A thin rod of length`(f)/(3)`is placed along the optic axis of a concave mirror of focal length f such that its image which is real and elongated, just touches the rod. The magnification is …..

Answer

A

Explanation

(##JMA_RWO_C16_008_S01##)
`Since the image formed is real and elongated, the situation is as shown in the figure. Since the image of B is formed at B itself
:' B is situated at the centre of curvature that is at a distance at 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)/(v)=(1)/(f) implies `(1)/((-5f)/(3))+(1)/(v)=(1)/(-f)`
`implies (1)/(v)=-(1)/(5f)` implies v=-2.5f Image length`=2.5f-2f=0.5f`
:' Magnification `=(0.5f)/((f)/(3))=1.5`