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Get Full Access to Essential Calculus - 2 Edition - Chapter 11 - Problem 4
Get Full Access to Essential Calculus - 2 Edition - Chapter 11 - Problem 4

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# Solved: Sketch the graph of the function

ISBN: 9781133112297 149

## Solution for problem 4 Chapter 11

Essential Calculus | 2nd Edition

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Essential Calculus | 2nd Edition

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Problem 4

Sketch the graph of the function

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Short Assignment By 3/25/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Short Assignment By 3/25/2016 Due: 11:00am on Friday, March 25, 2016 To understand how points are awarded, read therading Policy for this assignment. Constructive and Destructive Interference Conceptual Question Two sources of coherent radio waves broadcasting in phase are located as shown below. Each grid square is 0.5 square, and the radio sources broadcast at . Part A At Point A is the interference between the two sources constructive or destructive Hint 1. Path-length difference Since the two sources emit radio waves in phase, the only possible phase difference between the waves at various points is due to the different distances the waves have traveled to reach those points. The difference in the distances traveled by the two waves from source to point of interest is termed the path-length difference. If the path-length difference is an integer multiple of the wavelength of the waves, one wave will pass through an integer number of complete cycles more than the other wave, placing the two waves back in perfect synchronization, resulting in constructive interference. If the path-length difference is a half-integer multiple of the wavelength, one wave will be one-half of a cycle, or 180 , out of phase, resulting in destructive interference. Hint 2. Find the path-length difference What is the distance from the left source to Point A, What is the distance from the right source to Point A, Enter the distances in meters separated by a comma. ANSWER: , = 3,3 , ANSWER: Short Assignment By 3/25/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... constructive destructive Correct Part B At Point B is the interference between the two sources constructive or destructive Hint 1. Find the path-length difference What is the distance from the left source to Point B, What is the distance from the right source to Point B, Enter the distances in meters separated by a comma. ANSWER: , = 1.5,4.5 , ANSWER: constructive destructive Correct Part C At Point C is the interference between the two sources constructive or destructive ANSWER: constructive destructive Correct Part D At Point D is the interference between the two sources constructive or destructive ANSWER: Short Assignment By 3/25/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... constructive destructive Correct Problem 28.22 Light with a wavelength of 516 passes through two slits and forms an interference pattern on a screen 8.away. Part A If the linear distance on the screen from the central fringe to the first bright fringe above , what is the separation of the slits ANSWER: = 81.8 Correct Score Summary: Your score on this assignment is 100%. You received 2 out of a possible total of 2 points. Short Assignment By 3/21/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Short Assignment By 3/21/2016 Due: 11:00am on Monday, March 21, 2016 To understand how points are awarded, read the Grading Policy for this assignment. ± Find the Wavelength Assume the following waves are propagating in air. Part A 21 Calculate the wavelength $$\texttip{\lambda _{\rm 1}}{lambda_1}$$ for gamma rays of frequency $$\texttip{f_{\rm 1}}{f_1}$$ $${\rm Hz}$$ . Express your answer in meters. Hint 1. How to set up the problem Recall the formula $$c=\lambda f$$. ANSWER: −14 $$\texttip{\lambda _{\rm 1}}{lambda_1}$$4.11×10 $$\rm{m}$$ Correct Part B Now express this gamma-ray wavelength in nanometers. Express your answer in nanometers. Hint 1. Relation between meters and nanometers $$1\;{\rm meter} = 10^{9}\;{\rm nanometers}$$. ANSWER: $$\texttip{\lambda _{\rm 1}}{lambda_1}$$4.11×10−5 $$\rm nm$$ Correct Part C Calculate the wavelength $$\texttip{\lambda _{\rm 2}}{lambda_2}$$ for visible light of frequency $$\texttip{f_{\rm 2}}{f_2\({\rm Hz}$$ . Express your answer in meters. Hint 1. How to set up the problem Recall the formula $$c=\lambda f$$. ANSWER: $$\texttip{\lambda _{\rm 2}}{lambda_2}$$4.84×10−7 $$\rm{m}$$ Correct Part D Now express this visible wavelength in nanometers. Short Assignment By 3/21/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Express your answer in nanometers. Hint 1. Relation between meters and nanometers $$1\;{\rm meter} = 10^{9}\;{\rm nanometers}$$. ANSWER: $$\texttip{\lambda _{\rm 2}}{lambda_2}$$ = 484 Correct Electromagnetic Waves Ranking Task Part A Rank these electromagnetic waves on the basis of their speed (in vacuum). Rank from fastest to slowest. To rank items as equivalent, overlap them. Hint 1. Relating speed, frequency, and wavelength Like all waves, the relationship among wave speed, frequency, and wavelength is . ANSWER: Correct Part B Rank these electromagnetic waves on the basis of their wavelength. Rank from longest to shortest. To rank items as equivalent, overlap them. Hint 1. Electromagnetic spectrum Different wavelength electromagnetic waves have historically been given different names. The traditional names for the various wavelengths are listed below. Hint 2. Radio waves By examining a radio dial, you will discover that FM radio stations broadcast with frequencies between 88 and 108 (megahertz, or millions of cycles per second) and AM radio stations broadcast between 520 and 1720 (kilohertz, or thousands of cycles per second). Short Assignment By 3/21/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Hint 3. Visible light The lowest frequency visible light is red, and the frequencies of visible light increase in the order: red, orange, yellow, green, blue, and then violet. ANSWER: Correct Part C Rank these electromagnetic waves on the basis of their frequency. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Correct Score Summary: Your score on this assignment is 100%. You received 2 out of a possible total of 2 points. Short Assignment By 3/16/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Short Assignment By 3/16/2016 Due: 11:00am on Wednesday, March 16, 2016 To understand how points are awarded, read theg Policy for this assignment. Problem 27.25 Two concave lenses, each with \texttip{f}{f} = -13 {\rm cm} , are separated by 4.7 {\rm cm} . An object is placed 23 {\rm cm} in front of one of the lenses. Part A Find the location of the final image produced by this lens combination. ANSWER: in front of the lens closest to the object beyond the lens farest to the object between the two lenses Correct Part B Express your answer using two significant figures. ANSWER: d_{\rm i} = 1.8 \rm cm in front of the lens closest to the object All attempts used; correct answer displayed Part C Find the magnification of the final image produced by this lens combination. Express your answer using two significant figures. ANSWER: m = 0.18 All attempts used; correct answer displayed Score Summary: Your score on this assignment is 33.3%. You received 0.67 out of a possible total of 2 points. Short Assignment By 3/23/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Short Assignment By 3/23/2016 Due: 11:00am on Wednesday, March 23, 2016 To understand how points are awarded, read theGrading Policy for this assignment. Polarization of Light and Malus's Law Learning Goal: To understand polarization of light and how to use Malus's law to calculate the intensity of a beam of light after passing through one or more polarizing filters. The two transverse waves shown in the figure both travel in the +z direction. The waves differ in that the top wave oscillates horizontally and the bottom wave oscillates vertically. The direction of oscillation of a wave is called the polarization of the wave. The upper wave is described as polarized in the +x direction whereas the lower wave is polarized in the +y direction. In general, waves can be polarized along any direction. Recall that electromagnetic waves, such as visible light, microwaves, and X rays, consist of oscillating electric and magnetic fields. The polarization of an electromagnetic wave refers to the oscillation direction of the electric field, not the magnetic field. In this problem all figures depicting light waves illustrate only the electric field. A linear polarizing filter, often just called a polarizer, is a device that only transmits light polarized along a specific transmission axis direction. The amount of light that passes through a filter is quantified in terms of its intensity. If the polarization angle of the incident light matches the transmission axis of the polarizer, 100 of the light will pass through, so the transmitted intensity will equal the incident intensity. More generally, the intensity of light emerging from a polarizer is described by Malus's law: , where is the intensity of the polarized light beam just before entering the polarizer, is the intensity of the transmitted light beam immediately after passing through the polarizer, and is the angular difference between the polarization angle of the incident beam and the transmission axis of the polarizer. After passing through the polarizer, the transmitted light is polarized in the direction of the transmission axis of the polarizing filter. In the questions that follow, assume that all angles are measured counterclockwise from the +x axis in the direction of the +y axis. Part A A beam of polarized light with intensity and polarization angle strikes a polarizer with transmission axis . What angle should be used in Malus's law to calculate the transmitted intensity This process is illustrated in the figure , where the polarization of the light wave is visually illustrated by a magenta double arrow oriented in the direction of polarization, the transmission axis of the polarizer is represented by a blue double arrow, and the direction of motion of the wave is illustrated by a purple arrow. Short Assignment By 3/23/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... ANSWER: Correct Part B What is the polarization angle of the light emerging from the polarizer ANSWER: Correct Part C If = 20.0 , = 25.0 , and = 40.0 , what is the transmitted intensit Express your answer numerically in watts per square meter . ANSWER: = 18.7 Correct If the polarization axis of the incident light and the transmission axis of the filter are aligned (so ) or if they point in exactly opposite directions (so ), then in Malus's law, indicating that 100 of the incident intensity will be transmitted. Short Assignment By 3/23/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Most natural light sources emit "unpolarized" light, perhaps better described as "randomly polarized" light. These light sources emit numerous brief bursts of light whose polarization directions are unrelated, so on average the resulting beam has all polarization angles equally represented. When unpolarized light with intensity passes through a polarizer, its intensity is cut in half, regardless of the orientation of the transmission axis of the polarizer: . Part D One way to produce a beam of polarized light with intensity and polarization angle would be to pass unpolarized light with intensity through a polarizer whose transmission axis is oriented such that . How large must be if the transmitted light is to have intensity Express your answer as a decimal number times the symbol . For example, if , enter 0.25 * I. ANSWER: = Correct Part E A beam of unpolarized light with intensity falls first upon a polarizer with transmission axis then upon a second polarizer with transmission axis , where (in other words the two axes are perpendicular to one another). What is the intensity of the light beam emerging from the second polarizer Express your answer as a decimal number times the symbol . For example, if , enter 0.25 * I_0. Short Assignment By 3/23/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Hint 1. How to approach the problem Consider doing the problem in two steps. First, determine the intensitand polarization angle of the light emerging from the first polarizer. Then calculate the intensiof the light emerging from the second polarizer based on those answers as well as on the transmission axis angle of the second polarizer, . Hint 2. Determine the intensity of light between the two polarizers What is the intensity of the light emerging from the first polarizer, before it attempts to pass through the second polarizer Express your answer as a decimal number times the symbol . For example, if , enter 0.25 * I_0. ANSWER: = Hint 3. Determine the angle to be used in Malus's law When Malus's law is applied to calculate the intensity of light that emerges from the second polarizer, what angle should be used Recall that all angles are measured counterclockwise from the +x axis in the direction of the +y axis. Hint 1. Find the orientation of the light between the two polarizers What is , the polarization of the light after it has passed through the first polarizer Express your answer numerically in degrees. ANSWER: = 0 ANSWER: 0 degrees 45 degrees 90 degrees 180 degrees ANSWER: = Short Assignment By 3/23/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Correct Adjacent polarizers with perpendicular transmission axes are said to be "crossed." No light can get through both polarizers, regardless of the light's initial polarization state. Part F Notice that a polarizer modifies the light intensity according to Malus's law and also reorients the polarization angle of the beam to match its own transmission axis. Hence it is possible for light to pass through a pair of crossed polarizers if a third polarizer is inserted between them with an intermediate transmission axis direction. What is the new intensity of the light emerging from the final polarizer in Part E if a third polarizer (Polarizer A in the figure ), whose transmission axis is offset 45 from each of the others, is inserted between the original two polarizers Express your answer as a decimal number times the symbol . For example, if , enter 0.25 * I_0. Hint 1. How to approach the problem Besides the intensities , , and used in Part E, notice in the figure the new intensity immediately following the inserted polarizer. Use the rules learned in the previous parts to relate intensities across each polarizer individually. First relateto , then to , and finally to . Combine your expressions to relate to directly. Hint 2. Relate to What is the value of in terms of Express your answer as a decimal number times the symbol . For example, if , enter 0.25 * I_0. ANSWER: = Hint 3. Relate to What is the value of in terms of Express your answer as a decimal number times the symbol . For example, if , enter 0.25 * I_1. ANSWER: Short Assignment By 3/23/2016 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... = Hint 4. Relate to What is the value of in terms of Express your answer as a decimal number times the symbol . For example, if , enter 0.25 * I_A. ANSWER: = ANSWER: = Correct Polarizing filters for visible light are made of Polaroid, which contains long molecular chains that have been aligned by stretching the material during production. Polaroid is commonly used in sunglasses because it reduces the intensity of unpolarized sunlight light b. Glare is often at least partially polarized, so Polaroid sunglasses, when properly oriented, can selectively reduce glare by even more than .0 Problem 25.74 Unpolarized light passes through two polarizers whose transmission axes are at an angle of with respect to each other. Part A What fraction of the incident intensity is transmitted through the polarizers ANSWER: = 0.375 Correct Score Summary: Your score on this assignment is 100%. You received 2 out of a possible total of 2 points.

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