Question

An object of length $10cm$ is placed at right angles to the principal axis of a mirror of radius of curvature $60cm$ such that its image is virtual, erect and has a length $6cm$ . What kind of mirror it is and also determine the position of the object?

• A. $-20cm$
• B. $20cm$
• C. $-30cm$
• D. $30cm$

Answer 1

Answer: (A)

Since the image is virtual, erect and of a smaller

size, the given mirror is 'convex' (concave mirror does not form an image with the said description).
Given R = + 60 cm
∴ $f=\frac{R}{2}=30\text{cm}$
Transverse magnification,
$m=\frac{I}{O}=\frac{6}{10}=+\frac{3}{5}$
Further $m=-\frac{v}{u}=\frac{3}{5}$
∴ $v=-\frac{3u}{5}$
Using $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{-5}{3u}+\frac{1}{u}=\frac{1}{30}$
$\frac{-5+3}{3u}=\frac{1}{30}$
∴ u = - 20 cm
Thus the object is at a distance 20 cm ( from the pole) in front of the mirror.