Question

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

• A. $8/3$
• B. $2.5$
• C. $2$
• D. $-1.5$

Answer 1

The correct option is D $-1.5$

Let the object distance be $u$ then image distance is $u+10$

$u=-u$ ; $v=-(u+10)$ ; $f=-12$

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

$\frac{-1}{u+10}+\frac{-1}{u}=\frac{-1}{12}$

$\frac{2u+10}{u(u+10)}=\frac{1}{12}$

$u=20$cm

$v=-30$cm

Magnification is $-\frac{v}{u}=-\frac{30}{20}=-1.5$