p35_060

# p35_060 - Â âˆ’ 1 = 6 0 m to the left of the lens(that is 2...

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Â¶ âˆ’ 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 = Âµ âˆ’ i p Â¶Âµ âˆ’ i p Â¶ = Âµ âˆ’ 1 . 0 m 1 . 0 m Â¶Âµ âˆ’ . 60 m 3 . 0 m Â¶ = +0 . 20 ....
View Full Document

Ask a homework question - tutors are online