Φ21 Fall 2006
1
HW25 Solutions
Problem K23.32
Find the focal length of the glass lens (n
= 1.50)
in Figure 1.
The
distances shown are the radii of curvature.
Solution: The sign convention for radii of lenses in our book is negative
for the center to the left and positive for the center to the right, like a
number line. (Another explanation is on p. 743 of the book.) So, in this
case,
R1 = −40 cm
1
f
f
R2 = +40 cm. The focal length is found
)
(
)
(
1
1
1
1
−
= (n − 1)
−
= 0.5
R1
R2
−40 40
= −40 cm
and
from:
The focal length is negative so this is a diverging lens. (The fact that
of the value of
2
Figure 1: A thin diverging lens.
f
is equal to the radii is a coincidence
n.)
Problem YF 34.7
7
The diameter of Mars is 6790 km, and its minimum distance from the earth is 5.58×10 km. When Mars
is at this distance, nd the diameter of the image of Mars formed by a spherical, concave, telescope mirror
with a focal length of 1.95 m.
5.58 × 1010 m,
Solution: The object distance is
which is eectively innity for the purpose of nding the
image position (try it). So, the image is formed at
di = f = 1.95 m.
The height of the image comes from
the magnication formula.
m=
hi
di
1.95
=− =−
= −3.49 × 10−11
ho
do
5.58 × 1010
The diameter of Mars in the image would be scaled by this amount. (Remember that diameter is a positive
quantity, so the minus sign just means the image gets ipped.)
(
)(
)
|hi | = m · ho = 3.49 × 10−11 6790 × 103 m = 2.37 × 10−4 m = 0.237 mm
3
Problem YF 34.49
The focal length of a simple magnier is 7.00 cm. Assume the magnier to be a thin lens placed very close
to the eye. Part A. How far in front of the magnier should an object be placed if the image is formed at
the observer's near point, a distance of 25.0 cm in front of her eye?
f = 7.00 cm. The image position is on the
di = −25.0 cm. The object position is then
Solution: A magnier is a converging lens, so
the light is coming, so it is a virtual image and
1
1
+
do
di
=
do
=
1
f
(
1
1
−
f
di
)−1
= 5.47 cm
Part B. If the object has a height of 3.00 mm, what is the height of the image?
Solution: The magnication is
m = − ddoi ,
hi = −
so:
25.0
di
ho = −
(3.00 mm) = 13.7 mm
do
5.47
1
side from which
4
Problem YF 34.99
When an object is placed at the proper distance to the left of a converging lens, the image is focused on a
screen which is placed a distance of 30.5 cm to the right of the lens. A diverging lens is now placed a distance
of 14.7 cm to the right of the converging lens, and it is found that the screen must be moved a distance of
19.5 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens?
Solution: The image from the rst lens becomes the object for the second lens. Since the second lens is
placed between the rst lens and the rst image, it will be a virtual object with
do < 0.
This distance is
d0 = − (30.5 − 14.7) = −15.8 cm
The nal image is 19.5 cm further and is projected onto a screen, so it is a real image with
distance is
di = 15.8 + 19.5 = +35.3 cm
With these two values, it is easy to see that the focal length is
(
f=
1
1
+
do
di
(
)−1
=
1
1
+
−15.8 35.3
2
)−1
= −28.6 cm
di > 0.
The