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Version 123 – quiz – markert – (58710)
1
This printout should have 10 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Consider a divergent lens with a Focal length
f
.
An upright object is placed within the
interval
f
and 2
f
to the leFt oF the lens.
2
f
2
f
f
f
The corresponding image is
1.
real, upright, and reduced.
2.
real, upright, and enlarged.
3.
nonexistant.
4.
virtual, inverted, and the same size.
5.
virtual, upright, and enlarged.
6.
virtual, upright, and reduced.
correct
7.
virtual, inverted, and reduced.
8.
virtual, upright, and the same size.
9.
real, upright, and the same size.
10.
virtual, inverted, and enlarged.
Explanation:
2
f
f
The Focal length For a diverging lens is neg
ative.
f
=

f

<
0
Then the lens equation reads
1
p
+
1
q
=

1

f

where
p >
0 and
q
are the object and image
distances respectively. Solving For
q
yields
q
=


f

p
p
+

f

<
0
Since
q <
0, by convention the image is vir
tual. Since the magnifcation is less than one
the image is reduced.
002
10.0 points
A person in a boat sees a fsh in the water oF
index 1
.
33 at an angle oF 59
◦
relative to the
water’s surFace.
What is the true angle relative to the wa
ter’s surFace?
1. 66.5196
2. 53.0959
3. 53.6662
4. 59.1508
5. 57.8824
6. 51.4398
7. 67.2167
8. 51.9824
9. 49.8728
10. 54.245
Correct answer: 67
.
2167
◦
.
Explanation:
θ
i
=
π
2

θ
water
sin
θ
i
=
n
2
sin
θ
r
θ
fish
=
π
2

θ
r
003
10.0 points
A concave mirror has a Focal length oF 27 cm.
What is the position oF the resulting image
iF the image is inverted and 4 times smaller
than the object?
1. 13.5
2. 17.1429
3. 61.2
4. 53.7143
5. 9.12
6. 24.0
7. 18.6667
8. 62.5
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2
9. 40.8
10. 33.75
Correct answer: 33
.
75 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
= 27 cm
and
n
= 4
.
Since the image is inverted,
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 Spring '09
 MARKERT

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