Image location and magnification with mirror and diverging lens

[floridianfisher]

Homework Statement

The object in Figure P23.50 is midway between the lens and the mirror. The mirror's radius of curvature is 20.4 cm, and the lens has a focal length of -16.9 cm.
http://www.webassign.net/sf5/p23_54.gif

Figure P23.50
A. Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system.


B.What is the overall magnification of the image?

Homework Equations

1/p+ 1/q = 2/R and M=-q/p

The Attempt at a Solution

I tried the problem this way
1/12.5 +1/q = 2/20.4 so q=55.43 then I found M for the mirror 55.43/12.5 = 4.43
then I found q lens =25-55.43= -30.43 and M = -(-16.9/-30.43) = 0.56
So final image is 30.43-25= 5.43cm behind the mirror and M tot=4.43*0.56= 2.5
These are not the right answers though. Can anyone help?

 

[G01]

It seems to me that everything you have in the parts of the problem concerning the mirror is correct. You seem to be confused when you consider the light going through the lens.

Think of this as two separate problems. You solved the first one by finding the image from the mirror. In the second problem, the image from the mirror becomes the object for the lens. You have the right object distance for the lens, that is, the distance of the mirror image from the lens is -30.43 . This is the object distance,p, for the lens, not the image distance, q, as you said.

So for the lens, the object distance p=-30.43, q=?, and f=-16.9?
You should be able to solve for the image formed by the lens, which is the final image.
Once you find the final image distance, you can solve for the magnification of the lens as you were doing before.

BTW: I remember being assigned this problem too when I did intro optics!

 

[floridianfisher]

Thanks for the help!

 

[G01]

No problem. Anytime!