of the material. ANSWER: Light travels fastest in the material with the highest index of refraction. Light travels fastest in the material with the lowest index of refraction. The speed of light is the same in both materials. ANSWER: The second material through which the light propagates has a higher index of refraction. The second material through which the light propagates has a lower index of refraction. As the light passes into the second material, its speed increases. As the light passes into the second material, its speed decreases. Correct When a ray of light passes into a material having a lower index of refraction, and hence a higher wave speed, the refracted ray bends away from the normal to the interface and will appear closer to the interface than the incident ray. ± Refraction through Glass and Water A plate of glass with parallel faces having a refractive index of 1.58 is resting on the surface of water in a tank. A ray of light coming from above in air makes an angle of incidence 35.0 $${\rm ^\circ}$$ with the normal to the top surface of the glass. Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Part A What angle $$\texttip{\theta _{\rm 3}}{theta_3}$$ does the ray refracted into the water make with the normal to the surface Use 1.33 for the index of refraction of water. Express your answer in degrees. Hint 1. How to approach the problem Use Snell's law to determine the angle of refraction that the ray makes as it goes from the air into the glass. Then repeat this process as you follow the light ray from the glass into the water. Hint 2. Snell's law Snell's law states that $$n_1 \sin(\theta_1) = n_2 \sin(\theta_2),$$ where $$\texttip{\theta _{\rm 1}}{theta_1}$$ is the angle the light ray makes with the normal to the surface as it propagates through a material of index of refraction $$\texttip{n_{\rm 1}}{n_1}$$, and $$\texttip{\theta _{\rm 2}}{theta_2}$$ is the angle with respect to the normal that the continuing beam makes as it propogates through the material of index of refraction $$\texttip{n_{\rm 2}}{n_2}$$. ANSWER: $$\texttip{\theta _{\rm 3}}{theta_3}$$ = 25.5$$^\circ$$ Correct Note that the index of refraction of the glass doesn't figure at all in finding the angle of refraction in water. The mathematics reduces to: $$n_1\sin(\theta_1) = n_2\sin(\theta_2) = n_3\sin(\theta_3)$$ or $$n_1\sin(\theta_1) = n_3\sin(\theta_3)$$. The final angle is the same as if the light ray were passing directly from the air into the water. Problem 26.59 Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... You have a semicircular disk of glass with an index of refraction of $$\texttip{n}{n}$$ = 1.47. Part A Find the incident angle $$\theta$$ for which the beam of light in the figure will hit the indicated point on the screen. ANSWER: $$\theta$$ = 20.9 $$^\circ$$ Correct Total Internal Reflection Conceptual Question Consider scenarios A to F in which a ray of light traveling in material 1 is incident onto the interface with material 2. Material 1 ($$\texttip{n_{\rm Material 2 (\(\texttip{n_{\rm 1}}{n_1}$$) 2}}{n_2}\)) Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro..https://session.masteringphysics.com/myct/assignmentPrintViewdispl... A air (1.00) water (1.33) B water (1.33) air (1.00) C diamond (2.42) air (1.00) D air (1.00) quartz (1.46) E benzene (1.50) water (1.33) F diamond (2.42) water (1.33) Part A For which of these scenarios is total internal reflection possible List all correct answers in alphabetical order. For example, if scenarios A and E are correct, enter AE. Hint 1. Total internal reflection Total internal reflection occurs when, mathematically, the refracted angle is undefined (meaning, the sine of the refracted angle is greater than 1, which has no solution). This means, physically, that there is no refracted ray at all! Therefore, all of the light is reflected. Hint 2. Angle of refraction Total internal reflection is only possible if the refracted ray is bent farther from the normal than the incident ray is from the normal. Hint 3. Determine the role of the indices of refraction Snell's law says that for a light ray passing from region 1 to region 2, $$n_1\sin(\theta_1)=n_2\sin(\theta_2)$$, where $$\texttip{\theta _{\rm 1}}{theta_1}$$ is the angle of the incident ray from the normal, and $$\texttip{\theta _{\rm 2}}{theta_2}$$ is the angle of the refracted ray from the normal. If $$\theta_2>\theta_1$$, meaning that the refracted ray is bent farther from the normal than the incident ray, what is the relationship between $$\texttip{n_{\rm 1}}{n_1}$$ and $$\texttip{n_{\rm 2}}{n_2}$$ ANSWER: $$n_1n_2$$ $$n_1=n_2$$ ANSWER: Correct Part B Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... For the scenarios in which total internal reflection is possible, rank the scenarios on the basis of the critical angle, the angle above which total internal reflection occurs. At this angle, the refracted ray is at 90 $${\rm degrees}$$ from the normal. Rank from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Apply Snell's law At the critical angle, the refracted ray is at 90 $${\rm degrees}$$ from the normal. Therefore, Snell’s law can be used to determine the critical angle. Snell's law states that $$n_1\sin(\theta_1)=n_2\sin(\theta_2)$$, where $$\texttip{\theta _{\rm 1}}{theta_1}$$ is the angle of the incident ray from the normal, and $$\texttip{\theta _{\rm 2}}{theta_2}$$ is the angle of the refracted ray from the normal. Solve this equation for the sine of the critical angle $$\texttip{\theta _{\rm c}}{theta_c}$$. Express your answer in terms of $$\texttip{n_{\rm 1}}{n_1}$$ and $$\texttip{n_{\rm 2}}{n_2}$$. ANSWER: $$\sin(\theta_{\rm c})$$ = $$\large{\frac{n_{2}}{n_{1}}}$$ ANSWER: Correct Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... ± A Laser and a Lens A laser is mounted as shown in the figure (distances $$\texttip{d}{d}$$ and $$\texttip{h}{h}$$ are given) above the axis of a converging lens with positive focal length $$\texttip{f}{f}$$. The laser beam travels parallel to the axis of the lens. A large screen is placed at a distance $$\texttip{d_{\rm s}}{d_s}$$ to the right of the lens. The laser beam passes through the lens and makes a dot on the screen at a distance $$\texttip{h_{\rm s}}{h_s}$$, measured upward from the axis of the lens. Assume that a positive value means that the dot is above the axis, while a negative value means that the dot is below the axis. Part A Find the distance $$\texttip{h_{\rm s}}{h_s}$$. Express your answer in terms of some or all of the variables $$\texttip{d}{d}$$, $$\texttip{h}{h}$$, $$\texttip{f}{f}$$, and $$\texttip{d_{\rm s}}{d_s}$$. Note that not all of the variables may be required for the answer. Hint 1. How to approach the problem Make your own diagram and draw the refracted ray. Then use the geometry of similar triangles to find the position of the dot on the screen. Hint 2. Drawing the refracted ray Since the incident ray is parallel to the axis of the lens, the refracted ray passes through the focal point on the right side of the lens. Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro... https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Hint 3. Find a pair of similar triangles Once you have drawn your own diagram with the refracted ray shown, similar to the picture shown here, consider the two similar triangles ACF and EFG that the refracted ray forms with the axis of the lens, one on each side of the focal point F. The triangle ACF has sides of length $$\texttip{h}{h}$$ and $$\texttip{f}{f}$$. What are the lengths of the corresponding sides of the triangle EFG ANSWER: $$\texttip{h_{\rm s}}{h_s}$$ and $$\texttip{f}{f}$$ $$\texttip{h_{\rm s}}{h_s}$$ and $$\texttip{d_{\rm s}}{d_s}$$ $$\texttip{h_{\rm s}}{h_s}$$ and $$d_s -f$$ $$\texttip{h}{h}$$ and $$\texttip{h_{\rm s}}{h_s}$$ ANSWER: $$\texttip{h_{\rm s}}{h_s}$$ = $$\large{\frac{h}{f}\left(f-d_{s}\right)}$$ Correct Notice that the sign of the height depends upon the sign of $$d_s-f$$. If the focal point is in front of the screen, the dot will be below the axis of the lens. If the focal point is on the screen, the image will be at the level of the axis of the lens, and if the focal point is behind the screen, the dot will be above the axis of the lens. Now consider a specific case. Let the laser be 50 centimeters to the left of the lens and at height 15 centimeters above the axis of the lens. The lens has focal length 30 centimeters, and the screen is 1 meter to the right of the lens. Part B What is the position $$\texttip{h_{\rm s}}{h_s}$$ of the dot on the screen Express your answer in centimeters, to two significant figures. ANSWER: $$\texttip{h_{\rm s}}{h_s}$$ = -35 $$\rm cm$$ Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro.https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Correct Part C If the laser is moved farther from the lens, what happens to the dot on the screen Hint 1. How to approach the problem Consider the expression found in Part A. Does the position of the dot on the screen depend on the distance of the laser from the lens ANSWER: It moves up. It moves down. It does not move. Correct Part D If the laser is moved up, farther above the axis of the lens, what happens to the dot on the screen Assume that the ray still strikes the lens. Hint 1. How to approach the problem Recall the expression found in Part A. If you double the value of $$\texttip{h}{h}$$, how is $$\texttip{h_{\rm s}}{h_s}$$ affected ANSWER: It moves up. It moves down. It does not move. Correct Problem 26.77 A concave lens has a focal length of -27 $${\rm {\rm cm}}$$ . Find the image distance and magnification that results when an object is placed 29 $${\rm {\rm cm}}$$ in front of the lens. Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 phttps://session.masteringphysics.com/myct/assignmentPrintViewdispl... Part A Express your answer using two significant figures. ANSWER: $$d_{\rm i}$$ = -14$${\rm cm}$$ Correct Part B Express your answer using two significant figures. ANSWER: $$m$$ = 0.48 Correct Problem 26.78 When an object is located 42 $${\rm {\rm cm}}$$ to the left of a lens, the image is formed 21 $${\rm {\rm cm}}$$ to the right of the lens. Part A What is the focal length of the lens Express your answer using two significant figures. ANSWER: $$f$$ = 14 $${\rm cm}$$ Correct Problem 26.79 An object with a height of 2.48 $${\rm {\rm cm}}$$ is placed 36.0 $${\rm {\rm mm}}$$ to the left of a lens with a focal length of 32.9 $${\rm {\rm mm}}$$ . Part A Where is the image located ANSWER: Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 prohttps://session.masteringphysics.com/myct/assignmentPrintViewdispl... $$d_{\rm i}$$ = 0.382$${\rm m}$$ Correct Part B Where is the image located ANSWER: The image is located to the right of the lens. The image is located to the left of the lens. Correct Part C What is the height of the image ANSWER: $$h_{\rm i}$$ = 26.3$${\rm cm}$$ Correct Problem 26.82 IP You have two lenses at your disposal, one with a focal length $$f_1 =$$ 43.5 $${\rm {\rm cm}}$$ , the other with a focal length $$f_2 =$$-43.5 $${\rm {\rm cm}}$$ . Part A Which of these two lenses would you use to project an image of a lightbulb onto a wall that is far away ANSWER: the lens with a focal length $$f_1$$ the lens with a focal length $$f_2$$ Correct Part B Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 phttps://session.masteringphysics.com/myct/assignmentPrintViewdispl... If you want to produce an image of the bulb that is enlarged by a factor of 2.05, how far from the wall should the lens be placed ANSWER: $$d_{\rm i}$$ = 1.33$${\rm m}$$ All attempts used; correct answer displayed Problem 27.4 Part A Approximating the eye as a single thin lens 2.55 $${\rm {\rm cm}}$$ from the retina, find the eye's near-point distance if the smallest focal length the eye can produce is 2.25 $${\rm {\rm cm}}$$ . ANSWER: $$d_{\rm 0}$$ = 19.1 $${\rm cm}$$ Correct Problem 27.28 A converging lens of focal length 8.060 $${\rm cm}$$ is 20.0 $${\rm cm}$$ to the left of a diverging lens of focal length -6.54 $${\rm cm}$$ . A coin is placed 12.2 $${\rm cm}$$ to the left of the converging lens. Part A Find the location of the coin's final image. ANSWER: to the right of the diverging lens between the lenses to the left of the converging lens Correct Part B Express your answer using two significant figures. ANSWER: Weekly Assignment 08 (First 5 problems included in Exam 2; Last 7 pro.https://session.masteringphysics.com/myct/assignmentPrintViewdispl... $$d_{\rm i}$$ = 8.8 $$\rm cm$$ to the right of the diverging lens Correct Part C Find the magnification of the coin's final image. Express your answer using two significant figures. ANSWER: $$m$$ = -4.6 Correct Score Summary: Your score on this assignment is 91.7%. You received 11 out of a possible total of 12 points.

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