p35_040 - 1 p 1 + 1 i 1 = 1 f 1 = p 1 = 3 . 00 cm . (b) The...

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40. (a) The “object” for the mirror which results in that box-image is equally in front of the mirror (4 cm). This object is actually the Frst image formed by the system (produced by the Frst transmission through the lens); in those terms, it corresponds to i 1 =10 4=6cm . Thus ,w ith f
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Unformatted text preview: 1 p 1 + 1 i 1 = 1 f 1 = p 1 = 3 . 00 cm . (b) The previously mentioned box-image (4 cm behind the mirror) serves as an object (at p 3 = 14 cm) for the return trip of light through the lens ( f 3 = f 1 = 2 cm). This time, Eq. 35-9 leads to 1 p 3 + 1 i 3 = 1 f 3 = i 3 = 2 . 33 cm ....
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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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