tovar (jdt436) – homework 36 – Turner – (59070)
1
This printout should have 12 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
An object located 34 cm in Front oF a lens
Forms an image on a screen 9
.
73 cm behind
the lens.
±ind the Focal length oF the lens.
Correct answer: 7
.
56506 cm.
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
∞
1
Diverging Lens
f <
0
∞
>p >
0
0
0
<m<
1
Solution:
The Formula relating the Focal
length to the image distance
s
′
and the object
distance
s
is
1
s
+
1
s
′
=
1
f
so
f
=
s s
′
s
+
s
′
=
(34 cm) (9
.
73 cm)
(34 cm) + (9
.
73 cm)
= 7
.
56506 cm
.
002
(part 2 oF 2) 10.0 points
What is the magnifcation oF the object?
Correct answer:

0
.
286176.
Explanation:
Magnifcation is
M
=

s
′
s
=

(9
.
73 cm)
(34 cm)
=

0
.
286176
.
003
(part 1 oF 2) 10.0 points
A convergent lens has a Focal length oF
15
.
2 cm
.
The object distance is 7
.
6 cm
.
f
f
q
h
′
p
h
Scale: 10 cm =
±ind the distance oF the image From the
center oF the lens.
Correct answer: 15
.
2 cm.
Explanation:
1
p
+
1
q
=
1
f
M
=
h
′
h
=

q
p
Convergent Lens
f >
0
f > p>
0
∞
0
∞
>M >
1
Note:
The Focal length For a convergent
lens is positive,
f
= 15
.
2 cm.
Solution:
Substituting these values into the
lens equation
q
=
1
1
f

1
p
=
1
1
(15
.
2 cm)

1
(7
.
6 cm)
=

15
.
2 cm

q

=
15
.
2 cm
.
004
(part 2 oF 2) 10.0 points
±ind the magnifcation.
Correct answer: 2.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documenttovar (jdt436) – homework 36 – Turner – (59070)
2
Explanation:
M
=

q
p
=

(

15
.
2 cm)
(7
.
6 cm)
=
2
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Work, Orders of magnitude

Click to edit the document details