Question

An image formed by a concave mirror is $5$ times greater than the object. If the radius of curvature is $40cm$, determine the distance of the object from the mirror?

• A. $4cm$
• B. $8cm$
• C. $16cm$
• D. $32cm$

Answer 1

The correct option is C $16cm$

The magnification of image $\left(m\right)=5$
The radius of curvature $\left(r\right)=-40cm$
The focal length $\left(f\right)=\frac{r}{2}=\frac{-40}{2}=-20cm$
Let the object distance be $u$
and the image distance be $v$

We know,
$m=\frac{-v}{u}5=\frac{-v}{u}-v=5uv=-5u$

Using mirror formula,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\frac{1}{-20}=\frac{1}{-5u}+\frac{1}{u}\frac{1}{-20}=\frac{1}{-5u}+\frac{1}{u}\times \frac{5}{5}\frac{1}{-20}=\frac{4}{5u}u=\frac{4}{5}\times (-20)u=-16cm$
The distance of object from the mirror is $16cm$.