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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
=+
11
121
20
qRp
2
1
54
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
( )( )
()
22
25.0 cm
16.7 cm
25.0 cm
16.7 cm
pf
−−
==
−
−
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
( )
12
50.3 cm
50.0 cm
12.5 cm
25.0 cm
MMM
−
== −
=
−
−
−
+8.05
=
qq
pp
−
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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This document was uploaded on 10/20/2011.
 Spring '09
 Physics, Work

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