Golshani (ag5873) – Homework 02 – Yao – (57045)
1
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14
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before answering.
001
10.0 points
Consider a concave mirror with radius
R
. An
upright object is placed between the interval
R
2
and
R
.
R
R
/2
The image is
1.
virtual, upright, same size.
2.
virtual, inverted, reduced.
3.
virtual, upright, reduced.
4.
real, upright, enlarged.
5.
real, upright, same size.
6.
virtual, upright, enlarged.
7.
virtual, inverted, enlarged.
8.
real, upright, reduced.
9.
real, inverted, enlarged.
correct
10.
real, inverted, reduced.
Explanation:
R
R
/2
Note:
f
=
R
2
,
where
f
is positive for a concave mirror by
convention. From the mirror equation
1
p
+
1
q
=
1
f
.
Solving for
q
we have
q
=
fp
p

f
.
Since
p > f
, the image is real and inverted.
The absolute value of the magnification is

M

=
q
p
=
f
p

f
.
Since
f < p <
2
f
, recall
parenleftbigg
f
=
R
2
parenrightbigg
,

M

>
1, or the image is enlarged.
This is also
confirmed by the ray diagram shown.
002
10.0 points
A concave mirror with a radius of curvature
of 0
.
9 m is illuminated by a candle located on
the symmetry axis 2
.
9 m from the mirror.
Where is the image of the candle?
Correct answer: 0
.
532653 m.
Explanation:
1
p
+
1
q
=
1
f
=
2
R
m
=
h
′
h
=

q
p
Concave Mirror
f >
0
∞
>p> f
f <q<
∞
0
>m>
∞
f >p>
0
∞
<q<
0
∞
>m>
1
Let :
f
= 0
.
45 m
and
p
= 2
.
9 m
.
From the mirror equation
1
p
+
1
q
=
1
f
=
2
R
f
=
R
2
⇒
q
=
bracketleftbigg
1
f

1
p
bracketrightbigg
−
1
=
bracketleftbigg
1
(0
.
45 m)

1
(2
.
9 m)
bracketrightbigg
−
1
=
0
.
532653 m
.
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Golshani (ag5873) – Homework 02 – Yao – (57045)
2
003
10.0 points
A concave mirror has a focal length of 51 cm.
What is the position of the resulting image
if the image is inverted and 6 times smaller
than the object?
Correct answer: 59
.
5 cm.
Explanation:
1
p
+
1
q
=
1
f
=
2
R
m
=
h
′
h
=

q
p
Concave Mirror
f >
0
∞
>p> f
f <q <
∞
0
>m>
∞
f >p>
0
∞
<q <
0
∞
>m>
1
Let :
f
= 51 cm
and
n
= 6
.
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 Fall '10
 YAO
 0.45 m

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