This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 26. (a) (b) (c) and (d) Our ﬁrst step is to form the image from the ﬁrst lens. With p1 = 10 cm and
f1 = −15 cm, Eq. 359 leads to
1
1
1
+
=
p1
i1
f1 =⇒ i1 = −6 cm . The corresponding magniﬁcation is m1 = −i1 /p1 = 0.6. This image serves the role of “object” for
the second lens, with p2 = 12 + 6 = 18 cm, and f2 = 12 cm. Now, Eq. 359 leads to
1
1
1
+
=
p2
i2
f2 =⇒ i2 = 36 cm with a corresponding magniﬁcation of m2 = −i2 /p2 = −2, resulting in a net magniﬁcation of
m = m1 m2 = −1.2. The fact that m is positive means that the orientation of the ﬁnal image is
inverted with respect to the (original) object. The height of the ﬁnal image is (in absolute value)
(1.2)(1.0 cm) = 1.2 cm. The fact that i2 is positive means that the ﬁnal image is real. ...
View
Full
Document
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

Click to edit the document details