lenses

Apply the thin lens formula to lens 1: 1/u + 1/v = 1/f
with u = 6 cm, f = 24 cm, v = ?
hence, 1/6 + 1/v = 1/24
solve for v gives v = -8 cm
The image formed is thus virtual and 8 cm in front of the lens on the same side as the object.

Distance of image from lens 2 = (10 + 8) cm = 18 cm
Apply the thin lens formula to lens 2,
with u = 18 cm, f = 9 cm, v = ?
hence, 1/18 + 1/v = 1/9
solve for v gives v = 18 cm

The final image is therefore real, inverted, located at distance 18 cm behind lens 2 and of the same size as the object.

2010-03-26 20:16:59 補充：
Just a little clarification. The final image is of the same size as the first image (object of lens 2) . Because the first image is larger in size than the object. The final image is thus larger than the object.
The "object" in the last sentence in the post thus refers to the "object of lens 2".