Rodriguez, Brandon – Homework 23 – Due: Nov 9 2006, 11:00 am – Inst: Andrew J Rader
1
This
printout
should
have
15
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
Please notice that for your homework to be
considered towards your grade, it needs to be
submitted one hour before the corresponding
recitation starts.
Work submitted after this
time, but before the DUE DATE on top of
this page, will be accepted but not graded.
PLEASE
REMEMBER
THAT
YOU
MUST CARRY OUT YOUR CALCULA
TIONS TO AT LEAST THREE SIGNIFI
CANT FIGURES. YOUR ANSWER MUST
BE
WITHIN
ONE
PERCENT
OF
THE
CORRECT RESULT TO BE MARKED AS
CORRECT BY THE SERVER.
001
(part 1 of 1) 4 points
A converging lens of focal length 0
.
232 m
forms a virtual image of an object. The image
appears to be 0
.
897 m from the lens on the
same side as the object.
What is the distance between the object
and the lens?
Correct answer: 0
.
184326 m.
Explanation:
Basic Concepts:
1
p
+
1
q
=
1
f
m
=
h
0
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
.
232 m
+
1
0
.
897 m
=
1
5
.
42517 m

1
p
= 0
.
184326 m
.
keywords:
002
(part 1 of 1) 4 points
A
converging
lens
has
a
focal
length
of
34
.
4 cm.
If the object is 82
.
8 cm from the lens, what
is the image distance?
Correct answer: 58
.
8496 cm.
Explanation:
Basic Concepts:
1
p
+
1
q
=
1
f
m
=
h
0
h
=

q
p
Converging Lens
f >
0
∞
> p > f
f < q <
∞
0
> m >
∞
f > p >
0
∞
< q <
0
∞
> m >
1
f
is positive for a converging lens:
1
q
=
1
f

1
p
s
i
=
•
1
f

1
p
‚

1
=
•
1
(34
.
4 cm)

1
(82
.
8 cm)
‚

1
= 58
.
8496 cm
.
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 Spring '08
 pedon
 Work, Photographic lens, Thin lens

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