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Homework_22 - Nguyen Thanh Homework 22 Due 7:00 pm Inst...

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Nguyen, Thanh – Homework 22 – Due: Apr 25 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points The left face of a biconvex lens has a radius of curvature of 5 . 5 cm, and the right face has a radius of curvature of 21 cm. The index of refraction of the glass is 1 . 44. Calculate the focal length of the lens. Correct answer: 9 . 90566 cm. Explanation: Given : R 1 = 5 . 5 cm , R 2 = - 21 cm , and n = 1 . 44 . Using the lens maker’s equation, 1 f = ( n - 1) 1 R 1 - 1 R 2 = ( n - 1)( R 2 - R 1 ) R 1 R 2 f = R 1 R 2 ( n - 1)( R 2 - R 1 ) = (5 . 5 cm)( - 21 cm) (1 . 44 - 1)( - 21 cm - 5 . 5 cm) = 9 . 90566 cm . 002 (part 2 of 2) 10 points Calculate the focal length if the radii of cur- vature of the two faces are interchanged. Correct answer: 9 . 90566 cm. Explanation: Given : R 1 = 21 cm and R 2 = - 5 . 5 cm . 1 f = ( n - 1) 1 R 1 - 1 R 2 = ( n - 1)( R 2 - R 1 ) R 1 R 2 f = R 1 R 2 ( n - 1)( R 2 - R 1 ) = (21 cm)( - 5 . 5 cm) (1 . 44 - 1)( - 5 . 5 cm - 21 cm) = 9 . 90566 cm . keywords: 003 (part 1 of 3) 10 points A converging lens of focal length 19 . 2 cm is separated by 49 . 8 cm from a converging lens of the focal length 4 . 2 cm. Find the position of the final image with respect to the second lens of an object placed 38 . 5 cm in front the first lens. Correct answer: 6 . 61661 cm. Explanation: Given : L = 49 . 8 cm , f 1 = 19 . 2 cm , f 2 = 4 . 2 cm , and p 1 = 38 . 5 cm . The thin lens equation gives the image dis- tance for the first lens: 1 f 1 = 1 p 1 + 1 q 1 = q 1 + p 1 p 1 q 1 p 1 q 1 = f 1 ( q 1 + p 1 ) q 1 = p 1 f 1 p 1 - f 1 = (38 . 5 cm) (19 . 2 cm) 38 . 5 cm - 19 . 2 cm = 38 . 3005 cm .
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