Chapter 23 Solution
23.5
Since the mirror is convex,
0
R
<
. Thus,
0.550 m
R
= −
. With a real object,
0
p
>
,
so
. The mirror equation then gives the image distance as
10.0 m
p
=+
121
2
1
0.550
m1
−
0
.
0 m
qRp
=−=
−
, or
0.268 m
q
= −
Thus, the image is
virtual
and located
0.268
m behind the
mirror
The magnification is
0.026 8
q
M
p
−
=− =−
=
Therefore, the image is
upright
( )
since
0
M
>
and
diminished in size
()
1
M
<
23.7
The radius of curvature of a concave mirror is positive, so
20.0 cm
R
= +
. The
mirror equation then gives
10.0 cm
1
1
p
p
p
−
−=
, or
( )
p
q
p
=
−
(a) If
40.
,
0 cm
p
=
13.3 cm
q
and
0.333
40.0 cm
q
M
p
= −
The image is
13.3 cm in fron
inverted
t of the
mirror, real, and
(b) When
20.
,
p
=
q
= +
and
1.00
q
M
p
The image is
20.0 cm in fron t of
the
(c) If
10
,
.0 cm
p
=
( )
q
=→
−
∞
and
no image is form ed.
Parallel rays leave the
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View Full Document23.8
(a)
Since the object is in front of the mirror,
0
p
>
. With the image behind the
mirror,
. The mirror equation gives the radius of curvature as
0
q
<
211
1
1
1
1.00 cm
10.0 cm
10.0
Rpq
=
+
=−=
0

1
cm
or
22
.
22 cm
9
R
⎛⎞
==
+
⎜⎟
⎝⎠
23.11
The
magnified
,
virtual
images formed by a concave mirror are upright, so
0
M
>
.
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 Fall '08
 Abdulrazzaq
 Physics

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