qz9sp09sol - April 14, 2009 PHY2054 Discussion-Spring 09...

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April 14, 2009 PHY2054 Discussion-Spring ‘09 Quiz 9 (Chapter 23.6-24.5) Name: UFID: **1. (2.5pts) An object is placed 40.0 cm to the left of a converging lens with a focal length of 30.0 cm. A spherical mirror with a radius of curvature of -40.0 cm is located 60.0 cm to the right of the lens. Locate the final image. The light ray from the object passes the lens from left to right, hits the mirror and then passes the lens from right to left before forming the final image. The position of the image due to the lens is 1/p+1/q 1 = 1/f q 1 = (1/f-1/p) -1 = (1/30-1/40 -1 = 120 cm This image serves as an object for the mirror. The object distance and the image distance are |p 2 | = q 1 -60 = 60 cm p 2 = -60 cm (in the back of the mirror) 1/p 2 + 1/q 2 = 2/R q 2 = (2/R-1/p 2 ) -1 = [2/(-40)-1/(-60)]-1 = -30 cm (behind the mirror) This image serves as an object for the lens. Now the front side is to the right of the lens and the back is to the left. The final image is located p 3 = |q
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This note was uploaded on 12/13/2011 for the course PHY 2054 taught by Professor Avery during the Fall '08 term at University of Florida.

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qz9sp09sol - April 14, 2009 PHY2054 Discussion-Spring 09...

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