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Mirrors and Lenses
285
(b) The virtual image case is shown in the second
diagram. Note that in this case,
( )
12.9 cm
qp
=
−+
,
so the thin lens equation gives
11
1
12.9 cm
2.44 cm
pp
−=
+
(
or
)
22
12.9 cm
31.5 cm
+−
0
=
2.10 cm
or
The quadratic formula then gives
15.0 cm
=
=−
Since the object is real, the negative solution must be rejected leaving
2.10 cm
p
=
.
V
±²³´µl
¶·µ¸¹
±
º»º¼½¾¾º¿·
À¼½9º¿·
ÁÂÃ¹¿³
²
³
23.34
We must first realize that we are looking at an upright, magnified, virtual image. Thus,
we have a real object located between a converging lens and its frontside focal point, so
.
The magnification is
0,
0, and
0
f
<>
>
2
q
M
p
=+
, giving
2
= −
. Then, from the thin lens equation,
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This document was uploaded on 10/20/2011.
 Spring '09
 Physics, Work

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