Question

An object is placed 6.0 cm in front of a lens of focal length 5.0 cm. Another lens of focal length 4.0 cm is placed 2.4 cm behind the first lens.

(a) Where is the final image?

distance	cm
location	---Select--- behind the second lens in front of the second lens, but between lenses in front of the first lens

Is the image real or virtual?

virtualreal    

(b) What is the overall magnification?

Answer 1

The key to this kind of problem is that the 1st (near) lens's
(intermediate) image is the 2nd (far) lens's object. 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 = 0.06 m; f1 = 0.05 m; lens separation (sep) = 0.024 m; f2 = 0.04 m; v2 = TBD
Solving,
v1 = 1/(1/f1-1/u1) = 0.3 m
u2 = sep-v1 = -0.276 m
v2 = 1/(1/f2-1/u2) = 0.035 m

Final Image location relative to lens 2 = v2 = 0.035 m
Final Image location relative to lens 1 = v2+sep = 0.042220404 m
Final Image location relative to object = v2+sep+u1 = 0.119 m
Magnification = v1v2/(u1u2) = -0.634