Question

An object of length $10 \, cm$ is placed at right angles to the principal axis of a mirror of radius of curvature $60 \, cm$ such that its image is virtual, erect and has a length $6 \, cm$ . 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} = 3 0 \text{cm}$
Transverse magnification,
$m = \frac{ I ⁡}{ O ⁡} = \frac{6}{1 0} = + \frac{3}{5}$
Further $m = - \frac{ v ⁡}{ u ⁡} = \frac{3}{5}$
∴ $ v = - \frac{3 u ⁡}{5}$
Using $\frac{1}{ v } + \frac{1}{ u ⁡} = \frac{1}{ f ⁡}$
$\frac{- 5}{3 u } + \frac{1}{ u ⁡} = \frac{1}{3 0}$
$\frac{- 5 + 3}{3 u } = \frac{1}{3 0}$
∴ u = - 20 cm
Thus the object is at a distance 20 cm ( from the pole) in front of the mirror.