Question

A very small linear object of size $1\text{mm}$ is kept at a distance of $40\text{cm}$ from a concave mirror of focal length $30\text{cm}$. The length of the image when the object is placed along the principal axis will be

• A. $9\text{mm}$
• B. $6\text{mm}$
• C. $4\text{mm}$
• D. $2\text{mm}$

Answer 1

The correct option is A $9\text{mm}$

Given:
$L_o=1\text{mm}$
$u=-40\text{cm}$
$f=-30\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{40}=\frac{-1}{30}$
$\Rightarrow \frac{1}{v}=\frac{-1}{120}$
$\Rightarrow v=-120\text{cm}$
For longitudinal magnification,
$m_L=-\frac{v^2}{u^2}$
$\Rightarrow m_L=-\frac{(-120{)}^2}{(-40{)}^2}=-9$
$\Rightarrow \frac{L_i}{L_o}=|-9|$
$\Rightarrow \frac{L_i}{1}=9$
$\Rightarrow L_i=9\text{mm}$
$\begin{array}{c}\text{Why this question?}\\ \text{Since the length of object is small as compared to the the distance at which it is kept, therefore the formula for longitudinal magnification can be used.}\end{array}$