Question

An object is placed at (i) $10cm$, (ii) $5cm$ in front of a concave mirror of radius of curvature $15cm$. Find the position, nature and magnification of the image in each case.

Answer 1

(a) In concave mirror, using sign convention

$u=-10cm$

$R=-15cm$

$f=\frac{-15}{2}cm$

Using mirror’s relation

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}=-\frac{2}{15}+\frac{1}{10}$

$\frac{1}{v}=\frac{-1}{30}$

$v=-30cm$

Image is real and it forms at a distance of 30 cm in front of mirror.

Magnification

$m=\frac{-v}{u}$

$m=\frac{30}{-10}=-3$

This shows image is inverted and 3 times magnified.

(b) again using sign convention

$u=-5cm$

$f=\frac{-15}{2}cm$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}=\frac{1}{15}$

$v=+15cm$

Image is formed at a distance of 15 cm behind the mirror.

Magnification

$m=\frac{-v}{u}$

$=\frac{-15}{-5}$

$=3$

+3 magnification shows image is virtual and erect.