Why is a chain reaction more likely in a big piece of uranium than it is in a small piece?

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...