Question

A concave mirror produces a real image twice the size of the object when placed at a distance of 22.5 cm from it. Where should be object be placed so that the size of the image becomes three times that of the object?

Answer 1

Since image is real, it is inverted.

So $m=-2=\frac{-v}{u}$ , here $u=-22.5$

So $-2=\frac{v}{22.5}$

$v=-45cm$

so $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{-1}{45}-\frac{1}{22.5}$

$f=-15cm$

Now we want three times larger image so

So $m=-3=\frac{-v}{u}$ ,

So $3=\frac{v}{u}$

So $v=3u$

So $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{3u}+\frac{1}{u}$

$\frac{-1}{15}=\frac{4}{3u}$

$u=-20cm$