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# solution_pdf4 - McPartlin(anm464 assignment 7...

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McPartlin (anm464) – assignment 7 – frommhold – (58025) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 16.6 points Hint: The convergent lens in this problem is a part of a lens system so the object in this problem may be either real or virtual. Construct a ray diagram. Given: A real object is located to the left of a divergent lens. The object’s distance and image’s distance from the lens and the lens’ focal length are shown in the Fgures below. 0 13 6 f f f Which diagram correctly shows the image? 1. 0 13 6 f f f 13 7 f 2. 0 13 6 f f f 13 7 f correct 3. 0 13 6 f f f 13 7 f 4. 0 13 6 f f f 13 7 f Explanation:

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McPartlin (anm464) – assignment 7 – frommhold – (58025) 2 0 13 6 f f f p q 13 7 f Basic Concepts: 1 p + 1 q = 1 f , where 0 < f for a convergent lens. Solution: 1 q = 1 p + 1 f = (6) 13 f + 1 f = (6) + (13) 13 f = 7 13 f q = 13 7 f . The magniFcation m of this lens is m = q 1 p 1 = 13 7 f 13 6 f = 6 7 = 6 7 . 002 16.6 points A coin is at the bottom of a beaker. The beaker is Flled with 4 . 3 cm of water ( n 1 = 1 . 33) covered by 1 . 8 cm of liquid ( n 2 = 1 . 4) ±oating on the water.
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solution_pdf4 - McPartlin(anm464 assignment 7...

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