• question_answer An infinitely long rod lies along the axis of concave mirror of focal length$f$. The near end of the rod is at a distance$x>f$from the mirror. Then the length of the image of the rod is: A)  $\frac{{{f}^{2}}}{x+f}$               B)         $\frac{{{f}^{2}}}{x}$                   C)  $\frac{xf}{x-f}$             D)         $\frac{xf}{x+f}$E)  $\frac{{{f}^{2}}}{x-f}$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ and      $m=\frac{v}{u}=-\frac{I}{O}$ Using above relation, the length of image                 $I=\frac{{{f}^{2}}}{x-f}$