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# 37 - Kapoor(mk9499 homework37 Turner(60230 This print-out...

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Kapoor (mk9499) – homework37 – Turner – (60230) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A reFracting telescope has a 0 . 999 m diameter objective lens with Focal length 12 . 5 m and an eyepiece with Focal length 2 . 09 cm. What is the angular magnifcation oF the telescope? Correct answer: - 598 . 086. Explanation: The relation between the angular magnifca- tion and Focal lengths oF the lenses is: m = - f object f eye so the answer is: m = - 12 . 5 m 0 . 0209 m = - 598 . 086 002 (part 2 oF 2) 10.0 points Describe the fnal image. 1. inverted and virtual correct 2. inverted and real 3. None oF these. 4. erect and real 5. erect and virtual Explanation: Since the magnifcation m is negative, the image is inverted . The image is virtual due to the eyepiece lens. 003 10.0 points Assume that a camera has a fxed Focal length oF 64 . 1 mm and is adjusted to properly Focus the image oF a distant object. How Far must the lens be moved to Focus the image oF object that is 1 . 73 m away? Correct answer: 2 . 46642 mm. Explanation: Given : p 2 = 1 . 73 m and f = 64 . 1 mm . To Focus on a very distant object, the original distance From the lens to the flm was q 1 = f = 64 . 1 mm. To Focus on an object 1 . 73 m away, the thin lens equation gives 1 f = 1 p 2 + 1 q 2 1 q 2 = 1 f - 1 p 2 = p 2 - f fp 2 q 2 = p 2 f p 2 - f = (1730 mm) (64 . 1 mm) 1730 mm - 64 . 1 mm = 66 . 5664 mm Thus, the lens should be moved Farther From the flm by the distance Δ q = q 2 - q 1 = 66 . 5664 mm - 64 . 1 mm = 2 . 46642 mm . 004 (part 1 oF 4) 10.0 points A real object is located at 9 5 f to the leFt oF a convergent lens with a Focal length f as shown in the fgure below. 0

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37 - Kapoor(mk9499 homework37 Turner(60230 This print-out...

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