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Ch. 26 Solution
PHY 205
H.S.
Chapter 26: Geometrical Optics
Answers Conceptual Questions
4.
The concave side of the dish collects the parallel rays coming from a geosynchronous
satellite and focuses them at the focal point of the dish. The convex side of the dish would
send the parallel rays outward on divergent paths. The situation is analogous to that of light
in an optical telescope.
10.
The Sun is already well below the horizon when you see it setting. The reason is that as the
Sun’s light enters the atmosphere from the vacuum of space, it is bent toward the normal;
that is, toward the surface of Earth. Therefore, the light from the Sun can still reach us even
when a straight line from our eyes to the Sun would go below the horizon.
Answers Conceptual Exercises
4.
(a) Looking at the front side of a spoon means we are looking at a concave mirror. In
addition, holding the spoon at arm’s length means that we are outside the focal point of the
mirror—clearly, the focal length of the front side of a spoon is only a few centimeters. The
situation, then, is like that illustrated in Figure 26-18 (a). It follows that our image is inverted.
(b) Continuing to refer to Figure 26-18 (a), we see that our image is also reduced. (c) Light
rays pass through the image, and hence the image is real.
18.
(a) Over the period of a year on an airless Earth, the number of daylight hours would be
equal to the number of nighttime hours. (b) Because Earth's atmosphere bends light toward
the normal as it enters the atmosphere—along a more vertical direction—we see the Sun as
being higher in the sky than it actually is. As a result, we see the Sun even after a straight
line to the Sun is below the horizon—therefore, the number of daylight hours, when
averaged over a year, is greater than the number of nighttime hours. In fact, the atmosphere
gives us more than an extra 24 hours of daylight over the course of a year.
Solutions to Problems
6. Picture the Problem: The image shows you observing the
image of your feet in a vertical mirror mounted on a wall that
is 1.50 meters away.
Strategy: To observe your eyes the mirror should be tilted
by an angle equal to the angle of reflection of the light from
the shoes. Calculate the angle of reflection from the inverse
tangent of half your height divided by the distance to the
mirror.
Solution:
Calculate θ r :
θ r = tan −1
⎡ 1 (1.75 m ) ⎤
h
= tan −1 ⎢ 2
⎥ = 30.3°
d
⎣ 1.50 m ⎦
Insight: Rotating the mirror by any angle between 0° and 30.3° will enable you to see different parts of
your body.
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Ch. 26 Solution
PHY 205
H.S.
14. Picture the Problem: The image shows two rays
that approach each other at an angle of 27°. A
mirror of diameter 11 cm is placed in the path of
the rays causing them to reflect before they
intersect.
Strategy: The distance d will be greatest when the
rays are incident at the edges of the mirror, 11 cm
apart. The reflected rays and the diameter of the
mirror form an isosceles triangle. The distance d
bisects this triangle forming two right triangles
where d is one leg, (11 cm)/2 is the other leg, the
reflected ray from the mirror to the point of
intersection is the hypotenuse, and θ is 27°/2. Use
this triangle to calculate the distance d.
27° y 11 cm 2
= =
2
x
d
Solution: 1.
Set the tangent
equal to the
ratio y x :
tan θ = tan
2. Solve for
dmax:
d max = 5.5 cm tan13.5° = 23 cm
Insight: The distance d from the mirror to the point that the rays intersect is always equal to the
distance between the mirror and the point that the incident rays would intersect (dashed lines in the
figure).
18. Picture the Problem: The image shows parallel rays
from the Sun converging at the focal point of a
concave piece of glass.
Strategy: Solve equation 26-3 for the radius of
curvature, given that the rays converge at the focal
point.
Solution: Calculate the R = 2 f = 2 (15 cm ) = 30 cm
radius:
Insight: If the glass were turned around so that the
light was reflecting off the convex portion, a virtual
image of the Sun would appear to be 15 cm behind the
glass.
40. Picture the Problem: A light ray refracts as it travels from air to liquid benzene.
Strategy: Solve Snell’s Law (equation 26-11) for the index of refraction of benzene.
Solution: Calculate the index of refraction:
nbenzene = nair
sin θ i
sin 43°
= 1.000
= 1.5
sin θ 2
sin 27°
Insight: This solution agrees with the index of refraction given in table 26-2.
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Ch. 26 Solution
PHY 205
H.S.
55. Picture the Problem: The figure shows a horizontal beam
refracting into a right-isosceles prism. The beam makes a θ 4 = 34°
angle with the horizontal after exiting the prism. The angles θ2 and
θ3 are labeled as 2 and 3 due to limited space in the diagram.
Strategy: Use Snell’s Law (equation 26-11) to write an equation
for the refracted angle θ2 in terms of the index of refraction n of
the glass. Use Snell’s law again to write an equation for the
incident angle θ3 in terms of n and θ4. Set the sum of the angles in
the highlighted triangle equal to 180° to create a relationship
between the θ2 and θ3, and then use the relations to calculate θ2.
Finally, use θ2 to calculate n.
Solution: 1. Write an expression for n in terms of nair sin θ1 = n sin θ 2
θ2:
n = sin 45° sin θ 2
nair sin θ 4 = n sin θ 3
2. Write an expression for n in terms of θ3:
n = sin 34° sin θ 3
3. Sum the interior angles of the triangle and
solve for θ3:
θ 2 + (θ 3 + 90° ) + 45° = 180°
4. Use a trigonometric identity to write sin θ 3 :
sin θ3 = sin ( 45° − θ 2 ) = sin 45° cos θ 2 − cos 45° sin θ 2
5. Set the two equations from
steps 1 and 2 equal and write
sin θ 3 in terms of θ2:
n=
6. Multiply both sides by sin θ 2 and divide by
sin 34° :
θ 3 = 45° − θ 2
sin 45° sin 34°
sin 34°
=
=
sin θ 2
sin θ3
sin 45° cos θ 2 − cos 45° sin θ 2
sin 45°
1
=
sin 34° sin 45° cot θ 2 − cos 45°
⎡ cos 45° + sin 34° sin 45° ⎤
⎥ = 25°
sin 45°
⎣
⎦
7. Solve for θ2:
θ 2 = cot −1 ⎢
8. Calculate the index of refraction:
n = sin 45° sin 25° = 1.7
Insight: It would require an index of refraction greater than 1.7 in order to have a beam exit at an angle
greater than 34°. For example, if the index of refraction were 2.1, the beam would exit at θ4 = 60°.
60. Picture the Problem: The figure shows an
object located a distance 2 f from a convex
lens.
Strategy: Sketch the P ray, F ray, and M ray
for the object. Use the diagram to determine
the approximate image position, image
orientation, and image type.
Solution 1. (a) Sketch the three rays on the
diagram:
2. Note the location of the image:
3. (b) Note the image orientation:
The image is located on the right side of the lens at about 2f
.
The image is inverted.
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Ch. 26 Solution
4. (c) Note the image type:
PHY 205
H.S.
The image is real, because the image is on the opposite side
of the lens from the object, and light passes through it.
Insight: When the object distance is greater than the focal length of a convex lens, the image is real
and inverted.
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