HW#10-solutions - nanni(arn437 – HW#10 – Erskine –(56905 1 This print-out should have 21 questions Multiple-choice questions may continue on

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Unformatted text preview: nanni (arn437) – HW #10 – Erskine – (56905) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A certain blue-green light has a wavelength of 550 nm in air. What is its wavelength in water, where light travels at 75% of its speed in air? Correct answer: 412 . 5 nm. Explanation: Frequency doesn’t change when light trav- els from air to water, but the speed and the wavelength do: λ = 0 . 75(550 nm) = 412 . 5 nm . 002 (part 2 of 2) 10.0 points What is its wavelength in Plexiglas, where light travels at 67% of its speed in air? Correct answer: 368 . 5 nm. Explanation: λ = 0 . 67(550 nm) = 368 . 5 nm . 003 10.0 points The leaves of a tree are bright green. What does a leaf’s absorption spectrum look like? 1. The leaf would absorb all colors except green. correct 2. The leaf would reflect sunlight. 3. The leaf would reflect all colors except green. 4. None of these 5. The leaf would absorb sunlight. Explanation: The leaf would absorb all colors except green. 004 (part 1 of 4) 10.0 points The Fizeau experiment is performed such that the round-trip distance for the light is 28 . 7 m. Given: The wheel has 441 teeth / rev and the speed of light is 2 . 99792 × 10 8 m / s. Find the lowest speed of rotation that allow the light to pass through a notch between the teeth of the wheel. Correct answer: 23686 . 5 rev / s. Explanation: Basic Concept: Angular motion θ = ω t . Solution: For a total path of d = 2 l , the time traveled between notches on the rotating wheel is t = 2 l c . During this time the wheel rotates through an angle of θ = ω t = ω parenleftbigg 2 l c parenrightbigg . To calculate the lowest speed of rotation, note that also θ = 1 n rev, where n is the total number of notches on the wheel. Therefore ω 1 = c d θ = (2 . 99792 × 10 8 m / s) (28 . 7 m) × (0 . 00226757 rev) = 23686 . 5 rev / s . 005 (part 2 of 4) 10.0 points Find the second-lowest speed of rotation that allow the light to pass through a notch. Correct answer: 47372 . 9 rev / s. Explanation: To calculate the second-lowest speed of ro- tation, note that this time, the light beam will pass through the next notch to the notch one in Part 1! That is the wheel rotates through nanni (arn437) – HW #10 – Erskine – (56905) 2 an angle of θ = 2 parenleftbigg 1 n rev parenrightbigg during this time (where n is the total number of notches be- tween the teeth on the wheel). So, the speed should be two times the lowest one; i.e. , ω 2 = c (2 θ ) d = 2 c d θ = 2 ω 1 = 2 (23686 . 5 rev / s) = 47372 . 9 rev / s . 006 (part 3 of 4) 10.0 points Repeat the calculation in Part 1 for a round- trip distance of 2200 m....
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This note was uploaded on 11/02/2011 for the course PHYSICS 317L taught by Professor Erskine during the Fall '11 term at University of Texas at Austin.

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HW#10-solutions - nanni(arn437 – HW#10 – Erskine –(56905 1 This print-out should have 21 questions Multiple-choice questions may continue on

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