Homework set # 16
Problems: 23.13, 11, 5, 28
Solutions:
23.5
Since the mirror is convex,
0
R
. Thus,
0.550 m
R
. With a real object,
0
p
,
so
10.0 m
p
. The mirror equation then gives the image distance as
1
2
1
2
1
0.550 m
10.0 m
q
R
p
, or
0.268 m
q
Thus, the image is
virtual
and located
0.268 m behind the mirror
The magnification is
0.268 m
0.026 8
10.0 m
q
M
p
Therefore, the image is
upright
since
0
M
and
diminished in size
since
1
M
23.11
The image is upright, so
0
M
, and we have
2.0
q
M
p
, or
2.0
2.0
25 cm
50 cm
q
p
The radius of curvature is then found to be
2
1
1
1
1
2
1
25 cm
50 cm
50 cm
R
p
q
, or
0.50 m
2
1.0 m
+1
R
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23.13
The focal length of the mirror may be found from the given object and image
distances as
1
1
1
f
p
q
, or
152 cm
18.0 cm
16.1 cm
152 cm
18.0 cm
pq
f
p
q
For an upright image twice the size of the object, the magnification is
2.00
q
M
p
giving
2.00
q
p
Then, using the mirror equation again,
1
1
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 Spring '08
 SEATON
 Work, #, 0.50 m

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