Platt, David – Quiz 4 – Due: Dec 6 2005, 10:00 pm – Inst: Ken Shih
1
This
printout
should
have
28
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
In our everyday environment, diffraction is
much more evident for sound waves than for
light waves.
Why is this so?
1.
We see light more often than hear sound
in our everyday environment.
2.
None of these.
3.
Light waves travel much faster than sound
waves.
4.
Light waves have much shorter wave
length than sound waves.
correct
Explanation:
Diffraction around ordinarysized objects is
most pronounced for waves with a wavelength
as big or bigger than the objects. The wave
length of sound waves is relatively long, and
for light extremely short.
Hence the diffrac
tion of sound is more evident in our everyday
environment.
002
(part 1 of 1) 10 points
The thin composite lens has

R
1

=
R
and

R
2

=
R
2
,
as shown in the figure.
R
1
R
2
n
2
n
1
The lens has focal length
1.
f
=
n
1
4 (
n
2

n
1
)
R .
2.
f
=
n
1
4 (
n
1

n
2
)
R .
3.
f
=
n
1
2 (
n
1

n
2
)
R .
4.
f
=
n
2
n
1

n
2
R .
5.
f
=
n
2
2 (
n
2

n
1
)
R .
6.
f
=
n
1
2 (
n
2

n
1
)
R .
7.
f
=
n
2
n
2

n
1
R .
8.
f
=
n
1
n
2

n
1
R .
9.
f
=
n
1
n
1

n
2
R .
correct
10.
f
=
n
2
2 (
n
1

n
2
)
R .
Explanation:
Use the lens makers equation
1
f
=
n
2
n
1

1
¶
1
R
1

1
R
2
¶
=
n
2

n
1
n
1
¶
1
R

2
R
¶
=
n
1

n
2
n
1
¶
1
R
¶
f
=
n
1
n
1

n
2
R
.
003
(part 1 of 1) 10 points
Assume 515 nm light in the eye and a pupil
diameter of 2
.
5 mm.
What are the approximate dimensions of
the smallest object on Earth that astronauts
can resolve by eye when they are orbiting
219 km above the Earth?
Correct answer: 55
.
0391 m.
Explanation:
If the dimensions of the smallest object on
Earth that astronauts can resolve by eye is
x
,
then the Rayleigh’s criterion yields
x
D
= 1
.
22
λ
d
,
where
D
is the distance from the astronaut
to the Earth,
λ
is the wavelength of the light,
and
d
is the diameter of the astronaut’s pupil.
Therefore
x
=
1
.
22
λD
d
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Platt, David – Quiz 4 – Due: Dec 6 2005, 10:00 pm – Inst: Ken Shih
2
x
=
(1
.
22)(515 nm )(219 km )
2
.
5 mm
= 55
.
0391 m
.
004
(part 1 of 2) 10 points
A beam of light with a wavelength 660 nm in
air travels in water.
The index of refraction of water is 1
.
33.
What is the wavelength in water?
Correct answer: 496
.
241 nm.
Explanation:
Let :
n
w
= 1
.
33
,
and
λ
0
= 660 nm
.
The wavelength of light in water is
λ
w
=
λ
0
n
w
=
660 nm
1
.
33
=
496
.
241 nm
.
005
(part 2 of 2) 10 points
Does a swimmer underwater observe the same
color or a different color for this light?
1.
It is not possible to know what a swimmer
underwater will observe.
2.
A swimmer underwater observes a differ
ent color because the wavelength changes.
3.
A swimmer underwater observes a the
same color because the wavelength does not
change.
4.
A swimmer underwater observes a differ
ent color because the frequency changes.
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 Fall '08
 Turner
 Physics, Light, Correct Answer, Platt, Ken Shih

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