# A rod of length f/2 is placed along the axis of a concave mirror of focal length f.IF the near end of the REAL IMAGE formed by the rod just touches the far end of the ROD,what is its magnification?

## dont get confused ,question is absolutely correct,and read it twice,ans anyways.

Nov 12, 2016 A rod AB of length $\frac{f}{2}$ is placed along the axis of a concave mirror of focal length $f$ in such a way that the near end of the REAL IMAGE formed by the rod just touches the far end B of the ROD.
The near end of the real image of the rod is the image of far end B of the the rod AB. It is possible only if the far end B of the rod is placed at the center of curvature C of the concave mirror (as shown in the figure).
The image of near end A of the rod is A'

Now PC = the radius of curvature $= 2 f$

The object distance for near point of the rod A is $u$

So $u = P A = P C - A B = 2 f - \frac{f}{2} = \frac{3 f}{2}$

The conjugate foci relation of spherical mirror is

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \ldots \ldots \left(1\right)$

Where

v->"Image distance"=?

$u \to \text{Object distance} = - \frac{3 f}{2}$

$f \to \text{Focal length} = - f$

So inserting the values in equation (1)

$\frac{1}{v} - \frac{2}{3 f} = - \frac{1}{f}$

$\implies \frac{1}{v} = - \frac{1}{f} + \frac{2}{3 f}$

$\implies \frac{1}{v} = \frac{- 3 + 2}{3 f} = - \frac{1}{3 f}$

$\implies v = - 3 f$
Negative sign indicates the position of image point A' is at the same side of object point A

So length/size of the image

$B A ' = C A ' = P A ' - P C = 3 f - 2 f = f$

Now Magnification

$m = \text{Size of image"/"Size of object} = \frac{f}{\frac{f}{2}} = 2$