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

One end of image coincides with one end of rod i.e., either A or B is centre of curvature.
Since image is elongated i.e., point B is centre of curvature. Image of B is formed on B itself.
image of A: `u=-(2f-(f)/(3))=(-5f)/(3),f=-f,v=?`
`(1)/(v)+(1)/(u)=(1)/(f)implies(1)/(v)+(1)/((-5f)/(3))=(1)/(-f)`
`(1)/(v)=-(1)/(f)+(3)/(5f)=(-5+3)/(5f)=-(2)/(5f)`
`v=-(5f)/(2)`
Image length `A^(')B^(')=(5f)/(2)-2f=(f)/(2)`
Longitudinal magnification `m_L= - (A^(')B^('))/(AB)=((f)/(2))/((f)/(3))=-(3)/(2)`