Ch. 27 HW Solutions
1.) Because the lens takes the parallel rays from the sun and focuses them on a point, the
lens is converging. The point at which parallel rays converge is the focus of the lens, 12 cm.
3.) (a) q = 80 cm: real, inverted
(b) q = 200 cm: real, inverted
(c) q = 40 cm: virtual, upright
4.) (a) q = 100 cm: real, inverted
(b) q = 133 cm: real, inverted
(c) q = 33.3 cm: virtual, upright
8.)
p = 12 cm, q = 4.0 m.
f
= (1
/p
+ 1
/q
)

1
= (1
/
12 + 1
/
400)

1
= 11
.
65
cm.
M
=

q/p
=

400
/
12 =

33
.
3
. The image size is magnified 33.3 times, so it is 83.3 cm
×
116.7
cm (inverted).
14.) q = 80 cm, p = 12 cm.
f
= (1
/p
+ 1
/q
)

1
= (1
/
12 + 1
/
80)

1
= 10
.
4
cm.
15.)
f = 80 cm,
h
i
= 2
m
= 200
cm,
h
o
= 50
cm.
M
=

h
i
/h
o
=

200
/
50 =

4
(real images from converging lenses are inverted). Since
M
=

q/p
=

4
,
q
= 4
p
. Then
1
/f
= 1
/
80 = 1
/p
+ 1
/
4
p
= 5
/
4
p
. So (a)
p
= (5)(80)
/
4 = 100
cm. (b)
q
= 4
p
= 400
cm.
20.) The object is 150 cm from the screen (the image location) so
p
+
q
= 150
cm. With f
= 25 cm, we use the lens equation
1
25
=
1
p
+
1
150

p
=
150

p
p
(150

p
)
+
p
p
(150

p
)
=
150

p
+
p
p
(150

p
)
=
150
p
(150

p
)
.
25 =
p
(150

p
)
150
=

p
2
+ 150
p
150
p
2

150
p
+ 3750 = 0
Using the quadratic equation,
150
±
p
150
2

(4)(3750)
2
and solving yields
p
= 118
.
3
cm,
31
.
7
cm. The lens must be placed this far from the object.
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 Summer '08
 Lee
 Applied Physics, Quadratic equation, Harshad number, 1.2 cm

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