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Unformatted text preview: s larger (not smaller) than the object (see Section
26.7).
14. 2.5 cm
15. 9.17 cm
16. 417 cm
17. (c) (Statement A) The refractive power in diopters is the reciprocal of the focal length in
meters, according to Equation 26.8. Thus, a positive refractive power means that the focal
length is positive, which means that the lens is a converging lens. Converging lenses can
produce images that are reduced in size. (Statement D) Two lenses with different refractive
powers cannot have the same focal length. This follows directly from Equation 26.8.
(Statement E) Since refractive power and focal length are inversely proportional according
to Equation 26.8, it follows that lens A, with twice the refractive power of lens B, must have
a focal length that is onehalf that of lens B. 1350 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 18. (a) For small angles the angular size in radians is given by the object height divided by the
distance from the eye. Thus, when an object has the greatest angular size and is located the
farthest away, it must have the greatest height. This means that object B has the greatest
height. Objects A and C have the same angular size, but object C is twice as far away, which
means that object C must have a greater height than object A. The ranking in descending
order is, then, B, C, A.
19. 8.00 × 10−3
20. (d) Both dispersion and chromatic aberration occur because the refractive index depends on
the wavelength of light, as discussed in Sections 26.5 and 26.14. Chapter 26 Problems 1351 CHAPTER 26 THE REFRACTION OF LIGHT: LENSES AND
OPTICAL INSTRUMENTS
PROBLEMS
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1. SSM REASONING The substance can be identified from Table 26.1 if its index of
refraction is known. The index of refraction n is defined as the speed of light c in a vacuum
divided by the speed of light v in the substance (Equation 26.1), both of which are known.
SOLUTION Using Equation 26.1, we find that n= c 2.998 ×108 m/s
=
= 1.362
v 2.201×108 m/s Referring to Table 26.1, we see that the substance is ethyl alcohol .
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2. c
(Equation 26.1),
v
where n is the index of refraction of the medium, and c = 3.00×108 m/s is the speed of light
in a vacuum. Equation 26.1 is sufficient to determine the speed of light in this collection of
atoms.
REASONING The speed v of light in any medium is found from n = SOLUTION Solving Equation 26.1 for v, we obtain
c 3.00 × 108 m/s
=
= 19.1 m/s
n
1.57 ×107
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v= 3. REASONING The refractive index n is defined by Equation 26.1 as n = c/v, where c is the
speed of light in a vacuum and v is the speed of light in a material medium. We will apply
this definition to both materials A and B, and then form the ratio of the refractive indices.
This will allow us to determine the unknown speed.
SOLUTION Applying Equation 26.1 to both materials, we have nA = c
vA and nB = c
vB 1352 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS Dividing the equation for material A by that for material B gives nA c / vA vB
=
=
nB c / vB vA
Solving for vB, we find that
n
vB = vA A
n
B 4. 8
8 = 1.25 × 10 m/s (1.33 ) = 1.66 × 10 m/s ( ) REASONING The wavelength λ is related to the frequency f and speed v of the light in a
material by Equation 16.1 (λ =v/f ). The speed of the light in each material can be
expressed using Equation 26.1 (v = c/n) and the refractive indices n given in Table 26.1.
With these two equations, we can obtain the desired ratio.
SOLUTION Using Equations 16.1 and 26.1, we find λ= v c/ n c
=
=
f
f
fn Using this result and recognizing that the frequency f and the speed c of light in a vacuum
do not depend on the material, we obtain the ratio of the wavelengths as follows: c f n alcohol λalcohol
=
λ disulfide c c 1 n
f n alcohol
1.632
=
= disulfide =
= 1.198
c 1
nalcohol
1.362 f n disulfide f n disulfide
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5. SSM WWW REASONING Since the light will travel in glass at a constant speed v, the
time it takes to pass perpendicularly through the glass is given by t = d / v , where d is the
thickness of the glass. The speed v is related to the vacuum value c by Equation 26.1:
n = c / v.
SOLUTION Substituting for v from Equation 26.1 and substituting values, we obtain d nd (1.5) ( 4.0 ×10 –3 m )
=
=
= 2.0 ×10 –11 s
v
c
3.00 ×108 m/s
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t= Chapter 26 Problems 6. 1353 REASONING We can identify the substance in Table 26.1 if we can determine its index of
refraction. The index of refraction n is equal to the speed of light c in a vacuum divided by
the speed of light v in the substance, or n = c/v. According to Equation 16.1, however, the
speed of light is related to its wavelength λ and frequency f via v = f λ. Combining these two...
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 Spring '13
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 Physics, The Lottery

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