Question

The image of an object placed on the principal axis of a concave mirror of focal length 12 cm is formed at a point which is 10 cm more distant from the mirror than the object. The magnification of the image is

• A. 2
• B. 1.5
• C. 1
• D. 2.5

Answer 1

The correct option is B. 1.5.

Let $u$ be the object distance and $v$ be image distance, then from mirror formula we have $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

(sign convention: Real$\to$+ve and virtual$\to$-ve.)
Given,

Focal length = $f=-12cm$

Image distance = $v=(u+10)$
$\therefore \frac{1}{-u}+\frac{1}{-(u+10)}=\frac{1}{-12}$
$\Rightarrow \frac{u+10+u}{(u+10)u}=\frac{1}{12}$
$\Rightarrow u^2+10u=24u+120$
$\Rightarrow u^2-14u-120=0$
$\Rightarrow u=20$
$\Rightarrow u=-20cm$ and $v=-30cm$
Magnification = $m=-\frac{v}{u}=-\frac{-30}{-20}=-\frac{3}{2}$
Hence, $m=-1.5$.

Negative sign indicates that the image is inverted.