Question

A particle goes is in the form of a circle of radius $2.0\text{cm}$. A concave mirror of focal length $20\text{cm}$ is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is $30\text{cm}$. Calculate the radius of the circle formed as the image.

• A. $2\text{cm}$
• B. $8\text{cm}$
• C. $4\text{cm}$
• D. $5\text{cm}$

Answer 1

The correct option is C $4\text{cm}$

Given $u=-30\text{cm},f=-20\text{cm}$
From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}+(-\frac{1}{30})=(-\frac{1}{20})$
$\Rightarrow v=-60\text{cm}$.

Image of the circle is formed at a distance $60\text{cm}$ in front of the mirror.
$\therefore m=-\frac{v}{u}=\frac{R_{\text{image}}}{R_{\text{object}}}$
$\Rightarrow m=-\frac{-60}{-30}=\frac{R_{\text{image}}}{2}$

$\Rightarrow R_{\text{image}}=4\text{cm}$

Radius of image of the circle is $4\text{cm}$

Why this Question? Tip: when object is placed at a distance from principle axis, image position is calculated using magnification formula.