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Chapter 36
377
P36.74
(a)
Start with the second lens: This lens must form a virtual image located 19.0 cm to the left of
it (i.e.,
q
2
19 0
=−
. cm ). The required object distance for this lens is then
p
qf
2
22
19 0
20 0
19 0
20 0
380
39 0
=
−
=
−
−−
=
..
.
cm
cm
cm
cm
cm
a
fa
f
.
The image formed by the first lens serves as the object for the second lens. Therefore, the
image distance for the first lens is
qp
12
50 0
50 0
380
1570
=−=
cm
cm
cm
39.0
cm
39.0
.
The distance the original object must be located to the left of the first lens is then given by
111
1
10 0
39 0
157 39 0
118
pf
q
−
=
−
=
.
cm
cm
cm
1 570 cm
or
p
1
13 3
==
cm
118
cm
(b)
MMM
q
p
q
p
−
F
H
G
I
K
J
−
F
H
G
I
K
J
=
F
H
G
I
K
J
F
H
G
I
K
J
L
N
M
M
O
Q
P
P
−
L
N
M
O
Q
P
1
1
2
2
118
19 0
39 0
380
590
cm
39.0
cm
cm
cm
.
af
a
f
(c)
Since
M
<
0 , the final image is
inverted
.
P36.75
(a)
P
fpq
==+=
+
∞
=
1
00224
1
44 6
.
.
m
diopters
bg
(b)
P
+
∞
=
1
0330
1
303
.
.
m
diopters
P36.76
The object is located at the focal point of the upper mirror. Thus, the
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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