p35_041 - image), so that our result for focal length...

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41. (a) In this case m> +1 and we know we are dealing with a converging lens (producing a virtual image), so that our result for focal length should be positive. Since | p + i | =20cmand i = 2 p , we Fnd p =20cmand i = 40 cm. Substituting these into Eq. 35-9, 1 p + 1 i = 1 f leads to f = +40 cm, which is positive as we expected. (b) In this case 0 <m< 1 and we know we are dealing with a diverging lens (producing a virtual
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Unformatted text preview: image), so that our result for focal length should be negative. Since | p + i | = 20 cm and i = p/ 2, we Fnd p = 40 cm and i = 20 cm. Substituting these into Eq. 35-9 leads to f = 40 cm, which is negative as we expected....
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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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