On the axis of a spherical mirror of focal length `f` a short linear object of length `L` lies on the axis at a distance `mu` from the mirror. Its image has an axial length `L'` equal to

Options

A. `L[(f)/(u-f)]^(1//2)`
B. `L[(u+f)/(f)]^(1//2)`
C. `L[(u-f)/(f)]^(2)`
D. `L[(f)/(u-f)]^(2)`

Answer

D

Explanation

From mirror formula `(1)/(F)=(1)/(v)+(1)/(u)…(ii)`
On differentiating Eq. (i) we obtain `0=(1)/(v^(2))dv-(1)/(v^(2))du`
`dv=-((v)/(u))^(2)du ....(ii)`
Also, from Eq. (i) we get `(v)/(u)=(f)/(u-f)...(iii)`
From Eqs. (ii) and (iii) we get `dv=-((f)/(u-f))^(2)L`
Therefore, size of image is `((f)/(u-f))^(2)L`