Question

When an object is placed at a distance of $0.18m$ in front of a convex lens, a real image, double in size is produced. Where must be the object be placed so that its real image is three times as big as the object?

Answer 1

Magnification in lens

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

$-2=\frac{v}{-18}$

$v=36cm$

From lens formula

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

$\frac{1}{36}-\frac{1}{-18}=\frac{1}{f}$

$f=\frac{36}{1+2}$

$f=12cm$

Similarly

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

$v=-3u$

From lens formula we have

$\frac{1}{-3u}-\frac{1}{u}=\frac{1}{12}$

$\frac{1+3}{-3u}=\frac{1}{12}$

$u=\frac{12\times 4}{-3}$

$u=-16cm$

Thus the object distance will be$16cm$ on the left side of the lens.