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Unformatted text preview: Rodriguez, Brandon Homework 23 Due: Nov 9 2006, 11:00 am Inst: Andrew J Rader 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA TIONS TO AT LEAST THREE SIGNIFI CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 1) 4 points A converging lens of focal length 0 . 232 m forms a virtual image of an object. The image appears to be 0 . 897 m from the lens on the same side as the object. What is the distance between the object and the lens? Correct answer: 0 . 184326 m. Explanation: Basic Concepts: 1 p + 1 q = 1 f m = h h = q p Converging Lens f > > p > f f < q < > m > f > p > < q < > m > 1 The image appears on the same side as the object, so q is negative: 1 p = 1 f 1 q 1 p = 1 . 232 m + 1 . 897 m = 1 5 . 42517 m 1 p = 0 . 184326 m . keywords: 002 (part 1 of 1) 4 points A converging lens has a focal length of 34 . 4 cm. If the object is 82 . 8 cm from the lens, what is the image distance? Correct answer: 58 . 8496 cm. Explanation: Basic Concepts: 1 p + 1 q = 1 f m = h h = q p Converging Lens f > > p > f f < q < > m > f > p > < q < > m > 1 f is positive for a converging lens: 1 q = 1 f 1 p s i = 1 f 1 p  1 = 1 (34 . 4 cm) 1 (82 . 8 cm)  1 = 58 . 8496 cm ....
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.
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