Question

The radius of curvature of concave mirror is $12cm$ and the virtual image is magnified by $1.5$times. The object distance is

• A. $10cm$
• B. $8cm$
• C. $16cm$
• D. $4cm$

Answer 1

Answer: (A)

$10cm$

Given,

The radius of curvature $R=-12cm$ .

The focal length is given as $f=\frac{R}{2}=\frac{-12}{2}=-6cm$ .

Magnification is given as $1.5$, hence

$M=\frac{v}{u}$

$1.5=\frac{v}{u}$

$v=1.5u$

Now according to the mirror formula

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

$-\frac{1}{6}=\frac{1}{1.5u}+\frac{1}{u}$

$-\frac{1}{6}=\frac{2.5}{1.5u}$

$u=\frac{-2.5\times 6}{1.5}$

$u=-10cm$ .

Hence the object distance obtained is $10cm$ .