Question

A rod of length $\frac{f}{2}$ is placed along the axis of a concave mirror of focal length ${}^{\prime }{f}^{\prime }.$ if the near end of the real image formed by the mirror just touches the far end of the rod, find its magnification.

• A. $4$
• B. $3$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

Given: A rod of length $\frac{f}{2}$ is placed along the axis of a concave mirror of focal length 'f'. if the near end of the real image formed by the mirror just touches the far end of the rod

To find its magnification

Solution:

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 =$2f$

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

So, $u=PA=PC-AB=2f-\frac{f}{2}=\frac{3f}{2}$

The conjugate foci relation of spherical mirror is

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

where $v$ is image distance, $u$ is object distance = $-\frac{3f}{2}$

and $f$ is focal length = $-f$

substituting the values in eqn (i),we get

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

Negative sign indicates the position of image point A' is at the same side of object point A

So length/size of the image

$B{A}^{\prime }=C{A}^{\prime }=P{A}^{\prime }-PC=3f-2f=f$

Now magnification is

$m=\frac{\text{size of the object}}{\text{size of the image}}=\frac{f}{\frac{f}{2}}=2$