Question

A point object is placed at $60\text{cm}$ from the pole of a concave mirror of focal length $10\text{cm}$ on the principal axis. If the object is shifted $1\text{mm}$ towards the mirror along the principal axis, then find the shift in image.

• A. $\frac{1}{15}\text{mm}$ away from the pole
• B. $\frac{1}{15}\text{mm}$ toward the pole
• C. $\frac{1}{25}\text{mm}$ away from the pole
• D. $\frac{1}{25}\text{mm}$ toward the pole

Answer 1

The correct option is C $\frac{1}{25}\text{mm}$ away from the pole

Assuming the length shifted as a linear object, the length of the image will be the shift in the image of the point object.
Given:
$u=-60\text{cm}$
$f=-10\text{cm}$
$L_o=1\text{mm}$ $[$assumed as object$]$
From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{60}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-1}{12}$
$\Rightarrow v=-12\text{cm}$
For longitudinal magnification,
$m_L=-\frac{v^2}{u^2}$
$\Rightarrow m_L=-\frac{(-12{)}^2}{(-60{)}^2}=-\frac{1}{25}$
$\Rightarrow \frac{L_i}{L_o}=|-\frac{1}{25}|$
$\Rightarrow \frac{L_i}{1}=\frac{1}{25}$
$\Rightarrow L_i=\frac{1}{25}\text{mm}$
Thus, image gets shifted by $\frac{1}{25}\text{mm}$ away from pole as object is shifted toward the mirror.
$\begin{array}{c}\text{Why this question?}\\ \text{In concave mirror, as object move towards the mirror till focus,}\\ \text{the image shift away from the mirror.}\end{array}$