26.114 An arrow 2.10 cm long is located 74.0 cm from a lens that has a focal len

Question

26.114 An arrow 2.10 cm long is located 74.0 cm from a lens that has a focal length f = 31.0 cm

Part A

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?

Part B

Suppose, instead, that the arrow lies along the principal axis, extending from 75.0 cm to 77.1 cmfrom the lens, as indicated in the figure (b). What is the longitudinal magnification of the arrow, defined as Li/Lo? (Hint: Use the thin-lens equation to locate the image of each end of the arrow.)

Explanation / Answer

part A)

hi = image height

ho = object height = 2.10 cm

do = onject distance = 74 cm

f = focal length = 31 cm

using lens equation

1/f = 1/do + 1/di

1/31 = 1/74 + 1/di

di = 53.35 cm

hi/ho = - di/do = - 53.35 / 74 = - 0.72

Part B)

For nearer end :

do = object distance = 75 cm

f = focal length = 31 cm

using lens equation

1/f = 1/do + 1/di

1/31 = 1/75 + 1/di

di = 52.84 cm

For farther end :

d'o = object distance = 77.1 cm

f = focal length = 31 cm

using lens equation

1/f = 1/d'o + 1/d'i

1/31 = 1/77.1 + 1/d'i

d'i = 51.85 cm

Li = 52.84 - 51.85 = 0.99 cm

Lo = 77.1 - 75 = 2.1

Li/Lo = 0.99 / 2.1 = 0.47

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