An object is 16 m to the left of a lens of focal length 076

An object is 1.6 m to the left of a lens of focal length 0.76 m. A second lens of focal length -3.8 m is 0.72 m to the right of the first lens.

Find the distance between the object and the final image formed by the second lens. (m)

What is the overall magnification (with sign)?

Is the final image real or virtual? real virtual

Is it upright or inverted?

Solution

The key to this kind of problem is that the 1st (left) lens\'s
(intermediate) image is the 2nd (right) lens\'s object. (+uN is object distance to left of lens N and +vN is image distance to right of lens N), we have initially:
u1 = 1.6 m; f1 = 0.76 m; sep = 0.72 m; f2 = -3.8 m; v2 = ?
Solving,

v1 = 1/(1/f1-1/u1)

v1 = 1/(1/0.76-1/1.6) = 1.4476 m
u2 = sep-v1 = 0.72 - 1.4476 =-0.7276 m

v2 = 1/(1/f2-1/u2) = 1/(1/0.76 + 1/0.7276) = 0.3717 m

distance between the object and the final image formed by the second lens.

v2+sep+u1= 0.3717+0.72+1.6 = 2.691 m

overall magnification

Magnification = v1v2/(u1u2) = 2.6917*1.4476)/(-0.7276*1.6) =   - 3.34