An arrow 1.90 cm long is located 74.0 cm from a lens that has a focal length f =

Question

An arrow 1.90 cm long is located 74.0 cm from a lens that has a focal length f = 30.5 cm . (Figure 1)

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 73.0 cm to 74.9 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

(a) Lateral magnification is
m = -di/do = I/O
So we know that
do object distance = 74 cm
focal length f = 30.5 cm
therefore
using lens equation
1/d0 + 1/di = 1/f
So di = 51.88 cm
So lateral magnification = 51.88/74 = 0.7
(b) In this case we have to assume both end seperately
assuming left end do = 73 cm
so di = 52.38 cm
and Right end do = 74.9 cm
di = 51.45 cm
so the image length = 52.38 - 51.45 = 0.93 cm
Object length Lo = 74.9 - 73 = 1.9 cm
so longittudinal magnification = Li / Lo = 0.48

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