16homework12 - Husain, Zeena – Homework 12 – Due: Apr...

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Unformatted text preview: Husain, Zeena – 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 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.

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16homework12 - Husain, Zeena – Homework 12 – Due: Apr...

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