Unformatted text preview: magnification of the image inside the tube; (4) determine the overall angular magnification of the final image, as compared to the original object if viewed at a typical near point of 25 cm. 1/f = 1/d + 1/(10f e ) ; solve for d = 1.167 cm (1) How far to the left of the objective lens should the object be placed? __ 1.167 ___cm (2) Sketch the ray diagram, using several principle rays, showing the image in the tube. ____________ (3) Is the image in the tube ( REAL ) or VIRTUAL ? (circle one) ___________ Is the image in the tube ERECT or ( INVERTED ) ? (circle one) m =  s' / s =  (103) / 1.167 = 6 What is the magnification of the image in the tube? ____6 _________ M = m 25 / fe = 6* 25 / 3 = 50 (4) What is the overall angular magnification of the final image, as compared to the original object if viewed at a typical near point of 25 cm? ______50 ___________...
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 Sherman,DouglasF
 Physics, objective lens, overall angular magnification

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