Question

An object placed in front of a concave mirror of focal length $0.15\text{m}$ produces a virtual image which is twice the size of the object. The position of the object from the mirror as per sign convention is

• A. $-6.5\text{cm}$
• B. $+6.5\text{cm}$
• C. $-7.5\text{cm}$
• D. $+7.5\text{cm}$

Answer 1

The correct option is C $-7.5\text{cm}$


Here, $f=-0.15\text{m}$ and $m=+2$ as mirror is concave and image is virtual.
We know, $m=\frac{f}{f-u}$
$\Rightarrow +2=\frac{-0.15}{-0.15-u}$
$\Rightarrow -0.3-2u=-0.15$
$\Rightarrow 2u=-0.15$
$\Rightarrow u=\frac{-0.15}{2}=-0.075\text{m}=-7.5\text{cm}$