qz10f08sol - December 5, 2008 PHY2054 Discussion-Fall ‘08...

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Unformatted text preview: December 5, 2008 PHY2054 Discussion-Fall ‘08 Quiz 10 (Chapter 23.6-24.7) Name: UFID: **1. (5pts) A lens with a focal length of -20.0 cm and a mirror with a radius of curvature of 25.0 cm are placed 30.0 cm apart. An object is placed 15.0 cm to the left of the mirror. Considering only the light that leaves the object and travels first towards the mirror, find the overall magnification of the image. Applying the thin-lens equation, first we calculate the image formed by the mirror. (1/15)+(1/q 1 ) = (2/25) ⇒ q 1 = [(2/25)-(1/15)]-1 = 75 cm This image serves as the object for the lens. The object distance is |p 2 | = 75-30 = 45 cm ⇒ p 2 = -45 cm (Note p 2 is negative, because it is in the back of the lens.) We apply the lens-equation again and calculate the position of the final image. (1/-45)+(1/q2) = (1/-20) ⇒ q 2 = [(1/-20)+(1/45)]-1 = -36 cm The overall magnification is the product of the magnifications of the lens and the mirror....
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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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qz10f08sol - December 5, 2008 PHY2054 Discussion-Fall ‘08...

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