Homework 10 solutions
Physics 204, morning summer 2008
Chapter 25
Problems 20,33,19
Chapter 26
Problems 11,24
20.
REASONING
Since the focal length
f
and the object distance
d
o
are known, we will use the mirror equation
to determine the image distance
d
i
.
Then, knowing the image distance as well as the object distance, we will
use the magnification equation to find the magnification
m
.
SOLUTION
a.
According to the mirror equation (Equation 25.3), we have
()
i
io
o
111
1
1
or
4.3 m
11
1
1
7.0 m
11 m
d
df
d
fd
=−
=
=
=
−
−−
−
The image distance is negative because the image is a virtual image behind the mirror.
b. According to the magnification equation (Equation 25.4), the magnification is
( )
i
o
4.3 m
0.39
11 m
d
m
d
−
=
33.
REASONING AND SOLUTION
a.
The focal length is found from
1
1
27
1
65
46
d
f
=+=
+
−
=+
oi
cm
cm
so
cm
b.
The magnification is
m
d
d
=
−
=
=
i
o
cm
27 cm
65
24
bg
.
19.
REASONING
a. We are dealing with a concave mirror whose radius of curvature is 56.0 cm. Thus, the focal length of the
mirror is
1
2
28.0 cm
fR
==
(Equation 25.1) The object distance is
d
o
= 31.0 cm. With known values for
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 Spring '09
 rollino
 Physics, Snell's Law

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