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Unformatted text preview: Husain, Zeena – Homework 12 – Due: Apr 20 2004, 4:00 am – Inst: Sonia Paban 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Albert Michelson used an improved version of the technique developed by Fizeau to measure the speed of light. In one of Michelson’s ex periments, the toothed wheel was replaced by a wheel with eight identical mirrors mounted on its perimeter, with the plane of each mirror perpendicular to a radius of the wheel. The total light path was obtained by multiple re flections of a light beam within an evacuated tube 0.5 miles long. Below is a schematic drawing with an eight sided mirrored cylinder (for example only, since a 12 sided mirrored cylinder is used for this question) and five internal reflec tions from mirrors within the evacuated tube. Light passes from the source to the detector in short pulses at the times when a mirror on the rotating cylinder is at 45 ◦ to the inci dent light beam. Initially, the light is reflected from a mirror on the rotating cylinder and af ter traveling the total light path of 7 mi, the light is then reflected from an adjacent mir ror (which had moved into the location of the original mirror). mirror light detecter light source ϖ 0.5 miles For what minimum angular speed ( ω > rad/s) of the mirror would Michelson have calculated the speed of light to be 2 . 99792 × 10 8 m / s? Correct answer: 13936 . 9 rad / s. Explanation: Basic Concept: Angular motion: θ = ω t Solution: For a total path of 7 mi, the time traveled between faces on the rotating wheel is t = d c . During this time the wheel rotates through an angle of θ = ω t . Therefore ω = c d θ = c d 2 π 12 = (2 . 99792 × 10 8 m / s) (11263 m) × (0 . 523599 rad) = 13936 . 9 rad / s . 002 (part 1 of 2) 10 points Laser scientists have succeeded in generating pulses of light lasting as short as 10 15 s. Imagine that you create a pulse which lasts 2 . 25 × 10 15 s. What is the spatial length of this pulse along its direction of travel in a vacuum? Correct answer: 6 . 75 × 10 7 m. Explanation: Assume: The pulse started at t = 0 sec and the pulse lasts for 2 . 25 × 10 15 s, so the leading edge of the pulse which had started at t = 0 will have traveled a distance given by l = ct = (3 × 10 8 m / s) (2 . 25 × 10 15 s) = 6 . 75 × 10 7 m . Therefore the distance between the leading edge and the rear edge of the pulse is 6 . 75 × 10 7 m which is the required spatial length. 003 (part 2 of 2) 10 points What is the spatial length of this pulse as it travels through glass with index of refraction n = 1 . 39?...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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