Unformatted text preview: 47. (a) (b) and (c) Our ﬁrst step is to form the image from the ﬁrst lens. With p1 = 4 cm and f1 = −4 cm,
Eq. 359 leads to
1
1
1
+
=
=⇒ i1 = −2 cm .
p1
i1
f1
The corresponding magniﬁcation is m1 = −i1 /p1 = 1/2. This image serves the role of “object” for
the second lens, with p2 = 10 + 2 = 12 cm, and f2 = −4 cm. Now, Eq. 359 leads to
1
1
1
+
=
p2
i2
f2 =⇒ i2 = −3.00 cm with a corresponding magniﬁcation of m2 = −i2 /p2 = 1/4, resulting in a net magniﬁcation of
m = m1 m2 = 1/8. The fact that m is positive means that the orientation of the ﬁnal image is the
same as the (original) object. The fact that i2 is negative means that the ﬁnal image is virtual. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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