Daniels, Matthew – Quiz 4 – Due: Dec 7 2006, midnight – Inst: Ditmire
1
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printout
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23
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
What is the frequency of an electromagnetic
wave if it has a wavelength of 1.2 km?
The
speed of light is 3
×
10
8
m
/
s.
Correct answer: 250000 Hz.
Explanation:
Let :
λ
= 1
.
2 km
and
c
= 3
.
00
×
10
8
m
/
s
.
The speed is
c
=
fλ
f
=
c
λ
=
3
×
10
8
m
/
s
1
.
2 km
·
1 km
10
3
m
=
250000 Hz
.
keywords:
002
(part 1 of 2) 10 points
The secondorder bright fringe (
m
= 2) is
4
.
45 cm from the center line.
4
.
45 cm
1
.
11 m
0
.
0307 mm
S
1
S
2
θ
viewing
screen
Determine the wavelength of the light.
Correct answer: 615
.
383 nm.
Explanation:
Let
:
y
= 4
.
45 cm
,
L
= 1
.
11 m
,
and
d
= 0
.
0307 mm
,
r
2
r
1
y
L
d
S
1
S
2
θ
= tan

1
‡
y
L
·
viewing
screen
δ
≈
d
sin
θ
≈
r
2

r
1
P
O
6
S
2
Q S
1
≈
90
◦
Q
r
2
r
1
d
S
1
S
2
θ
= tan

1
‡
y
L
·
θ
δ
≈
d
sin
θ
≈
r
2

r
1
6
S
2
Q S
1
≈
90
◦
Q
For constructive interference
y
bright
=
λ L
d
m ,
(1)
with
m
= 2,
y
2
= 0
.
0445 m,
L
= 1
.
11 m, and
d
= 3
.
07
×
10

5
m
λ
=
d y
2
m L
=
(3
.
07
×
10

5
m) (0
.
0445 m)
(2) (1
.
11 m)
= 6
.
15383
×
10

7
m
= 615
.
383 nm
.
003
(part 2 of 2) 10 points
Calculate
the
distance
between
adjacent
bright fringes.
Correct answer: 2
.
225 cm.
Explanation:
From equation (1) and the results of the
first part of the problem, we get
y
m
+1

y
m
=
λ L
(
m
+ 1)
d

λ L m
d
=
λ L
d
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Daniels, Matthew – Quiz 4 – Due: Dec 7 2006, midnight – Inst: Ditmire
2
=
(6
.
15383
×
10

7
m) (1
.
11 m)
(3
.
07
×
10

5
m)
= 0
.
02225 m
= 2
.
225 cm
.
keywords:
004
(part 1 of 1) 10 points
The reflecting surfaces of two intersecting flat
mirrors are at an angle of 64
◦
, as shown in
the figure.
A light ray strikes the horizon
tal mirror, reflects off the horizontal mirror,
impinges on the raised mirror, reflects off the
raised mirror, and proceeds in the righthand
direction.
64
◦
φ
Figure is not drawn to scale.
Calculate the angle
φ
.
Correct answer: 52
◦
.
Explanation:
Basic Concept:
θ
incident
=
θ
reflected
Solution:
θ
1
θ
2
φ
θ
Figure is to scale.
The sum of the angles in a triangle is 180
◦
.
In the triangle on the left we have angles
θ ,
180
◦

θ
1
2
,
and
180
◦

θ
2
2
,
so
180
◦
=
θ
+
180
◦

θ
1
2
+
180
◦

θ
2
2
,
or
θ
1
+
θ
2
= 2
θ .
(1)
In the triangle on the right we have angles
θ
1
,
θ
2
,
and
φ .
180
◦
=
θ
1
+
θ
2
+
φ ,
so
θ
1
+
θ
2
= 180
◦

φ .
(2)
Combining Eq. 1 and 2, we have
φ
= 180
◦

2
θ
= 180
◦

2 (64
◦
)
=
52
◦
.
As a matter of interest,
in the upperhalf
of the figure the angles (clockwise) in the
triangles from left to right are
47
◦
,
47
◦
,
and 86
◦
;
94
◦
,
26
◦
,
and 60
◦
;
120
◦
,
17
◦
,
and 43
◦
;
137
◦
,
17
◦
,
and 26
◦
;
and in the lowerhalf of the figure the angles
(counterclockwise) in the triangles from left
to right are
17
◦
,
17
◦
,
and 146
◦
;
34
◦
,
26
◦
,
and 120
◦
;
60
◦
,
47
◦
,
and 73
◦
;
107
◦
,
47
◦
,
and 26
◦
.
keywords:
005
(part 1 of 1) 10 points
An object 8
.
6 cm high is placed 7
.
9 cm in
front of a convex mirror with a focal length of

18 cm.
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 Spring '08
 Turner
 Light, Wavelength, Correct Answer, Total internal reflection

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