How Does Lens Placement Affect Image Height and Orientation?

[Rawr]

A 1.5-cm tall object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. What is the height and orientation with respect to the original object of the final image.

Okay, first, I find the distance of the image with respect to the diverging lens first.

-1/.2 = 1/di + 1/.5
di = -0.1428...

Then, I add that distance image to the converging lens distance to get the object distance for the converging lens.

0.08 + -.1428.. = .2228 m

Then I use the lens equation again except with that object distance and the focal length of the converging lens.

1/.17 = 1/di + 1/.2228
di = 1.395...

Then I take the magnification equation and use hi/h0 = -di/d0

hi / (1.5) = -(1.395)/(.5)
Solve for hi.. and get -4.185.

Apparently, that is not the answer. The answer key says that it is 1.4 cm with an inverted image. I don't really know how they get 1.4. Where did I go wrong in my work? Or am I wwwaaayy off with what I'm doing?

 

[hage567]

For your final di of 1.395 I think you forgot to invert it. It should be 1/1.395.
For the magnification, find the magnification due to the first lens, and then the second lens. The product of the two will give you the final magnification. Your way won't work because after the first lens, the size of the object is not 1.5 cm anymore. Also the object distance changes after the first lens. Work that out and see what you get.