# Finding length and longitudinal magnification

## Homework Statement

A small object of length dL is placed on the axis of a converging mirror of focal length f. The center of the object is a distance o from the mirror. Determine the length of the image and thus effective determine a longitudinal m' = dL'/dL for the lens.

## The Attempt at a Solution

I know that the distance from the center to the object to the mirror is o so one end of the mirror is o+dL/2 and the other end is o - dL/2. I will call x = o + dL/2 and y = o -dL/2

I know 1/f = 1/o + 1/i

1/f = 1/x + 1/i
1/f - 1/x = 1/i
i = (x - f)/xf
x' = (x-f)/xf

For the other one

i = (y-f)/yf
y' (y-f)/(yf)

I know dL =|x-y|
which is just dL

dL'=|x'-y'|
=[(x-f)*yf - (y-f)*xf]/xyf^2
[xyf - yf^2 - xyf -xf^2] / xyf^2
[-yf^2 - xf^2] / (xyf^2)
(-y - x) /xy

[-(o - dL/2) - (o+dL/2) ]/ [(o-dL/2)(o+dL/2)]
dL'=-2o / (o-dL^2/4)

m' = -2o/(o-dL^2/4)/dL
m' = -2o/(dL*o - dL^3/4)

The correct answer is m'=-m^2. By definition, m = -i/o. I don't see how to get this. Did I approach the problem incorrectly or am I overlooking some kind of algebra manipulations? How do I get to -m^2?