A thin rod of length `f//3` is placed along the optical axis of a concave mirror of focal length f such that its image whichis real and elongated just touches the rod. Calculate the magnification.

Explanation

Image is real and enlarged, the object must be between `C` and `F`.
One end `A'` of the image coincides with the end `A` of rod itself.
So `v_(A) = u_(A), (1)/(V_(A)) + (1)/(v_(A)) = (1)/(-f)` i.e. , `v_(A) = u_(A) = -2f`
so it clear that the end `A` is at `C` . `because ` the length of rod is `(f)/(3)`
`therefore` Distance of the other end `B` from `P` is `u_(B)` = 2f - (f)/(3) = (5)/(3)f`
if the distance of image of end `B` from `P` is `v_(B)` then `(1)/(v_(B)) + (1)/(-(5)/(3)f) = (1)/(-f) implies v_(B) = -(5)/(2)f`
`therefore` the length of the image `|v_(B)| - |v_(A)| = (5)/(2)f = (1)/(f)` and magnification `m = (|v_(B)| - |v_(A)|)/(|(u_(B)| - |u_(A)|) = ((1)/(2)f)/(-(1)/(3)f) = -(3)/(2)`
Negative sign implies that image is inverted with respect to object and so it is real.