p35_048 - 48. (a) (b) (c) and (d) Our first step is to...

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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. 35-9 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. 35-9 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.

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