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286
CHAPTER 23
23.36
We are given that
. Then, the thin lens equation,
12.5 cm and
30.0 cm
fq
=+
=−
111
p
qf
+=
, gives
( )( )
30.0 cm
12.5 cm
8.82 cm
30.0 cm
12.5 cm
−
=
+
−−
−
p
==
and the lateral magnification is
30.0 cm
3.40
8.82 cm
q
M
p
−
= +
Since
0
M
>
, the image is
upright
23.37
All virtual images formed by diverging lenses are upright images. Thus,
0
M
>
, and the
magnification gives
1
3
q
M
p
, or
3
p
q
Then, from the thin lens equation,
13 21
p
pp
f
−
=
or
22
p
ff
=
The object should be placed at distance
2
in front of the lens
f
23.38
(a) The total distance from the object to the real image is the objecttoscreen distance,
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.
 Fall '10
 STAFF
 Physics

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