p35_060 - 1 = 6 . 0 m to the left of the lens (that is, 2 ....

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60. (a) Suppose that the lens is placed to the left of the mirror. The image formed by the converging lens is located at a distance i = µ 1 f 1 p 1 = µ 1 0 . 50 m 1 1 . 0m 1 =1 . 0m to the right of the lens, or 2 . 0m 1 . 0m=1 . 0 m in front of the mirror. The image formed by the mirror for this real image is then at 1 . 0 m to the right of the the mirror, or 2 . 0m+1 . 0m=3 . 0m to the right of the lens. This image then results in another image formed by the lens, located at a distance i 0 = µ 1 f 1 p 0 1 = µ 1 0 . 50 m 1 3 . 0m
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Unformatted text preview: 1 = 6 . 0 m to the left of the lens (that is, 2 . 6 cm from the mirror). (b) The Fnal image is real since i > 0. (c) It also has the same orientation as the object, as one can verify by drawing a ray diagram or Fnding the product of the magniFcations (see the next part, which shows m > 0). (d) The lateral magniFcation is m = i p i p = 1 . 0 m 1 . 0 m . 60 m 3 . 0 m = +0 . 20 ....
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