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ID: 2124392 • Letter: H
Question
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An arrow 2.00cm long is located 76.0cm from a lens that has a focal length f = 31.0cm . If the arrow is perpendicular to the principal axis of the lens, as in the figure (a), what is its lateral magnification, defined as hi/ho? Suppose, instead, that the arrow lies along the principal axis, extending from 74.0cm to 76.0cm from the lens, as indicated in the figure (b). What is the longitudinal magnification of the arrow, defined as L i /L o ?Explanation / Answer
(1/f) = (1/xo) + (1/xi)
(1/31) = (1/76) + ( 1/xi)
xi=52.35
Lateral Magnification = hi/ho = -xi/xo = -76/52.35 = -1.45 cm
for 74 cm
(1/31) = (1/74) + (1/xi')
xi' = 53.35
Longitudinal magnification = Li/Lo = 53.35 - 52.35 = 1 cm
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