An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is distance `u gt f` from the mirror. Its image will have length

Options

A. `f^(2)/(u-f)`
B. `(uf)/(u-f)`
C. `f^(2)/(u+f)`
D. `(uf)/(u+f)`

Answer

A

Explanation

(a) From the relation,
`1/v+1/u=1/for1/v-1/u=1/(-f)`
`1/v=1/u-1/forv=((uf)/(u-f))`
Since,`ugtf`,v is negative
or `abs(v)=((uf)/(u-f))gtf`
The end which is at infinity will have its image at focus.
`therefore` Length of image,
`L=abs(v)-f=f^(2)/(u-f)`