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Mirrors and Lenses
297
Combining fractions gives
()
11
1
1.50 m
1.50 m
1.50 m
0.900 m
0.600 m
1
p
pp
p
=
−+
−
(
or
)( ) ( )
1
0.900 m
0.600 m
1.50 m
1
p
p
+−
=
−
(a) Simplifying yields
( ) ( )
2
m
0.600 m
0.900 m
1.50 m
0.540
p
p
=
or
2
1
0.540 m
1.80 m
p
==
0.300 m and
21
0.900 m
1.20 m
=+
=
(b) Then,
q
and the thin lens equation gives
1.50 m
1.20 m
p
=−
=
( )
0.300 m
1.20 m
0.240 m
1.50 m
pq
f
=
+
(c) The second image distance is
0.600 m
0.300 m
qp
=
−=
+
and the magnification for
this configuration is
2
2
2
0.300 m
1.20 m
q
M
p
= −
0.250
Thus, the second image is
real, inverted, and diminished
23.57
From the thin lens equation, the image distance for the first lens is
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This document was uploaded on 10/20/2011.
 Spring '09
 Physics, Work

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