Question

A linear thin rod of length $\frac{f}{3}$ is placed along the optical axis of a concave mirror of focal length $f$ such that its image which is real and elongated just touches the rod. Find the magnification.

• A. $\frac{3}{2}$
• B. $2$
• C. $\frac{5}{2}$
• D. $3$

Answer 1

The correct option is A $\frac{3}{2}$

Given that image of the object touches the object. It means one end of the object is at centre of curvature and also image is enlarged , that means orientation of object is towards focus only then we can get an enlarged image. (Details are as shown in the image below)




Let $l$ be the length of image.

Magnification: $m=\frac{l}{\left(\frac{f}{3}\right)}$
$\Rightarrow l=\frac{mf}{3}$

From the above diagram
$u^{\prime }=u+\frac{f}{3}$

Since the one end of the object is at centre of curvature we can say that, $2f=u^{\prime }$

$\Rightarrow 2f=u+\frac{f}{3}$

$\Rightarrow u=\frac{5f}{3}$

Now,
$v=(u^{\prime }+l)$

$\Rightarrow v=(u+\frac{f}{3}+l)$

$\Rightarrow v=(\frac{5f}{3}+\frac{f}{3}+\frac{mf}{3})$

$\Rightarrow v=\left(\frac{6f+mf}{3}\right)$

Using mirror formula:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{3}{6f+mf}+\frac{3}{5f}=\frac{1}{f}$

$\Rightarrow \frac{3}{6+m}=\frac{2}{5}$

$\Rightarrow 12+2m=15$

$\Rightarrow m=\frac{3}{2}$

Hence, option (a) is the correct answer.

Why this question: To understand the concept of magnification for the longitudinal objects.