Physics - Geometrical Optics Question with Solution | TestHub

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.