# 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?

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

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

##### 1 Answer

A rod AB of length **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

The object distance for near point of the rod A is

So

The conjugate foci relation of spherical mirror is

#1/v+1/u=1/f......(1)#

Where

So inserting the values in equation (1)

#1/v-2/(3f)=-1/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**

Now Magnification