Mirrors and Lenses
293
23.50
Since the object is midway between the lens and mirror, the object distance for the
mirror is
. The mirror equation gives the image position as
1
12.5 cm
p
= +
1
1
1
2
1
20
q
R
p
2
1
5
4
1
.0 cm
12.5 cm
50.0 cm
50.0 cm
−
=
−
=
−
=
=
2
25.0 cm
p
, or
This image serves as the object for the lens, so
1
50.0 cm
q
= +
1
25.0 cm
q
=
−
= −
. Note that
since
, this is a virtual object. The thin lens equation gives the image position for
the lens as
2
0
<
p
(
)(
)
(
)
2
2
2
2
25.0 cm
16.7 cm
25.0 cm
16.7 cm
p f
p
f
−
−
=
=
−
−
− −
2
q
2
0
q
<
50.3 cm
= −
Since
, this is a
virtual
image that is located 50.3 cm in front of the lens or
25.3
hind the
cm be
mirror
. The overall magnification is
(
)
(
)
1
2
1
2
1
2
50.3 cm
50.0 cm
12.5 cm
25.0 cm
M
M M
−
=
=
−
=
−
−
−
+8.05
=
q
q
p
p
−
Since
0
M
>
, the final image is
upright
23.51
As light passes lefttoright through the lens, the image position is given by
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 Spring '09
 Physics, Work

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