Question

A small object $ab$ of size $1\text{mm}$ is kept at a distance of $40\text{cm}$ from a concave mirror of focal length $30\text{cm}$. Then, the size of the image is :

• A. $2\text{mm}$
• B. $5\text{mm}$
• C. $6\text{mm}$
• D. $9\text{mm}$

Answer 1

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

Given:
$u=-40\text{cm};f=-30\text{cm}$
[since object size is very small]

Applying the mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f},$ we have

$\frac{1}{v}+\frac{1}{-40}=\frac{1}{-30}$

$\Rightarrow v=-120\text{cm}$

Further, applying magnification formula,
$m=-\frac{v}{u}=-\frac{(-120)}{(-40)}=-3$

For small size of object ($ab$), Longitudinal magnification will be $m^2=9$

Hence, Image length
$L_I=m^2{L}_O=\left(9\right)\left(1\text{mm}\right)=9\text{mm}$