Problem 35E

Based on the molecular formula, determine whether each compound is an alkane, alkene, or alkyne. (Assume that the hydrocarbons are noncyclical and there is no more than one multiple bond.)

a. C5H12

b. C3H6

c. C7H12

d. C11H22

weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... weekly Assignment 10 Due: 9:00pm on Monday, April 4, 2016 To understand how points are awarded, read the Grading Policy for this assignment. ± Fringes from Different Interfering Wavelengths Coherent light with wavelength 610 passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 from the slits. The first-order bright fringe is a distance of 4.84 from the center of the central bright fringe. Part A For what wavelength of light will the first-order dark fringe (the first dark fringe next to a central maximum) be observed at this same point on the screen Express your answer in micrometers (not in nanometers). Hint 1. How to approach the problem For this problem we can use the wavelength of the first beam of light, as well as the dimensions of the interference pattern that it creates, to determine the separation of the two slits. Using this information and the dimensions of the interference pattern of the second beam of light, we can then determine the second beam's wavelength. Hint 2. Interference pattern equation The equation for the constructive interference fringes from two slits projected on a screen is , where is the distance between the two slits, is the wavelength of light, and is the angle between the constructive peak and the centerline. Note that . For the destructive interference pattern, one can use the equation , where all the variables are the same as for the case of constructive interference. Using the approximation , which is valid for small , will be helpful. Hint 3. Correct order to use for destructive interference You might have some confusion about whether to use or for the "first-order" destructive interference fringe. The correct way to look at the situation is to use the main equation for destructive interference, , and note that if we use , we get the same answer as if we use (just with a minus sign). This is weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... due to the fact that we have arbitrarily defined the equation with instead of , which would also have worked just fine. Since we are looking for the first dark fringe, we can use and . Both give the same answer for the magnitude of the angle off the centerline. ANSWER: 1.22 Correct Notice that the answer is twice the first wavelength. This makes sense, because we are dealing with the same point on the screen, so the path difference, given by , is the same for each wavelength. Since the first wavelength experiences constructive interference, the path difference must equ. Therefore, for light of wavelength , this same path difference is exactly half of its wavelength, giving destructive interference. Problem 28.21 Light from a He-Ne laser ( = 632.8 ) strikes a pair of slits at normal incidence, forming a double-slit interference pattern on a screen located 1.40 from the slits. The figure shows the interference pattern observed on the screen. Part A What is the slit separation ANSWER: = 154 Correct Problem 28.25 weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... A physics instructor wants to produce a double-slit interference pattern large enough for her class to see. For the size of the room, she decides that the distance between successive bright fringes on the screen should be at least 2.70 . Part A If the slits have a separation= 0.0250 , what is the minimum distance from the slits to the screen when 632.8- light from a He-Ne laser is used ANSWER: = 107 Correct Antireflective Coating A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). The index of refraction of the polystyrene is 1.49, and the index of refraction of the fabulite is 2.409. Part A What is the minimum thickness of film required Assume that the wavelength of the light in air is 520 nanometers. Express your answer in nanometers. Hint 1. How to approach the problem An antireflective coating works because destructive interference occurs between the light that is reflected off the surface of the coating and the light that is reflected off the coating/substrate interface. As a result, there is a specific thickness of coating in which a certain wavelength of reflected light will experience destructive interference. weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Hint 2. What is the phase shift Which statement accurately describes the phase shift of the light reflected off the coating Hint 1. Phase shift of reflection off different materials Recall that if light is reflected off a surface that has a higher index of refraction than that in which the light ray is being propagated, there will be a half-cycle phase shift. If the light reflects off a surface with lower index of refraction than that in which it is being propagated, it will not experience a phase shift. ANSWER: The light reflected off the top of the coating has no phase shift, while the light that reflects off the bottom has a half-wave phase shift. The light reflected off the top of the coating has a half-wave phase shift, and the light that reflects off the bottom also has a half-wave phase shift. The light reflected off the top of the coating has a half-wave phase shift, while the light that reflects off the bottom has no phase shift. The light reflected off the top of the coating has no phase shift, and the light that reflects off the bottom also has no phase shift. Hint 3. Find the wavelength of light in the coating Find the wavelength of the light as it propagates in the antireflective coating. Express your answer in nanometers. Hint 1. Speed of light in a material Recall that the speed of light in a material with index of refraction is given by , where is the speed of light in the material and is the speed of light in vacuum. Hint 2. Relation between speed and frequency The speed of a wave is related to its frequency and wavelength via the equation . The frequency of the light cannot change as it passes into a material (or else there would be discontinuites over time), so the wavelength of the light must decrease by a factor of the index of refraction, relative to the wavelength in vacuum, in order for the correct velocity of light in the material to be obtained. weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... ANSWER: = 349 ANSWER: 87.2 Correct Interference from Reflection off a Soap Film Part A What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 565 The index of refraction of the film is 1.33, and there is air on both sides of the film. Express your answer in nanometers. Hint 1. How to approach the problem You can obtain a black soap bubble if the soap is at just the thickness for which destructive interference occurs between the light that is reflected off the top surface of the soap and the light that is reflected off the bottom surface. As a result, there is a specific thickness of soap in which a certain wavelength of light will experience destructive interference. Hint 2. What is the phase shift Which of the following best describes the phase shifts of the light reflecting off the soap bubble weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Hint 1. Phase shift of a reflected wave Recall that if light is reflected off a surface that has a higher index of refraction than that in which the light ray is propagating, there will be a half-cycle phase shift. If it reflects off a surface with lower index of refraction than that in which it is propagating, it will not have a phase shift. ANSWER: The light reflected off the top of the soap has no phase shift, while the light that reflects off the bottom has a half-wave phase shift. The light reflected off the top of the soap has a half-wave phase shift, and the light that reflects off the bottom also has a half-wave phase shift. The light reflected off the top of the soap has a half-wave phase shift, while the light that reflects off the bottom has no phase shift. The light reflected off the top of the soap has no phase shift, and the light that reflects off the bottom also has no phase shift. Hint 3. The wavelength of light in soap Find the wavelength of the light as it propagates in the soap film. Express your answer in nanometers. Hint 1. The speed of light in a material Recall that the speed of light in a material with index of refractiois given by , where is the speed of light in the material andis the speed of light in vacuum. Hint 2. The relation between wave speed and frequency The speed of a wave is related to its frequency and wavelength via the equation . The frequency of the light cannot change as it passes into a material (or else there would be discontinuites over time), so the wavelength of the light must decrease by a factor of the index of refraction relative to the wavelength in vacuum for the velocity of light in the material to decrease correspondingly. ANSWER: = 425 ANSWER: 212 weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Correct Problem 28.40 White light is incident normally on a thin soap fi= 1.33) suspended in air. Part A What are the two minimum thicknesses that will constructively reflect yellow= 581 ) light Enter your answers in ascending order separated by a comma. ANSWER: = 109,328 Correct Part B What are the two minimum thicknesses that will destructively reflect yello= 581 ) light Enter your answers in ascending order separated by a comma. ANSWER: = 218,437 Correct ± Thin Film (Oil Slick) A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off the surface. She aims a spectrometer at a particular spot and measures the wavelength to be 750 (in air). The index of refraction of water is 1.33. Part A The index of refraction of the oil is 1.20. What is the minimum thicknof the oil slick at that spot Express your answer in nanometers to three significant figures. Hint 1. Thin-film interference In thin films, there are interference effects because light reflects off the two different surfaces of the film. In this problem, the scientist observes the light that reflects off the air-oil interface and off the oil-water interface. Think about the phase difference created between these two rays. The phase difference will arise from differences in path length, as well as differences that are introduced by certain types of reflection. Recall that if the phase difference between two waves is (a full wavelength) then the waves interfere constructively, whereas if the weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... phase difference is (half of a wavelength) the waves interfere destructively. Hint 2. Path-length phase difference The light that reflects off the oil-water interface has to pass through the oil slick, where it will have a different wavelength. The total "extra" distance it travels is twice the thickness of the slick (since the light first moves toward the oil-water interface, and then reflects back out into the air). Hint 3. Phase shift due to reflections Recall that when light reflects off a surface with a higher index of refraction, it gains an extra phase shift of radians, which corresponds to a shift of half of a wavelength. What used to be a maximum is now a minimum! Be careful, though; if two beams each reflect off a surface with a higher index of refraction, they will both get a half-wavelength shift, canceling out that effect. ANSWER: = 313 Correct Part B Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now Express your answer in nanometers to three significant figures. Hint 1. Phase shift due to reflections Keep in mind that when light reflects off a surface with a higher index of refraction, it gains an extra shift of half of a wavelength. What used to be a maximum is now a minimum! Be careful, though; if two beams reflect, they will both get a half-wavelength shift, canceling out that effect. Also, reflection off a surface with a lower index of refraction yields no phase shift. ANSWER: = 125 Correct Part C Now assume that the oil had a thickness of 200 and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver Express your answer in nanometers to three significant figures. Hint 1. How to approach the problem weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... For transmission of light, the same rules hold as before, only now one beam travels straight through the oil slick and into the water, while the other beam reflects twice (once off the oil-water interface and once again off the oil-air interface) before being finally transmitted to the water. Hint 2. Determine the wavelength of light in air Find the wavelength of the required light in air. Express your answer numerically in nanometers. ANSWER: = 600 Hint 3. Relationship between wavelength and index of refraction There is a simple relationship between the wavelength of light in one medium (with one index of refraction ) and the wavelength in another medium (with a different index of refraction ): . ANSWER: = 451 Correct This problem can also be approached by finding the wavelength with the minimum reflection. Conservation of energy ensures that maximum transmission and minimum reflection occur at the same time (i.e., if the energy did not reflect, then it must have been transmitted to conserve energy), so finding the wavelength of minimum reflection must give the same answer as finding the wavelength of maximum transmission. In some cases, working the problem one way may be substantially easier, so you should keep both approaches in mind. ± Single-Slit Diffraction You have been asked to measure the width of a slit in a piece of paper. You mount the paper centimeters from a screen and illuminate it from behind with laser light of wavelength nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance between them to be millimeters. weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Part A What is the width of the slit Express your answer in micrometers, to three significant figures. Hint 1. The equation for single-slit diffraction The equation for the angle between a line perpendicular to the screen and a line connecting the slit to thth dark fringe is , where is a nonzero integer, is the wavelength of the light, and is the width of the slit. Use this equation, together with some small-angle approximations, to solve this problem. Hint 2. Small-angle approximations Recall that for a small angle , . For this problem, find the value of , in terms of the distance to the screen and the distance from the central maximum to the dark fringe, and then substitute this expression for in the equation for location of the dark fringes. ANSWER: = 170 Correct Part B If the entire apparatus were submerged in water, would the width of the central peak change Hint 1. How to approach the problem If you are puzzled, look at each variable in your equations and ask yourself whether it would be changed by placing the whole apparatus underwater. If you identify any of the variables as changing, think about how such a change would affect the width of the central fringe (if at all). ANSWER: The width would increase. The width would decrease. The width would not change. Correct Problem 28.46 weekly Assignment 10 https://session.masteringphysics.com/myct/assignmentPrintViewdispl... Green light (= 536 ) strikes a single slit at normal incidence. Part A What width slit will produce a central maximum that is 2.00wide on a screen 1.70 from the slit ANSWER: = 91.1 All attempts used; correct answer displayed Problem 28.67 The yellow light from a helium discharge tube has a wavelength of 587.5. When this light illuminates a certain diffraction grating it produces a first-order principal maximum at an angle of 1.350 Part A Calculate the number of lines per centimeter on the grating. Express your answer using four significant figures. ANSWER: = 401.0 Correct Score Summary: Your score on this assignment is 90.0%. You received 9 out of a possible total of 10 points.