Nguyen, Thanh – Homework 20 – Due: Apr 18 2008, 7:00 pm – Inst: Weathers
1
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printout
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have
13
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A 1
.
73 m long pole stands vertically in a lake
having a depth of 0
.
81 m.
When the Sun is 35
.
7
◦
above the horizon,
find the length of the pole’s shadow on the
bottom of the lake.
The index of refraction
for water is 1
.
33.
Correct answer: 1
.
90482 m.
Explanation:
θ
φ
2
s
2
pole
s
1
l
φ
1
d
Given :
‘
= 1
.
73 m
,
d
= 0
.
81 m
,
n
= 1
.
33
,
and
θ
= 35
.
7
◦
.
Light just passing the top of the pole has
an angle of incidence of
φ
1
= 90
◦

θ
= 90
◦

35
.
7
◦
= 54
.
3
◦
.
This light falls on the water’s surface at dis
tance
s
1
= (
‘

d
) tan
φ
1
= (1
.
73 m

0
.
81 m) tan 54
.
3
◦
= 1
.
28031 m
from the pole. As the light enters the water,
Snell’s Law becomes
sin
φ
1
=
n
sin
φ
2
φ
2
= arcsin
sin
φ
1
n
¶
= arcsin
sin 54
.
3
◦
1
.
33
¶
= 37
.
6321
◦
.
Then distance between the extremity of the
pole’s shadow on the water’s surface and the
extremity of the pole’s shadow on the bottom
of the lake is
s
2
=
d
tan
φ
2
= (0
.
81 m) tan 37
.
6321
◦
= 0
.
624507 m
.
The whole shadow’s length is
s
=
s
1
+
s
2
= 1
.
28031 m + 0
.
624507 m
=
1
.
90482 m
.
Note: The shadow’s length without the water
would be
s
‘
=
‘
tan
φ
1
so its length is shortened by the water by
Δ
s
=
s
‘

s
= (1
.
73 m) tan 54
.
3
◦

1
.
90482 m
= 0
.
502726 m
.
keywords:
002
(part 1 of 4) 10 points
Light of wavelength 396 nm is incident on the
face of a silica prism at an angle of
θ
1
= 56
.
2
◦
(with respect to the normal to the surface).
The apex angle of the prism is
φ
= 72
.
6
◦
.
Given:
The value of the index of refraction
for silica is
n
= 1
.
455.
θ
2
θ
3
θ
1
φ
θ
4
Find the angle of refraction at this first
surface.
Correct answer: 34
.
8286
◦
.
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Nguyen, Thanh – Homework 20 – Due: Apr 18 2008, 7:00 pm – Inst: Weathers
2
Explanation:
Basic Concept:
Snell’s Law
n
1
sin
θ
1
=
n
2
sin
θ
2
.
Solution:
Applying Snell’s law, the refrac
tion angle at the first surface
θ
2
= sin

1
sin
θ
1
n
¶
= sin

1
sin 56
.
2
◦
1
.
455
¶
= 34
.
8286
◦
.
003
(part 2 of 4) 10 points
Find the angle of incidence at the second sur
face.
Correct answer: 37
.
7714
◦
.
Explanation:
The incident angle at the second surface is
θ
3
=
φ

θ
2
= (72
.
6
◦
)

(34
.
8286
◦
)
= 37
.
7714
◦
,
where
φ
is the apex angle of the prism.
004
(part 3 of 4) 10 points
Find the angle of refraction at the second
surface.
Correct answer: 63
.
0252
◦
.
Explanation:
The incident angle at the second surface is
θ
3
= 37
.
7714
◦
, so the refraction angle is
θ
4
= sin

1
[(
n
sin
θ
3
)]
= sin

1
[(1
.
455) sin(37
.
7714
◦
)]
= 63
.
0252
◦
.
005
(part 4 of 4) 10 points
Find the angle between the incident and
emerging rays.
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 Spring '00
 Littler
 Work, Angle of Incidence, Snell's Law, Correct Answer, Total internal reflection

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