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Unformatted text preview: aljabr (faa335) Hw23 Ross (89251) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A diverging lens has a focal length of 24 cm. An object 2 . 82 cm in height is placed 170 cm in front of the lens. Locate the position of the image. Correct answer: 21 . 0309 cm. Explanation: Basic Concepts: 1 p + 1 q = 1 f m = h h = q p Diverging Lens > f >p> f <q < <m< 1 Solution: Using the thin lens equation 1 p + 1 q = 1 f , with p = 170 cm and f = 24 cm , we get q = pf p f = (170 cm) ( 24 cm) (170 cm) ( 24 cm) = 21 . 0309 cm . The negative sign tells us that the image is virtual. 002 (part 2 of 3) 10.0 points What is the magnification? Correct answer: 0 . 123711. Explanation: The magnification is given by M = q p = ( 21 . 0309 cm) 170 cm = 0 . 123711 . The positive sign of M means that the image is upright and on the same side of the lens as the object. 003 (part 3 of 3) 10.0 points Find the height of the image. Correct answer: 0 . 348866 cm. Explanation: From the formula M = H h , where H is the height of the image and h is the height of the object, we obtain H = M h = (0 . 123711) (2 . 82 cm) = 0 . 348866 cm . 004 10.0 points A converging lens of focal length 0 . 216 m forms a virtual image of an object. The image appears to be 0 . 912 m from the lens on the same side as the object. What is the distance between the object and the lens? Correct answer: 0 . 174638 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:...
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 Gavrine

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