aljabr (faa335) – Hw23 – Ross – (89251)
1
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printout
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15
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001
(part 1 of 3) 10.0 points
A diverging lens has a focal length of

24 cm.
An object 2
.
82 cm in height is placed 170 cm
in front of the lens.
Locate the position of the image.
Correct answer:

21
.
0309 cm.
Explanation:
Basic Concepts:
1
p
+
1
q
=
1
f
m
=
h
′
h
=

q
p
Diverging Lens
0
> f
∞
>p>
0
f <q <
0
0
<m<
1
Solution:
Using the thin lens equation
1
p
+
1
q
=
1
f
,
with
p
= 170 cm and
f
=

24 cm , we get
q
=
pf
p

f
=
(170 cm) (

24 cm)
(170 cm)

(

24 cm)
=

21
.
0309 cm
.
The negative sign tells us that the image is
virtual.
002
(part 2 of 3) 10.0 points
What is the magnification?
Correct answer: 0
.
123711.
Explanation:
The magnification is given by
M
=

q
p
=

(

21
.
0309 cm)
170 cm
= 0
.
123711
.
The positive sign of
M
means that the image
is upright and on the same side of the lens as
the object.
003
(part 3 of 3) 10.0 points
Find the height of the image.
Correct answer: 0
.
348866 cm.
Explanation:
From the formula
M
=
H
h
,
where
H
is the height of the image and
h
is
the height of the object, we obtain
H
=
M h
= (0
.
123711) (2
.
82 cm)
= 0
.
348866 cm
.
004
10.0 points
A converging lens of focal length 0
.
216 m
forms a virtual image of an object. The image
appears to be 0
.
912 m from the lens on the
same side as the object.
What is the distance between the object
and the lens?
Correct answer: 0
.
174638 m.
Explanation:
Basic Concepts:
1
p
+
1
q
=
1
f
m
=
h
′
h
=

q
p
Converging Lens
f >
0
∞
>p> f
f <q<
∞
0
>m>
∞
f >p>
0
∞
<q<
0
∞
>m>
1
The image appears on the same side as the
object, so
q
is negative:
1
p
=
1
f

1
q
1
p
=
1
0
.
216 m
+
1
0
.
912 m
=
1
5
.
72612 m
−
1
p
= 0
.
174638 m
.
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 Spring '10
 Gavrine
 Thin lens

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