Solution to set 9
Physics 122
Chapter27:
21. Picture the Problem: A nearsighted person with a far point of 1.8 meters looks at himself in the
mirror.
Strategy: Because the student is using a plane mirror, the image distance is equal to the object
distance. The image is therefore twice the distance he is standing from the mirror. Divide the student’s
far point in half to determine how far he can stand from the mirror.
Solution: Calculate the maximum
distance from the mirror:
d=
1.6 m
= 0.80 m
2
Insight: If the person had been hyperopic (farsighted) with a near point of 1.6 m, then the closest he
could stand to the mirror would be 80 cm.
82. Picture the Problem: An optical system consists of two lenses, one with a focal length of 50 cm and
the other with a focal length of 2.5 cm. The separation between the lenses is 52.5 cm.
Strategy: Recall the principles of the construction of microscopes and telescopes when answering the
question.
Solution: If a telescope were constructed from these lenses, the separation between the lenses would
equal the sum of the focal lengths, or 52.5 cm. This matches the given separation, so we conclude that
the instrument is a telescope.
Insight: The magnifying power of this telescope would be f o fe = 50 cm 2.5 cm = 20 × . If a
microscope were constructed from these lenses with a “barrel length” of 16 cm, the shorter focal
length would be the objective and the magnifying power would be
M total = −
(
(
)(
)(
)
)
16 cm 25 cm
di N
=−
= − 3.2 × . The lenses make a better telescope than
f objective feyepiece
2.5 cm 50 cm
microscope!
96. Picture the Problem: The figure shows a farsighted
person wearing glasses with refractive power 3.6
diopters. The glasses create an image at the person’s
near point of an object that is 25.0 cm from the
person’s eye (the corrected near point).
Strategy: Solve equation 26-16 for the image
distance, where the inverse of the focal length is the
refractive power (R.P.) of the lens and the object
distance is the distance between the lens and the
corrected near point. Add the eye-to-lens distance D
= 2.5 cm to the absolute value of the image distance
to find the uncorrected near point.
Solution: 1. Calculate the image distance:
"1 1%
di = $ − '
# f do &
−1
"
1%
= $ R.P. − '
do &
#
−1
"
%
1
= $ 3.6 diopters −
0.25 m − 0.025 m '&
#
−1
= −1.18 m
2. Calculate the near point:
Insight: This person could alternatively use R.P. =
contact lenses.
N = 1.18 m + 0.025 m = 1.21 m
1 1
1
1
+ =
+
= +3.2 diopter
do di 0.25 m −1.18 m