What is the image height formed by the new diverging lens?

[matt72lsu]

Homework Statement

An object is located 27.5 cm from a certain lens. The lens forms a real image that is twice as high as the object.
b) Now replace the lens used in Part A with another lens. The new lens is a diverging lens whose focal points are at the same distance from the lens as the focal points of the first lens. If the object is 5.00 cm high, what is the height of the image formed by the new lens? The object is still located 27.5 cm from the lens.

Homework Equations

hi/ho = di/do 1/f = 1/di + 1/di

The Attempt at a Solution

for part A, i found the focal length to be 18.3 cm. For part B i really don't know how to start. I was thinking that i had to solve for either di or do and plug it into the first equation but I don't think that is possible with the info given.

 

[AtticusFinch]

Ok you know converging lenses form real images that are inverted.

So -2 = -q/27.5

q = 55 cm

1/27.5 + 1/55 = 1/f

f = 18.3 cm so your part A is correct.

For part B, you know f = -18.3 cm and p = 27.5 cm so just find q (the image distance) and then find the magnification (-q/p) and multiply that to the object height.

 

[matt72lsu]

Ok I'm getting 2 cm. Could you double-check that please?

 

[AtticusFinch]

Yep that checks out with what I got. Good job.