HW-25Solutions-04-07-08

# HW-25Solutions-04-07-08 - Lehigh University HW-25 Solutions...

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Lehigh University Physics 21, Spring 2008 April 7, 2008 HW-25 Solutions 25-1. (HRW 34-107) A fruit fly of height H sits in front of lens 1 on the central axis through the lens. The lens forms an image of the fly at a distance from the fly; the image has the fly's orientation and height . What are (a) the focal length f 1 of the lens and (b) the object distance p 1 of the fly? The fly then leaves lens 1 and sits in front of lens 2, which also forms an image at that has the same orientation as the fly, but now . What are (c) f 2 and (d) p 2 ? Solution: 107. (a) In this case m > +1 and we know that lens 1 is converging (producing a virtual image), so that our result for focal length should be positive. Since |P + i 1 | = 20 cm and i 1 = – 2 p 1 , we find p 1 = 20 cm and i 1 = – 40 cm. Substituting these into Eq. 34-9, 11 1 111 p if + = leads to f 1 = +40 cm, which is positive as we expected. (b) The object distance is p 1 = 20 cm, as shown in part (a). (c) In this case 0 < m < 1 and we know that lens 2 is diverging (producing a virtual image), so that our result for focal length should be negative. Since |p + i 2 | = 20 cm and i 2 = – p 2 /2, we find p 2 = 40 cm and i 2 = – 20 cm. Substituting these into Eq. 34-9 leads to f 2 = – 40 cm, which is negative as we expected. (d) p 2 = 40 cm, as shown in part (c).

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HW-25Solutions-04-07-08 - Lehigh University HW-25 Solutions...

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