Physics help: a two lens magnifying system uses lenses of <span title="...................................................">............................</span>

Question

focal lengths 2.5cm and 9.5cm for the objective and eyepiece respectively. The two lenses are positioned 23cm apart. An object for study is placed 3.0cm in front of the objective lens. Find the position of the final image relative to the eyepiece lens.

hint: the answer is -50.7cm. (i don't know how to get to it)

Answer 1

See the refs. The key to this kind of problem is that the 1st (objective) lens&#039;s (intermediate) image is the 2nd (eyepiece) lens&#039;s object. Assume the objective is on the left and the eyepiece on the right. Then using notation from ref. 1 (+uN is object distance to left of lens N and +vN is image distance to right of lens N), we have initially:

u1 = .03 m; f1 = .025 m; sep = .23 m; f2 = .095 m; v2 = unknown

Solving,

v1 = 1/(1/f1-1/u1) = .15 m

u2 = sep-v1 = 0.08 m

v2 = 1/(1/f2-1/u2) = -.506666666666664 m

Final Image location relative to lens 2 = v2 =-.506666666666664 m

Final Image location relative to lens 1 = v2+sep =-.276666666666664 m

Final Image location relative to object = v2+sep+u1 =-.246666666666664 m

Magnification = v1v2/(u1u2) =-31.6666666666665