Unformatted text preview: 1.333 for water, we
find that the angles of refraction are sin θ 2 = n1 sin θ1
n2 or n1 sin θ1 n2 θ 2 = sin −1 (1.000 ) sin 60.0° = 41.4°
1.309 Ice θ 2, ice = sin −1 Water θ 2, water = sin −1 (1.000 ) sin 60.0° = 40.5°
1.333 The difference in the angles of refraction is θ 2, ice − θ 2, water = 41.4° − 40.5° = 0.9°
______________________________________________________________________________ 1356 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 12. REASONING AND SOLUTION The angle of incidence is found from the drawing to be 8.0 m = 73° 2.5 m θ1 = tan −1 Snell's law gives the angle of refraction to be
sin θ2 = (n1/n2) sin θ1 = (1.000/1.333) sin 73° = 0.72 or θ2 = 46° The distance d is found from the drawing to be
d = 8.0 m + (4.0 m) tan θ2 = 12.1 m
______________________________________________________________________________ 13. SSM REASONING We will use the geometry of the situation to determine the angle of
incidence. Once the angle of incidence is known, we can use Snell's law to find the index of
refraction of the unknown liquid. The speed of light v in the liquid can then be determined.
SOLUTION From the drawing in the text, we see that the angle of incidence at the liquidair interface is 5.00 cm θ1 = tan –1 = 39.8° 6.00 cm The drawing also shows that the angle of refraction is 90.0°. Thus, according to Snell's law
(Equation 26.2: n1 sin θ1 = n2 sin θ 2 ), the index of refraction of the unknown liquid is n1 = n2 sinθ 2
sinθ1 = (1.000) (sin 90.0°)
= 1.56
sin 39.8° From Equation 26.1 ( n = c / v ), we find that the speed of light in the unknown liquid is c 3.00 × 10 8 m/s
=
= 1.92 × 10 8 m/s
n1
1.56
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v= 14. REASONING The index of refraction of the oil is one of the factors that determine the
apparent depth of the bolt. Equation 26.3 gives the apparent depth and can be solved for
the index of refraction. Chapter 26 Problems 1357 SOLUTION According to Equation 26.3, the apparent depth d ′ of the bolt is
n d′ = d 2 n 1 where d is the actual depth, n1 is the refractive index of the medium (oil) in which the
object is located, and n2 is the medium (air) in which the observer is located directly above
the object. Solving for n1 and recognizing that the refractive index of air is n2 = 1.00, we
obtain
d 5.00 cm n1 = n2 = (1.00 ) = 1.47 d′ 3.40 cm 15. REASONING The drawing shows a ray of
sunlight reaching the scuba diver (drawn as a
black dot). The light reaching the scuba diver
makes of angle of 28.0° with respect to the
vertical. In addition, the drawing indicates that
this angle is also the angle of refraction θ2 of the
light entering the water. The angle of incidence
for this light is θ1. These angles are related by
Snell’s law, n1 sin θ1 = n2 sin θ 2 (Equation 26.2),
where n1 and n2 are, respectively the indices of refraction of the air and water. Since θ2, n1, and
n2 are known, the angle of incidence can be
determined. Sun θ1 Air
Water
28.0° θ2 = 28.0° Scuba
diver SOLUTION The angle of incidence of the light is given according to sin θ1 = n2 sin θ 2
n1 or n2 sin θ 2 −1 1.333sin 28.0° = 38.7° = sin n1 1.000 θ1 = sin −1 where the values n1 = 1.000 and n2 = 1.333 have been taken from Table 26.1.
______________________________________________________________________________
16. REASONING When the light ray passes from a into b, it is bent toward the normal.
According to the discussion in Section 26.2, this happens when the index of refraction of b
is greater than that of a, or nb > na . When the light passes from b into c, it is bent away
from the normal. This means that the index of refraction of c is less than that of b, or
nc < nb . The smaller the value of nc, the greater is the angle of refraction. As can be seen
from the drawing, the angle of refraction in material c is greater than the angle of incidence 1358 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS at the ab interface. Applying Snell’s law to the ab and bc interfaces gives
na sin θa = nb sin θb = nc sin θc. Since θc is greater than θa, the equation na sin θa = nc sin θc
shows that the index of refraction of a must be greater than that of c, na > nc . Thus, the
ordering of the indices of refraction, highest to lowest, is nb, na, nc.
SOLUTION The index of refraction for each medium can be evaluated from Snell’s law,
Equation 26.2:
n sin θ a (1.20 ) sin 50.0°
nb = a
=
= 1.30
ab interface
sin θ b
sin 45.0°
bc interface nc = nb sin θ b
sin θ c = (1.30 ) sin 45.0° = 1.10
sin 56.7° As expected, the ranking of the indices of refraction, highest to lowest, is
nb = 1.30, na = 1.20, nc = 1.10
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17. REASONING When the incident light is in a vacuum, Snell’s law, Equation 26.2, can be
used to express the relation between the angle of incidence (35.0°), the (unknown) index of
refraction n2 of...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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