Question

An object is kept in front of a concave mirror of focal length 15 cm. The image formed is three times of size of object. Calculate two possible distance of the object from the mirror.

Answer 1

f=-15
3u=v
1/v+1/u=1/-15
1/3u+1/u=-1/15
u+3/3u=-1/15
3u=-15u-45
u=-45/18

Answer 2

From the mirror, the two possible distance of objects are 10 cm and 20 cm.

Solution:

The possible two distances of the object from the mirror be calculated as following –

Given: Focal length f = 15 cm

Magnification m = Thrice of object size = can either be -3 or +3

As we know, $\bold{\begin{aligned} \mathrm{m} &=\frac{-v}{u}} \\-3 &=\frac{-v}{u} \\ v &=3 u \end{aligned}$

From the lens maker formula,

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

$\begin{array}{l}{\frac{1}{-15}=\frac{1}{3 u}+\frac{1}{u}} \\\\ {\frac{1}{-15}=\frac{4}{3 u}} \\\\ {3 u=4 \times(-15)} \\\\ {3 u=-60} \\\\ {u=-\frac{60}{3}} \\\\ \bold{{u=-20 \mathrm{cm}}}\end{array}$

Also when m taken as +3

$\begin{array}{l}{\frac{1}{-15}=\frac{1}{-3 u}+\frac{1}{u}} \\\\ {\frac{1}{-15}=\frac{2}{3 u}} \\\\ {3 u=2 \times-15} \\\\ {3 u=-30} \\\\ {u=-\frac{30}{3}} \\\\ \bold{{\mathrm{u}=-10 \mathrm{cm}}}\end{array}$