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hw12sol - Flores Michael Homework 12 Due 4:00 am Inst Sonia...

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Flores, Michael – Homework 12 – Due: Apr 20 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 19 questions. Multiple-choice 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 48 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 4 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 ω 0.5 miles For what minimum angular speed ( ω > 0 rad/s) of the mirror would Michelson have calculated the speed of light to be 2 . 99792 × 10 8 m / s? Correct answer: 6097 . 38 rad / s. Explanation: Basic Concept: Angular motion: θ = ω t Solution: For a total path of 4 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 48 = (2 . 99792 × 10 8 m / s) (6436 m) × (0 . 1309 rad) = 6097 . 38 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 7 . 43 × 10 - 15 s. What is the spatial length of this pulse along its direction of travel in a vacuum? Correct answer: 2 . 229 × 10 - 6 m. Explanation: Assume: The pulse started at t = 0 sec and the pulse lasts for 7 . 43 × 10 - 15 s, so the leading edge of the pulse which had started at t = 0 will have traveled a distance given by l = c t = (3 × 10 8 m / s) (7 . 43 × 10 - 15 s) = 2 . 229 × 10 - 6 m . Therefore the distance between the leading edge and the rear edge of the pulse is 2 . 229 × 10 - 6 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 . 26? Correct answer: 1 . 76905 × 10 - 6 m. Explanation: The speed of light decreases when it travels through glass. Its velocity becomes v glass = c n
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Flores, Michael – Homework 12 – Due: Apr 20 2004, 4:00 am – Inst: Sonia Paban 2 Therefore the spatial length in glass will be reduced by a factor 1 . 26.
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