honea (cth632) – H9Refraction – avram – (1956)
1
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001
10.0 points
A shallow pool of a liquid 7
.
4 m deep is not
the depth one expects when viewed from over
head. The index of refraction of the liquid is
1
.
35.
How deep does it appear to be?
Correct answer: 548
.
148 cm.
Explanation:
Basic Concepts
n
1
s
1
+
n
2
s
2
=
n
2

n
1
R
Solution:
Consider a point on the bottom
of the pool as our object. In the refraction
surface formula
n
1
s
1
+
n
2
s
2
=
n
2

n
1
R
.
Because the water surface of a pool is a plane,
R
=
∞
, thus
n
2

n
1
R
= 0. The object
s
1
is under the water,
n
1
= 1
.
35 is the index of
refraction of water, and
n
2
= 1 is the index of
refraction of air. Hence,
s
2
=

s
1
n
2
n
1
=

(7
.
4 m)(1)
(1
.
35)
=

5
.
48148 m
.
The image is virtual and on the same side
of the interface as the object at a distance
d
apparent
=

5
.
48148 m

= 548
.
148 cm.
Note: One may also determine the apparent
depth in an alternative approach. ±irst sketch
a ray diagram for a Fnite incident angle where
the refacted ray hits the bottom of the pool
at P. Draw a vertical line OP, where O is at
the surface of the water, OP is the real depth.
The apparent depth is deFned by the depth
OP’, where P’ is the intersection between OP
and the extrapolated line of the incident ray.
We leave it an exercise for the reader to show
that
OP
′
/OP
=
n
1
/n
2
.
002
10.0 points
Batman and Robin are attempting to escape
that dastardly villain, the Joker, by hiding
in a large pool of water (refractive index
n
water
= 1
.
333). The Joker stands gloating at
the edge of the pool. (His makeup is water
soluble.) He holds a powerful laser weapon
y
1
= 1
.
33 m above the surface of the water
and Fres at an angle of
θ
1
= 29
.
9
◦
to the hor
izontal. He hits the Boy Wonder squarely on
the letter “R”, which is located
y
2
= 3
.
92 m
below the surface of the water.
θ
x
y
y
1
1
2
R
J
Batplastic
surface
Mirrored
Surface
water
B
How far (horizontal distance) is Robin from
the edge of the pool? (±ear not, Batfans. The
“R” is made of laserre²ective material.)
Correct answer: 5
.
66886 m.
Explanation:
Basic Concepts:
Snell’s law, Total inter
nal re²ection.
Solution:
Let
x
1
be the horizontal distance
from the laser to where the laser beam strikes
the water and
x
2
the horizontal distance from
that point to Robin (see the following Fgure).
1
x
y
1
x
y
2
2
air
water
J
R
θ
90θ
ο
1
1
θ
2
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Then we have
x
1
=
y
1
tan
θ
1
=
(1
.
33 m)
tan(29
.
9
◦
)
= 2
.
31294 m
.
From Snell’s law,
n
air
sin(90
◦

θ
1
) =
n
water
sin
θ
2
.
So
sin
θ
2
=
sin(90
◦

29
.
9
◦
)
(1
.
333)
= 0
.
650335
θ
2
= arcsin(0
.
650335)
= 40
.
5669
◦
.
Hence,
x
2
=
y
2
tan
θ
2
= (3
.
92 m) tan 40
.
5669
◦
= 3
.
35592 m
.
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 Avram
 Physics, Work, Correct Answer, Avram –

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