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

Since the image formed is real and elongate, the situation is as shown in the figure. Since the image of B is formed at `B^(')` itself, therefore,
B is situtated at the center of curvature, that is, at a distance 2f from the pole.
`:.PA=2f-(f)/(3)=(5f)/(3)`
Let us find the image of A. For point A
`u=-(5f)/(3),v=?`
Applying `(1)/(u)+(1)/(upsilon)=(1)/(f)`
`rArr (1)/((-5f)/(3))+(1)/(upsilon)=(1)/(-f)`
`rArr (1)/(upsilon)=-(1)/(f)+(3)/(5f)rArr(1)/(upsilon)=(-5+3)/(5f)=(-2)/(5f)rArrv=-2.5f`
`:.` Image length `=2.5f-2f=0.5f=(f)/(2)`
`:.` Magnification `=((f)/(2))/((f)/(3))=1.5`