Find the location and length of the final image

[hraghav]

Homework Statement

An object of length 3.23cm is located at the origin of a coordinate system. A concave reflecting surface is located at 30.5cmi^, and has a radius of 45.4cm. A converging lens of focal length 42cm is located at −83cmi^. This arrangement is shown in the diagram below. The light from the object that reflects off the mirror then refracts through the lens. The image from the mirror acts as an object for the lens, and the lens produces the final image. HINT: This question uses a coordinate system, but your lens and mirror formulas use scalar quantities with a set of rules. Use the coordinate system to convert and calculate the scalar quantities, then draw out the lens/mirror system and convert back into the coordinate system. The more accurate your drawing, the better your analysis will be. What is the location of the final image of the object as produced by the lens? What is the length of the final image of the object as produced by the lens? (A negative value means the image is inverted relative to the object orientation.)

Relevant Equations

1/f = 1/do + 1/di
Mm = -di/do
Ml = di/do

So I found the location of the image of the object produced by the mirror to be -58.2628 cmi^ and the length of the image of the object produced by the mirror to be -9.40012 cm. (these are correct values).

To find the location of the final image of the object as produced by the lens I used 1/f = 1/do + 1/di where
f = 45.4/2 = 22.7
do = 30.5
and found di = 88.7628
Then I added 30.5 - 88.7628 = -58.2628 (mirror image)
For object distance from lens: do = -58.2628 - (-83) = 24.7372
As Lens's focal length is 42 cm, using lens equation:
1 / 42 = 1 / 24.7372 + 1 /di
where di = -60.18504 as the final answer

For the length:
I used the mirror magnification formula ie Mm = -di/do = - 88.7628/ 30.5 = -2.91
Lens magnification formula ie Ml = di/do = -60.18504/ 24.7372 = -2.4329
total magnification is M = Mm * Ml
M = -2.91 * -2.4329 = 7.07996

Image length is 3.23 * 7.07996 = 22.8682

Both my answers are incorrect and I am not sure where I am making an error. Could someone please look at this and help me out?
Thank you!

 

[Steve4Physics]

The more accurate your drawing, the better your analysis will be.

You need a good ray diagram with distances and x-coordinates both marked. Without that, things can go horribly wrong!

where di = -60.18504 as the final answer

Are you expected to convert this back to an x-coordinate? Maybe that's the problem.
(Also, units are missing and there are far too many significant figures.)

From my (distant) memories the negative sign here indicates that the image is virtual. In that case, the lens's image will be upright, not inverted - like a simple magnifying glass.

 

[hraghav]

Thank you for your help I got both the answers :)