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Unformatted text preview: 48. (a) (b) (c) and (d) Our ﬁrst step is to form the image from the ﬁrst lens. With p1 = 3 cm and
f1 = +4 cm, Eq. 359 leads to
1
1
1
+
=
p1
i1
f1 =⇒ i1 = −12 cm . The corresponding magniﬁcation is m1 = −i1 /p1 = 4. This image serves the role of “object” for
the second lens, with p2 = 8 + 12 = 20 cm, and f2 = −4 cm. Now, Eq. 359 leads to
1
1
1
+
=
p2
i2
f2 =⇒ i2 = −3.33 cm with a corresponding magniﬁcation of m2 = −i2 /p2 = 1/6, resulting in a net magniﬁcation of
m = m1 m2 = 2/3. 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 (and
therefore to the left of the second lens). ...
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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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