CHAPTER
26
THE REFRACTION OF
LIGHT: LENSES AND
OPTICAL INSTRUMENTS
CONCEPTUAL QUESTIONS
____________________________________________________________________________________________
1.
REASONING AND SOLUTION
Since the index of refraction of water is greater than that
of air, the ray in Figure 26.1
a
is bent toward the normal at the angle
q
w1
when it enters the
water.
According to Snell's law (Equation 26.2), the sine of
w1
is given by
air
1
1
w1
water
water
sin
sin
sin
n
nn
θ
==
(1)
where we have taken
n
air
=
1.000 .
When a layer of oil is added on top of the water, the
angle of refraction at the air/oil interface is
oil
and, according to Snell's law, we have
air
1
1
oil
oil
oil
sin
sin
sin
n
(2)
But
oil
is also the angle of incidence at the oil/water interface.
At this interface the angle of
refraction is
w2
and is given by Snell's law as follows:
oil
oil
oil
1
w2
water
water
oil
water
sin
sin
sin
sin
n
n
1
== =
(3)
where we have substituted Equation (2) for
oil
sin
.
According to Equation (1), this result is
equal to
w1
sin
.
Therefore, we can conclude that the angle of refraction as the ray enters
the water does
not
change due to the presence of the oil.
____________________________________________________________________________________________
2.
REASONING AND SOLUTION
When light travels from a material with refractive index
n
1
into a material with refractive index
n
2
, the angle of refraction
2
is related to the angle of
incidence
1
by Equation 26.2:
11
2
sin
2
=
or
1
21
2
sin
(
/
)sin
1
−
=
⎡
⎤
⎣
⎦
.
When
n
1
<
n
2
, the angle of refraction will be less than the angle of incidence.
The larger the value
of
n
2
, the smaller the angle of refraction for the same angle of incidence.
The angle of
refraction is smallest for slab B; therefore, slab B has the greater index of refraction.
____________________________________________________________________________________________
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THE REFRACTION OF LIGHT:
LENSES AND OPTICAL INSTRUMENTS
3.
REASONING AND SOLUTION
When an observer peers
over the edge of a deep empty bowl, he does not see the entire
bottom surface, so a small object lying on the bottom is
hidden from view.
However, when the bowl is filled with
water, the object can be seen.
When the object is viewed from the edge of the bowl, light
rays from the object pass upward through the water.
Since
n
air
<
n
water
, the light rays from the object refract away from
the normal when they enter air. The refracted rays travel to
the observer, as shown in the figure at the right.
When the
rays entering the air are extended back into the water, they
show that the observer sees a virtual image of the object at an
apparent depth that is less than the actual depth, as indicated in the drawing.
Therefore, the
apparent position of the object in the water is in the line of sight of the observer, even
though the object could not be seen before the water was added.
Imag
Object
____________________________________________________________________________________________
4.
SSM
REASONING AND SOLUTION
Two identical containers, one filled with water
(
n
= 1.33) and the other filled with ethyl alcohol (
n
= 1.36) are viewed from directly above.
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 Spring '08
 holland
 Light, Snell's Law, refractive index, Total internal reflection, Geometrical optics

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