(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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?

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# Homework Help: Finding length and longitudinal magnification

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