1023_Physics ProblemsTechnical Physics

1023_Physics ProblemsTechnical Physics - 364 Image...

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364 Image Formation (c) To find the area, first find q R and q L , along with the heights h R and h L , using the thin lens equation. 11 1 pq f RR += becomes 1 20 0 13 3 .. cm cm q R or q R = 40 0 . cm == F H G I K J =− = hh Mh q p R R 10 0 2 00 20 0 ... cm cm af a f 1 30 0 13 3 cm cm q L or q L = 24 0 = M LL 10 0 0 800 8 00 cm cm a f Thus, the area of the image is: Area +− ′ − ′ = qq h RL L 1 2 224 cm 2 . P36.41 (a) The image distance is: qdp . Thus, 111 pq f becomes pdp f + = . This reduces to a quadratic equation: pd p f d 2 0 + = which yields: p dd f d fd = ±− 2 2 4 22 4 . Since f d < 4 , both solutions are meaningful and the two solutions are not equal to each other. Thus, there are two distinct lens positions that form an image on the screen. (b) The smaller solution for p gives a larger value for q , with a real, enlarged, inverted image .
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