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Unformatted text preview: The image, virtual in this case, is located at a distance of: The magnification for the eyepiece is: So the height of the final image is 1.8 mm x 3.85 = 6.9 mm. The overall magnification of the two lens system is: This is equal to the final height divided by the height of the object, as it should be. Note that, applying the sign conventions, the final image is virtual, and inverted compared to the object. This is consistent with the ray diagram....
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This note was uploaded on 11/22/2011 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.
 Fall '10
 DavidJudd
 Physics, Work

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