Question

An object is placed at a distance of $40\text{cm}$ from a concave mirror of focal length $15\text{cm}$. If the object is displaced through a distance of $20\text{cm}$ towards the mirror, the displacement of the image will be :

• A. $30\text{cm}$ away from the mirror
• B. $36\text{cm}$ away from the mirror
• C. $36\text{cm}$ towards the mirror
• D. $30\text{cm}$ towards the mirror
  

Answer 1

The correct option is B $36\text{cm}$ away from the mirror

Given that,
Focal length of the mirror, $f=-15\text{cm}$ (concave mirror)
Initial position of the object, $u_1=-40\text{cm}$ (concave mirror)


Let, when the object is at $u_1$ image is formed at $v_1$.
From mirror formula,
$\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f}$
$\Rightarrow \frac{1}{v_1}+\frac{1}{(-40)}=\frac{1}{(-15)}$
$\Rightarrow \frac{1}{v_1}=\frac{1}{40}-\frac{1}{15}=\frac{3-8}{120}$
$\therefore v_1=-24\text{cm}$

When object is displaced $20\text{cm}$ towards mirror,
$u_2=-(40-20)=-20\text{cm}$
Using mirror formula,
$\frac{1}{v_2}=\frac{1}{20}-\frac{1}{15}=\frac{3-4}{60}=\frac{-1}{60}$
$\therefore v_2=-60\text{cm}$
Hence, displacement of image,
$\Delta v=v_2-v_1=-60-(-24)$
$\Rightarrow \Delta v=-36\text{cm}$
Thus, image has shifted in -ve direction i.e., away from mirror along principal axis.