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Unformatted text preview: the magnification equation that hi
di
m
{= h = − d
o
o
1 so that di = − 1 d o
4 (25.4) 4 We can substitute this relation into the mirror equation to find the ratio do/f.
SOLUTION Substituting di = − 1 do into the mirror equation gives
4 1
1
1
+
=
d o di
f or 1
1
1
+1
=
d o − 4 d0
f Solving this equation for the ratio do/f yields do/f = −3 . or −3 1
=
do f (25.3 ) 1314 THE REFLECTION OF LIGHT: MIRRORS 46. REASONING AND SOLUTION
We know that do − di = 45.0 cm. We also have
1/do + 1/di = 1/f. Solving the first equation for di and substituting the result into the second
equation yields,
1
1
1
+
=
d o do − 45.0 cm
30.0 cm
Cross multiplying gives do2 − 105 do + 1350 = 0, which we can solve using the quadratic
equation to yield two roots, do = (105 ± 75)/2.
a. When the object lies beyond the center of curvature we have
do+ = (1.80 × 102 cm)/2 = +9.0 × 10 1 cm and di+ = +45 cm b. When the object lies within the focal point
do– = (3.0 × 101 cm)/2 = +15 cm , and di– = −3.0 × 10 1 cm 47. SSM REASONING AND SOLUTION We
can see from the upper triangle in the drawing
that
tan θ = (L − x)/L L−x L We also see from the lower triangle that
tan θ = x/(L/2) x Equating these two expressions gives L−x
=
L x
L 1
2 or L /2 x=1L
3 Therefore,
x
1 −1 3 L
θ = tan 1 = tan 1 = L 2 2L
−1 θ
θ 33.7° CHAPTER 26 THE REFRACTION OF LIGHT: LENSES AND
OPTICAL INSTRUMENTS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (c) When the light is refracted into liquid B it is bent away from the normal, so that nA > nB.
When the light is refracted into liquid C it is bent toward the normal, so that nC > nA.
Therefore, we conclude that nC > nA > nB.
2. 1.41
3. (b) When the light is refracted into liquid B it is bent away from the normal, so that nA > nB.
When the light is refracted into liquid C it is also bent away from the normal, so that nA > nC.
However, the bending is less than that in liquid B. Thus, the index of refraction of liquid C is
closer to that of liquid A than the index of refraction of liquid B is. In fact, if nA and nC were
equal, the ray would not be bent at all upon entering liquid C. This means that nC > nB.
Therefore, we conclude that nA > nC > nB.
4. 4.41 cm 1.00 5. (d) The apparent depth d ′ is given by d ′ = d , according to Equation 26.3, where d is
n
the actual depth and n is the refractive index of the liquid. Thus, the apparent depth and the
refractive index are inversely proportional, which means that the ranking in descending order
is nC, nB, nA.
6. (c) Material B has the smaller refractive index. Therefore, a light ray that begins in material
B and travels toward material A, with its greater refractive index, is bent toward the normal
(not away from it) when it crosses the interface.
7. (a) When light travels from a material with a greater refractive index toward a material with a
smaller refractive index, total internal reflection occurs at the interface when the angle of
incidence is greater than the critical angle. Here, Equation 26.4 reveals that the critical angle 1.00 for the glassair interface is θ c = sin −1 = 41.1° . At point A the angle of incidence is 1.52 35º and is less than the critical angle. Therefore, some light passes into the air at point A. At
point B, however, total internal reflection occurs, because the angle of incidence is 55º and
exceeds the critical angle. Chapter 26 Answers to Focus on Concepts Questions 1349 8. 67.5 degrees
9. (d) The displacement occurs because of the refraction or bending of the ray that occurs when
the ray enters and leaves the glass. The bending is greater when the refractive index is
greater. Hence, the displacement is greatest for violet light.
10. (b) Equation 26.4 reveals that the critical angles for the glassair interface are 1.000 −1 1.000 The angle of
θ c, red = sin −1 = 41.14° and θ c, violet = sin = 40.56° . 1.520 1.538 incidence shown is 40.85º, which is greater than the critical angle for violet light, but less
than the critical angle for red light. Therefore, only red light enters the air, the violet light
being totally internally reflected.
11. (e) For a converging lens, rays that are parallel to and close to the principal axis are bent
toward the axis by the lens and pass through the focal point on the far side of the lens. For a
diverging lens, rays that are parallel to and close to the principal axis are bent away from the
axis by the lens. After being bent they appear as if they originated at the focal point on the
near side of the lens. They do not actually pass through that focal point, however.
12. (b) A converging lens can produce a virtual image if the object is within the focal point of
the lens. However, the image is upright. A diverging lens produces only upright virtual
images, no matter where the object is located with respect to the lens.
13. (d) A converging lens can produce an upright virtual image, if the object is within the focal
point of the lens. However, this image i...
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 Spring '13
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 Physics, The Lottery

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